Introduction to IEEE STANDARDS and its different types.pptx
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Energy balance (leacture 6)
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3. LECTURE 6
Energy Balance
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4. ο FORMS OF ENERGY: THE FIRST LAW OF THERMODYNAMICS
1. Kinetic Energy
2. Potential Energy
3. Internal Energy
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5. Suppose a process system is closed, meaning that no mass is transferred
across its boundaries while the process is taking place. Energy may be
transferred between such a system and its surroundings in two ways:
1. As heat, or energy that flows as a result of temperature difference between
a system and its surroundings. The direction of flow is always from a higher
temperature to a lower one. Heat is defined as positive when it is
transferred to the system from the surroundings.
2. As work, or energy that flows in response to any driving force other than a
temperature difference, such as force, a torque, or a voltage.
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6. - The principle that underlies all energy balances is the law of conservation of
energy, which states that energy can neither be created nor destroyed. This
law is also called as βfirst law of thermodynamicsβ.
- In its most general form, the first law states that the rate at which energy
(K.E.+P.E.+I.E.) is carried into a system by the input streams, plus the rate at
which it enters as heat, minus the rate at which it leaves as work, equals the
rate of accumulation of energy in the system (accumulation = input β output)
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7. ο Kinetic & Potential Energy
πΈπ = 1/2ππ£2 πΈπ = πππ§
1. Water flows into a process unit through a 2-cm ID pipe at a rate of 2.00
π3/h. Calculate πΈπ for this stream in joules/sec.
2. Crude oil is pumped at a rate of 15 kg/s from a point 220 m below the
earthβs surface to a point 20 m above ground level. Calculate the
attendant rate of increase of potential energy.
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8. ο ENERGY BALANCES ON CLOSED SYSTEM
- A system is termed as open or close according to whether or not mass
crosses the system boundary during the period of time covered by the
energy balance.
- A batch process is by definition, closed, and semi batch and continuous
are open.
ππππππππππππ = πππππ β ππππππ
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9. πππππ π π¦π π‘ππ energy β ππππ‘πππ π π¦π π‘ππ ππππππ¦
= πππ‘ ππππππ¦ π‘ππππ ππππππ π‘π π‘βπ π π¦π π‘ππ ππ β ππ’π‘
ππππ‘πππ π π¦π π‘ππ ππππππ¦ = ππ + πΈππ + πΈππ
πππππ π π¦π π‘ππ ππππππ¦ = ππ + πΈππ + πΈππ
ππππππ¦ π‘ππππ ππππππ = π β π
Now the equation becomes,
ππ βππ + πΈππ β πΈππ + πΈππ β πΈππ = π β π
Or
βπ + βπΈπ + βπΈπ = π β π
This equation is the basic form of the first law of thermodynamics for a closed
system.
10. When we apply this equation to a given process, we should know the following
point:
1. The internal energy of the system depends almost entirely on the chemical
composition, state and temperature of the system materials. It is independent
of pressure for ideal gases and nearly independent of pressure for liquids and
solids. If no temperature changes, phase changes or chemical reactions
occur in a closed system and if pressure changes are less than a few
atmospheres, then βπΌ β π.
2. If a system is not accelerating, then βπ¬π = π. If a system is not rising or
falling then βπ¬π = π.
3. If a system and its surroundings are at the same temperature or the
system is perfectly insulated then πΈ = π. The process is then termed as
adiabatic.
13. 2.
βπ + βπΈπ + βπΈπ = π β π
βπΈπ = 0 (π π¦π π‘ππ π π‘ππ‘ππππππ¦)
βπΈπ = 0 (ππ π π’πππ ππππππππππ)
βπ = 0
0 = Q β W
π = +100 π½
π π, π = 100 π½
Thus an additional 100 J of heat is transferred to the gas as it expands and re-
equilibrates at 100ΛC.
14. ο ENERGY BALANCES ON OPEN SYSTEMS AT STEADY STATE
- An open system by definition has mass crossing its boundaries as the process
occurs.
- Work must be done on such system to push mass in and work is done on the
surroundings by mass that emerges.
