1. EGR 334 Thermodynamics Chapter 4: Section 6-8
Lecture 16:
Control Volume Applications:
Day 1
Quiz Today?

2. Today’s main concepts:
Be able to set up mass and energy balance models for Turbines Pumps Compressors Boilers Heat Exchangers Nozzles Diffusers Throttle
Reading Assignment:
Read Chapter 4, Sections 10-12
Homework Assignment:
Problems from Chap 4: 36, 43, 52, 66

3. Review: For a Control Volume:
Mass Rate Balance:
Energy Rate Balance:

4. Modeling applications with Control Volumes:
Many important applications involve one inlet, one exit control volumes at steady state. Today a number of these useful models will be developed using the one inlet, one outlet, steady state forms of the mass balance and energy balance given below.
Mass Rate Balance: 1 path, steady state
Energy Rate Balance: 1 path, steady state

5. Control Volume Applications:
Nozzles
Diffuser
Turbine
Boiler
Compressor
Pump
Throttling Valve
Heat Exchanger

6. Common Modeling assumptions:
Application models generally make use of simplifying assumptions to reduce the complexity of the Energy Balance. By removing terms that do not apply to a particular application or whose impact on the application is generally only minor the models take on simplified, useful, and easy to use forms.
Assumption:
If:
a) Outer surface of CV is well insulated…
b) Small change of elevation…
c) No mechanical mechanisms present…
d) CV maintains same shape and volume…
e) No electrical effects act on CV…
f) Inlet and Outlet have same physical size…
g) Outer surface of CV is small…
h) Small ΔT between CV and environment…
i) flow passes through CV in short time…
j) inlet size much larger than outlet size…

7. Nozzles and Diffusers
Nozzle: a flow passage of varying cross-sectional area in which the velocity of a gas or liquid increases in the direction of flow.
Diffuser: a flow passage of varying cross-sectional area in which the velocity of a gas or liquid decreases in the direction of flow.

8. Sec 4.6: Nozzles and Diffusers
Nozzles and Diffusers are used to change the speed of the mass flow through the control volume. Vi Ve Ve Vi
For continuous flow, changing the size of the cross section alters the speed of the flow. if
continuous if
incompressible  
What common assumptions may be used to simplify the energy balance?

9. Typical Energy Balance simplifications,
Sec 4.6: Nozzles and Diffusers
Typical Energy Balance simplifications,
No pump/turbines
Horizontal Section
(or very short vertical)
Steady State
Even though there is no insulation, the V is high so there may be little heat transfer.
Therefore,

10. Sec 4.6: Nozzles and Diffusers
Example: (4.34) Air with a mass flow rate of 5 lb/s enters a horizontal nozzle operating at steady state at 800°R, 50 psi and a velocity of 10 ft/s. At the exit, the temperature is 570°R and the velocity is 1510 ft/s. Using the ideal gas model for air determine (a) the area of the inlet, in ft2, and (b) the heat transfer between the nozzle and its surroundings in BTU/lb of air flowing. .
m = 5 lbm/s
50 psi
800°R
10 ft/s
570°R
1510 ft/s
Ain = ?, Q = ?

11. Sec 4.6: Nozzles and Diffusers
Example: (4.34) Using the ideal gas model for air determine (a) the area of the inlet, in ft2, and (b) the heat transfer between the nozzle and its surroundings in BTU/lb of air flowing.
mass balance
50 psi
800°R
10 ft/s
570°R
1510 ft/s
Ain = ?, Q = ?
continuity
ideal gas

12. reduced using assumptions:
Sec 4.6: Nozzles and Diffusers
Example: (4.34) Using the ideal gas model for air determine (a) the area of the inlet, in ft2, and (b) the heat transfer between the nozzle and its surroundings in BTU/lb of air flowing.
Energy balance at SS
reduced using assumptions:
50 psi
800°R
10ft/s
570°R
1510ft/s
from Table A-22E

13. Turbines
Turbine: a device in which power is developed as a result of a gas or liquid passing through a set of blades attached to a shaft free to rotate.

14. Use Mass and Energy Balances still hold:
Sec 4.7: Turbines
A turbine is a device that develops power from a gas or liquid passing through a set of blades which are attached to a shaft free to rotate.
Use Mass and Energy Balances still hold:

15. Typical Energy Balance simplifications,
Sec 4.7: Turbines
Typical Energy Balance simplifications,
V  0
Horizontal Section
(or very short vertical)
Steady State
Even though there is no insulation, the V is high so there is no heat transfer.
WCV .
min
mexit
Therefore,
Then,

16. Sec 4.7: Turbines
Example: (4.50) Steam enters the first stage of a turbine at 40 bar and 500 °C with a volumetric flow rate of 90 m3/min. Steam exits the turbine at 20 bar and 400°C. The steam is then reheated at constant pressure to 500 °C before entering the second stage turbine. Steam leaves the second stage as saturated vapor at 0.6 bar. For operation at steady state, and ignoring stray heat transfer and KE and PE effects, determine the
Mass flow rate of steam, in kg/hr
Total power produced by both turbines, in kW
The rate of heat transfer to the steam flowing through the reheater, in kW.
40 bar
500°C
90 m3/min
WCV,1
WCV,2
Sat. vapor
0.6 bar
Assumptions
KE=  PE=0
QTurbine = 0
20 bar
400°C
20 bar
500°C
Reheater

