1. EGR 334 Thermodynamics Chapter 4: Section 4-5
Test #1 Next Monday
Lecture 13:
Control Volumes and Energy Balance
Quiz Today?

2. Today’s main concepts:
Be able to draw a graphic representation of a control volume with energy balance.
Explain what flow work.
Write the energy balance equation
Identify some commons assumptions that simplify the energy balance equation.
Use mass balance and energy balance to solve thermodynamic problems.
Test #1 Next Monday
Reading Assignment:
Read Chap 4: Sections 6-9 (for next Wed.)
Homework Assignment:
From Chap 4: 25, 28, 31, 34

3. For a Control Volume:
Mass Rate Balance:
The rate at which mass accumulates in the control volume is the difference between the
rate of mass flow in and flow out.
Energy Rate Balance:
The rate at which energy accumulates in the control volume is the difference between the
heat flow rate in and the
power out of the system
and
the difference between the
rate at which mass carries
energy with it either into and
out of the control volume.

4. [ ] [ ] [ ] [ ] [ ] + = = E within the system net Q input +
Sec 4.4: Conservation of Energy for a Control Volume
Energy Balance
Net E from
Mass transfer +
[ ]
Add a term to the energy balance.
E within
the system
net Q input
net W output
[ ] =
[ ] +
Mass Balance
m within
the system
net m input
net m output
[ ] =
[ ] +
where then energy due to mass transfer is
mass transfer energy expressed as a rate equation
then the full 1st law of Thermodynamics is given as

5. [ ] Two Types of Work Flow Work: work done BY the flow
Sec 4.4: Conservation of Energy for a Control Volume
Two Types of Work
Flow Work: work done BY the flow
Control Volume Work: work done BY the control volume (WCV)
- work done by PV, turning a shaft, electricity (types of work we have dealt with thus far)
Define  Flow Work
[ ]
Time rate of Energy transfer from the
control volume at exit due to mass transfer
Work done by system due to rotating shafts, electrical effects, or boundary displacement (∫pdV)
(+) Work
Done On environment as mass exits
(-) Work
Done ON system as mass enters

6. h  “heat energy + flow work”
Sec 4.4: Conservation of Energy for a Control Volume
Putting all these terms together including the separate work terms, W = WCV + WPAv hi he
Recall that enthalpy is defined
h = u + p v
then the form that will usually be your starting equation is
Thus u  “heat energy”
h  “heat energy + flow work”

7. For Steady State Conditions:
Sec 4.5: Analyzing Control Volumes at Steady State
For Steady State Conditions:
and
Mass Balance at steady state:
Energy Balance at steady state:
Common Simplifying Assumptions:
Often Q = 0 if
Small surface area
System is insulated
T between system & environment
is small
Often W = 0 if
No change in system volume
(fixed container) No turbines/pumps or
electrical devices

8. 600 kPa 120 kPa 330 K 300 K d =1.2 cm Example 1: (Problem 4.27)
Sec 4.5: Analyzing Control Volumes at Steady State
Example 1: (Problem 4.27)
Air at 600 kPa, 330 K enters a well-insulated, horizontal pipe having a diameter of 1.2 cm and exits at 120 kPa, 300 K. Applying the ideal gas model for air, determine at steady state
(a) the inlet and exit velocities, each in m/s
(b) the mass flow rate, in kg/s.
600 kPa
330 K
120 kPa
300 K
d =1.2 cm
Identify State Properties
state
Inlet
Exit
P (kPa)
600
120
T (K)
330
300
h (kJ/kg)
330.34
300.19
state
Inlet
Exit
P (kPa)
600
120
T (K)
330
300
h (kJ/kg)
state
Inlet
Exit
p [kPa]
T [K]
h [kJ/kg]
Look up values for h on Table A-22

9. No significant heat loss No work done by system
Sec 4.5: Analyzing Control Volumes at Steady State
Assumptions:
System at steady state
No significant heat loss
No work done by system
No large change in elevation
state
Inlet
Exit
P (kPa)
600
120
T (K)
330
300
h (kJ/kg)
330.34
300.19
Table A-22
Energy balance:
is reduced with assumptions to:
Steady state also implies that mass flow rate is equals mass flow rate out.

