1. EGR 334 Thermodynamics Chapter 4: Section 9-10
Lecture 17:
Control Volume Applications:
Day 2
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 11-12
Homework Assignment:
Problems from Chap 4: 75, 83, 87, 90

3. Modeling applications with Control Volumes:
Many important applications involve one inlet, one exit control volumes at steady state. Other systems include feature additional inlet and outlet streams. One very common device, the heat exchanger, requires multiple streams of mass flow to be modeled.
Mass Rate Balance: multi-path, steady state
Energy Rate Balance: 1 path, steady state

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

5. Heat Exchangers
Direct contact: A mixing chamber in which hot and cold streams are mixed directly.
Tube-within-a-tube counterflow: A gas or liquid stream is separated from another gas or liquid by a wall through which energy is conducted. Heat transfer occurs from the hot stream to the cold stream as the streams flow in opposite directions.

6. Direct Contact Heat Exchangers:
Sec 4.9: Heat Exchangers
Direct Contact Heat Exchangers:
In direct contact heaters, mass streams combine
and exchange heat by mass transport as well as by convection and conduction.
Examples:
Swamp Coolers
Hot-Cold water
faucet
Cooling Towers

7. Separate Flow (Tube) Heat Exchangers:
Sec 4.9: Heat Exchangers
Separate Flow (Tube) Heat Exchangers:
Heat transfer occurs by conduction through a material that separates individual flow streams
Parallel Flow: Streams enter and leave along same directions
Cross Flow: Stream pass along direction perpendicular to each other
Counter Flow: Streams enter and leave opposite each other.

8. Tube Heat Exchanger are all around you.
Still
Radiator
Home air conditioning
Industrial Heating Boiler
Geothermal
Heating
Refrigerator
condenser
Industrial
Systems
Power Plant

9. Designed to maximize “contact” between the two fluids
Sec 4.9: Heat Exchangers
A heat exchanger attempts to maximize the heat transfer between separate fluid streams.
Designed to maximize “contact” between the two fluids
Maximize time of contact and Maximize area of contact

10. For Separate Stream Heat Exchangers:
Sec 4.9: Heat Exchangers
For Separate Stream Heat Exchangers:
Mass Balance Model:
and
Energy Balance Model:
Steady State
V0
Horizontal Section
(or very short vertical)
Insulated
from surroundings
No work
First Fluid
Second Fluid

11. Assumptions Steady State KE=  PE = 0 QCV= 0 Sec 4.9: Heat Exchangers
Example: (4.81) A feedwater heater operates at steady state with liquid water entering at inlet #1 at 7 bar, 42 °C and a mass flow rate of 70 kg/s. A separate stream of water enters at inlet 2 as a two phase liquid-vapor mixture at 7 bar with a quality of 98%. Saturated liquid at 7 bar exits the feedwater heater at #3. Ignoring heat transfer with surroundings and neglecting KE and PE effects, determine the mass flow rate in kg/s at inlet #2.
state 1 2 3
m (kg/s) 70 x
0.98
p (bar) 7
T (°C) 42
165
h (kJ/kg)
state 1 2 3
m (kg/s) x
p (bar)
T (°C)
h (kJ/kg)
state 1 2 3
m (kg/s) 70 x
0.98
p (bar) 7
T (°C) 42
165
h (kJ/kg)
175.9
697.22 #2
7 bar
98% #1
7 bar
42 °C
70 kg/s #3
7 bar
Sat. Liq.
Look up h values on
Table A-2 and A-3
Assumptions
Steady State
KE=  PE = 0
QCV= 0

12. Example: (4.81) Determine the mass flow rate in kg/s at inlet #2.
Sec 4.9: Heat Exchangers
Example: (4.81) Determine the mass flow rate in kg/s at inlet #2.
From Table A-3
state 1 2 3
m (kg/s) 70 x
0.98
p (bar) 7
T (°C) 42
165
h (kJ/kg)
175.9
697.22
Mass Balance:
Energy Balance:
Out In
Combining

13. Using energy balance: for the heat exchanger
Sec 4.9: Heat Exchangers
Example: (4.85) A parallel flow heat exchanger has separate streams of air and water. Streams undergo no significant change of pressure. Ignore heat loss to the surroundings and changes in KE and PE. Apply ideal gas to the air stream. If both streams leave at the same temperature, determine the exit temperature.
state 1 2 3 4
H2O
Air
m (kg/s) 10 5
p (bar) T
99.63 oC
1200 K
h (kJ/kg)
2675.5
state 1 2 3 4
H2O
Air
m (kg/s) 10 5
p (bar) T
1200 K
h (kJ/kg)
from Table A-3 at psat = 1 bar, T1= Tsat = C and h1=hg= kJ/kg
Using energy balance: for the heat exchanger
Reducing this with assumptions

14. Sec 4.9: Heat Exchangers
Example: (4.85) A parallel flow heat exchanger has separate streams of air and water. Streams have no significant change of pressure. Ignore heat loss to the surroundings and changes in KE and PE. Apply ideal gas to the air stream. If both stream leave at the same temperature, determine the exit temperature.
state 1 2 3 4
H2O
Air
m (kg/s) 10 5
p (bar) T
99.63oC
1200 K
h (kJ/kg)
2675.5
Using ideal gas model
Note: since h2 depends upon T2=T4 and p2 = 1 bar, it could be found from Table A-4 if the exit temperature were known. This problem can be solved by picking a value of T2 and then looking up h2 and checking to see if it satisfies the equation. It can also be easily solved using IT’s iterative solver.

15. Then the heat transfer between the two fluids is
Sec 4.9: Heat Exchangers
Solution using IT
Exit Temperature
Then the heat transfer between the two fluids is
state 1 2 3 4
H2O
Air
m (kg/s) 10 5
p (bar) T
99.63oC
288
1200 K
h (kJ/kg)
2675.5
3049

16. Throttling Devices
Throttling Device: a device that achieves a significant reduction in pressure by introducing a restriction into a line through which a gas or liquid flows. Means to introduce the restriction include a partially opened valve or a porous plug.

17. Throttling Device: Reduces Pressure
Sec 4.10: Throttling Devices
Throttling Device: Reduces Pressure
Mass Balance:
Typical Energy Balance simplifications,
Steady State
v0
Horizontal Section
(or very short vertical)
No heat transfer
No work

18. Enthalpy of saturated liquid at 80°F
Sec 4.10: Throttling Devices
Example: (4.92) Refrigerant 134a enters the expansion valve of an air conditioning unit at 140 psi, 80 °F and exits at 50 psi. If the refrigerant undergoes a throttling process, what are the temperature, in °F, and the quality at the exit valve?
state
inlet
exit x ?
P (psi)
140 50
T (°F) 80
h (BTU/lb)
37.27
Assumptions
Steady State
KE= PE = 0
QCV= WCV = 0
Enthalpy of saturated liquid at 80°F
140 psi
80 °F
50 psi

19. Example: (4.92) Refrigerant 134a
Sec 4.10: Throttling Devices
Example: (4.92) Refrigerant 134a
state
inlet
exit x ?
P (psi)
140 50
T (°F) 80
40.27
h (BTU/lb)
37.27
At 140 psi, the saturation T is °F, so this is a super-cooled liquid. There are no tables for super-cooled liquid, so neglect pressure effects.
From Table A-10E

20. end of Lecture 17 Slides