The document contains solutions to math problems using Mathcad. Section 1 covers topics including: converting between Celsius and Fahrenheit, calculating pressure, force, work, energy, and cost analysis problems. Section 2 covers additional thermodynamics problems calculating things like internal energy, enthalpy, heat transfer, and phase changes.
Introduction to chemical engineering thermodynamics, 6th ed [solution]Pankaj Nishant
This document contains solutions to math problems involving concepts of thermodynamics, including calculations of work, heat, internal energy, enthalpy, and phase changes. Problem 1 calculates the work done in lifting a mass and the resulting internal energy change. Problem 2 determines the heat transferred and final temperature when water gains a small amount of heat. Problem 3 is a series of thermodynamic steps where the initial and final internal energies must sum to zero.
Heat transfer is the transfer of thermal energy between objects of differing temperatures. There are three main modes of heat transfer: conduction, convection, and radiation. Conduction involves the transfer of heat through direct contact of objects. Convection involves the transfer of heat by the circulation of fluids like gases and liquids. Radiation involves the transfer of heat through electromagnetic waves between objects not in contact.
1) The aluminum block absorbs more energy than the copper block when dropped into the calorimeter containing water. This is because aluminum has a higher specific heat than copper, so it takes more energy to raise the temperature of the aluminum block.
2) The final temperature of the water after the horseshoe and water reach equilibrium is 38°C. Using conservation of energy, the temperature of the horseshoe can be calculated.
3) It takes 63.97 minutes for a cup of water to boil in the microwave, assuming 50% of the microwave's 1200W power goes to heating the water. This estimate matches everyday experience of how long it takes for water to boil in a microwave.
Thermodynamics Assignment 02 contains calculations for various cycles of a steam power plant operating between 40 bar and 0.04 bar:
1) Carnot, simple Rankine, and modified Rankine cycles are analyzed. The modified Rankine cycle with superheat has the highest efficiency of 40.86% and lowest SSC of 2.4820 kg/kWh.
2) "Metallurgical limit" refers to the maximum safe pressures and temperatures a power plant's components can withstand without damage.
3) Implementing reheating in the Rankine cycle increases efficiency to 41.05% and lowers SSC to 2.4663 kg/kWh by utilizing the steam's initial high temperature again
This document contains a 16 question multiple choice mechanical engineering review problem set. It covers topics including: specific weight calculations, changes in weight due to elevation, pressure and force calculations for scuba diving, determining height using barometer readings, properties of gas mixtures, heat transfer between materials, gas turbine processes, combustion calculations, and thermodynamic processes including changes in temperature, pressure, volume, entropy and heat/work.
This document contains a tutorial for CHE2163 with 5 questions. It provides the relevant equations, assumptions, and step-by-step working to calculate various heat transfer rates. For question 1, it calculates the heat transfer rate from a module with air flowing over it. For question 2, it finds the initial and changing temperature of a heating plate. For question 3, it determines the heat loss from an uninsulated steam pipe. For questions 4 and 5, it analyzes heat transfer from pin fins and an underwater instrument pod respectively.
This document contains 20 multiple choice questions regarding concepts of thermodynamics and the second law of thermodynamics. It includes calculations of efficiency, COP, temperatures, enthalpies, and entropy involving heat engines and refrigerators. Key concepts covered are the Carnot cycle, Carnot efficiency, ideal gas processes, heat transfer, work, and isentropic processes.
The document contains solutions to math problems using Mathcad. Section 1 covers topics including: converting between Celsius and Fahrenheit, calculating pressure, force, work, energy, and cost analysis problems. Section 2 covers additional thermodynamics problems calculating things like internal energy, enthalpy, heat transfer, and phase changes.
Introduction to chemical engineering thermodynamics, 6th ed [solution]Pankaj Nishant
This document contains solutions to math problems involving concepts of thermodynamics, including calculations of work, heat, internal energy, enthalpy, and phase changes. Problem 1 calculates the work done in lifting a mass and the resulting internal energy change. Problem 2 determines the heat transferred and final temperature when water gains a small amount of heat. Problem 3 is a series of thermodynamic steps where the initial and final internal energies must sum to zero.
Heat transfer is the transfer of thermal energy between objects of differing temperatures. There are three main modes of heat transfer: conduction, convection, and radiation. Conduction involves the transfer of heat through direct contact of objects. Convection involves the transfer of heat by the circulation of fluids like gases and liquids. Radiation involves the transfer of heat through electromagnetic waves between objects not in contact.
1) The aluminum block absorbs more energy than the copper block when dropped into the calorimeter containing water. This is because aluminum has a higher specific heat than copper, so it takes more energy to raise the temperature of the aluminum block.
2) The final temperature of the water after the horseshoe and water reach equilibrium is 38°C. Using conservation of energy, the temperature of the horseshoe can be calculated.
3) It takes 63.97 minutes for a cup of water to boil in the microwave, assuming 50% of the microwave's 1200W power goes to heating the water. This estimate matches everyday experience of how long it takes for water to boil in a microwave.
Thermodynamics Assignment 02 contains calculations for various cycles of a steam power plant operating between 40 bar and 0.04 bar:
1) Carnot, simple Rankine, and modified Rankine cycles are analyzed. The modified Rankine cycle with superheat has the highest efficiency of 40.86% and lowest SSC of 2.4820 kg/kWh.
2) "Metallurgical limit" refers to the maximum safe pressures and temperatures a power plant's components can withstand without damage.
3) Implementing reheating in the Rankine cycle increases efficiency to 41.05% and lowers SSC to 2.4663 kg/kWh by utilizing the steam's initial high temperature again
This document contains a 16 question multiple choice mechanical engineering review problem set. It covers topics including: specific weight calculations, changes in weight due to elevation, pressure and force calculations for scuba diving, determining height using barometer readings, properties of gas mixtures, heat transfer between materials, gas turbine processes, combustion calculations, and thermodynamic processes including changes in temperature, pressure, volume, entropy and heat/work.
This document contains a tutorial for CHE2163 with 5 questions. It provides the relevant equations, assumptions, and step-by-step working to calculate various heat transfer rates. For question 1, it calculates the heat transfer rate from a module with air flowing over it. For question 2, it finds the initial and changing temperature of a heating plate. For question 3, it determines the heat loss from an uninsulated steam pipe. For questions 4 and 5, it analyzes heat transfer from pin fins and an underwater instrument pod respectively.
This document contains 20 multiple choice questions regarding concepts of thermodynamics and the second law of thermodynamics. It includes calculations of efficiency, COP, temperatures, enthalpies, and entropy involving heat engines and refrigerators. Key concepts covered are the Carnot cycle, Carnot efficiency, ideal gas processes, heat transfer, work, and isentropic processes.
This document discusses heat engines, refrigerators, and the second law of thermodynamics. It contains the following key points:
1. A heat engine transforms heat partly into work by a working substance undergoing a cyclic process, but is not 100% efficient due to the second law of thermodynamics.
2. The Carnot cycle involves two reversible isothermal processes and two reversible adiabatic processes, and represents the maximum possible efficiency for a heat engine.
3. Refrigerators and heat pumps move heat from a cold to a hot reservoir, but also require work input. Their performance is measured by the coefficient of performance.
The document proposes replacing pressure reducing valves (PRVs) with backpressure steam turbine-generators at a heat plant to increase efficiency. Currently, PRVs reduce steam pressure from 180 psig to 25 psig without recovering work. Installing a turbine would capture this pressure reduction as electricity. Calculations show the turbine cycle would achieve an efficiency of 77.2% versus 76.6% for the PRV cycle. Case studies at other universities found turbines reduced emissions by 1,200-2,000 metric tons annually and saved $120,000-275,000 per year. A pre-design analysis identified space in the plant basement as optimal for a turbine, which could fit through double doors and utilize existing steam piping.
Module 5 (properties of pure substance)2021 2022Yuri Melliza
This document discusses properties of pure substances and steam. It defines key terms like saturation temperature, saturation pressure, subcooled liquid, compressed liquid, saturated mixture, and superheated vapor. It also describes temperature-specific volume, temperature-entropy, and enthalpy-entropy diagrams. Sample problems are provided to calculate properties like quality, enthalpy, specific volume, power output, and mass flow rate using steam tables and the concepts introduced.
Module 6 (ideal or perfect gas and gas mixture) 2021 2022Yuri Melliza
This document summarizes key concepts related to ideal gases. It defines the characteristic gas equation and gas constant. It describes Boyle's, Charles', Avogadro's, and combined gas laws. It also covers specific heats at constant pressure and volume, the ratio of specific heats, entropy change, non-ideal gas behavior, compressibility factor, and includes sample problems applying these concepts.
1) The document defines various forms of energy including work, heat, internal energy, kinetic energy, potential energy, and enthalpy. It provides equations to calculate changes in these energy forms.
2) Two sample problems are included, one calculating potential energy change and velocity when a hammer is dropped, the other calculating drop height using given kinetic and potential energy changes.
This document contains 20 multiple choice problems related to mechanical engineering. The problems cover topics such as fluid mechanics, thermodynamics, heat transfer, and other mechanical engineering principles. They involve calculations related to things like tank volumes, pressure differences, flow rates, heat transfer between substances, and more. The questions provide relevant equations, known values, and ask the reader to determine unknown values or temperatures based on the given information.
This document discusses reversible work, irreversibility, and availability in thermodynamics. It defines reversible work as the maximum possible work for a process, calculated by assuming the process is reversible with no entropy production. Irreversibility is defined as the difference between reversible and actual work. Availability is the maximum reversible work that can be extracted from a system as it moves to equilibrium with its surroundings. The document provides equations for calculating reversible work, irreversibility, and availability. It also introduces exergy as a measure analogous to availability. Several examples apply these concepts to problems involving turbines, compressors, and nozzles.
[W f stoecker]_refrigeration_and_a_ir_conditioning_(book_zz.org)Mike Mentzos
- The document describes thermal principles and psychrometric concepts.
- It provides solutions to example problems involving state changes of water, heat transfer calculations, psychrometric chart readings, and enthalpy/humidity ratio determinations.
- Key concepts covered include the use of steam tables, Bernoulli's equation, psychrometric equations, and heat transfer relationships for convection and radiation.
An isobaric process is a constant pressure process where pressure (P) remains constant. Work (W) done is equal to pressure (P) times the change in volume (ΔV). For an ideal gas, the change in internal energy (ΔU) is equal to heat (Q) added.
An isometric process is a constant volume process where volume (V) remains constant. Work (W) done is zero since there is no change in volume. For any substance, the change in internal energy (ΔU) is equal to the heat (Q) added.
An isothermal process is a constant temperature process where temperature (T) remains constant. For an ideal gas, the ratio
1) The document discusses different types of thermodynamic processes including reversible, irreversible, isobaric, isochoric, isothermal, and adiabatic processes.
2) It provides examples of an ideal gas cycle consisting of isobaric, isochoric, and isothermal processes and calculates the work, internal energy change, heat, and efficiency for each process.
3) The heat capacities of ideal monoatomic and diatomic gases are derived from their internal energy changes, showing the relationship between heat capacity at constant volume and constant pressure.
solution manual to basic and engineering thermodynamics by P K NAG 4th editionChandu Kolli
- There are three main temperature scales: Celsius (C), Fahrenheit (F), and Kelvin (K)
- Celsius and Kelvin have the same increments but different reference points, with 0°C being the freezing point of water and 100°C being the boiling point, while 0K is absolute zero
- Fahrenheit has a different increment than the other two scales, with 32°F being the freezing point of water and 212°F being the boiling point
- Conversions between the scales can be done using the following relationships:
(C − 0°C) = (F − 32°F) × 5/9 = (K − 273
This document provides links to download university textbooks and solutions manuals for many books for free. It encourages following the website on social media and visiting the site to download files freely. The rest of the document consists of sample math problems and solutions.
1. Thermodynamics covers basic concepts including open and closed systems, state functions, and the laws of thermodynamics.
2. The zeroth law defines thermal equilibrium and allows for the definition of temperature.
3. The first law concerns the conservation of energy and establishes the concept of internal energy. Heat and work are both means of transferring energy.
This document provides an overview of basic thermodynamics concepts including:
- The objectives of understanding the laws of thermodynamics and their constants.
- Definitions of perfect gases and their properties of pressure, volume, and temperature.
- Explanations of Boyle's Law, Charles' Law, and the Universal Gas Law.
- Introduction of specific heat capacity at constant volume and constant pressure.
- Examples demonstrating applications of the gas laws and calculations involving specific heat.
The document describes various thermodynamic processes including:
1) Isometric processes which occur at constant volume with changes in temperature and pressure.
2) Isobaric processes which occur at constant pressure with changes in volume and temperature.
3) Isothermal processes which occur at constant temperature with changes in pressure and volume.
4) Isentropic processes which are reversible adiabatic processes with no heat transfer and constant entropy.
Several examples are provided to calculate temperature, pressure, volume, work, heat, internal energy and entropy changes during these different thermodynamic processes.
The document describes the second law of thermodynamics and reversible processes involving perfect gases on temperature-entropy (T-s) diagrams. It discusses:
1) Constant pressure, volume, temperature, and adiabatic processes on T-s diagrams, with constant pressure lines sloping more steeply than constant volume lines.
