APPLIED THERMODYNAMICS PAMPHLET

BPE 211 APPLIED THERMODYNAMICS
Musadoto IWR/D/2016/0011
SOKOINE UNIVERSITY OF AGRICULTURE
Department of Engineering sciences and Technology
BPE213/AE211 :APPLIED THERMODYNAMICS
A COMPREHESIVE SIMPLIFIED TO THE APPLIED THERMODYNAMICS
1st
EDITION
2017
Musadoto

About the author:
musadoto is a student at Sokoine University Of Agriculture undertaking bachelor of
science in Irrigation and Water Resources engineering second year with full registration number
iwr/d/2016/0011.Samuye and Njombe secondary were his ordinary and advanced education
level respectively.
i

Forewords
This pamphlet aim to help all engineering students at Sokoine University of Agriculture in
attempting the BPE211/AE213 exams because as curriculum instructs that,this kind of course
should be an open system. The main aim is to reduce the amount materials in the examination
room that may cause confusion to individual student and time consuming due to dis organization
of materials. This pamphlet consists of 9 Topics as given out by the course instructor and atleast
1000 solved questions at which at least 100 from each Topic. All past papers.tutorials,summaries
from many books are included inside.Dont forget to study carefuly by your own to improve your
understanding .
ii

Acknowledgement
Much thanks to my God who laid all the following;
Bsc .of irrigation and water resources engineering students for their infinity support both
materially and time. Never the less my friend Kitowe Mariam for her inspiration to complete this
pamphlet. BSc.Age/Bpe for their intention. I appreciate too Mr Deus,Heri,Donath,Ole David
and Mashauri (Age) for their materials as far as many supporters participated in the preparation
of this pamphlet.
God bless you.
iii

Dedication
This pamphlet is dedicated to all Engineering students at Sokoine University of Agriculture
taking BPE211/AE213(Applied thermodynamic)-2017 and all upcoming youngs at this
institution.
iv

Table of content
About the author…………………………………………………………………….i
Foreword………………………………………………………………………..….ii
Acknowledgement………………………………………………………………....iv
Dedication…………………………………………………………………………v
1. Fundamental concept of thermodynamics…………………………………..1
2. Thermodynamics properties of liquids,property diagrams………………….8
3. First law of Thermodynamics………………………………………………
4. Second law of thermodynamics……………………………………………...
5. Carnot cycle and Gas power system…………………………………………
6. Steam/Vapour power…………………………………………………………
7. Refrigeration cycles………………………………………………………….
8. Properties of mixtures……………………………………………………….
9. Psychometry…………………………………………………………………
V

TOPIC 1
FUNDAMENTAL CONCEPT OF THERMODYNAMICS
Thermodynamic System
Thermodynamics is the science relating heat and work transfers and the related changes in the
properties of the working substance. The working substance is isolated from its surroundings in
order to determine its properties.
System - Collection of matter within prescribed and identifiable boundaries. A system may be
either an open one, or a closed one, referring to whether mass transfer or does not take place
across the boundary.
Types of Thermodynamic Systems
There are three mains types of system: open system, closed system and isolated system. All these
have been described below:
1) Open system: The system in which the transfer of mass as well as energy can take place
across its boundary is called as an open system. Our previous example of engine is an open
system. In this case we provide fuel to engine and it produces power which is given out, thus
there is exchange of mass as well as energy. The engine also emits heat which is exchanged with
the surroundings. The other example of open system is boiling water in an open vessel, where
transfer of heat as well as mass in the form of steam takes place between the vessel and
surrounding.
2) Closed system: The system in which the transfer of energy takes place across its boundary
with the surrounding, but no transfer of mass takes place is called as closed system. The closed
system is fixed mass system. The fluid like air or gas being compressed in the piston and
cylinder arrangement is an example of the closed system. In this case the mass of the gas remains
constant but it can get heated or cooled. Another example is the water being heated in the closed
vessel, where water will get heated but its mass will remain same.
3) Isolated system: The system in which neither the transfer of mass nor that of energy takes
place across its boundary with the surroundings is called as isolated system. For example if the
piston and cylinder arrangement in which the fluid like air or gas is being compressed or
expanded is insulated it becomes isolated system. Here there will neither transfer of mass nor
that of energy. Similarly hot water, coffee or tea kept in the thermos flask is closed system.
However, if we pour this fluid in a cup, it becomes an open system.
2

Surroundings - Is usually restricted to those particles of matter external to the system which
may be affected by changes within the system, and the surroundings themselves may form
another system.
Boundary - A physical or imaginary surface, enveloping the system and separating it from the
surroundings.
THERMODYNAMIC PROPERTIES.
Property - is any quantity whose changes are defined only by the end states and by the process.
Examples of thermodynamic properties are the Pressure, Volume and Temperature of the
working fluid in the system above.Thermodynamic properties are related to the energy of the
system, i.e. temperature, pressure, mass, volume.
Extensive properties depend on the size or extent of the system, e.g. volume, mass, total energy.
Intensive properties are independent of size, e.g. temperature, pressure, entropy, density,
specific volume.
3.

Thermodynamic Process
When the system undergoes change from one thermodynamic state to final state due change in
properties like temperature, pressure, volume etc, the system is said to have undergone
thermodynamic process. Various types of thermodynamic processes are: isothermal process,
adiabatic process, isochoric process, isobaric process and reversible process. These have been
described below:
1) Isothermal process: When the system undergoes change from one state to the other, but
its temperature remains constant, the system is said to have undergone isothermal
process. For instance, in our example of hot water in thermos flask, if we remove certain
quantity of water from the flask, but keep its temperature constant at 50 degree Celsius,
the process is said to be isothermal process.
Another example of isothermal process is latent heat of vaporization of water. When we
heat water to 100 degree Celsius, it will not start boiling instantly. It will keep on
absorbing heat at constant temperature; this heat is called latent heat of vaporization.
Only after absorbing this heat water at constant temperature, water will get converted into
steam.
4