10. Next find mass flow rate
Sec 4.5: Analyzing Control Volumes at Steady State
Next find mass flow rate
state
Inlet
Exit
P (kPa)
600
120
T (K)
330
300
h (kJ/kg)
330.34
300.19
600 kPa
330 K
120 kPa
300 K
d =1.2 cm
Apply continuity equation and mass balance:
therefore:
applying Ideal Gas Equation:

11. Finally combine what we know
Sec 4.5: Analyzing Control Volumes at Steady State
Finally combine what we know
state
Inlet
Exit
P (kPa)
600
120
T (K)
330
300
h (kJ/kg)
330.34
300.19
600 kPa
330 K
120 kPa
300 K
d =1.2 cm
from Energy Balance: from Mass balance:
therefore:
solving for velocities:

12. Sec 4.5: Analyzing Control Volumes at Steady State
Inlet
Exit
P (kPa)
600
120
T (K)
330
300
h (kJ/kg)
330.34
300.19
600 kPa
330 K
120 kPa
300 K
d =1.2 cm
to find the mass flow rate: from table 3.1: R = kJ/kg-K

13. 2 bar d= 4 cm Example 2: (Problem 4.29)
Sec 4.5: Analyzing Control Volumes at Steady State
Example 2: (Problem 4.29)
Refrigerant 134a flows at a steady state through horizontal pipe with an inside diameter of 4cm. It enters as -8 deg C saturated vapor with a mass flow rate of 17 kg/min. It exits at a pressure of 2 bar. If the heat transfer rate to the refrigerant
is 3.4 kW, determine
a) the exit temperature
b) the inlet and outlet velocities
-8 oC
17 kg/min
2 bar
d= 4 cm .
Q = 3.4 kW
Identify State Properties
statec
Inlet
(sat. vapor)
Exit
p [bar]
2.1704 2
T [oC] -8
v [m3/kg]
0.0919
h [kJ/kg]
242.54
statec
Inlet
(sat. vapor)
Exit
p [bar]
T [oC]
v [m3/kg]
h [kJ/kg]
Look up values for h
on Table A-10

14. Sec 4.5: Analyzing Control Volumes at Steady State
-8 oC
17 kg/min
2 bar
d= 4 cm .
Q = 3.4 kW
state
Inlet
(sat. vapor)
Exit
p [bar]
2.1704 2
T [oC] -8
v [m3/kg]
0.0919
h [kJ/kg]
242.54
Using continuity equation at the inlet
Find the volumetric flow rate
Find the inlet velocity

15. Sec 4.5: Analyzing Control Volumes at Steady State
-8 oC
17 kg/min
2 bar
d= 4 cm .
Q = 3.4 kW
state
Inlet
(sat. vapor)
Exit
p [bar]
2.1704 2
T [oC] -8
v [m3/kg]
0.0919
h [kJ/kg]
242.54
for steady state: mass balance is given as
For steady state the energy balance is given as:
for steady state, no work done, and a horizontal pipe:

16. Sec 4.5: Analyzing Control Volumes at Steady State
-8 oC
17 kg/min
2 bar
d= 4 cm .
Q = 3.4 kW
state
Inlet
(sat. vapor)
Exit
p [bar]
2.1704 2
T [oC] -8
v [m3/kg]
0.0919
h [kJ/kg]
242.54
combining the mass balance and energy balance equations:
gives or

17. Sec 4.5: Analyzing Control Volumes at Steady State
-8 oC
17 kg/min
2 bar
d= 4 cm .
Q = 3.4 kW
state
Inlet
(sat. vapor)
Exit
p [bar]
2.1704 2
T [oC] -8
v [m3/kg]
0.0919
h [kJ/kg]
242.54
This equation contains two unknowns, he and ve.
Note that both are properties of the exit state which can be both considered functions of an unknown temperature, TB and a known pressure, pB = 2 bar.
One way to solve this would be to guess a temperature of the outlet….then look up the corresponding values of h and v….put them back in the equation and see if they give the correct value.
This iterative approach is exactly the type of solution that IT is very good at….so anther way to solve this is to use IT to define the h and v in terms of an unknown TB and let it iteratively find the temperature for us.

18. c Results: Outlet temperature Te = -10.09 oC Inlet velocity
Vi = 20.7 m/s
Outlet velocity

19. end of Lecture 13 Slides
Test #1 Next Monday