2) Analyzing a example problem involving a constant pressure expansion of nitrogen gas, calculating work, heat, entropy change, and sketching the process on a T-s diagram.
3) The relationships between pressure, volume, temperature and entropy for perfect gases during various reversible thermodynamic processes.
The document discusses the second law of thermodynamics and various reversible processes on a temperature-entropy (T-s) diagram for a perfect gas. It defines:
1) Constant pressure, volume, temperature, adiabatic, and polytropic processes on a T-s diagram.
2) Equations to calculate work, heat, and entropy change for constant pressure, volume, and temperature processes.
3) Provides an example problem calculating properties of air undergoing two processes - constant volume heating and constant pressure cooling.
This document contains 57 comments on the textbook "Thermodynamics: An Engineering Approach, 7th ed." by Yunus Cengel and Michael Boles. The comments note typos, suggest clarifications or corrections to figures, equations, or explanations. They also recommend revising some terminology for consistency or accuracy. The reviewer aims to improve the quality and precision of information provided in the textbook.
Module 4 (first law of thermodynamics) 2021 2022Yuri Melliza
The document discusses the first law of thermodynamics, also known as the law of conservation of energy. It states that energy cannot be created or destroyed, only converted from one form to another. The first law is expressed mathematically as: Energy In - Energy Out = Change in Stored Energy. The document provides examples of applying the first law to closed and open systems, including deriving equations relating heat, work, internal energy, and other properties. It also includes examples solving thermodynamics problems for systems like turbines, compressors, and nozzles.
This document contains a thermodynamics tutorial with 11 questions covering concepts like states of water and steam, properties from steam tables, processes involving heat and work. Key concepts covered include determining the state of water/steam based on pressure and other properties, calculating heat, work and internal energy change using the first law of thermodynamics for various processes involving water and steam.
This document contains multiple problems involving ideal gas processes. The first problem describes a steady flow compressor handling nitrogen with known intake conditions and discharge pressure. It asks to determine the final temperature and work for two process types. The second problem involves air in a cylinder being compressed in a polytropic process with known initial and final pressures and temperatures. It asks to determine the work and heat transfer. The third problem describes a gas turbine expanding helium polytropically and asks to determine the final pressure, power produced, heat loss, and entropy change.
The document summarizes an experiment on a multi-pass counter-flow liquid-to-liquid heat exchanger. It includes readings from 4 trials measuring inlet and outlet temperatures on the hot and cold sides. Calculations were shown to determine the heat transfer (q) and overall heat transfer coefficient (U) for each trial. While q transferred was expected to be equal between the hot and cold sides, some trials showed a deviation, possibly due to heat loss. The heat exchanger performance improved with increasing flow rate as both q and U increased.
This document discusses heat engines, refrigerators, and the second law of thermodynamics. It contains the following key points:
1. A heat engine transforms heat partly into work by a working substance undergoing a cyclic process, but is not 100% efficient due to the second law of thermodynamics.
2. The Carnot cycle involves two reversible isothermal processes and two reversible adiabatic processes, and represents the maximum possible efficiency for a heat engine.
3. Refrigerators and heat pumps move heat from a cold to a hot reservoir, but also require work input. Their performance is measured by the coefficient of performance.
The document proposes replacing pressure reducing valves (PRVs) with backpressure steam turbine-generators at a heat plant to increase efficiency. Currently, PRVs reduce steam pressure from 180 psig to 25 psig without recovering work. Installing a turbine would capture this pressure reduction as electricity. Calculations show the turbine cycle would achieve an efficiency of 77.2% versus 76.6% for the PRV cycle. Case studies at other universities found turbines reduced emissions by 1,200-2,000 metric tons annually and saved $120,000-275,000 per year. A pre-design analysis identified space in the plant basement as optimal for a turbine, which could fit through double doors and utilize existing steam piping.
Module 5 (properties of pure substance)2021 2022Yuri Melliza
This document discusses properties of pure substances and steam. It defines key terms like saturation temperature, saturation pressure, subcooled liquid, compressed liquid, saturated mixture, and superheated vapor. It also describes temperature-specific volume, temperature-entropy, and enthalpy-entropy diagrams. Sample problems are provided to calculate properties like quality, enthalpy, specific volume, power output, and mass flow rate using steam tables and the concepts introduced.
Module 6 (ideal or perfect gas and gas mixture) 2021 2022Yuri Melliza
This document summarizes key concepts related to ideal gases. It defines the characteristic gas equation and gas constant. It describes Boyle's, Charles', Avogadro's, and combined gas laws. It also covers specific heats at constant pressure and volume, the ratio of specific heats, entropy change, non-ideal gas behavior, compressibility factor, and includes sample problems applying these concepts.
1) The document defines various forms of energy including work, heat, internal energy, kinetic energy, potential energy, and enthalpy. It provides equations to calculate changes in these energy forms.
2) Two sample problems are included, one calculating potential energy change and velocity when a hammer is dropped, the other calculating drop height using given kinetic and potential energy changes.
This document contains 20 multiple choice problems related to mechanical engineering. The problems cover topics such as fluid mechanics, thermodynamics, heat transfer, and other mechanical engineering principles. They involve calculations related to things like tank volumes, pressure differences, flow rates, heat transfer between substances, and more. The questions provide relevant equations, known values, and ask the reader to determine unknown values or temperatures based on the given information.
This document discusses reversible work, irreversibility, and availability in thermodynamics. It defines reversible work as the maximum possible work for a process, calculated by assuming the process is reversible with no entropy production. Irreversibility is defined as the difference between reversible and actual work. Availability is the maximum reversible work that can be extracted from a system as it moves to equilibrium with its surroundings. The document provides equations for calculating reversible work, irreversibility, and availability. It also introduces exergy as a measure analogous to availability. Several examples apply these concepts to problems involving turbines, compressors, and nozzles.
[W f stoecker]_refrigeration_and_a_ir_conditioning_(book_zz.org)Mike Mentzos
- The document describes thermal principles and psychrometric concepts.
- It provides solutions to example problems involving state changes of water, heat transfer calculations, psychrometric chart readings, and enthalpy/humidity ratio determinations.
- Key concepts covered include the use of steam tables, Bernoulli's equation, psychrometric equations, and heat transfer relationships for convection and radiation.
An isobaric process is a constant pressure process where pressure (P) remains constant. Work (W) done is equal to pressure (P) times the change in volume (ΔV). For an ideal gas, the change in internal energy (ΔU) is equal to heat (Q) added.
An isometric process is a constant volume process where volume (V) remains constant. Work (W) done is zero since there is no change in volume. For any substance, the change in internal energy (ΔU) is equal to the heat (Q) added.
An isothermal process is a constant temperature process where temperature (T) remains constant. For an ideal gas, the ratio
1) The document discusses different types of thermodynamic processes including reversible, irreversible, isobaric, isochoric, isothermal, and adiabatic processes.
2) It provides examples of an ideal gas cycle consisting of isobaric, isochoric, and isothermal processes and calculates the work, internal energy change, heat, and efficiency for each process.
3) The heat capacities of ideal monoatomic and diatomic gases are derived from their internal energy changes, showing the relationship between heat capacity at constant volume and constant pressure.
solution manual to basic and engineering thermodynamics by P K NAG 4th editionChandu Kolli
- There are three main temperature scales: Celsius (C), Fahrenheit (F), and Kelvin (K)
- Celsius and Kelvin have the same increments but different reference points, with 0°C being the freezing point of water and 100°C being the boiling point, while 0K is absolute zero
- Fahrenheit has a different increment than the other two scales, with 32°F being the freezing point of water and 212°F being the boiling point
- Conversions between the scales can be done using the following relationships:
(C − 0°C) = (F − 32°F) × 5/9 = (K − 273
This document provides links to download university textbooks and solutions manuals for many books for free. It encourages following the website on social media and visiting the site to download files freely. The rest of the document consists of sample math problems and solutions.
1. Thermodynamics covers basic concepts including open and closed systems, state functions, and the laws of thermodynamics.
2. The zeroth law defines thermal equilibrium and allows for the definition of temperature.
3. The first law concerns the conservation of energy and establishes the concept of internal energy. Heat and work are both means of transferring energy.
This document provides an overview of basic thermodynamics concepts including:
- The objectives of understanding the laws of thermodynamics and their constants.
- Definitions of perfect gases and their properties of pressure, volume, and temperature.
- Explanations of Boyle's Law, Charles' Law, and the Universal Gas Law.
- Introduction of specific heat capacity at constant volume and constant pressure.
- Examples demonstrating applications of the gas laws and calculations involving specific heat.
The document describes various thermodynamic processes including:
1) Isometric processes which occur at constant volume with changes in temperature and pressure.
2) Isobaric processes which occur at constant pressure with changes in volume and temperature.
3) Isothermal processes which occur at constant temperature with changes in pressure and volume.
4) Isentropic processes which are reversible adiabatic processes with no heat transfer and constant entropy.
Several examples are provided to calculate temperature, pressure, volume, work, heat, internal energy and entropy changes during these different thermodynamic processes.
The document describes the second law of thermodynamics and reversible processes involving perfect gases on temperature-entropy (T-s) diagrams. It discusses:
1) Constant pressure, volume, temperature, and adiabatic processes on T-s diagrams, with constant pressure lines sloping more steeply than constant volume lines.
2) Analyzing a example problem involving a constant pressure expansion of nitrogen gas, calculating work, heat, entropy change, and sketching the process on a T-s diagram.
3) The relationships between pressure, volume, temperature and entropy for perfect gases during various reversible thermodynamic processes.
The document discusses the second law of thermodynamics and various reversible processes on a temperature-entropy (T-s) diagram for a perfect gas. It defines:
1) Constant pressure, volume, temperature, adiabatic, and polytropic processes on a T-s diagram.
2) Equations to calculate work, heat, and entropy change for constant pressure, volume, and temperature processes.
3) Provides an example problem calculating properties of air undergoing two processes - constant volume heating and constant pressure cooling.
This document contains 57 comments on the textbook "Thermodynamics: An Engineering Approach, 7th ed." by Yunus Cengel and Michael Boles. The comments note typos, suggest clarifications or corrections to figures, equations, or explanations. They also recommend revising some terminology for consistency or accuracy. The reviewer aims to improve the quality and precision of information provided in the textbook.
Module 4 (first law of thermodynamics) 2021 2022Yuri Melliza
The document discusses the first law of thermodynamics, also known as the law of conservation of energy. It states that energy cannot be created or destroyed, only converted from one form to another. The first law is expressed mathematically as: Energy In - Energy Out = Change in Stored Energy. The document provides examples of applying the first law to closed and open systems, including deriving equations relating heat, work, internal energy, and other properties. It also includes examples solving thermodynamics problems for systems like turbines, compressors, and nozzles.
This document contains a thermodynamics tutorial with 11 questions covering concepts like states of water and steam, properties from steam tables, processes involving heat and work. Key concepts covered include determining the state of water/steam based on pressure and other properties, calculating heat, work and internal energy change using the first law of thermodynamics for various processes involving water and steam.
This document contains multiple problems involving ideal gas processes. The first problem describes a steady flow compressor handling nitrogen with known intake conditions and discharge pressure. It asks to determine the final temperature and work for two process types. The second problem involves air in a cylinder being compressed in a polytropic process with known initial and final pressures and temperatures. It asks to determine the work and heat transfer. The third problem describes a gas turbine expanding helium polytropically and asks to determine the final pressure, power produced, heat loss, and entropy change.
The document summarizes an experiment on a multi-pass counter-flow liquid-to-liquid heat exchanger. It includes readings from 4 trials measuring inlet and outlet temperatures on the hot and cold sides. Calculations were shown to determine the heat transfer (q) and overall heat transfer coefficient (U) for each trial. While q transferred was expected to be equal between the hot and cold sides, some trials showed a deviation, possibly due to heat loss. The heat exchanger performance improved with increasing flow rate as both q and U increased.
This document describes experiments conducted to test the feasibility of using a thermoelectric cooler powered by a bicycle dynamo to heat and cool drinks. It was found that the power output of 60W from cycling at 15km/h was not sufficient to cool a drink to the desired temperature within an optimum time. Additional experiments showed heating 350ml of water would take a very long time and cooling was hindered by lack of forced convection. Insulation would be needed for practical application of this thermoelectric cooler system.
The document discusses different heat engines and their thermodynamic cycles, including:
- Stirling engines, which use a gas that remains inside the engine during the cycle. Their efficiency can approach that of a Carnot engine.
- Internal combustion engines that operate on the Otto cycle, using combustion to heat the gas rather than a separate heat reservoir. Their theoretical efficiency is lower than the Carnot limit.
- The third law of thermodynamics, which states that the entropy of a perfect crystal at 0K is zero, implying it is impossible to reach absolute zero temperature or 100% efficiency. Entropy is also discussed from statistical and experimental perspectives.
1) The document discusses the processes involved in a Carnot cycle for an ideal gas, including isothermal expansion and compression and adiabatic processes.
2) It examines the efficiencies of Carnot engines and refrigerators, noting that engines are more efficient when the temperature difference is large, while refrigerators are more efficient when the temperature difference is small.
3) It then shows how assuming the heat engine statement of the second law is false would allow using a refrigerator to violate the refrigerator statement of the second law by creating a perpetual motion machine.