2) Adiabatic process: The process, during which the heat content of the system or certain
quantity of the matter remains constant, is called as adiabatic process. Thus in adiabatic
process no transfer of heat between the system and its surroundings takes place. The wall
of the system which does not allows the flow of heat through it, is called as adiabatic
wall, while the wall which allows the flow of heat is called as diathermic wall.
3) Isochoric process: The process, during which the volume of the system remains
constant, is called as isochoric process. Heating of gas in a closed cylinder is an example
of isochoric process.
4) Isobaric process: The process during which the pressure of the system remains constant
is called as isobaric process. Example: Suppose there is a fuel in piston and cylinder
arrangement. When this fuel is burnt the pressure of the gases is generated inside the
engine and as more fuel burns more pressure is created. But if the gases are allowed to
expand by allowing the piston to move outside, the pressure of the system can be kept
constant.
The constant pressure and volume processes are very important. The Otto and diesel
cycle, which are used in the petrol and diesel engine respectively, have constant volume
and constant pressure processes. In practical situations ideal constant pressure and
constant pressure processes cannot be achieved.
5) Reversible process: In simple words the process which can be revered back completely
is called a reversible process. This means that the final properties of the system can be
perfectly reversed back to the original properties. The process can be perfectly reversible
only if the changes in the process are infinitesimally small. In practical situations it is not
possible to trace these extremely small changes in extremely small time, hence the
reversible process is also an ideal process. The changes which occur during reversible
process are in equilibrium with each other
6) Irreversible process A process that is not reversible—duh. There will be entropy
generation in an irreversible process. Factors that can cause a process to be irreversible
include friction, unrestrained expansion, mixing of two gases, heat transfer across a finite
temperature difference, electric resistance, inelastic deformation of solids, and chemical
reactions
7) Throttling process A throttling is defined as a process in which there is no change in
enthalpy from state one to state two, h1=h2; no work is done, W=0; and the process is
adiabatic, Q=0. To better understand the theory of the ideal throttling process let’s
compare what we can observe theoretical assumption. An example of a throttling process
is an ideal gas flowing through a valve in mid position
5.

.
THERMODYNAMICS EQUILIBRIUM:PROCESSES
6.

7.

APPLICATION AREA OF THERMODYNAMICS
Application Areas of Thermodynamics summary
Some examples include the electric or gas range, the heating and air-conditioning systems, the
refrigerator, the humidifier, the pressure cooker, the water heater, the shower, the iron, and even
the computer and the TV. On a larger scale, thermodynamics plays a major part in the design and
analysis of automotive engines, rockets, jet engines, and conventional or nuclear power plants,
solar collectors, and the design of vehicles from ordinary cars to airplanes.
8.

QUESTIONS FOR TOPIC 1
All questions from this chapter are based on explanation (in words) so when meet with any
question refer the notes above…….,
9.

TOPIC 2
THERMODYNAMICS PROPERTIES OF LIQUIDS,PROPERTY
DIAGRAMS
Pure Substance
A substance that has a fixed chemical composition throughout is called pure substance. Water,
helium carbon dioxide, nitrogen are examples. It does not have to be a single chemical element
just as long as it is homogeneous throughout, like air. A mixture of phases of two or more
substance is can still a pure substance if it is homogeneous, like ice and water (solid and liquid)
or water and steam (liquid and gas)
.
Phases of a Pure Substance
There are three principle phases
Solid, liquid and gas, but a substance can have several other phases within the principle phase.
Examples include solid carbon (diamond and graphite) and iron (three solid phases).
Nevertheless, thermodynamics deals with the primary phases only. In general:
 Solids have strongest molecular bonds.
 Solids are closely packed three dimensional crystals.
 Their molecules do not move relative to each other
 Intermediate molecular bond strength
 Liquid molecular spacing is comparable to solids but their molecules can float about
in groups. - There is molecular order within the groups
 Weakest molecular bond strength.
 Molecules in the gas phases are far apart, they have no ordered structure
 The molecules move randomly and collide with each other.
 Their molecules are at higher energy levels, they must release large amounts of
energy to condense or freeze. Vapor Liquid Vapor Liquid Water (Pure substance)
Air (Not a pure substance because the composition of liquid air is different from the
Phase – Change Processes Of Pure Substances
At this point, it is important to consider the liquid to solid phase change process. Not
so much solid to liquid because thermodynamics deals only with liquid to gases (or
vice versa) to generate power.
Consider water at room temperature (20°C) and normal atmospheric pressure (1 atm)
in a piston-cylinder device. The water is in liquid phase, and it is called compressed
liquid or sub cooled liquid (not about to vaporize).
9.

If we add heat to water, its temperature will increase; let us say until 50°C. Due to the
increase in temperature, the specific volume v will increase. As a consequence, the
piston will move slightly upward therefore maintaining constant pressure (1 atm).
Now, if we continue to add heat to the water, the temperature will increase further
until 100°C. At this point, any additional addition of heat will vaporize some water.
This specific point where water starts to vaporize is called saturated liquid. (Point 2)
If we continue to add heat to water, more and more vapor will be created, while the
temperature and the pressure remain constant (T = 100°C and P = 1 atm). The only
property that changes is the specific volume. These conditions will remain the same
until the last drop of liquid is vaporized. At this point, the entire cylinder is filled with
vapor at 100°C. This state is called saturated vapor (Point 4)
The state between saturated liquid (only liquid) and saturated vapor (only vapor)
where two phases exist is called saturated liquid-vapor mixture. (Point 3)
10.

After the saturated vapor phase, any addition of heat will increase the temperature of
the vapor, this state is called superheated vapor (Point 5)
11.

12.

Saturation Temperature And Saturation Pressure
Recall that during a phase change, pressure and temperature are not independent
intensive properties. As a consequence, the temperature at which water starts boiling
depends on the pressure. In other words, water starts boiling at 100 ºC but only at
1 atm. At different pressures, water boils at different temperatures.
At a given pressure, the temperature at which a pure substance changes phase is
called the saturation temperature (Tsat).
Likewise, at a given temperature, the pressure at which a pure substance changes
phase is called the saturation pressure (Psat)
Saturation is defined as a condition in which a mixture of vapor and liquid can exist
together at a given temperature and pressure
Saturation pressure is the pressure at which the liquid and vapor phases are in
equilibrium at a given temperature
For a pure substance there is a definite relationship between saturation pressure and
saturation temperature. The higher the pressure, the higher the saturation temperature
Property Diagrams for Phase Change Processes
T-v Diagram
If we increase the pressure of water in the piston-cylinder device, the process from
compressed liquid to superheated vapor will follow a path that looks like the process
for P = 1 atm, the only difference is that the width of the mixture region will be
shorter. Then, at a certain pressure, the mixture region will be represented only by one
point. This point is called the critical point. It is defined as the point at which the
saturated liquid and saturated vapor states are identical. At the critical point, the
properties of a substance are called critical properties (critical temperature (Tcr),
critical pressure (Pcr) and critical specific volume (vcr))
13.

Figure showing T-v diagram.
If we connect all the points representing saturated liquid we will obtain the saturated
liquid line. If we connect all the points representing saturated vapor we will obtain the
saturated vapor line. The intersection of the two lines is the critical point.
T-v diagram and saturation lines
14.
P-v Diagram

If we consider the pressure-cylinder device, but with some weights above the piston,
if we remove the weights one by one to decrease the pressure, and we allow a heat
transfer to obtain an isothermal process, we will obtain one of the curves of the P-v
diagram.
The P-v diagram can be extended to include the solid phase, the solid liquid and the
solid-vapor saturation regions. As some substances, as water, expand when they
freeze, and the rest (the majority) contracts during freezing process, we have two
configurations for the P-v diagram with solid phase.
P-v diagram for a substance that contracts during freezing (left) and for a
substance that expends during freezing (right).
15.