- Both work term must be included in the energy balance.
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15. ο§ Flow work and Shaft work
- Net rate of work done by an open system on its surroundings may be written as:
π = ππ + πππ
where,
ππ = shaft work, or rate of work done by the process fluid on a moving part within
the system.
πππ = flow work, or rate of work done by the fluid at the system outlet minus the
rate of work done on the fluid at the system inlet.
16. To derive an expression for πππ, we initially consider single-inlet-single-outlet
system shown here,
πππ = πππ πππ
(π. π/π ) = (π/π2) (π3/π )
πππ’π‘ = πππ’π‘ πππ’π‘
πππ = πππ’π‘ πππ’π‘ β πππ π
ππ
Process Unit
π½ππ(ππ/π)
π·ππ(π΅/ππ)
π½ πππ(ππ/π)
π·πππ(π΅/ππ)
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17. ο§ Specific Properties & Enthalpy
- A specific property is an intensive quantity obtained by dividing an extensive
property (or its flow rate) by the total amount (or flow rate) of the process
material.
- Example:
i) volume of fluid is 200ππ3 and mass of the fluid is 200g, then specific
volume of the fluid is 1πππ/g.
ii) If the rate at which kinetic energy is transported by a stream is 300 J/min,
having mass flow rate 100 kg/min, then the specific kinetic energy of the
stream material is 3J/kg.
18. - This property can be denoted by a capital letter with a cap, example: π will
denote specific volume, π will denote specific internal energy and so on.
- If the temperature and pressure of a process material are such that the specific
internal energy of the material is π (J/kg), then mass (kg) of this material has a
total internal energy:
U(J) = m(kg) πΌ(J/kg)
Similarly, πΌ (J/s) = π (kg/s) πΌ(J/kg)
- A property that occurs in the energy balance equation for open systems is
specific enthalpy, defined as:
π― = πΌ + π·
π½
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19. οTHE STEADY-STATE OPEN-SYSTEM ENERGY BALANCE
βπ»+ βπΈ π + βπΈ π = πβ π
π
We will use this equation as the starting point for most energy balance
calculation on open systems at steady state.
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20. Example 7.4-2
Five hundred kilograms per hour of steam drives a turbine. The steam enters
the turbine at 44 atm and 450β at a linear velocity of 60 m/s and leaves at a
point 5m below the turbine inlet at atmospheric pressure and a velocity of 360
m/s. The turbine delivers the shaft work at a rate of 70 kW, and the heat loss
from the turbine is estimated to be 104 kcal/h. Calculate the specific enthalpy
change associated with the process.
Solution
500 kg/h
44 atm, 450β
60 m/s
500 kg/h
1 atm,
360 m/s
5 m
πΈ = βπππ ππππ/π πΎπ = ππππΎ
21. Solution:
βπ» = π β ππ β βπΈ π β βπΈ π β (
π΄)
For units consistency,
π = 500
ππ
β
3600 = 0.139
π ππ
β π
1
2
βπΈ π= π 2 1
π£2 β π£2 =
1 ππ
2 π
0.139 3602 β 602
π2
π 2
= 8757 π β π/π
βπΈπ= 8.75 π₯ 103 π ππ 8.75 ππ
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23. The question says we have to calculate specific enthalpy,
So,
βπ» = π π»2 β π»1
ππ
π
βπ»
= β
π»
β
π» =
β90.3 ππππ
= β650
ππ½
0.139 ππ/π ππ
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24. Example 7.6-1
Two streams of water are mixed to form the feed to a boiler. Process data are as
follows:
πΉπππ π π‘ππππ 1
πΉπππ π π‘ππππ 2
π΅πππππ ππππ π π’ππ
120 ππ/ min @ 30β
175 ππ/min @ 65β
17 πππ πππ πππ’π‘π
The exiting steam emerges from the boiler through a 6 cm ID pipe. Calculate the
required heat input to the boiler in kilojoules per minute if the emerging steam is
saturated at the boiler pressure. Neglect the kinetic energies of the liquid inlet
streams.
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