This document provides the solutions to homework problems assigned in a thermodynamics course. It summarizes the key steps and conclusions for 6 problems involving concepts like approximations for enthalpy of compressed water, properties of water at different temperatures and pressures, heat transfer for a refrigerant, the Rankine cycle diagram, and properties of propane using different equations of state. The last problem calculates about 700 kJ/kg of work done by steam expanding adiabatically between two states.
This chemistry problem set covers topics in thermodynamics including:
1) Calculating the isothermal compressibility of ideal gases and van der Waals gases.
2) Finding work, heat, internal energy and enthalpy change for ideal gas processes including isothermal compression/expansion and cooling.
3) Deriving an expression for work during isothermal reversible expansion of a van der Waals gas.
4) Calculating enthalpy changes for hydrogenation reactions of unsaturated hydrocarbons.
5) Deriving a relationship between initial and final temperatures for adiabatic ideal gas processes.
This document discusses the first law of thermodynamics for closed systems. It defines the first law as the law of conservation of energy, where energy can be transformed but not created or destroyed within a closed system. The energy balance equation for closed systems is presented, accounting for heat, work, internal energy, kinetic energy and potential energy. Several examples are worked through applying the first law to calculate changes in internal energy, heat transfer and entropy change for closed thermodynamic processes.
Methods of handling Supply air in HVAC Yuri Melliza
The document provides examples of calculations for air conditioning systems that use outside air and recirculated air. It determines parameters such as mass flow rates, cooling and heating capacities, and condensate removal. For the last example, it calculates:
a) The supply air mass flow rate is 14.2 kg/sec
b) The recirculated air mass flow rate is 8.51 kg/sec
c) The outside air mass flow rate is 5.7 kg/sec
d) The condensate removal rate is 0.056 kg/sec
e) The refrigeration capacity of the AC unit is 90.21 tons
The document contains 10 examples solving thermodynamics problems involving concepts like the first law of thermodynamics, ideal gas law, steady state conditions, heat transfer, work done, efficiency and refrigeration cycles. The last example involves a reversible heat engine and refrigerator operating between different temperature reservoirs. It is determined that the heat rejection to the 40°C reservoir is 5539 kJ.
This document provides information about a lecture on reheat and intercooling in gas turbine systems. It includes:
- An explanation of the concepts and purposes of using reheat and intercooling in gas turbines.
- An example problem calculating efficiency and mass flow rate for a gas turbine cycle with reheat and intercooling.
- Diagrams of gas turbine cycles with the different enhancements labeled.
The document summarizes key concepts from thermodynamics including:
- The Zeroth Law of Thermodynamics states that if two systems are each in thermal equilibrium with a third system, they are in thermal equilibrium with each other.
- The First Law of Thermodynamics states that energy cannot be created or destroyed, only changed in form. For a closed system, the net energy transferred as heat and work equals the net change in internal energy.
- Derivations from the First Law relate internal energy change, enthalpy change, and specific heat capacities for constant volume and constant pressure processes. Relationships between specific heat capacities are also derived for ideal gases.
- Several examples show applying the
Jamuna Oil Comany Ltd recruitment question & ans (5th july 2018)Alamin Md
This document contains information about an exam for Jamuna Oil Company Ltd, including:
- The exam date is July 5th and will take place from 3:30-5:00 PM at BUET.
- The exam contains 50 multiple choice questions worth 25 marks and 12 departmental questions worth 50 marks, for a total of 75 marks.
- Sample multiple choice and departmental questions are provided.
- Solutions to two of the departmental questions are provided, calculating the thickness required for a pressure vessel and the velocity of a pendulum bob.
- The document requests feedback on any mistakes found in the solutions.
The document provides examples and problems related to thermodynamics. Example 1 involves calculating the final temperature of a system consisting of a brass block and ice water. Example 2 calculates the root mean square speeds of nitrogen and oxygen molecules in air. Example 3 determines the new temperature of an ideal gas that is compressed.
The document discusses the third law of thermodynamics and the entropy of perfect crystals. It states that the entropy of a pure perfect crystal is zero at absolute zero temperature (0 K). A perfect crystal is one where every molecule is identical and aligned perfectly throughout. At higher temperatures, molecular motion causes disorder and entropy increases from zero. The document provides an example calculation of the thermodynamic properties of an ideal diesel cycle using air as the working fluid.
Thermodynamic Chapter 3 First Law Of ThermodynamicsMuhammad Surahman
This document provides an overview of the first law of thermodynamics for closed systems. It defines key terms like internal energy, kinetic energy, and potential energy. It presents the general energy balance equation for closed systems undergoing various processes like constant volume, constant pressure, or adiabatic. Example problems demonstrate applying the first law to calculate changes in internal energy or heat transfer. The document also discusses thermodynamic cycles and how the first law applies to systems that return to their initial state.
Solution Manual for Physical Chemistry – Robert AlbertyHenningEnoksen
https://www.book4me.xyz/solution-manual-physical-chemistry-alberty/
Solution Manual for Physical Chemistry - 6th Edition
Author(s) : Robert A. Alberty
This solution manual include all chapters of textbook (1 to 21).
Let's Integrate MuleSoft RPA, COMPOSER, APM with AWS IDP along with Slackshyamraj55
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thermal science Ch 04
1. FOUR
The First Law of Thermodynamics
4.1FE
B
4.2FE
A
4.3FE
C
Because the container is insulated we assume the heat transfer is zero. For a rigid container:
Q electric v
W
mc T
( 400 0 60) 10 10.085(T2
20). T2
29 C
Table B.2 was used to find cv .
4.4FE
A
For a rigid container the boundary work is zero:
q w 2 1 2961
u u u
2553.6
407 kJ/kg
where we used v2 1 and interpolated for
v
u2 in Table C.3 as follows:
v2 v1
0.4625
2961 kJ/kg
u2
T2 400
C
4.5FE
C
For this adiabatic process we use
(k / k
1)
p
= m(u2
W
u
and T2 1 2
T
293 0.2857
6
488.9 K
p1
v (T2 1 )
W mc
T
2 0.717
(489
293) kJ
281
4.6FE
C
This is an adiabatic process so that
Then
= m(u2 u v (T2 1 )
W
W mc
T
.
( 10 100 10 50) (5 60) 2 717 T
T 313.8
C
V2
1 0.287 ln 0.5
373
74.2 kJ .
V1
The minus sign indicates that heat is rejected.
Q
W mRT ln
4.7FE
D
4.8FE
D
mcv
W
T 5 0.717
100 358.5 kJ
4.9FE
A
p2 = p1 = 400 kPa. Use Table C.3 for this superheated steam:
V1
1 0.7726
0.7726 m 3 .
V
2
V 1
2 0.05
0.6726 m 3 and v2
0.6726 m 3 / kg.
Interpolate at 400 kPa and 0.6726 m3/kg:
0.7726
0.6726
T2 400
(400
300)
315 C
0.7726
0.6548
4.10FE D
Use Table C.3 at 315 and interpolate to find h2:
C
Q
m h 1 (3098
3273) kJ
175
31
2. 4.11FE A
Use Tables C.3 and C.2 with W = 0:
Q (u3 2 )
m
u
1 (1245
2829) = 1584 kJ
We interpolated at 315 in Table C.3 to find u2 = 2829 kJ/kg and used v2 = v3 =
C
0.672 m3/kg which is in the quality region. Then using Table C.2,
x3 = 0.3963 and u3 = 1245 kJ/kg.
4.12FE D
If T = const,
Q
W mRT ln
4.13FE C
p1
100
10 0.287
333 ln
1987 kJ
p2
800
For p = const,
q h2 1
h 3273
604.7
2668 kJ/kg
4.14FE C
For p = const,
Q (h2 1 ).
m
h
170 h2
1 (
3108).
h2
3278 and T2
402 C
We interpolated at 320 in Table C.3 to find h1 and then to find T2.
C
4.15FE D
For p = const,
W ( v2 1 )
mp
v
1 400 (0.7749
0.6784)
38.6 kJ
We interpolated in Table C.3 to find v1 and
v2 and the two temperatures.
4.16FE D
In Table C.3 we observe that u =2949 kJ/kg is between p = 1.6 MPa and 1.8 MPa but
closer to 1.6 MPa so the closest answer is 1.7 MPa.
4.17FE C
Interpolate in Table C.3 and find h = 3459 kJ/kg.
4.18FE C
Use a central difference for accuracy:
3213.6
h
2960.7
cp
2.53 kJ/kg
蚓
T
400
300
4.19FE C
We use q because the mass is not known:
q p (T2 1 )
c
T
2.254
300 676.2 kJ/kg
4.20FE B
The copper gains heat and the water losses an equal amount of heat:
mc c p ,c (T2 w c p , w (30 2 ).
0) m
T
4.21FE B
20
0.38 2
T 10 4.18 2 ). T2
(30 T
25.4
C
First the ice melts and then the ice water heats up:
mi ( i p , w i ) w c p , w w .
h c
T
m
T
10
[333
4.18(T2
0)] 60 4.18 2 ) .
(20 T
4.22FE C
For process 1:
Q = W =100 = a.
For process 2:
b = 40 + 60 = 100.
W or 100 .
100 40 100 60 c
Q
32
c
80
T2 = 5.85蚓
3. 4.23FE B
For the T = const process:
v
0.8
w1-2 RT1 ln 2
0.287
278.7
ln
166.4 kJ/kg
v1
0.1
We used pv = RT to find T1 = (800)(0.1)/0.287 = 278.7
C.
4.24FE C
For the adiabatic process:
3-1 v (T1 3 )
w
c
T
0.717(278.7
121.3) kJ/kg
113
k
1
0.4
v
0.1
where T3 1 1
T
278.7
121.3 K .
v
0.8
3
4.25FE D
The net heat transfer is equal to the net work. The work for the 2-3 process is zero.
Therefore, using the results from the above two problems, the net q is
qnet w1-2 w2-3 3-1 = 166.4
w
113 = 53 kJ/kg
4.26FE A
k k
1/
4.27FE D
p
T2 1 2
T
p
1
4.28FE B
For the Q = 0 process:
293 0.2857
8
530.7 K or 258 .
C
W mcv (T2 1 )
T
2 0.717
(622
373) 358 kJ
k k
1/
p
where T2 1 2
T
p
1
4.29FE A
0.2857
373 6
622.3 K .
100 400
Q p or
mc T
15 T
0.3 60 (10 20 3 1.2) 1.00
3600
T 14.3 C
We assumed a p = const process since there are openings where air is allowed to escape
or enter.
4.30FE D
4.31FE D
This term includes the flow work rate term due to the pressure force moving due to the
velocity.
4.32FE B
4.33FE C
4.34FE D
V 2 12
V
202 2
200
0 p 2
c T
.
1.00
T
.
.
T 19.8 C
2
2
1000
The factor 1000 is needed to convert J to kJ since cp includes the kJ unit.
h2 h1
3500.9 kJ/kg.
T2 = 508
C
p2 MPa. Interpolate in Table C.3 at 0.8 MPa and find
0.8
33
4. k
1/1
4.35FE D
p
T2 1 2
T
p
1
4.36FE B
p
1
m A1V1 1 A1V1 .
RT1
4.37FE D
The heat that leaves the steam enters the cooling water:
0.2857
293 8
530.7 K or 258 .
C
100
2
2 1
0.1 V
0.287
293
V1
53.5 m/s
h mw
ms s c p w
T
Use hfg from Table C.2:
10
2373
400 4.18 T .
4.38FE C
.
T 14.2 C
Assume p = const in the tube:
T 100 1.00 (25 20) 500 kJ/min
Q mc p
4.39FE C
For liquid:
p
100
m 4 6000
WP
23.6 kW
1000
4.40FE B
The enthalpy of the fluid when it enters the tank is assumed equal to the enthalpy
of the fluid in the pipe on the upstream side of a valve connecting the tank and pipe.
4.41FE C
Q1 c1 A1 1 1
h
T t
and Q2 c 2 A2 2 2 . But, Q2 1 , h2 h1 , 2 1 , and A2 A1.
h
T t
Q
2
T
T
So,
t2 Q2 h1 A1 1 1
T
t
. t2 1 s
20
t1 Q1h2 A2 2 2
T
2
4.42FE C
q
q B or k A
A
TA2 A1
T
T
T
B B 2 B1
k
LA
LB
or
If LB LA / 2 and TB1 TA1 and TB 2 TA2
2
2
TA2 A1 2(TA2 A1 )
T
T
LA
2 LA
and 1 = 1 so C is the answer.
34
TA2 A1 TB 2 B1
T
T
LA
4LB
then
5. 4.1
Qnet net
W
100 m
9.81 m
3.
3.398 kg
4.2
1
1
Qnet net . Qnet mV 2
W
1500 2
30
675000 J
2
2
4.3
0.
F
pA patm A
mg
0
p 2
0.05 10 9.81 000 2. p 490 Pa
100
0.05
112
Q .
W
U
U 300 (112 490 2 )
0.05
0.2 123.3 J
4.4
24)] ft-lbf or
U Q W
2 778 [ 600 (2 1/
381
0.49 Btu
4.5
a) 20 kJ E2 E2 kJ
E Q W
5 15
7.
22
b) Q 6 kJ. E2 E2 kJ
E W
3 3
8 6.