Triple point
Until now, we have defined the equilibrium between two phases. However, under
certain conditions, water can exist at the same time as ice (solid), liquid and vapor.
These conditions define the so called triple point. On a P-T diagram, these conditions
are represented by a point. Example Water T = 0.01°C = 273.16 K and P = 0.6113kP
Figure showing P-T diagram and the triple point
16.

16.

18.

T-V diagam showing the saturated liquid and saturated water vapour point
Latent heat
Latent heat: The amount of energy absorbed or released during a phase-change process.
Latent heat of fusion: The amount of energy absorbed during melting. It is equivalent to
the amount of energy released during freezing
Latent heat of vaporization: The amount of energy absorbed during vaporization and it
is equivalent to the energy released during condensation.
 At 1 atm pressure, the latent heat of fusion of water is 333.7 kJ/kg and the latent
heat of vaporization is 2256.5 kJ/kg.
Quality and Saturated Liquid-Vapor Mixture
Now, let’s review the constant pressure heat addition process for water shown in
Figure 3 (pg 4). Since state 3 is a mixture of saturated liquid and saturated vapor, how
do we locate it on the T-v diagram? To establish the location of state 3 a new
parameter called the quality x is defined as:
The quality is zero for the saturated liquid and one for the saturated vapor ( 0  x  1).
The average specific volume at any state 3 is given in terms of the quality as follows.
Consider a mixture of saturated liquid and saturated vapor. The liquid has a mass mf
and occupies a volume Vf. The vapor has a mass mg and occupies a volume Vg.
19.

19.

20.

PROPERTY TABLE CONCEPTS
Superheated Water Table
A substance is said to be superheated if the given temperature is greater than the
saturation temperature for the given pressure. State is a superheated state. In the
superheated water Table A-6, T and P are the independent properties. The value of
temperature to the right of the pressure is the saturation temperature for the pressure.
The first entry in the table is the saturated vapor state at the pressure.
Superheated vapor is characterized by:
 Lower pressure (P < Psat at a given T)
 Higher temperature (T > Tsat at a given P)
 Higher specific volumes (v > vg at a given P or T)
 Higher internal energies (u > ug at a given P or T)
 Higher enthalpies (h > hg at a given P or T)
Compressed Liquid Water Table
A substance is said to be a compressed liquid when the pressure is greater than the
saturation pressure for the temperature. It is now noted that state 1 (above pages) is
called a compressed liquid state because the saturation pressure for the temperature T1
is less than P1. Data for water compressed liquid states are found in the compressed
liquid tables, Table A-7. Table A-7 is arranged like Table A-6, except the saturation
states are the saturated liquid states. Note that the data in Table A-7 begins at 5 MPa or
50 times atmospheric pressure. At pressures below 5 MPa for water, the data are
approximately equal to the saturated liquid data at the given TEMPERATURE. We
approximate intensive parameter y, that is v, u, h, and s data as
y  y f@T
The enthalpy is more sensitive to variations in pressure; therefore, at high pressures the
enthalpy can be approximated by
h  h f@T  vf(P  Psat)
For our work, the compressed liquid enthalpy may be approximated by h  hf@T
Compressed liquid is characterized by:
 Higher pressure (P > Psat at a given T)
 Lower temperature (T < Tsat at a given P)
 Lower specific volumes (v < vg at a given P or T)
 Lower internal energies (u < ug at a given P or T)
 Lower enthalpies (h < hg at a given P or T)
21.

How to Choose the Right Table
The correct table to use to find the thermodynamic properties of a real substance can always be
determined by comparing the known state properties to the properties in the saturation region.
Given the temperature or pressure and one other property from the group v, u, h, and s, the
following procedure is used. For example if the pressure and specific volume are specified, three
questions are asked: For the given pressure,
Is v<vf ?
Is vf<v<vg ?
Is v>vg ?
The answer to one of these questions must be yes. If the answer to the first question is yes, the
state is in the compressed liquid region, and the compressed liquid tables are used to find the
properties of the state. If the answer to the second question is yes, the state is in the saturation
region, and either the saturation temperature table or the saturation pressure table is used to find
the properties. Then the quality is calculated and is used to calculate the other properties, u, h,
and s. If the answer to the third question is yes, the state is in the superheated region and the
superheated tables are used to find the other properties.
Some tables may not always give the internal energy. When it is not listed, the internal energy is
calculated from the definition of the enthalpy as u  h  Pv
22.

Equations of State( IDEAL GAS LAW)
The tables with which we have been working provide an accurate relationship between
temperature, pressure and other important thermodynamic properties. If we restrict consideration
to only the vapor state, then we may often find a simple algebraic relationship between
temperature, pressure and specific volume; such a relationship is called an equation of state. For
gases at low pressure it has been observed that pressure, P, is directly proportional to
temperature, T, (this is known as Charles’ law) and inversely proportional to specific volume, v
(Boyle’s law).
P =R.T/v
Here, R is known as the gas constant.
It is further found that there is a relationship between the various gas constants and the molecular
weight of the particular gas. Specifically, it is found that the product of the gas constant and the
gas molecular weight yields the same constant for all gases. This product is known as the
Universal Gas Constant.
23.

By combining the results of Charles' and Boyle's experiments, the following relationship can
be obtained
PV/T =constant
 The constant in the above equation is called the ideal gas constant and is designated by
R; thus the ideal gas equation becomes
Pv = RT or PV = mRT
 In order to make the equation applicable to all ideal gas, a universal gas constant RU is
introduced
R RU/M
24.

.
25.

Important Definitions
 Critical point - the temperature and pressure above which there is no distinction between
the liquid and vapor phases.
 Triple point - the temperature and pressure at which all three phases can exist in
equilibrium.
 Sublimation - change of phase from solid to vapor.
 Vaporization - change of phase from liquid to vapor.
 Condensation - change of phase from vapor to liquid Fusion or melting - change of phase
from solid to liquid
26.

Qn 1/2
Determine the saturated pressure, specific volume, internal energy and enthalpy for saturated
water vapor at 45o
C and 50o
C.
Qn 2/2
Determine the saturated pressure, specific volume, internal energy and enthalpy for saturated
water vapor at 47⁰ C
27.

28.
Qn 3/2 Fill in the blank using R-134a

Qn 4/2
Calculatethespecificvolumeandthespecificenthalpyofa35%qualitysteamatapressureof20kPa
Qn 5/2
Qn 6/2

29.
Qn 7/2

Qn 8/2
Four kg of water is placed in an enclosed volume of 1m3
. Heat is added until the temperature is
150°C. Find ( a ) the pressure, ( b )the mass of vapor, and ( c ) the volume of the vapor.