17. Example: (4.50) 40 bar 500°C 90 m3/min WCV,1 WCV,2 Sat’d vapor 0.6 bar
Sec 4.7: Turbines
Example: (4.50)
Mass flow rate of steam, in kg/hr
Total power produced by both turbines, in kW
The rate of heat transfer to the steam flowing through the reheater, in kW.
40 bar
500°C
90 m3/min
WCV,1
WCV,2
Sat’d vapor
0.6 bar
Assumptions
KE=  PE=0
QTurbine = 0
20 bar
400°C
20 bar
500°C
Reheater
state
In T1
Ex T1
Ex RH
Ex T2
sat. vapor
P (bar) 40 20
0.6
T (C)
500
400
v (m3/kg)
h (kJ/kg)
state
In T1
Ex T1
Ex RH
Ex T2
phase
P (bar)
T (C)
v (m3/kg)
h (kJ/kg)
state
In T1
Ex T1
Ex RH
Ex T2
sat. vapor
P (bar) 40 20
0.6
T (C)
500
400
36.16
v (m3/kg)
0.1512
0.1757
23.739
h (kJ/kg)
3445.3
3247.6
3467.6
2567.4
1) Identify state properties given in problem statement.
2) Find intensive
properties from
Table A-4

18. Example: (4.50) Sec 4.7: Turbines Mass flow rate of steam, in kg/hr
Total power produced by both turbines, in kW
The rate of heat transfer to the steam flowing through the reheater, in kW.
state
In T1
Ex T1
Ex RH
Ex T2
Sat’d vap
P (bar) 40 20
0.6
T (C)
500
400
36.16
v (m3/kg)
0.1512
0.1757
23.739
h (kJ/kg)
3445.3
3247.6
3467.6
2567.4
To find mass flow rate from volumetric flow rate:

19. Example: (4.50) Sec 4.7: Turbines Mass flow rate of steam, in kg/hr
Total power produced by both turbines, in kW
The rate of heat transfer to the steam flowing through the reheater, in kW.
state
In T1
Ex T1
Ex RH
Ex T2
Sat’d vap
P (bar) 40 20
0.6
T (C)
500
400
36.16
v (m3/kg)
0.1512
0.1757
23.739
h (kJ/kg)
3445.3
3247.6
3467.6
2567.4
To find the power produced in both Turbines:

20. Example: (4.50) Sec 4.7: Turbines Mass flow rate of steam, in kg/hr
Total power produced by both turbines, in kW
The rate of heat transfer to the steam flowing through the reheater, in kW.
state
In T1
Ex T1
Ex RH
Ex T2
Sat’d vap
P (bar) 40 20
0.6
T (C)
500
400
36.16
v (m3/kg)
0.1512
0.1757
23.739
h (kJ/kg)
3445.3
3247.6
3467.6
2567.4
Heat transferred in the reheater…starting with energy balance
Simplified with assumptions: 

21. Compressors and Pumps
Compressors and Pumps: devices in which work is done on the substance flowing through them to change the state of the substance, typically to increase the pressure and/or elevation.
Compressor : substance is gas
Pump: substance is liquid

22. min WCV WCV mexit Sec 4.8: Compressors and Pumps . Compressor Model:
Device where work is used to increase pressure and/or elevation of the flow substance.
WCV
min
mexit .
Compressor Model:
used for gases
Pump Model: used for liquids
WCV .

23. Typical Energy Balance simplifications,
Sec 4.8: Compressors and Pumps
Typical Energy Balance simplifications,
v0
Horizontal Section
(or very short vertical)
Steady State
Even though there is no insulation, the V is high so there is no heat transfer.
WCV
min
mexit .
Therefore,
Then,

24. Sec 4.8: Compressors and Pumps
Example: (4.60) Air is compressed at steady state from 1 bar, 300 K, to 6 bar with a mass flow rate of 4 kg/s. Each unit of mass passing from the inlet to the exit undergoes a process described by pV1.27= constant. Heat transfer occurs at a rate of kJ/kg of air flowing to the cooling water circulating in a water jacket enclosing the compressor. If ΔKE and ΔPE of the air are negligible, calculate the compressor power in kW.
1 bar
300 K
4 kg/s
Assumptions
Steady State
KE= PE=0
pV1.27= constant
Ideal gas
WCV .
QCV /m= kJ/kg
6 bar
state In Ex
P (bar) 1 6
T (K)
300
h (kJ/kg)
state In Ex
P (bar)
T (K)
h (kJ/kg)
state In Ex
P (bar) 1 6
T (K)
300
h (kJ/kg)
300.19
Table A-22 (Ideal Gas Properties of Air)
h is independent of p

25. Example: (4.60) Sec 4.8: Compressors and Pumps . state In Ex P (bar) 1
WCV .
1 bar
300 K
4 kg/s
Example: (4.60)
state In Ex
P (bar) 1 6
T (K)
300
h (kJ/kg)
300.19
6 bar
QCV = kJ/kg
From ideal gas equation and polytropic eq.
Rearranging:
Combining 

26. Example: (4.60) Sec 4.8: Compressors and Pumps . state In Ex P (bar) 1
WCV .
1 bar
300 K
4 kg/s
state In Ex
P (bar) 1 6
T (K)
300
439.1
h (kJ/kg)
300.19
440.7
state In Ex
P (bar) 1 6
T (K)
300
h (kJ/kg)
300.19
6 bar
QCV = kJ/kg
Therefore the exit temperature is
(Since T2 and p2 are now
known, h2 may be found
on table A22)
The energy balance can then find the pump work

27. end of Lecture 16 Slides