14
c) W E 40 25 kJ. kJ
Q
(30 15)
E 30 15 15
d) W E kJ. 20 1 . E1 kJ
Q
10 20
30
10 E
10
e) Q kJ. kJ
E W
8 6 10
4
E
8 6
14
4.6
a 1-2 1-2 U kJ
W
Q
200 0
200
( )cycle 0. 0
U
c 400 1200 c
0.
800 kJ
b 2-3 2-3 U kJ
W
Q
800 800 0
d 3-4 3-4 400
Q
U W
600 1000 kJ
e 4-1 4-1 U
W
Q
0 ( 1200)
1200 kJ or we could use Qnet net
W
4.7
a 1-2 1-2 200 kJ
Q
E W
100 100
( )cycle 100
E
0.
c 200 c kJ
0.
100
b 2-3 2-3 kJ
Q
E W
100 50 50
d 3-1 3-1 E
W
Q
100 ( 200)
300 kJ or we could use Qnet net
W
35
6. 4.8
First, find the initial pressure in the cylinder:
W
60
9.81
p patm
000 000 Pa
100
119
A
2
0.1
Now, apply the 1st law adding the work of the pressure force and the work to
compress the spring:
1
Q .
W
U
Q pA Kx 2 )
(
h
U
2
1
200
U
(119 000 2
0.1 0.05 000 2 ) J
50
0.05
49
2
4.9
W1-2 6 000 J.
VI t
5 (20 60) 36
Q1-2 1-2 2-1 W2-1 )cycle 0. Q2-1 1-2 kJ
W
Q
( U
W
36
4.10
Q 000 600 J or 377.6 kJ
U W
400
12 3 (6 60 60) 377
4.11
Q U 300 000 84 000 J or kJ
W
12 10 (30 60)
84
4.12
Q
U W 8000 /1055
110 15 (2 60 60)
3260 Btu
w e t f t 15 cne s tBus
hr h a o 05 ovr Jo t .
e e cr
t
’
4.13
The only transfer of energy across the boundary of the system is via the
electrical wires leading to the refrigerator. The 1st law is
Q . ) 2
W
U
U
W
( VI t
0.746 2686 kJ
(30 60)
V
0.3
(u2 1 )
u
(3655.3
2621.9)
1505 kJ
v
0.206
4.14
Q W
m u
4.15
a) Q W
m u
V
(u2 1 )
u
v
0.2
1000
{u2
[669.9
0.8(2567.4
669.9)]}
0.0011
0.8(.3157
.0011)
u2
3452 kJ/kg and v2
0.2528 m 3 /kg . We must find where in Table C.3
this state exists. After checking the table we interpolate for the following:
p2 MPa, v2
1.5
0.2528, u2
3216 : T2
556 C
686
C
T2
p2 MPa, v2
2.0
0.2528, u2
3706 : T2
826 C
b) TK solution:
Rule Sheet
;This is a closed, constant-volume system so no work is done and the first law is
Q = m * (u2 -u1)
; First law
V1 = m * v1
; Definition of specific volume
v2 = v1 ; For a constant-volume system
; Steam tables based on NBS/NRC Steam Tables by Lester Haar, ; John S. Gallagher, and George S.
Kell, Hemisphere Publishing Corp., 1984.
; STEAM8SI.TKW
36
8. At T2 F :
150
vg
96.99.
slightly superheat
975
118 941.3 x2 .
x2
0.912
State 2 lies between 140 F and 150 F . Since the quality is insensitive to the
internal energy, we find T2 such that v2
101.5 ft 3 /lbm:
101.5
96.99
T2
150
F
10 148
122.88
96.99
A temperature slightly less than this provides the answer: T2 F .
147
b) TK solution:
Rule Sheet
;This is a closed, constant-volume system so no work is done and the first law is
Q = m * (u2 -u1)
; First law
V1 = m * v1
; Definition of specific volume
v2 = v1
; For a constant-volume system
; Steam tables based on NBS/NRC Steam Tables by Lester Haar, ; John S. Gallagher, and George
S. Kell, Hemisphere Publishing Corp., 1984.
; STEAM8SI.TKW
Variable Sheet
Input Name Output
Unit
Comment
`
Thermal Sciences, Potter & Scott
*STEAM8E.TKW Steam, 1-8 States, English units
P4-16.tkw
Problem 4.16
120
T1
F
Temperature
p1
1.69
psi
Pressure
h1
601
B/lbm
Enthalpy
s1
1.05
B/(lbm*R) Entropy
v1
101
ft^3/lbm
Specific Volume
0.5
x1
Quality
phase1 'SAT
Phase
101
8
2
T2
p2
h2
s2
v2
x2
phase2
Q
m
u2
u1
V1
144
3.21
1030
1.73
F
psi
B/lbm
B/(lbm*R)
ft^3/lbm
0.913
'SAT
0.0197
975
569
B
lbm
B/lbm
B/lbm
ft^3
Temperature (starting guess needed)
Pressure
Enthalpy
Entropy
Specific Volume
Quality
Phase
*thunits.tkw Units for thermodynamics
Heat transfer
Mass of steam
Internal energy
Internal energy
Initial volume
38
9. V 100 / 1728
0.01633 lbm
v
3.544
120
144
100
Q m(u2 1 )
u
W 0.01633{1371.5
[312.3
0.95(796)]}
(7.208
)
778
1728
6.28 Btu
4.17 v1
0.0179
0.95(3.73
0.0179)
3.544. m
4.18
First, find the mass: m
V 1 0.004
0.0302 kg . The 1st law is Q m(u2 1 )
W
u
v1
0.1325
which takes the form
40
1500
0.03019( v2
0.1325)
0.0302(u2
2598) or 124.4 v2
45.3
0.0302u2
The above equation has two variables but the steam tables represent the 2nd equation.
'
This requires trial-and-error (let u2 be the u2 from the above equation):
'
At T2
600, p2 : v2
1.5
.2668, u2
3294. u2
3720
'
At T2
780, p2 : v2
1.5
.3230, u2
3622. u2
3636 T2
785 C
'
At T2
800, p2 : v2
1.5
.3292, u2
3627. u2
3720
4.19
100 /1728
a) Q m(h2 1 )
h
{1531.5
[312.7
0.95(878.5)]}
6.27 Btu
3.544
b) TK solution:
Rule Sheet
;This is a closed system, so the first law, Q =m* (u2 - u1 + INT pdv), at constant pressure
becomes
; Q = m * (u2 - u1 + p1 * (v2 - v1) = m * (u2 - u1 + p2*v2 - p1 *v1) or
Q = m * (h2 - h1)
v1 = V1/m
; Definition of v1
; Steam tables based on NBS/NRC Steam Tables by Lester Haar, ; John S. Gallagher, ;and George
S. Kell, Hemisphere Publishing Corp., 1984. ; STEAM8E.TKW
Variable Sheet
Input Name
`
120
0.95
1000
120
T1
p1
h1
s1
v1
x1
phase1
T2
p2
Output
341
1150
1.53
3.54
'SAT
Unit
Comment
Thermal Sciences, Potter & Scott
*STEAM8E.TKW Steam, 1-8 States, English units
P4-19.tkw Problem 4.19
F
Temperature
psi
Pressure
B/lbm
Enthalpy
B/(lbm*R) Entropy
ft^3/lbm
Specific Volume
Quality
Phase
F
Temperature
psi
Pressure
39
10. h2
s2
v2
x2
phase2
100
4.20
1530
B/lbm
1.9
B/(lbm*R)
7.2
ft^3/lbm
'mngless
'SH
Q
V1
m
6.28
B
in^3
lbm
0.0163
Enthalpy
Entropy
Specific Volume
Quality
Phase
*THUNITS.TKW Units for thermo
Heat transfer
Initial volume
Mass of steam
0.004
40
(h2
2797).
h2 4120 kJ/kg .
0.1325
At p2 MPa we interpolate T2 787
1.5
C.
b) TK solution:
a) Q m .
h
Rule Sheet
;This is a closed system, so the first law, Q =m* (u2 - u1 + INT pdv) constant pressure becomes
; Q = m * (u2 - u1 + p1 * (v2 - v1) = m * (u2 - u1 + p2*v2 - p1 *v1) or Q = m * (h2 - h1)
v1 = V1/m
; Definition of v1
; Steam tables based on NBS/NRC Steam Tables by Lester Haar, ; John S. Gallagher, and George
S. Kell, Hemisphere Publishing Corp., 1984.
; STEAM8SI.TKW
Variable Sheet
Input
Name
Output
Unit
`
200
1500
4
1500
40
T1
p1
h1
s1
v1
x1
phase1
V1
m
T2
p2
h2
s2
v2
x2
phase2
Q
2800
6.45
0.132
'mngless
'SH
0.0302
785
4120
8.28
0.324
'mngless
'SH
C
kPa
kJ/kg
kJ/(kg*K)
m^3/kg
L
kg
C
kPa
kJ/kg
kJ/(kg*K)
m^3/kg
kJ
Comment
Thermal Sciences, Potter & Scott
*STEAM8SI.TKW Steam, 1-8 States, SI Units
P4-20.tkw Problem 4.20
Temperature
Pressure
Enthalpy
Entropy
Specific Volume
Quality
Phase
Initial volume
Mass of steam
Temperature (starting guess is needed)
Pressure
Enthalpy
Entropy
Specific Volume
Quality
Phase
*THUNITS.TKW Units for thermo
Heat transfer
40
11. 4.21
From Table D.1 we find h1
62 0.8(190.3) 214.2 kJ/kg. Then
q h2 1 .
h
80 h2
214.2.
h2 294.2 kJ/kg
We interpolate at p2 0.4 MPa in Table D.3:
294.2
291.8
T2
(10)
50 52.5
C
301.5
291.8
4.22
a) Q m(h2 1 ) 2(3278
h
209)
6138 kJ
b) Q m(h2 1 ) 2(4044
h
3278)
1531 kJ
c) TK solution:
Rule Sheet
;This is a closed system, so the first law, Q =m* (u2 - u1 + INT pdv), at constant pressure becomes
Q = m * (u2 - u1 + p1 * (v2 - v1) = m * (u2 - u1 + p2*v2 - p1 *v1) for each process or
Q12 = m * (h2 - h1)
; for the process 1-2
Q23 = m * (h3 - h2)
; for the process 2-3
; Steam tables based on NBS/NRC Steam Tables by Lester Haar, John S. Gallagher, and George S.
Kell, Hemisphere Publishing Corp., 1984.
; STEAM8SI.TKW
Variable Sheet
Input
Name
Output
Unit
`
50
100
400
100
750
100
T1
p1
h1
s1
v1
x1
phase1
T2
p2
h2
s2
v2
x2
phase2
T3
p3
h3
s3
v3
x3
phase3
209
0.704
0.00101
'mngless
'CL
3280
8.54
3.1
'mngless
'SH
4040
9.45
4.71
'mngless
'SH
C
kPa
kJ/kg
kJ/(kg*K)
m^3/kg
C
kPa
kJ/kg
kJ/(kg*K)
m^3/kg
C
kPa
kJ/kg
kJ/(kg*K)
m^3/kg
Comment
Thermal Sciences, Potter & Scott
*STEAM8SI.TKW Steam, 1-8 States, SI Units
P4-22.tkw Problem 4.22
Temperature
Pressure
Enthalpy
Entropy
Specific Volume
Quality
Phase
Temperature
Pressure
Enthalpy
Entropy
Specific Volume
Quality
Phase
Temperature
Pressure
Enthalpy
Entropy
Specific Volume
Quality
Phase
41
12. 2
4.23
m
Q12
Q23
6140
1530
*THUNITS.TKW Units for thermo
Mass of steam
Heat transfer, process 1-2
Heat transfer, process 2-3
kg
kJ
kJ
h1
1008
0.8(1796)
2445. v1
0.0012
0.8(0.06668
0.0012)
0.5359
V1
1.2
T
m
22.39 kg
v1
0.05359
a)
Q m(h2 1 )
h
(b)
3000 22.39(h2
2445). h2 2579
h2 2579
T2 234 C
p2 MPa
3
1 (a)
v
b) Q m(h2 1 ) . 30 000 22.39(h2
h
2445). h2
3785
h2
3785
T2 645 C
p2 MPa
3
4.24
v1
0.0012
0.9(0.1274
0.0012)
0.1148 m 3 /kg and
v2
3 0.1148
0.3444 m 3 /kg
u1
850.6
0.9(2595.3
850.6) 2421 kJ/kg
Interpolate in Table C.3:
v2
0.3444
0.3444
0.2608
0.8
(0.2)
0.617 MPa
p2
T2
200 C
0.3520
0.2608
T 200
C
0.617
0.6
Also, 2
(2638.9
2630.6) 2638 kJ/kg
u2 2638.9
0.8
0.6
p2
0.617
To find the heat transfer we must know the work. It is found by using graph paper
and plotting p vs. v and graphically integrating. The work is twice the area since
m = 2 kg. Doing this, we find
W 2 456 kJ.
228
4.25
Q m (u2 1 ) 2(2638
u
W
2421)
456
890 kJ
3489
h
3076.5
a) c p
2.06 kJ/kg
K
T
200
3488.1
h
3074.3
b) c p
2.07 kJ/kg
K
T
200
2821.4
h
2151
c) c p
13.4 kJ/kg
K
T
50
42
13. d) TK solution:
Rule Sheet
; cp = the slope of a constant pressure line on an hT diagram or approximately.