Qn 9/2
A piston-cylinder device contains 0.1 m3
of liquid water and 0.9 m3
of water vapor in equilibrium at
800 kPa.Heat is transferred at constant pressure until the temperature reaches 350°C.
(a) what is the initial temperature of the water,
(b) determine the total mass of the water,
(c) calculate the final volume, and
(d) show the process on a P-v diagram with respect to saturation lines.

Qn 10/2
For a specific volume of 0.2 m3
/kg, find the quality of steam if the absolute pressure is (a) 40 kPa
and ( b ) 630 kPa. What is the temperature of each case?

Qn 11/2
Water is contained in a rigid vessel of 5 m3
at a quality of 0.8 and a pressure of 2 MPa. If the a
pressure is reduced to 400 kPa by cooling the vessel, find the final mass of vapor mg and mass of
liquid mf.
Qn 12/2
Qn 13/2

Qn 14/2

Qn 15/2
The pressure in an automobile tire depends on the temperature of the air in the tire. When the air
temperature is 25°C, the pressure gage reads 210 kPa. If the volume of the tire is 0.025 m3,
determine the pressure rise in the tire when the air temperature in the tire rises to 50°C. Also,
determine the amount of air that must be bled off to restore pressure to its original value at this
temperature. Assume the atmospheric pressure is 100 kPa. [ 26 kPa, 0.007 kg]

Qn 16/2
A 1-m3
tank containing air at 25°C and 500 kPa is connected through a valve to another tank
containing 5 kg of air at 35°C and 200 kPa. Now the valve is opened, and the entire system is
allowed to reach thermal equilibrium with the surroundings, which are at 20°C. Determine the
volume of the second tank and the final equilibrium pressure of air. [ 2.21 m3
, 284.1 kPa]
Qn 17/2
A 1 m3
rigid tank has propane at 100 kPa, 300 K and connected by a valve to another tank of 0.5
m3
with propane at 250 kPa, 400 K. The valve is opened and the two tanks come to a uniform
state at 325 K. What is the final pressure? [ 139.9 kPa]

Qn 18/2
A cylindrical gas tank 1 m long, inside diameter of 20 cm, is evacuated and then filled with
carbon dioxide gas at 25°C. To what pressure should it be charged if there should be 1.2 kg of
carbon dioxide?
[ 2152 kPa]
Solution T= 298 k: m=1.2kg:

QN 19/2
QN 20/2

QN 21/2 Given: Property table for H2O
QN 23/2

QN24/2
QN 25/2

QN 26/2

QN 27/2
A mass of 12 kg of saturated refrigerant-134a vapor is contained in a piston-cylinder device
at 240 kPa. Now 300 kJ of heat is supplied to the refrigerant at constant pressure while a 110-
V source supplies current to a resistor within the cylinder for 6 min. If the final temperature
(a) Write the full energy balance equation of this system
(b) Find the final phase of the refrigerant-134a
(c) Evaluate the total enthalpy change (in kJ),
(d) Determine the current supplied (in A), [Hint: 1 V = 1 J/C
(e) Show the process on T-v diagram with respect to the saturation lines.
Assume the cylinder is stationary and the kinetic and potential energies are negligible

QN 28/2
A cylinder has a thick piston initially held by a pin as shown in fig below. The cylinder contains
carbondioxide at 200 Kpa and ambient temperature of 290 k. the metal piston has a density of 8000
Kg/m
3
and the atmospheric pressure is 101 Kpa. The pin is now removed, allowing the piston to
move and after a while the gas returns to ambient temperature. Is the piston against the stops?

Conclusion: Pressure is grater than this value. Therefore the piston is resting against the stops.
QN 29/2
A 1-m
3
tank is filled with a gas at room temperature 20
°
C and pressure 100 Kpa. How much mass is
there if the gas is
a) Air
b) Neon, or
c) Propane?

QN 30/2 A 1-m
3
rigid tank with air 1 Mpa, 400 K is connected to an air line as shown in fig: the
valve is opened and air flows into the tank until the pressure reaches 5 Mpa, at which point the
valve is closed and the temperature is inside is 450 K.
a. What is the mass of air in the tank before and after the process?
b. The tank is eventually cools to room temperature, 300 K. what is the pressure inside the tank
then?

QN 31/2
QN 32/2
Two tanks are connected as shown in fig, both containing water. Tank A is at 200 Kpa,ν=1m
3
and tank B contains 3.5 Kg at 0.5 Mp, 400
0
C. The valve is now opened and the two come to a
uniform state. Find the specific volume

QN 33/2
The valve is now opened and saturated vapor flows from A to B until the pressure in B Consider
two tanks, A and B, connected by a valve as shown in fig. Each has a volume of 200 L and tank
A has R-12 at 25
°
C, 10 % liquid and 90% vapor by volume, while tank B is evacuated has
reached that in A, at which point the valve is closed. This process occurs slowly such that all
temperatures stay at 25
°
C throughout the process. How much has the quality changed in tank A
during the process?

QN 34/2

QN 35/2
QN 36/2
QN37/2
QN 38/2

QN 39/2

QN 40/2
QN 41/2

QN 42/2 (figure above)
Qn 43/2

TOPIC 3
FIRST LAW OF THERMODYNAMICS
Introduction
The first law of thermodynamics is a version of the law of conservation of energy, adapted
for thermodynamic systems. The law of conservation of energy states that the total energy of
an isolated system is constant; energy can be transformed from one form to another, but can be
neither created nor destroyed. The first law is often formulated as
ΔE=Q-W.
It states that the change in the internal energy ΔE of a closed system is equal to the amount
of heat Q supplied to the system, minus the amount of work W done by the system on its
surroundings. An equivalent statement is that perpetual motion machines of the first kind are
impossible. Simply first law of Thermodynamics Used in piston-cylinder problems.