;cp = (h2 - h1)/(T2 - T1) ; where states 1 and 2 are close together, say T1 = 398C and T1 = 402C
; at 10 kPa. Likewise, we define states 3 and 4 at these temperatures on a 100-kpa line and
;states 5 and 6 at these temperatures ona 30,000-kPa line.. Then
cp1 = (h2 - h1) / (T2-T1) ; at 10 kPa, 400 kPa
cp2 = (h4 - h3) / (T4 - T3) ; at 100 kPa, 400 kPa
cp3 = (h6 - h5 ) / (T6 - T5) ; at 30000 kPa, 400 kPa
; Steam tables based on NBS/NRC Steam Tables by Lester Haar, ; John S. Gallagher, and ;George
S. Kell, Hemisphere Publishing Corp., 1984. ; STM8SI.tkw
Variable Sheet
Input Name Output
`
398
10
402
10
398
100
402
100
398
T1
p1
h1
s1
v1
x1
phase1
T2
p2
h2
s2
v2
x2
phase2
T3
p3
h3
s3
v3
x3
phase3
T4
p4
h4
s4
v4
x4
phase4
T5
3280
9.6
31
'mless
'SH
3280
9.61
31.2
'mless
'SH
3270
8.54
3.09
'mnless
'SH
3280
8.55
3.11
'mless
'SH
Unit
C
kPa
kJ/kg
kJ/(kg*K)
m^3/kg
C
kPa
kJ/kg
kJ/(kg*K)
m^3/kg
C
kPa
kJ/kg
kJ/(kg*K)
m^3/kg
C
kPa
kJ/kg
kJ/(kg*K)
m^3/kg
C
Comment
Thermal Sciences, Potter & Scott
*STEAM8SI.TKW Steam, 1-8 States, SI Units
P4-25.tkw Problem 4.25
Temperature
Pressure
Enthalpy
Entropy
Specific Volume
Quality
Phase
Temperature
Pressure
Enthalpy
Entropy
Specific Volume
Quality
Phase
Temperature
Pressure
Enthalpy
Entropy
Specific Volume
Quality
Phase
Temperature
Pressure
Enthalpy
Entropy
Specific Volume
Quality
Phase
Temperature
43
14. 30000 p5
h5
s5
kPa
2120 kJ/kg
4.43 kJ/(kg*K)
0.0026
v5
m^3/kg
9
x5
'mless
phase5 'SH
402 T6
C
30000 p6
kPa
h6
2180 kJ/kg
s6
4.51 kJ/(kg*K)
v6
0.0029 m^3/kg
x6
'mless
phase6 'SH
cp1
cp2
cp3
4.26
2.07
2.07
13.9
Pressure
Enthalpy
Entropy
Specific Volume
Quality
Phase
Temperature
Pressure
Enthalpy
Entropy
Specific Volume
Quality
Phase
*THUNITS.TKW Units for thermo
kJ/(kg*K) Approx cp at 10 kPa, 400 C
kJ/(kg*K) Approx cp at 100 kPa, 400 C
kJ/(kg*K) Approx cp at 30000 kPa, 400 C
1373.9
u
1219.3
a) cv
0.386 Btu/lbm-
R
T
400
1373.7
u
1218.6
b) cv
0.388 Btu/lbm-
R
T
400
1156.5
u
960.7
c) cv
1.96 Btu/lbm-
R
T
100
4.27
117.6
h
126.4
a) c p
0.22 Btu/lbm-
R
T
120
80
115.5
u
107.6
cv
0.20 Btu/lbm-
R
T
120
80
b) TK solution:
Rule Sheet
; cp = the slope of a constant-pressure line on an hT diagram or approximately:
; cp = (h2 - h1)/(T2 - T1) ; where states 1 and 2 are close together, say T1 = 98 F and T2 = 102 F
; at 30 psia. Then cp = (h2 - h1) / (T2-T1) ; at 30 psia, 100 F
; cv = the slope of a constant-volume line on a uT diagram or approximately
;cv = (u4 - u3) / (T4 - T3) where states 3 and 4 are close together, say T3 = 98 F and T4 = 102 F at
the specific volume of the state of 30 psia, 100 F: 1.89 ft^3/lbm. Then
cv = (u4 - u3) / (T4 - T3)
;*R1348e.tkw, R134a, 1-8 States, English units
; R134a tables based on 'Thermodynamic Properties of HFC-134a' DuPont Technical Information,
which is based upon the Modified Benedict-Webb-Rubin equation of state.
R1348E.tkw
44
16. d) The error in (a) is
2.6% and that in (b) is 1.1% assuming that of (c) is the
best answer. All three methods are acceptable for the temperature range of
this problem.
4.29
a) mc p (T2 1 ) 2
H
T
.006
(600
400) 402 kJ
.213
10 3
.031
10 6
b) 2
H
.946 200
(6002 2 )
400
(6003 3 ) 418 kJ
400
2
3
c) m(h2 1 ) 2(607
H
h
401) 412 kJ
4.30
water mc p 2
H
T
4.18(60 418 kJ
10)
ice mc p 2
H
T
1.86[ kJ
10 ( 60)] 186
where we averaged the (c p )ice between and in Table B.4.
10 C
60 C
4.31
Assume that all of the ice melts. The ice warms up to 0 , melts at 0 , then
C
C
warms up to the final temperature T2 . Fr, t f dh m so t ice:
it e si t as fh
s l’ n e
e
6
V 16
8 10
mi
0.1174 kg
v
0.00109
where we found v in Table C.5. Energy is conserved so that the heat gained by the
ice equals the heat lost by the water:
mi p )i ( )i if c p ) w ( )iw mw (c p ) w ( ) w
(c
T
h (
T
T
0.1174
2.1 10 330 4.18(T2 (1000 )
0)
10 3 4.18(20 2 )
T
T2
9.07
C
4.32
a) Assume the ice melts and then the water is heated:
Q m(c p )i ( )i if (c p ) w ( ) w
T
mh m
T
2000 2.3 T2 T2
1.89 60 330 4.18 .
102 C
b) Assume the ice melts and then the water is heated:
Q mhif (c p ) w ( ) w
m
T
2000 2.3 T2 T2
330 4.18 .
129.1
C
This is in the superheat region so the above temperature is too high. The water
must first vaporize at T2
as e n m a r
120.2 ,o ht t f at pr ue
C s t ’ h i le e t .
46
17. 4.33
First, the temperature of the ice is raised, then the ice is melted, then the
temperature is raised to the boiling point, then the water is vaporized, then it is
superheated. This requires the following:
T
Q mc p ( )i p ( )w fg (h2 g )
T
m h mc T
mh
m
h
10 0.486
32 143 1.0(250
32) 945 (1334
1164)
2
=14,900 Btu
where for ice c p
0.47
0.001T and we used an average
1
temperature of 16 .
C
4.34
v
The heat gained by the ice cubes is lost by the cola. The mass of the ice is
mc 4 0.0734 kg
2 2 5 10 6 917
a) Assume all of the ice melts:
mi c p ( )i i i c p ( ) w mc (c p )c ( )c
T
m h m
T
T
0.0734(2.0 T2 ) 2
20 330 4.18
10 3 4.18
1000(20 2 ). T2
T
16.1 C
b) Assume that all of the ice does not melt:
mi c p ( )i x mc (c p )c ( )c
T
h
T
0.0734 x
2.0 20
330 0.25
10 3 1000
4.18
20. x
0.0561 kg
0.0561
100 76.4%
0.0734
4.35
The heat lost by the copper is gained by the water:
mc (c p )c ( )c mw (c p ) w ( ) w
T
T
Use an average value of c p for copper from Table B.4:
5
0.39
(300 2 ) 20
T
10 3 1000
4.18 T2
(
0). T2
6.84
C
4.36
The heat lost by the copper is gained by the water:
mc (c p )c ( )c mw (c p ) w ( ) w
T
T
v (T2 1 ) 4
u c
T
0.717(400
100)
860 kJ
4.37
The heat lost by the iron is gained by the copper:
mc (c p )c ( )c mi (c p )i ( )i
T
T
50 T2
0.39
100 0.45(200 2 ). T2
T
139.5
C
4.38
p (T2 1 ) 4
h c
T
1.00(400
100)
1200 kJ
v (T2 1 ) 4
u c
T
0.717(400
100)
860 kJ
47
18. 4.39
a) 0, W kJ,
U
H
Q 60
60 0.4
0.287
373ln
T1
100 C,
0.4
0.287
373
0.856 m3 ,
50
V2
0.856
p1 p2
50
203 kPa
V1
0.211
V
0.856
. V 1
0.211 m3 ,
V1
2
b) If V const, W
0.
H 0.4 1.00(300 1 )
T
60. T1
150 C
0.4
U
0.717(300
150) 43 kJ, Q 43 kJ,
U
T
423
0.4
0.287
573
p1 p2 1 200
147.6 kPa, V 2 V 1
0.329 m 3
T2
573
200
c) Q mc p . kJ
T
H 100
0.4 1.00(T2
200). T2
450 C,
p2
500 kPa,
T
0.4
0.287
723
473
0.166 m3 , V 1 V 2 1
0.166
0.109 m3
500
T2
723
W p ( V 2 V 1 )
500(0.166
0.109) 28.5 kJ
Also, 0.4
U
0.717(450
200)
71.7 kJ
V
2
k
1
0.4
V
0.48
1
d) T2 1
T
323
605 K or 332
C,
V2
0.1
0.4
U
0.717(332
50) 80.9 kJ, W U
80.9 kJ,
0.4
0.287
323
0.4
H
1.00(332 kJ, p1
50) 113
77.3 kPa,
0.48
0.4
0.287
605
p2
695 kPa
0.1
3.42
4.40
a) Q ( ). 60
m u m pdv
0.4(2512 1 ) . Integrate graphically selecting
u
pdv
v1
various initial states. We find W 49.4 kJ,
U 10.2 kJ,
H 11.8 kJ,
T1
100 C,
p1 kPa,
100
V
2
1.37 m3 ,
V 1 0.662 m3
b) To locate state 1 using the t mt l ivr d f u . ol’asm t th
s a a e s e ii l S , t s eh t
e
bs
y fc t
es u
a e
steam acts almost as an ideal gas and use the constants from Table B.2:
If V const, W
0.
H 0.4 1.872(300 1 )
T
60. T1 220 .
C
0.4
U
1.411(300
220) 45 kJ, Q 45 kJ,
U
T
493
0.4
0.462
573
p1 p2 1 200
kPa, V 2 V 1
172
0.529 m3
T2
573
200
c) Q m(h2 1 ). kJ h2
H
h
H 100
0.4(
2855). h2
3105, p2 MPa.
0.5
Interpolate and find T2
320 C. Then V 2
0.4 0.542
0.217 m3 and
V 1 0.4
0.4249
0.170 m3 and W
500(0.217
0.170) 23.5 kJ
Also, 0.4(2835
U
2643)
76.8 kJ
48
19. 4.41
Q mRT ln
U W
4.42
m
p1
20
2
(53.5 / 778) ln
560
176.7 Btu
p2
200
p
pV
200
2
800
0.596 kg. T2 1 2
T
323
1292 K
RT
2.077
323
p1
200
Q W mcv 0.596
T
3.116(1292
323)
1800 kJ
4.43
Q m(h2 1 )
h
h2
1832
1551F
T2
p2 60 psia
h2
1939
b) 1000 2.0(h2
1439). h2
1939 kJ/kg.
1741
F
T2
p2 psia
60
a) 1000 2.0(h2
1332). h2
1832 kJ/kg.
c) TK solution:
Rule Sheet
;*P4-43.tkw Problem 4.43
;This is a closed system, so the first law, Q =m* (u2 - u1 + INT pdv) constant pressure becomes
; Q = m * (u2 - u1 + p1 * (v2 - v1) = m * (u2 - u1 + p2*v2 - p1 *v1) or, for part (a),
Q = m * (h2 - h1)
;Similarly, for part (b):
Q = m * (h4 - h3)
; Steam tables based on NBS/NRC Steam Tables by Lester Haar, ; John S. Gallagher, ;and
George S. Kell, Hemisphere Publishing Corp., 1984. ; STM8E.tkw
Input Name
`
600 T1
60 p1
h1
s1
v1
u1
x1
phase1
T2
60 p2
h2
s2
v2
u2
Output
1330
1.82
10.4
1220
'mngless
'SH
1830
Variable Sheet
Unit
Comment
Thermal Sciences, Potter & Scott
*Stm8e.tkw Steam, 1-8 States, English
P4-43.tkw Problem 4.43
F
Temperature
psi
Pressure
B/lbm
Enthalpy
B/(lbm*R) Entropy
ft^3/lbm Specific Volume
B/lbm
Internal Energy
Quality
Phase
F
Final temperature, part (a)
psi
Pressure
B/lbm
Enthalpy
B/(lbm*R) Entropy
ft^3/lbm Specific Volume
B/lbm
Internal Energy
49
20. x2
phase2
815 T3
60 p3
h3
s3
v3
u3
x3
phase3
T4
60 p4
h4
s4
v4
u4
x4
phase4
1440
1.91
12.6
1300
'mngless
'SH
1740
1940
2.2
21.7
1700
'mngless
'SH
1000 Q
2
m
4.44
Q mc p (T2 1 ) ,
T
Quality
Phase
F
Temperature
psi
Pressure
B/lbm
Enthalpy
B/(lbm*R) Entropy
ft^3/lbm Specific Volume
B/lbm
Internal Energy
Quality
Phase
F
Final temperature, part (b)
psi
Pressure
B/lbm
Enthalpy
B/(lbm*R) Entropy
ft^3/lbm Specific Volume
B/lbm
Internal Energy
Quality
Phase
*THUNITS.TKW Units for thermo
B
Heat transfer
lbm
Mass of steam
m
p1 V 1
400
0.2
1.021 kg
RT1
0.287
273
a) 50
1.021 T2
1.0(
0). T2 49
C
b) 50
1.021 T2
1.0(
200). T2 249
C
4.45
4.46
p1 V 1 800
8000
10 6
Q W mcv (T2 1 ). m
T
0.05978 kg
RT1
0.287
373
p
200
a) Q 0.05978
0.717(93.2
373)
11.99 kJ where T2 1 2
T
373
93.2 K
p1
800
p
3000
b) Q
T
373
1399 K
0.05978
0.717(1399
373) 43.98 kJ where T2 1 2
p1
800
We assume a quasiequilibrium process with q = 0 so that
k k
1/
p
10
000
T2 1 2
T
373
1390 K
p
100
1
q . w v (T2 1 )
w
u
c
T
0.717(1390
373) kJ/kg
729
0.4 / 1.4
50
21. n n
1/
4.47
p
100
For the polytropic process T2 1 2
T
373
276.7 K
p
600
1
p v 1v1 R(T2 1 ) 0.287(276.7
p
T
373)
w 2 2
138.2 kJ/kg
1
n
1
n
1
1.2
0.2 /1.2
q v
c T w 0.745(276.7
373)
138.2
66.5 kJ/kg
4.48
100
(3 5 2.4)
Q mc p (T2 1 )
T
1.00(27
10) 753 kJ
0.287
283
4.49
Q m .
W
u
pA paddle mcv (T2 1 )
h W
T
The pressure is found from a force balance on the piston:
W
175
p patm
14.7
18.18 psia
A
2
4
The mass is found from the ideal-gas law:
p1 V 1 (18.18
144) 2 /1728
4 10
m
0.0255 lbm
RT1
53.3
560
The temperature at state 2 is
p V 2 (18.18
144) 2 /1728
4 15
T2 2
R
840
mR
0.0255
53.5
Finally,
Wpaddle
18.18 2 /12
4 5
0.0255
0.171
778 (840
560)
1331 ft-lbf
4.50
a) v1
.0011
.5(.4625
.0011)
0.2318 m 3 /kg
u1
604.3
.5(2553.6
604.3)
1579 kJ/kg
0.15
Q W m(u2 1 ). 800
u
(u2
1579). u2 2815 kJ/kg
0.2318
Now, search Table C.3 for the location of state 2. It takes a couple interpolations:
u2
2815
314 and p2
C
1.14 MPa
T2
v2
0.2318
We used:
p2
1.0
248 and u2
C
2707 kJ/kg
T2
v2
0.2318
p2
1.5
489 and u2
C
3102 kJ/kg
T2
v2
0.2318
p
(a)
(b)
1
v
51
22. b) Use some values from part (a):
0.15
Q W m(u2 1 ). 200
u
(u2
1579). u2
1888 kJ/kg
0.2318
Try several guesses for T2 . We tried T2 and T2 . Finally
150 C
160 C
u2
1888
154 and p2
C
0.533 MPa
T
v2
0.2318 2
4.51
60
144 3
a) m
0.9925 lbm. Q W m(u2 1 )
u
0.9925(u2
83.5)
53.3
490
u2 285 Btu/lbm and T2
1595R or 1135F
600
144 3
b) m
3.80 lbm. Q W m(u2 1 )
u
3.80(u2
244)
53.3
1280
u2 297 Btu/lbm and T2
1655R or 1195F
4.52
a) Q mc p
T 5 1.00(313
20) 1465 kJ
V2
293
2 586 K or 313 .
C
V1
b) Q W mcv
T 5 0.717(313
20) 1050 kJ
where we used T2 1
T
p2
293
2 586 K or 313 .
C
p1
p
1
mRT ln 1
5 0.287
293ln kJ
291
p2
2
where we used T2 1
T
c) Q mRT ln
W
V2
V1
d) Q mc p
T 5 1.00(586
293)
1465 kJ
4.53
0.5
m
1.35 kg, h1
604.7
.8 2133.8 2312 kJ/kg
.00108
.8(0.4625
.00108)
a) Q m(h2 1 )
h
1.35(3485
2312)
1584 kJ
b) Q m(h2 1 )
h
1.35(3870
2312) 2104 kJ where h2 is found by interpolation.
c) TK solution:
Rule Sheet
This is a closed system, so the first law, Q =m* (u2 - u1 + INT pdv) at constant pressure becomes
Q = m * (u2 - u1 + p1 * (v2 - v1) = m * (u2 - u1 + p2*v2 - p1 *v1) or
Q = m * (h2 - h1)
v1 = V1/m
; Definition of v1
W = m * p1* (v2 - v1) ; from w = INT pdv for a constant pressure process
; Steam tables based on NBS/NRC Steam Tables by Lester Haar, John S. Gallagher, and George S.
Kell, Hemisphere Publishing Corp., 1984.
; STM8SI.tkw
52
23. Variable Sheet
Input
Name
Output
Unit
`
0.8
0.5
675
400
4.54
144
Q
m
W
400
T1
p1
h1
s1
v1
x1
phase1
V1
T2
p2
h2
s2
v2
x2
phase2
2110
1.35
390
2310
5.87
0.37
C
kPa
kJ/kg
kJ/(kg*K)
m^3/kg
'SAT
3870
8.64
1.09
'mngless
'SH
m^3
C
kPa
kJ/kg
kJ/(kg*K)
m^3/kg
kJ
kg
kJ
Comment
Thermal Sciences, Potter & Scott
*STEAM8SI.TKW Steam, 1-8 States, SI Units
P4-53.tkw
Problem 4.53(b)
Temperature
Pressure
Enthalpy
Entropy
Specific Volume
Quality
Phase
Initial volume
Temperature
Pressure
Enthalpy
Entropy
Specific Volume
Quality
Phase
*THUNITS.TKW Units for thermo
Heat transfer
Mass of steam
Work
0.005
1.495
1.688
mf
4.713 kg, mg
1.688 kg. x1
0.2637
0.001061
0.8857
1.688
4.713
v1 2
v 0.001061
0.2637(0.8857
0.001961)
0.234 m 3 / kg
u1
504.5
0.2637(2025)
1038 kJ/kg
a) Find u2 in Table C.2: Q m(u2 1 )
u
6.401(2578
1038)
9850 kJ
b)
T2
400 C
2953 and Q (u2 1 )
m
u
6.401(2953
1038) 260 kJ
12
u2
v2
0.234
c)
p2 MPa
0.8
2530 and Q (u2 1 )
m
u
6.401(2530
1038)
9550 kJ
u2
v2
0.234
4.55
Q mc p . 10 4
T
0.171 T .
T 14.62F and 4
H
0.24
14.62
14.04 Btu
4.56
Q mRT ln
W
p1
p
200
p1 V 1 ln 1 200
(8000 ) ln
10 6
2.22 kJ
p2
p2
800
53
24. k k
1/
4.57
p
a) T2 1 2
T
p
1
0.4 /1.4
15000
473
400
1332 K or 1059
C
W v
mc T
2 0.717(1059
200)
1230 kJ
k k
1/
p
15000
b) T2 1 2 623
T
1755 K or 1482
C
p
400
1
W v
mc T
2 0.717(1482
200)
1620 kJ
0.4 /1.4
k k
1/
4.58
p V 1 100 2
p
0.3 0.8
T1 1
394 K and T2 1 2
T
mR
0.2
0.287
p
1
0.2857
394 50
1205 K
W v (T2 1 )
mc
T
0.2 0.717(1205
394) kJ
116
4.59
15 100
10 75 150
a) Q mc p . 400
T
1000
0.24 T . 49.4F
T
60
53.3
530
15 100
10 75 150
b) Q mcv . 400
T
1000
0.171 T .
T 69.4
F
60
53.3
530
c) Probably the constant pressure assumption since there are air passages under the doors
and through the air vents so the volume is not constant. The passages allow the pressure
to be essentially constant.
4.60
200
2
Q mcv . 200
W
T
0.717(T2
100). T2
174.7
C
0.287
373
mRT 3.737
0.287
447.7
p2
240 kPa where m = 3.737 kg.
V2
2
4.61
Maximum increase would be for an insulated container so that Q = 0:
Q mc p .
W
T
( 2 10 9.81)
10 10 3 1000 T . 4.69
T
C
4.62
Q paddle ) mc p
( W
T
400
0.1
a) Q
1.00(1092
273) /1000
10 100 45
373.1 kJ
0.287
273
V2
where we used T2 1
T
273
4 1092 K . The factor of 1000 in the above
V1
equation changed J to kJ.
54
25. 400
0.1
b) Q
1.00(2292
573) /1000
10 100 45
373.1 kJ
0.287
573
V2
where we used T2 1
T
2292 K
573 4
V1
4.63
0.8
53.3
1260
V1
6.218 ft 3
60
144
V1
V3
6.218
60 144(10
6.218)
0.8 53.3
1260 ln
7150 kJ
10
7150
net
W
9.19 Btu
778
a) Wnet 1-2 W2-3 3-1 p1 ( V 2 V 1 )
W
W
mRT ln
Qnet
b) Wnet 1-2 W2-3 3-1 p1 ( V 2 V 1 ) v (T1 3 )
W
W
mc
T
60 144(10
6.218)
0.8 0.171(1042
1260)
9480 kJ
9480
Qnet net
W
12.2 Btu
778
k
1
0.4
V
6.218
1
where we used T3 1
T
1260
1042 K.
V
3
10
4.64
The temperatures and V 3 are
100
0.08
800
0.08
T1
278.7 K. T2 3
T
2230 K
0.1
0.287
0.1
0.287
p
800
V 3 V 2 2 0.8
0.64 m3
p3
100
p
Wnet W1-2 2-3 3-1 mRT ln 2 p1 ( V 1 V 3 )
W
W
p3
0.1 0.287
2230 ln
For a cycle Qnet net
W
77.1 kJ
4.65
800
100(0.08
0.64)
77.1 kJ
100
a) Wnet 1-2 W2-3 3-4 W4-1 1 ( v2 1 ) 3 ( v4 3 )
W
W
mp
v
mp
v
10.04
4000(0.04978
0.00125)
47.39(0.00125
0.04978) 1930 kJ
V2
0.5
where we used m
10.04 kg.
v2
0.04978
Qnet net
W
1930 kJ
55
26. b) TK solution
Rule Sheet
;For this cycle of a closed system, assume that all processes are frictionless. Then
Qnet = Wnet ; = W12 + W23 + W34 + W 41 =m * (INT pdv + 0 + INT pdv + 0) and
Wnet = m * (p1 * (v2 - v1) + p3 * (v4 - v3)) ; for the two constant pressure processes
m = V2 / v2
v3 = v2
v4 = v1
; Steam tables based on NBS/NRC Steam Tables by Lester Haar, John S. Gallagher, and
; George S. Kell, Hemisphere Publishing Corp., 1984.
STM8si.tkw
Input
Name
`
4000
0
4000
1
80
0.0498
80
0.00125
0.5
T1
p1
v1
x1
phase1
T2
p2
v2
x2
phase2
T3
p3
v3
phase3
T4
p4
v4
phase4
V2
m
Qnet
Wnet
Variable Sheet
Output
Unit
Comment
Thermal Sciences, Potter & Scott
*Stm8si.tkw, Steam, 1-8 States, SI Units
P4-65.tkw Problem 4.65
250
C
Temperature
kPa
Pressure
0.00125 m^3/kg Specific Volume
Quality
'SAT
Phase
250
C
Temperature (starting guess needed)
kPa
Pressure
0.0498
m^3/kg Specific Volume (transfer value to input)
Quality
'SAT
Phase
*THUNITS.tkw
Units for thermo
C
47.4
kPa
m^3/kg Specific volume (transfer value to input)
'SAT
C
47.4
kPa
m^3/kg Specific volume (transfer value to input)
'SAT
m^3
Volume at state 2
10
kg
Mass of steam
1930
kJ
Net heat transfer for the cycle
1930
kJ
Net work for the cycle
k
1
4.66
100
0.02
V
1
0.02
T3
232.3 K. T1 3 232.3
T
583.5 K
0.03
0.287
V
0.002
3
0.4
V2
v (T1 3 )
mc
T
V3
0.02
0.03 0.287
583.5ln
0.717(583.5
232.3) 4.014 kJ
0.002
Wnet 1-2 W2-3 3-1 mRT ln
W
W
56
27. 4.67
4.68
A1
2
5
100
25 m/s
A2
2
10
V2 1
V
0.02
1 1
m AV1 A2V2 . V2
0.0025 m/s
2
1000
(800
0.01
0.001)
L
0.60
t
240 s or 4 min
V 0.0025
A1
2
1
150
37.5 fps
A2
2
4 0.5
4.69
V2 1
V
4.70
4000
1 1
m AV1
)
(10 10 4 150 3.65 kg/s
0.287
573
m
3.65
V2
m/s
195
400
A2
4
2
)
(50 10
0.287
373
4.71
a) The continuity equation with one inlet is
dm
d
AV1 c.v. V
1 1
dt
dt
where V is the volume of the tank. Then, using v1 ,
1/
1
d A1V1 2
0.05 20
0.0409 kg/m 3
s
dt
V v1
10
0.3843
b) TK solution:
Rule Sheet
; State 1: State of entering steam in the pipe; State 2: State of steam in the tank. A mass
;balance on the tank is: Rate of mass flow in = rate of mass increase in the tank. This is
mdot1 = dm2%dt ; where dm2%dt = dm2/dt
mdot1 = A1* V1/ v1
A1 = pi() * d1^2/4
drho%dt = dm2%dt/Volume2
; Steam tables based on NBS/NRC Steam Tables by Lester Haar, John S. Gallagher, and
; George S. Kell, Hemisphere Publishing Corp., 1984.