NOTE: Since the piston-cylinder is a closed system, we normally use the Closed System version
of the law. An exception occurs when the piston is allowed to move as the gas expands under
constant pressure. In this case, there is boundary work Wb, which can be included on the right-
hand side of the equation by using the Open Systems version since ΔU + Wb = ΔH.
 Q = net heat transfer across system boundaries, positive when flowing inward [kJ]
 W = net work done in all forms, positive when flowing outward [kJ]
 ΔE = net change in the total energy of the system [kJ]
 ΔU = net change in the internal energy of the system [kJ]
 ΔKE = net change in the kinetic energy of the system [kJ]
 ΔPE = net change in the potential energy of the system [kJ]
 m = mass [kg]
 u = internal energy [kJ/kg]
 h = enthalpy [kJ/kg]
 Cp,avg = specific heat at constant pressure, averaged for the two temperatures
[kJ/(kg·°C)]
 Cv,avg = specific heat at constant volume, averaged for the two temperatures
[kJ/(kg·°C)]

Energy Balance for Closed System

Mass and Energy Analysis of Control Volumes (Open Systems)

Compressors and fans
Compressors and fans are essentially the same devices. However, compressors operate over
larger pressure ratios than fans. If we neglect the changes in kinetic and potential energies as
fluid flows through an adiabatic compressor having one entrance and one exit, the steady-state,
steady-flow first law or the conservation of energy equation becomes

Throttling devices
Consider fluid flowing through a one-entrance, one-exit porous plug. The fluid experiences a
pressure drop as it flows through the plug. No net work is done by the fluid. Assume the
process is adiabatic and that the kinetic and potential energies are neglected; then the
conservation of mass and energy equations become

This process is called a throttling process. What happens when an ideal gas is throttled?
When throttling an ideal gas, the temperature does not change.The throttling process is an
important process in the refrigeration cycle.
A throttling device may be used to determine the enthalpy of saturated steam. The steam is
throttled from the pressure in the pipe to ambient pressure in the calorimeter. The pressure drop
is sufficient to superheat the steam in the calorimeter. Thus, the temperature and pressure in the
calorimeter will specify the enthalpy of the steam in the pipe.
Mixing Chamber
The section where the mixing process takes place. An ordinary T-elbow or a Y-elbow in a
shower, for example, serves as the mixing chamber for the cold- and hot-water streams.
Energy Balance:↑
Heat Exchanger
Devices where two moving fluid streams exchange heat without mixing.Heat exchangers
typically involve no work interactions (w = 0) and negligible kinetic and potential energy
changes for each fluid stream
 
   
1 1 2 2 3 3
1 1 3 1 2 3 3
1 1 2 3 3 2
3 2
1 3
1 2
m h m h m h
m h m m h m h
m h h m h h
h h
m m
h h
 
  
  
 
  
 

.
Summary
• Conservation of mass
 Mass and volume flow rates
 Mass balance for a steady-flow process
 Mass balance for incompressible flow
• Flow work and the energy of a flowing fluid
 Energy transport by mass
• Energy analysis of steady-flow systems
• Some steady-flow engineering devices
 Nozzles and Diffusers
 Turbines and Compressors
 Throttling valves
 Mixing chambers and Heat exchangers
 Pipe and Duct flow

QN1/3
QN2/3

QN3/3

QN4/3

QN5/3 A piston cylinder device contains 0.2 kg of water initially at 800 kPa and 0.06 m3
. Now
200 kJ of heat is transferred to the water while its pressure is held constant. Determine the final
temperature of the water. Also, show the process on a T-V diagram with respect to saturation
lines. [ 721.1o
C]

QN6/3 A piston-cylinder device contains 6 kg of refrigerant-134a at 800 kPa and 50o
C. The
refrigerant is now cooled at constant pressure until it exist as a liquid at 24o
C. Show the process
on T-v diagram and determine the heat loss from the system. State any assumption made.
[1210.26 kJ]
QN7/3 A 0.5 m3
rigid tank contains refrigerant-134a initially at 200 kPa and 40 percent quality.
Heat is now transferred to the refrigerant until the pressure reaches 800 kPa. Determine (a) the
mass of the refrigerant in the tank and (b) the amount of heat transferred. Also, show the process
on a P-v diagram with respect to saturation lines.
solution
The mass of refrigerant may be found from the initial state using the equation that m = V / v
where V = 0.5 m3
and v is found from the temperature and the quality. Since we are given an
initial quality, x1 = 40%, we know that we are in the mixed region. The specific volume is found
from the specific volumes of the saturated liquid and vapor, which are found in a steam Table ;
vf(200 kPa) = 0.0007532 m3
/kg and vg(200 kPa) = 0.0993 m3
/kg. We then find the initial
specific volume as follows.
kg
m04017.0
kg
m0993.0)4.0(
kg
m000753.0)4.01(xvv)x1(v
333
gf 








With this specific volume, we then find the refrigerant mass as follows:
kg
m04017.0
m5.0
v
V
m 3
3
 = 12.45 kg
To compute the heat transfer we apply the first law, Q = DU + W. We assume that there is no
volume change in the ―rigid‖ tank. If there is no volume change, no work is done. With W = 0,
Q = DU. We find DU = m(u2 – u1) where the specific internal energies are found from the
property tables. At the initial state we find u from the quality in the same way that we found the
volume.
kg
kJ59.110
kg
kJ43.221)4.0(
kg
kJ69.36)4.01(xuu)x1(u gf 








The final state has the same specific volume as the initial state (0.04017 m3
/kg, because of the
constant volume process) and a given pressure of 800 kPa. From the superheat table, A-13, on
page 845, we see that the specific volume of 0.04017 m3
/kg occurs at a pressure of 800 kPa (0.8
MPa) between 140o
C and 150o
C. The internal energy at the final state is found by interpolation.

kg
kJ
kg
m
kg
m
kg
m
kg
m
kg
kJ
kg
kJ
kg
kJu 84.34903997.004017.0
03997.004113.0
09.34815.358
09.348
33
33




 



We can now find the heat transfer as follows.




 
kg
kJ59.110
kg
kJ84.349)kg45.12()uu(mQ 12
= 2,978 kJ
The P-v diagram for this problem is shown at the right. For the linear scale used here, the
saturated liquid line is too close to the v = 0 axis to be shown. The constant volume process is
seen to move from an initial state inside the mixed region to a final state in the superheated vapor
region
QN8/3 An insulated tank is divided into two parts by a partition. One part of the tank contains 6
kg of an ideal gas at 50°C and 800 kPa while the other part is evacuated. The partition is now
removed, and the gas expands to fill the entire tank. Determine the final temperature and the
pressure in the tank.

solution
The most important thing to recognize in this problem is that removing the partition is equivalent
to allowing the partition to move the to the right (in our diagram) until the ideal gas fills the entire
tank. The resisting force in this expansion process is zero because there is a vacuum in the right
chamber of the tank. Since the resisting force is zero, the work done by the expanding gas is also
zero. If we take the ideal gas to be our system, there is no heat transfer during the expansion
either, because the tank is insulated. The 1st Law tells us that ΔU = 0when no work or heat
transfer occur during a process on a closed system. Also, because U is a function of T only for an
ideal gas, T2 = T1. Then, all we need to do is apply the IG EOS to determine P2.
Given m 6 kg asked: P2 ??? kPa
P1 800 kPa
T1 50 °C
Assumptions:- The gas behaves as an ideal gas
- Changes in kinetic and potential energy are negligible.
- The tank is perfectly insulated, so the process is adiabatic: Q = 0.
For an ideal gas, internal energy depends only on the temperature of the gas. If the
internal energy is the same in state 2 as in state 1, then the temperature in state 2
must also be the same as the temperature in state 1 !
But, we know that T1 = T2 and because the system is
closed, n1 = n2. Also, because the left and right chambers
of the tank are equal in size, V2 = 2 V1
or 5