Stm8si.tkw
Variable Sheet
Input
Name
Output
Unit
Comment
Engineering Thermo, Jones & Dugan
*Stm8si.tkw Steam, 1-8 States, SI Units
P4-71.tkw Problem 4.71
`
57
28. 400
800
20
10
10
4.72
T1
p1
v1
x1
phase1
T2
p2
v2
x2
phase2
mdot1
A1
V1
d1
Volume2
dm2%dt
drho%dt
m3
m1 m2 .
C
kPa
m^3/kg
C
kPa
m^3/kg
0.384
'mngless
'SH
0.409
0.00785
Temperature in inlet pipe
Pressure in inlet pipe
Specific Volume in inlet pipe
Quality
Phase
Temperature
Pressure
Specific Volume
Quality
Phase
Mass flow rate in
Cross-sectinal area of pipe
Velocity in pipe
Diamter of pipe
Volume of tank
Rate of iuncreas of mass in tank
Rate of increase of density in tank
kg/s
m^2
m/s
cm
m^3
kg/s
kg/(m^3*s)
0.409
0.0409
2000
150
2
.04 125
2
.05 40 m2
.287
623
.287
423
6.64 kg/s, V2
m2
6.64
450
2
0.05
0.287
473
4.73
255 m/s
m
15
Vavg
1.442 fps
62.4 24
A
4 1/
m V dA
y2
15 max 2
V
1
h
h
h
4
62.4 Vmax
4
dy
h
y3
4 1/
48
y
62.4 Vmax
4
2
3
3h h
Vmax 2.16 fps
4.74
1
1
4.95
a) m A1V1
2
0.25 30 4.95 kg/s, V2
109 m/s
v1
0.5951
2
0.1252 / 1.08
b) TK solution:
Rule Sheet
;For steady flow, mdot1 = 2 * mdot2, with state 1 in large pipe, state 2 in each smaller one.
A1 * V1 / v1 = 2 * A2 * V2 / v2
A1 = pi() * d1^2/4
A2 = pi() * d2^2 /4
mdot2 = A2 * V2 / v2
; Steam tables based on NBS/NRC Steam Tables by Lester Haar, John S. Gallagher, and
; George S. Kell, Hemisphere Publishing Corp., 1984.
Stm8si.tkw
58
29. Variable Sheet
Input
Name
Output
Unit
`
250
400
30
50
200
200
25
T1
p1
h1
s1
v1
x1
phase1
V1
d1
T2
p2
h2
s2
v2
x2
phase2
V2
d2
A1
A2
mdot2
2960
7.38
0.595
'mngless
'SH
2870
7.51
1.08
'mngless
'SH
109
0.196
0.0491
4.95
C
kPa
kJ/kg
kJ/(kg*K)
m^3/kg
m/s
cm
C
kPa
kJ/kg
kJ/(kg*K)
m^3/kg
m/s
cm
m^2
m^2
kg/s
Comment
Thermal Sciences, Potter & Scott
*Stm8si.tkw Steam, 1-8 States, SI Units
Temperature
Pressure
Enthalpy
Entropy
Specific Volume
Quality
Phase
Velocity in large pipe
Diameter of large pipe
Temperature
Pressure
Enthalpy
Entropy
Specific Volume
Quality
Phase
Velocity in each small pipe
Diameter of each small pipe
*THUNITS.TKW Units for thermo
Area of large pipe
Area of each small pipe
Flow rate in each small pipe
4.75
1
1
10
2.18
m A1V1
800 2.18 lbm/sec, V1
fps
182
v1
25.43 144
(2 / 144) / 1.1583
4.76
a) V
d
d
1
1
AV1 2 A2V2 . 10
2
0.04 20
2
0.06 10
1 1
dt
dt
0.1996
0.3066
d
0.01348 kg/m3
s
dt
b) TK solution:
Rule Sheet
; State 1 is defined as the state of steam in the entrance.pipe. State 2 is defined as the
; state of steam in the exit pipe. m is the mass of steam in the tank at a specified instant.
; A mass balance on the tank is therefore dm/dt = mdot1 - mdot2 or
dm%dt = A1* V1 /v1 - A2* V2 / v2
A1 = pi() * d1^2/4
A2 = pi() * d2^2/4
drho%dt = dm%dt/Volumet
59
30. ; Steam tables based on NBS/NRC Steam Tables by Lester Haar, John S. Gallagher, and
; George S. Kell, Hemisphere Publishing Corp., 1984.
Stm8si.tkw
Variable Sheet
Input
Name
Output
Unit
`
600
2000
400
1000
20
8
10
10
12
4.77
T1
p1
v1
x1
phase1
T2
p2
v2
x2
phase2
A1
V1
d1
Volumet
A2
V2
d2
dm%dt
drho%dt
0.2
'mngless
'SH
0.307
'mngless
'SH
0.00503
0.0113
0.135
0.0135
C
kPa
m^3/kg
C
kPa
m^3/kg
m^2
m/s
cm
m^3
m^2
m/s
cm
kg/s
kg/(m^3*s)
Comment
Thermal Sciences, Potter & Scott
*Stm8si.tkw Steam, 1-8 States, SI Units
P4-76.tkw
Problem 4.76
Temperature in inlet pipe
Pressure in inlet pipe
Specific Volume in inlet pipe
Quality
Phase
Temperature in exit pipe
Pressure in exit pipe
Specific Volume
Quality
Phase
Cross-sectinal area of pipe
Velocity in pipe
Diamter of pipe
Volume of tank
Rate of iuncreas of mass in tank
Rate of increase of density in tank
Use Eq. 4.58 with :
1
2
0.006
r2
AV1 A2V2 .
0.0062 Vmax
0.8
1
1
2
0
0.006
2 r
1.6
dr. Vmax m/s
1 1
m AV1
1000
0.0062
0.8 0.0905 kg/s
4.78
1
1
1 4
m 2(1 2 )2 r dr
r
10 4
1000 10
4
0.314 kg/s
0
2
4
0.314
1000
0.00252 2 . V2 m/s
V
16
4.79
60
144 2
2
1 1
m AV1
2.526 lbm/sec
100
53.3
560 144
70
144 2
2
A2V2
V2 . V2
116.3 fps.
2
53.3
760 144
60
31. If we neglect , Q p (T2 1 )
T
2.526
0.24(300
100)
121.2 Btu/sec
KE mc
V 2 12
V
116.32 2
100
Note: m 2
KE
2.526
4453 ft-lbf/sec or 5.72 Btu/sec
2
2
4.80
a) Across a valve . From Table D.2 at 0.8 MPa we find h1
h 0
93.42 :
h2 h1
93.42
3.46 x2 (221.3). x2
0.4065
u2 f x2 (u g f )
u
u
3.41
0.4065(206.1
3.41)
85.8 kJ/kg
b) TK solution:
Rule Sheet
;This is a throttling process. w = 0. Assume q = 0 and change in ke is negligible. Therefore,
h2 = h1 ; from the first law.
; R134a tables based on 'Thermodynamic Properties of HFC-134a'
; DuPont Technical Information, which is based upon the Modified
; Benedict-Webb-Rubin equation of state.
R1348si.tkw
h2 = u2 + p2 * v2
Variable Sheet
Input
Name
Output
Unit
`
30
800
T1
p1
h1
s1
v1
x1
phase1
T2
60
p2
93.602 h2
s2
v2
x2
phase2
u2
4.81
93.602
0.34761
0.00084403
'mngless
'CLQ
-36.91
0.39624
0.12501
0.40047
'SAT
86.101
C
kPa
kJ/kg
kJ/(kg*K)
m^3/kg
C
kPa
kJ/kg
kJ/(kg*K)
m^3/kg
kJ/kg
Comment
Thermal Sciences, Potter & Scott
*R1348si.tkw, R134a, 1-8 States, SI units
P4-80.tkw
Problem 4.80
Temperature
Pressure
Enthalpy
Entropy
Specific Volume
Quality
Phase
Temperature
Pressure
Enthalpy (transfer value to input)
Entropy
Specific Volume
Quality
Phase
Internal energy
*THUNITS.TKW Units for thermo
Neglecting kinetic energy changes and using h1 h f
1344 kJ/kg (from Table C.1),
p2 0.6 MPa. h2 h1
1344
670.6 x2 (2086.3). x2
0.3228
u2 f x2 (u g f )
u
u
669.9
0.3228(2567.4
669.9)
1282 kJ/kg
61
32. 4.82
Neglecting kinetic energy changes and using h1
3634 kJ/kg (from Table C.3),
a)
p2 0.6 MPa
569
C
T2
h2
3634 kJ/kg
b) With V2 1 and m2 m1 we have
V
A
1
1
A2 V2
A1 V1 . 2 22.3
0.9695
0.04341
A1
c) For air T2 1
h 0.
T 600 C
4.83
4.84
W
Using the energy equation in the form of Eq. 4.65 we have, with Q and
0
S
V2 1 and z2 z1 ,
V
p
p
(450
30) 144
h2 h1 or u2 2 1 1 . u2
u
38.09
39.34 Btu/lbm
62.4
778
2
1
The factor 778 converts ft-lbf to Btu.
The energy equation in the form of Eq. 4.68 is used. First, the velocities are
m
200
200
V1
6.366 m/s. V2
17.68 m/s
2
1 1000
A
0.1
1000 2
0.06
2 12
V V
p
WP 2
m
g
z
2
2
2
4 000 000
17.68
6.366
200
827 000 W or 1110 hp
2
1000
4.85
W
0.021 kg/s
Q mc p (T2 1 ). m
T
( 5) 1.00(530.7
293). m
S
k k
1/
p
where we used T2 1 2
T
p
1
4.86
0.4 /1.4
400
293
50
530.7 K .
Fr, t f dh m s f x
it e si t asl :
s l’ n e
u
500
2
m A2V2
2 2.72 kg/s
0.05 100
0.287
503
0.4 /1.4
500
where T2 293
503 K .
80
The energy equation gives (the kinetic energy change is negligible)
( W
Q ) mc p (T2 1 ). WC 2.72
T
1.00(503
293)
558 kW
C
4.87
( W
a) Q ) m(h2 1 ). Q 0.01
h
(2839
2592)
6
3.53 kJ/s
C
The negative sign indicates a heat loss.
62
33. b) TK solution:
Rule Sheet
Qdot = mdot * (h2 - h1) + Wdot ; First law for steady flow through compressor with negligible
; changes in kinetic and potential energy.
; Steam tables based on NBS/NRC Steam Tables by Lester Haar, John S. Gallagher, and
; George S. Kell, Hemisphere Publishing Corp., 1984.
Stm8si.tkw
Variable Sheet
Input
Name
Output
Unit
`
50
1
200
800
0.01
-6
4.88
T1
p1
h1
x1
phase1
T2
p2
h2
x2
phase2
Qdot
mdot
Wdot
12.3
2590
C
kPa
kJ/kg
'SAT
2840
'mngless
'SH
-3.52
C
kPa
kJ/kg
kJ/s
kg/s
kW
Comment
Thermal Sciences, Potter & Scott
*Stm8si.tkw Steam, 1-8 States, SI Units
Temperature
Pressure
Enthalpy
Quality
Phase
Temperature
Pressure
Enthalpy
Quality
Phase
*THUNITS.TKW Units for thermo
Rate of heat addition, i.e., heat is removed
Mass flow rate
Power output
We use the energy equation in the form of Eq. 4.68:
p p
(2000
2) 144
1 2000
) 2
( WP m
2594 ft-lbf/sec or 4.72 hp
1/ v 3600 1/ 0.01623
4.89
Use Eq. 4.58 with :
2
1
A2V2 AV1 0.1 m3 /s
1
0.1
0.1
V1
50.93 m/s and V2
12.73 m/s
2
0.025
2
0.05
The volume flow rate is the product AV . Assuming a 100% efficient pump for
minimum power,
2 12 p2 p1
V V
) m 2
( WP
2
12.732
50.932 4000
200
0.1 1000
258.4 kW or 346 hp
1000
1000
2
T e os n “00 i t dnm nt o t k ec nrye cne s t k
h cnt t10”nh eo i o fh i t ee t m ovr W o W.
a
e
ar e ni
g r
t
63
34. 4.90
Assume a 100% efficient turbine:
2 12
V V
p p1
0
300
WT AV1 2
1
2
20
6000 kW
1000
2
1000
We neglected the kinetic energy change because information was not given and
because it is most often negligible. Since the pressure is in kPa the work rate is in kW.
4.91
Assume a 100% efficient turbine with the pressure of zero gage on the surface of the
backwater and at the turbine outlet:
2 12
V V
p p1
WT AV1 2
1
2
g ( z2 1 )
z
2
240 W or 39.24 kW
100 9.81 ( 40) 39
4.92
Assume a 100% efficient turbine with the pressure of zero gage on the surface of the
backwater and at the turbine outlet:
2 12
V V
p p1
WT AV1 2
1
2
g ( z2 1 )
z
2
1000 9.81 21200 W or 21.2 kW
(0.6 1.2) 1.5
( 2)
4.93
3414
W
10
a) Q m(h2 1 ). Q 000
h
30(1131
1512).
T
3600
1954 Btu/sec
Q
b) TK solution:
Rule Sheet
Wdot = mdot * (h1 - h2) + Qdot ; First law for steady flow with no changes in potential or
; kinetic energy
;*Stm8e.tkw Steam, 1-8 States, English units
; Steam tables based on NBS/NRC Steam Tables by Haar, Gallagher,
; and Kell, Hemisphere Publishing Corp., 1984. See Comment sheet.
Input
Name
`
1000
800
5
T1
p1
h1
v1
x1
phase1
T2
p2
h2
Variable Sheet
Output
Unit
Comment
Thermal Sciences, Potter & Scott
*Stm8e.tkw Steam, 1-8 States, English units
F
Temperature
psi
Pressure
1510
B/lbm
Enthalpy
1.05
ft^3/lbm Specific Volume
'mngless
Quality
'SH
Phase
162
F
Temperature
psi
Pressure
1130
B/lbm
Enthalpy
64
35. 1
10
30
4.94
4.95
v2
x2
phase2
Wdot
mdot
Qdot
73.5
ft^3/lbm
'SAT
MW
lbm/s
B/s
-1950
Specific Volume
Quality
Phase
*THUNITS.TKW Units for thermo
Power output
Mass flow rate
Heat is removed at a rate of 1950 B/s
1000
m(
WT h2 1 )
h
(1116.1
1623.8)
8462 Btu/sec or 11,970 hp
60
m
1000 / 60
V
230 fps
A 2 /173.75
2
where we used v.
1/
h1
3445.2 kJ/kg, h2 h f x2 h fg 251.4
0.9(2358.3) 2374 kJ/kg
m(
WT h2 1 )
h
6(2374
3445)
6430 kW
m
6
V2
82.2 m/s
2
A
0.4 /[0.001
0.9(7.649
0.001)]
4.96
4.97
600
2
0.05 100
0.287
373
A2V2 AV1 . V2
18.17 m/s
2
1 1
140
2
0.2
0.287
253
600
1 1
m AV1
2 4.402 kg/s
0.05 100
0.287
373
V 2 12
V
18.17 2 2
100
m
WT c p (T2 1 ) 2
T
4.402
1.00( 20 100)
539 kW
2
2
1000
T eat “00 it dnm nt o t k ec nrye cne s t k
h f o 10”n h eo i o fh i t ee t m ovr W o W.
cr
e
ar e ni
g r
t
800
1 1
m AV1
2
0.05 20 0.9257 kg/s
0.287
473
Q mc (T ) 0.9257
T W
1.0(200
20) ( 400) kJ/s
233
p
2
1
S
The negative sign indicates a heat loss, as expected.
4.98
450
Q mc p (T2 1 )
T WS
2
.06 150 1.0 (20 100)
500
70.5 kJ/s
.287
373
65
36. 4.99
Fr, t cl leh vl ie:
it e s a u tt e ci
s l’ c a e o ts
m
30
30
V1
5.509 fps, V2
137.7 fps
2
A1 62.4 /144
2
62.4 2 /144
0.4
1
2 V2
V 1 p2 p1
Q WS m 2
2
137.7 2
5.5092 (14.7 1 )
P 144
0
30
.
142
p1 psia
32.2
62.4
2
The factor 32.2 above converts slugs to lbm.
4.100 The energy equation: V22 12 c p (T1 2 ) 202
V
2
T
2 1.0(293 2 )
T
p1
AV1 A2V2
1
2
RT1
80
2
4
V2
20 5.384 kg/m 2
s
2
0.287
293 2
3
Assume an adiabatic expansion using v :
1/
Continuity equation: AV1 A2V2 .
1 1
2
k-1
T2 2
T2
293
298.9
or 0.4
T1 1
[80 / 0.287 0.4
293]
2
The above three equations include three unknowns , V2 , and T2 . They are solved
2
by trial-and-error to give
3.47 kg/m3 , V2
1.55 m/s and T2 219
C
2
k /k
1
p
4.101 a) Assume an adiabatic process: T2 1 2
T
p
1
0.4 /1.4
585
468
85
812 K or 539
C
2 12
2 2
V V
V 100
b) 0 2
p (T2 1 ) 2
c
T
1.0(812
468) V2
.
835 m/s
1000
2
2
p
p
585 2
0.1 100 812
c) 2 ( d 22 / 4)V2 1 12 1 . d22
r V
. d2
0.239 m
RT2
RT1
185 / 4
468 835
p
80
4.102 a) m AV
2
0.05 200 1.672 kg/s
RT
0.297
253
V22 12
V
152 2
200
b) 0
p (T2 1 ). 0
c
T
1.042(T2
20). T2
0.91 C
2
2
1000
V 2 12
V
502 2
600
4.103 0 2
2 1. 0
h h
2
h 1154.
2
2
1000
P2 20 psia
T2 238 F
h2
1161
66
h2
1161 Btu/lbm
37. 4.104
The heat transfer rate for the steam, assuming no pressure drop through the condenser, is
Q m (h ) 600
h
(94.02
1116.1) 200 Btu/min
613,
s
s
2s
1s
This equals the energy gined by the water:
40,900 lbm/min
Qw mw c p (T2 w 1w ). 613, 200
T
mw 1.00 (15). mw
4.105
a) The heat lost by the air is gained by the water:
ma c pa (T2 a 1a ) mw c pw (T2 w 1w ). 5 mw
T
T
1.0 200 4.18
10. mw 23.9 kg/s
b) Q m c 23.9
T
4.18
200 1000 kJ/s
w
4.106
w pw
With Q 0 and mi 0 there results u f h1 since m f m1 . Across a valve the enthalpy
is constant so that h1 hline
3674.4 kJ/kg . The final pressure is p f MPa and
4
uf
3674.4 kJ/kg. The temperature is interpolated from Table C.3 to be
3674.4
3650.1
Tf
50 800 813 C
3650.1
3555.5
The final specific volume is
812.8
800
vf
(0.1229
0.1169)
0.1229
0.1244 m 3 /kg
50
V f
4
The mass of steam in the tank is then m f
32.15 kg
vf
0.1244
4.107
a)
pf
800 psia
1587
F
T f
uf 1
h 1620
b) Interpolate and find v f
1.512 m 3 /kg. m f
V f
50
33.1 lbm
vf
1.512
pVi
250
3
80
3
4.108 mi i
8.769 kg. m f
2.806 kg
RTi
0.287
298
0.287
298
Q m u u h
m
m
f
f
i i
2 2
2.806
0.717
298 8.769
0.717
298 (8.769
2.806)
1.0 298 503 kJ
13937
90
5
800
5
4.109 mi
5.351 kg. m f
0.287
293
0.287 f
T
Tf
0 f m f i mi 1 h1 f m f
u
u
m
u
0.717
293 5.351 m f
(
5.351)
1.0 353
Then 0
0.717 f
T
13937
1124 m f
(
5.351)
353. m f
30.48 kg
Tf
13937
Finally, T f
457 or 184 m1
C.
30.48
5.351 25.1 kg
30.48
67
38. 4.110 Since Q and mi , u f h1
0
0
0.24
530 0.171T f . T f or 284F
744 R
p f V f 12
144 10
mf
4.36 lbm
RT f
53.3
744
k k
1/
0.4 /1.4
p f Tf
p
f
95
4.111 a)
. Tf i
T
303
127
146
K or C
V i pi RTi
piTi
2000
pi
T
303
b) For V
const p2 p1 2
95
227 kPa
T1
127
V f p f RT f
800
4
4.112 mi f 0.02 kg. mi
m
300 6
36.8 kg. m f
30.8 kg
0.287
303
k
1
0.4
m
f
30.8
a) T f i 282 or 9
T
303
C
m
36.8
i
m RT
30.8
0.287
282
b) p f f f
624 kPa
V f
4
1/ k
1
T
f
c) m f mi
T
i
2.5
253
36.8 23.44 kg
303
m 13.36 0.02 t 60.
t 11.13 min
1.8
0.2
4.113 mi
158.9
C
1726 kg . a) T
0.001043 1.694
1
1
b) m f
911.4 kg.
m 814.5 kg
0.001101 0.3157
0.5
1.5
d
c) m f
458.9 kg. Q (um) 2 m2 f m2 2 m2
h u h
0.001101 0.3157
dt
10000
10 000 (u f 2 )m2 . m2
h
1.878 kg/min
2567
2757
1.878
t 1725.9
458.9.
t 675 min
4.114 a) Assume steady state with the heat sink area equal to the transistor area:
Q Q
Q Q2
1
3
Tj
T
4
4
j ) j sur )
h A (T T A (T T
1
0.5
h
A
Tj
325
4
1
Tj Tj 4 )
15 10 4 (
325) 0.5 5.67 10 8 10 4 (
300
1
0.5
15
10 4
4
2.835 Tj
10 12
0.003Tj
2 0
68
39. By trial and error,
Tj = 568 K (295 >> 125
C)
C
The transistor exceeds its maximum allowable temperature of 125
C
b)
A (T 4 T 4
Qrad j surr ) 4 4 )
0.5 5.67 10 8 10 4 (568 300
0.27 W
0.27
fraction
0.27
1
4.115 Assume the temperature to be uniform at any instant, negligible radiation, and
constant properties.
Air
T = 20
C
h = 8 W/m2·K
3 cm
7.5 cm
h
L
Bi c
k
a)
where Lc is the effective length and is defined as Volume/Surfce area. For the
pancake,
(0.075)2
(0.003)
4
LC
0.0015 m
2
2
(0.075)
0.0075
0.003
4
Thus,
h 8
L
0.0015
Bi c
0.048 0.1
k
0.25
The lumped capacitance assumption is valid.
b) Energy balance:
out = Estored
E
s ) V
h A (T T
c
h
A
dT
dT
or
s dt
dt
T
T
V
c
Let T . Then,
=
T
h
A
h
A
d
T
T
s dt. ln
s
t
V
c
Ti
T
V
c
V T
c
T
500
3000
0.0015 25
20
t
ln
ln
647 s
h s Ti
A
T
8
80
20
69
40. 4.115 Assume steady-state, one-dimensional conduction, with negligible contact
resistance and constant properties.
1
1
1
1
Roconv
0.07
C/W, Riconv
0.21
C/W
ho w 20.8
A
0.6
hi w 8
A
0.6
t panel
0.005
R panel
0.07
C/W
k panel panel 0.12
A
0.6
t plaster
0.015
R plaster
0.015
C/W
k plaster plaster 0.17
A
0.6
t
t
0.2
0.2
Rstud stud
312.5
C/W, Rins ins
11.91
C/W
k stud stud 0.16
A
0.06
kins ins 0.03
A
0.56
tsiding
0.025
Rsidings
0.443
C/W
ksiding siding 0.094
A
0.6
R ins
R
Rtotal Roconv panel plaster stud
R
R
sidings iconv
R
R
Rstud ins
R
T
20
5
15
Q
16.67 W
312.5
11.91
Rtotal 0.07
0.9
0.07
0.015
0.443
0.21
312.5
11.91
4.117 Neglect radiation.
Tair, h
D2, D1
lpipe
Rpipe
Rair
Tinside
Tair
T
T
a) Q inside air
R pipe air
R
Tinside air
T
200
10
9.98 kW
ln r2 / r1
1
ln 4 / 3.75
1
16.5 2 .04
7 30
2 l k 2 r2 2 7
l h
Tinside air
T
200
10
536 W
ln 4 / 3.75
ln 9 / 4
1
R pipe insulation air
R
R
2 7
16.5 2 7
0.055 2 0.09
7 30
b) Q
Q
% reduction 1 insulation
Q
536
94%
100 1
100
9980
70
41. 4.118 Assume the system to be at steady state.
r2 = 3.5 cm
r1 = 2.5 cm
Q total = 500W
1 cm
Ti = 400oC
Rcond
Tamb
Rconv
T
T
T
T
Qtot i amb . Rcond i amb conv
R
Rcond conv
R
Qtot
T T
1 1 i amb
1
1
2
4 1 r2 Qtot
k r
4 h
r
1 1
1
1
r1 r2
0.025 0.035
k
8.23 W/m
K
20
1
i amb
T T
1 4 400
4
2
4
0.0352
100
500
4 h
r
Qtot
71