P2=800kpa/2
P2=400kpa
Now the final temperature is 50o
c and p =400kpa
QN9/3 Sketch a P-V diagram showing the following processes in a cycle
Process 1-2: isobaric work output of 10.5 kJ from an initial volume of 0.028 m3
and pressure
1.4 bar,
Process 2-3: isothermal compression, and
Process 3-1: isochoric heat transfer to its original volume of 0.028 m3
and
pressure 1.4 bar.
Calculate (a) the maximum volume in the cycle, in m3
, (b) the isothermal work, in kJ, (c) the net
work, in kJ, and (d) the heat transfer during isobaric expansion, in kJ.
The net work
 
31
12 23 31
0
10.5 18.78
8.2
1
8
3
net
W
W W
Section isochoric
W W
kJ

   
 
 


QN10/3 A fluid at 4.15 bar is expanded reversibly according to a law PV = constant to a pressure
of 1.15 bar until it has a specific volume of 0.12 m3
/kg. It is then cooled reversibly at a constant
pressure, then is cooled at constant volume until the pressure is 0.62 bar; and is then allowed to
compress reversibly according to a law PVn
= constant back to the initial conditions. The work
done in the constant pressure is 0.525 kJ, and the mass of fluid present is 0.22 kg. Calculate the
value of n in the fourth process, the net work of the cycle and sketch the cycle on a P-V diagram.

QN11/3 A mass of 0.15 kg of air is initially exists at 2 MPa and 350o
C. The air is first
expanded isothermally to 500 kPa, then compressed polytropically with a polytropic exponent of
1.2 to the initial state. Determine the boundary work for each process and the net work of the
cycle.
Solution 1. If air properties are calculated at 300 K, then from Table A-2a R = 0.287 kJ/kg .K
k = 1.4
If air properties are calculated at 350 K, then from Table A-2b
R = cp − cv = 1.008 − 0.721 = 0.287 kJ/kg ・ K
k = 1.398

QN12/3 0.078 kg of a carbon monoxide initially exists at 130 kPa and 120o
C. The gas is then
expanded polytropically to a state of 100 kPa and 100o
C. Sketch the P-V diagram for this
process. Also determine the value of n (index) and the boundary work done during this process.
[1.248,1.855 kJ]
QN13/3 A system contains 0.15 m3
of air pressure of 3.8 bars and 150⁰ C. It is expanded
adiabatically till the pressure falls to 1.0 bar. The air is then heated at a constant pressure till its
enthalpy increases by 70 kJ. Sketch the process on a P-V diagram and determine the total work
done. Use cp=1.005 kJ/kg.K and cv=0.714 kJ/kg.K
QN14/3 Two kg of air experiences the three-process cycle shown in Fig. 3-14. Calculate the net
work
.

QN15/3
QN16/3

QN17/3

QN18/3
Thus
QN19/3
 0.02 400.98 280.13 16
2.74
inW
kW
    


QN20/3
QN21/3 Air flows through the supersonic nozzle . The inlet conditions are 7 kPa and 420°C.
The nozzle exit diameter is adjusted such that the exiting velocity is 700 m/s. Calculate ( a ) the
exit temperature, ( b )the mass flux, and ( c ) the exit diameter. Assume an adiabatic
quasiequilibrium flow.

QN22/3 Steam at 5 MPa and 400°C enters a nozzle steadily velocity of 80 m/s, and it leaves at 2
MPa and 300°C. The inlet area of the nozzle is 50 cm2
, and heat is being lost at a rate of 120
kJ/s. Determine (a) the mass flow rate of the steam, (b) the exit velocity of the steam, and (c) the
exit area nozzle.

QN23/3 Steam enters a turbine at 4000 kPa and 500o
C and leaves as shown in Fig A below. For
an inlet velocity of 200 m/s, calculate the turbine power output. ( a )Neglect any heat transfer and
kinetic energy change ( b )Show that the kinetic energy change is negligible.
QN24/3 Consider an ordinary shower where hot water at 60°C is mixed with cold water at
10°C. If it is desired that a steady stream of warm water at 45°C be supplied, determine the ratio
of the mass flow rates of the hot to cold water. Assume the heat losses from the mixing chamber
to be negligible and the mixing to take place at a pressure of 150 kPa.

Substitute…………………………..data.
QN+
The electric heating systems used in many hoses consist of a simple duct with resistance
heaters. Air is heated as it flows over resistance wires. Consider a 15-kW electric heating system.
Air enters the heating section at 100 kPa and 17˚C with a volume flow rate of 150 m3
/min. If
heat is lost from the air in the duct to the surroundings at a rate of 200 W, determine the exit
temperature of air.

QN+
Air at 100 kPa and 280 K is compressed steadily to 600 kPa and 400 K. The mass flow rate
of the air is 0.02 kg/s, and the heat loss of 16 kJ/kg occurs during the process. Assuming the
changes in kinetic and potential energies are negligible. Determine the necessary power input to
the compressor.

QN A cylinder of total volume 1.2 m3
is divided into two compartments A and B
of equal volume by a thin, frictionless, adiabatic piston as shown in Fig. E4.14.
Initially, compartment A contains air and B helium, both at 30o
C and 100 kPa. The
cylinder is well-insulated except for one face of A which is a diathermal wall. Heat
is supplied to A in a quasi-static manner until a final equilibrium state where the
pressureis 220 kPa is attained. Calculate (i) the final temperature of the helium and
the air, (ii) the work done by the air and (iii) the heat input to the air. Assume that
helium and air are ideal gases with the following specificheat capacities. For
helium = 3.12 v c kJkg-1K-1, = 5.21 p c kJkg-1K-1 and for air v c = 0.72 kJkg-1K-
1 and p c = 1.00 kJkg-1K

QN25/3 Refrigerant-134a is to be cooled by water in a condenser. The refrigerant enters the
condenser with a mass flow rate of 6 kg/min at 1 MPa and 70ºC and leaves at 35°C. The cooling
water enters at 300 kPa and 15°C and leaves at 25ºC. Neglecting any pressure drops, determine
(a) the mass flow rate of the cooling water required and (b) the heat transfer rate from the
refrigerant to water.

h1 = hf@15C = 62.99 kJ/kg

Qn 26/3
In a non-flow process there is heat transfer loss of 1055 kJ and an internal energy increase
of 210 kJ. Determine the work transfer and state whether the process is an expansion or
compression. [Ans: -1265 kJ, compression]
solutions
closed system for which the first law of thermodynamics applies
Q-W=ΔU
-1055-W=210
Since the workdone can be found as W= -1265KJ
Since negative ,it must be work input,ie compression.
Qn 27/3
In a non-flow process carried out on 5.4 kg of a substance, there was a specific internal
energy decrease of 50 kJ/kg and a work transfer from the substance of 85 kJ/kg. Determine
the heat transfer and state whether it is gain or loss. [Ans: 189 kJ, gain]
Qn 28/3
During the working stroke of an engine the heat transferred out of the system was 150
kJ/kg of the working substance. If the work done by the engine is 250 kJ/kg, determine the
change in internal energy and state whether it is decrease or increase.[Ans: -400 kJ/kg, decrease]
solution
closed system for which the first law applies

since the sign is negative there is decrease in internal energy
Qn 29/3
Steam enters a cylinder fitted with a piston at a pressure of 20 MN/m2 and a temperature of
500 deg C. The steam expands to a pressure of 200 kN/m2 and a temperature of 200 deg C.
During the expansion there is a net heat loss from the steam through the walls of the
cylinder and piston of 120 kJ/kg. Determine the displacement work done by one kg of
steam during this expansion.

Qn 30/3
A closed rigid system has a volume of 85 litres contains steam at 2 bar and dryness fraction
of 0.9. Calculate the quantity of heat which must be removed from the system in order to
reduce the pressure to 1.6 bar. Also determine the change in enthalpy and entropy per unit
mass of the system. [Ans: 98.9 kJ]
solution

Qn 31/3
2kg of air is heated at constant pressure of 2 bar to 500 o
C. Determine the change in its
entropy if the initial volume is 0.8 m3. [Ans: 2.05 kJ/K]
solution

Qn 32/3
A boiler is designed to work at 14 bar and evaporate 8 kg/s of water. The inlet water to the
boiler has a temperature of 40 deg C and at exit the steam is 0.95 dry. The flow velocity at
inlet is 10 m/s and at exit 5 m/s and the exit is 5 m above the elevation at entrance.
Determine the quantity of heat required. What is the significance of changes in kinetic and
potential energy on the result? [Ans: 20.186 MW]

Qn 33/3
Steam flows along a horizontal duct. At one point in the duct the pressure of the steam is 1
bar and the temperature is 400°C. At a second point, some distance from the first, the
pressure is 1.5 bar and the temperature is 500°C. Assuming the flow to be frictionless and
adiabatic, determine whether the flow is accelerating or decelerating. [Ans: Decelerating]

Qn 34/3
Steam is expanded isentropically in a turbine from 30 bar and 400°C to 4 bar. Calculate
the work done per unit mass flow of steam. Neglect changes in Kinetic and Potential
energies. [Ans: 476 kJ/kg]

Qn 35/3
A compressor takes in air at 1 bar and 20°C and discharges into a line. The average air
velocity in the line at a point close to the discharge is 7 m/s and the discharge pressure is
3.5 bar. Assuming that the compression occurs isentropically, calculate the work input to
the compressor. Assume that the air inlet velocity is very small. [Ans: -126.6 kW/kg]

Qn 36/3
Air is expanded isentropically in a nozzle from 13.8 bar and 150°C to a pressure of 6.9 bar.
The inlet velocity to the nozzle is very small and the process occurs under steady-flow,
steady-state conditions. Calculate the exit velocity from the nozzle knowing that the nozzle
is laid in a horizontal plane and that the inlet velocity is 10 m/s. [Ans: 390.9 m/s]
solution

Qn 37/3
A rotary air compressor takes in air (which may be treated as a perfect gas) at a pressure of
1 bar and a temperature of 20°C and compress it adiabatically to a pressure of 6 bar. The
isentropic efficiency of the processes is 0.85 and changes in kinetic and potential energy
may be neglected. Calculate the specific entropy change of the air. Take R = 0.287 kJ/kg
K and Cp = 1.006 kJ/kg K. [Ans: 0.05 kJ/kg K]
solution

Qn 38/3
An air receiver has a capacity of 0.86m3 and contains air at a temperature of 15°C and a
pressure of 275 kN/m2. An additional mass of 1.7 kg is pumped into the receiver. It is then
left until the temperature becomes 15°C once again. Determine,
a) the new pressure of the air in the receiver, and
b) the specific enthalpy of the air at 15°C if it is assumed that the specific enthalpy of the
air is zero at 0°C. Take Cp = 1.005 kJ/kg, Cv = 0.715 kJ/kg K [Ans: 442 kN/m2, 15.075 kJ/kg]
solution
Qn 39/3
Oxygen has a molecular weight of 32 and a specific heat at constant pressure of 0.91 kJ/kg K.
a) Determine the ratio of the specific heats.
b) Calculate the change in internal energy and enthalpy if the gas is heated from 300 to 400 K.

Qn 40/3
A steam turbine inlet state is given by 6 MPa and 500°C. The outlet pressure is 10 kPa.
Determine the work output per unit mass if the process:-
a) is reversible and adiabatic (ie 100% isentropic),
b) such that the outlet condition is just dry saturated,
c) such that the outlet condition is 90% dry.
solution

Qn 41/3
Determine the volume for carbon dioxide contained inside a cylinder at 0.2 MPa, 27°C:-
a) assuming it behaves as an ideal gas
b) taking into account the pressure and volume associated with its molecules

Qn 42/3
A cylindrical storage tank having an internal volume of 0.465 m3
contains methane at 20°C
with a pressure of 137 bar. If the tank outlet valve is opened until the pressure in the cylinder is
halved, determine the mass of gas which escapes from the tank assuming the
tank temperature remains constant.

Qn 43/3
Find the specific volume for H20 at 1000 kN/m2 and 300°C by using:-
a) the ideal gas equation assuming R = 461.5 J/kg K

b) steam tables
Qn 44/3
Determine the specific volume of steam at 6 MPa using the steam tables for the following
conditions:-
a) dryness fraction x = 0
b) dryness fraction x = 0.5
c) dryness fraction x = 1
d) its temperature is 600o
C

Qn 45/3
Steam at 4 MPa, 400oC expands at constant entropy till its pressure is 0.1 MPa.
Determine:
a) the energy liberated per kg of steam
b) repeat if the process is 80% isentropic

Qn 46/3
Steam (1 kg) at 1.4 MPa is contained in a rigid vessel of volume 0.16350 m3
. Determine its
temperature.
a) If the vessel is cooled, at what temperature will the steam be just dry saturated?
b) If cooling is continued until the pressure in the vessel is 0.8 MPa; calculate the final dryness
fraction of the steam, and the heat rejected between the initial and the final states.

Qn 47/3
Steam (0.05 kg) initially saturated liquid, is heated at constant pressure of 0.2 MPa until its
volume becomes 0.0658 m3. Calculate the heat supplied during the process.

Qn 48/3
Steam at 0.6 MPa and dryness fraction of 0.9 expands in a cylinder behind a piston
isentropically to a pressure of 0.1 MPa. Calculate the changes in volume, enthalpy and
temperature during the process.

Qn 49/3
The pressure in a steam main pipe is 1.2 MPa; a sample is drawn off and throttled where its
pressure and temperature become 0.1 MPa, 140oC respectively. Determine the dryness
fraction of the steam in the main stating reasonable assumptions made!
Qn 50/3
A boiler receives feed water at 20 kPa as saturated liquid and delivers steam at 2 MPa and
500oC. If the furnace of this boiler is oil fired, the calorific value of oil being 42000 kJ/kg
determine the efficiency of the combustion when 4.2 tonnes of oil was required to process
42000 kg of steam.

Qn 51/3
10 kg/s steam at 6 MPa and 500o
C, expands isentropically in a turbine to a pressure of 100
kPa. If the heat transfer from the casing to surroundings represents 1 per cent of the overall
change of enthalpy of the steam, calculate the power output of the turbine. Assume exit is 5
m above entry and that initial velocity of steam is 100 m/s whereas exit velocity is 10 m/s.

Qn 52/3
Four kilograms of a certain gas is contained within a piston cylinder assembly.The gas undergoes
a process for which the pressure volume relationship is pv1.5
=constant.The initial pressure is 3
bar,the initial volume is 0.1m3
and the final volume is 0.2m3
.The change in specific internal
energy is U2-U1=-4.6kJ/kg.There are no significant changes in kinetic or potential
energy.Determine the net heat transfer for the process , in KJ.
Solution

Qn 53/3
Measured data for pressure versus volume during the expansion of gases within the cylinder of
an internal combustion engine are given in the table below.Using the data from the
table,complete the following:
I. Determine a value a value of n such that the data the data are fit by an equation of
the form PVn
=constant.
II. Using the graphical or numerical integration of the data ,evaluate the work done
by the gases ,in KJ
Data point P (bar) V (cm3
)

*The author preferred to use A MATLAB SOFTWARE to simplify solving but you insert
in a normal equation still will obtain same values ( ITS PRACTICAL ORIANTED)
1
2
3
4
5
6
15
12
9
6
4
2
300
361
459
644
903
1608

OR USE
Where Ai is the area of each rectangle interval, and n is the number of rectangle intervals

Qn 54/3
A 1kg of air expands reversely according to a liner law (pv=constant) from 4.0 bar to 1.0 bar:
initial and final volumes are 0.006m3
and 0.05m3 respectively .The fluid is then cooled
reversibly at a constant pressure and finally reversibly according to a lwa of pv1.2
=constant back
to the initial condiions of 4.0 bar and 0.006m3
.Calculate the total workdone in each process and
the net-work of the cycle.Sketch the cycle on p-v diagram.
Qn 55/3
Air enters a compressor and is compressed adiabatically from 0.1Mpa,27o
C to a state of
0.5Mpa.Find the work done on the air for a compressor drop adiabatically of 80%.
solution
System: The compressor control volume
Property Relation: Ideal gas equations, assume constant properties
2a
2s
1 s

QN 56/3 A gas in a piston–cylinder assembly undergoes an expansion process for which the
relationship between pressure and volume is given by
The initial pressure is 3 bar, the initial volume is 0.1 m3, and the final volume is 0.2 m3.
Determine the work for the process, in kJ, if (a) n = 1.5, (b) n = 1.0, and (c) n = 0.
Solution
Known: A gas in a piston–cylinder assembly undergoes an expansion for which pVn _ constant.
Find: Evaluate the work if (a) n _ 1.5, (b) n _ 1.0, (c) n _ 0.
Schematic and Given Data: The given p–V relationship and the given data for pressure and
volume can be used to construct
the accompanying pressure–volume diagram of the process
Assumptions:
1. The gas is a closed system.
2. The moving boundary is the only work mode.
3. The expansion is a polytropic process.
Analysis: The required values for the work are obtained by integration of Eq. 3.9 using the given pressure–volume relation.
(a) Introducing the relationship p _ constant_Vn into Eq. 3.9 and performing the integration
The constant in this expression can be evaluated at either end state: constant = p1V1n
= p2V2n
The work expression then becomes

QN 57/3 A 12-V automotive storage battery is charged with a constant current of 2 amp for 24 h.
If electricity costs $0.08 per kW·h, determine the cost of recharging the battery.
QN58/3 Each line in the following table gives about a process of a closed system. Each line in
the following table Every entry has the same energy units. Fill in the blank spaces in the table.

QN 65/3

from which h2 = 2573 kJ/kg , The turbine specific work (m=1 kg/s) is =W = m *(h2 – h1) = 1 x (
3422.2 – 2573) = 849 kJ/kg.
QN 66/3
A steam turbine receives steam at 2 MPa and 250 oC, and exhausts at 0.1 MPa, 0.85 dry.
a) Neglecting heat losses and changes in ke and Pe, estimate the work output per kg steam.
b) If, when allowance is made for friction, radiation, and leakage losses, the actual work
obtained is 80% of that estimated in (a), calculate the power output of the turbine when
consuming 600 kg of steam per minute.

QN 67/3
30 kg/s steam at 3 MPa,300o
C expands isentropically in a turbine to a pressure of 100 kPa. If the
heat transfer from the casing to surrounding air represents 1 per cent of the overall change of
enthalpy of the steam, calculate the power output of the turbine. Assume exit is 2 m above entry
and that initial velocity of steam is 10 m/s whereas exit velocity is 1 m/s.

QN 68/3
200 gram of air initially at 0/6 bar and 20oc undergoes the cycle consinsting of the three quasi
static process shown in a fegure

200g of air initially at 0.6 bar and 20oC undergoes the cycle consisting of the three quasi-static processes
shown in Fig. E4.11. The air is compressed isothermally to a pressure of 1.0 bar during the process 1-2.
From 2 to 3 the air is heated in a constant volume process until the pressure is 2 bar. The adiabatic
expansion from 3 to 1 returns the air to the initial state, completing the cycle. Calculate (i) the work done,
the change in internal energy and the heat transfer during the processes 1-2, 2-3 and 3-1 and (ii) the net
work done during the cycle. Assume that air is an ideal gas with constant specific heat capacities, = .0 718
vc kJkg-1
K -1
and = 1. 005 p c kJkg-1
K -1

APPLIED THERMODYNAMICS PAMPHLET

Recommended

Recommended

More Related Content

What's hot

What's hot (20)

Similar to APPLIED THERMODYNAMICS PAMPHLET

Similar to APPLIED THERMODYNAMICS PAMPHLET (20)

More from musadoto

More from musadoto (20)

Recently uploaded

Recently uploaded (20)

APPLIED THERMODYNAMICS PAMPHLET