### Control Volumes and Mass Balance

```EGR 334 Thermodynamics
Chapter 4: Section 1-3
Lecture 12:
Control Volumes and
Conservation of Mass
Quiz Today?
Today’s main concepts:
• Be able to explain what a control volume is
• Be able to write mass balance and mass rate balance equations
for a control volume.
• Be able to explain the continuity of mass flow equation.
• Explain the difference between mass flow rate and volumetric
flow rate.
• Be able to set up problems involving mass balance
• Read Chap 4: Sections 4-5
Homework Assignment:
From Chap 4: 1, 6, 11, 22
Sec 4.1: Conservation of Mass
3
So far we have looked at closed systems
Only Q and W pass the boundary
Energy Balance
Q and W
Now move to open system,
mass can also pass the boundary.
Mass
From Chapter 2: Energy Balance
[
E within
the system
][ ] [ ]
Mass Balance
[
m within
the system
=
net Q
input
+
net W
output
][ ] [ ]
=
net m
input
+ net m
output
Sec 4.1: Conservation of Mass
Mass Balance
[
m within
the system
][ ] [ ]
dm CV
dt
4
=
net m
input
+ net m
output
 m i  m e
CV = control volume
i = inlet, e = exit
dmCV
dt
mi
General form for multiple inlets/exits
dm CV
dt

 m
i

 m
e
me
Sec 4.1.2: Mass Flow Rate
5
Continuity Principle:
mass flow is steady and continuous
m   V   ( V n  tA )
where ρ = density
Vn = normal velocity component
A = cross sectional area
V = volume
Mass flow rate:
dm
dt
 m   Vn A
m 

A
Mass Flux
  Vn
Vn d A
for constant Area
for variable Area
Sec 4.2 Mass Rate Balance
6
One-Dimensional Flow (Continuity Equation)
• Flow is normal to the boundary
• The fluid is homogeneous (intensive properties
are uniform with position).
m   V   AV 
AV
v
where:
ρ = density
V = velocity
A = area
m = mass flow rate
V = volumetric flow rate
v = specific volume
General form for multiple inlets/exits
dmCV
Ai V i
 

dt
vi
dm CV
dt
 0
and

Ae V e

ve
Ai V i
vi


Ae V e
ve
Sec 4.3 : Applications of the Mass Rate Balance
Example: (4.16) Ammonia enters a control volume operating at steady
state at pA = 14 bar, TA = 28oC, with a mass flow rate of 0.5 kg/s.
Saturated vapor at pB = 4 bar leaves through one exit, with a
volumetric flow rate of 1.036 m3/min and saturated liquid at pC=4 bar
leaves through a second exit. Determine
(a) the minimum diameter of the inlet pipe, in cm, so the ammonia
velocity does not exceed 20 m/s
(b) the volumetric flow rate of the second exit stream in m3/min.
pA = 14 bar,
TA = 28oC
0.5 kg/s
Saturated vapor
pB = 4 bar,
1.036 m3/min
pC= 4 bar
Saturated liquid
7
Sec 4.3 : Applications of the Mass Rate Balance
pA= 14 bar
TA= 28oC
mA=0.5 kg/s
Saturated vapor
pB = 4 bar,
1.036 m3/min
pC= 4 bar
Saturated liquid
Ammonia
state
Inlet A
Exit B
Exit C
P (bar)
14
4
4
T (°C)
28
v (m3/kg)
8
What else can you determine
State 1: from table A-14 (state A is compressed liquid…
let vA≈ vf @28oC = 1.6714x10-3 m3/kg)
State 2: from table A-14 ( vB = vg @4bar) = 0.3094 m3/kg and TB=Tsat = -1.9oC)
State 3: from table A-14 ( vC = vf @4bar)= 0.0015597 m3/kg and TC=Tsat = -1.9oC)
Sec 4.3 : Applications of the Mass Rate Balance
Ammonia
9
Saturated vapor
V B  1.036 m / m in
3
state
Liquid
m A  0.5 kg / s
Saturated
liquid
Inlet A
Exit B
Exit C
P (bar)
14
4
4
T (°C)
28
-1.90
-1.90
1.6714x10-3
0.3094
1.5597x10-3
v (m3/kg)
Consider mass flows:
State A: already known… m A  0.5 kg / s
State B: can be found from volumetric flow rate
m B   B VB 
VB
vB
3

1.036 m / m in
3
0.0016714 m / kg
1 m in
 0.0559 kg / s
60 s
State C: can be found from mass rate balance
m A  m B  mC
m C  m A  m B  0.5  0.0559  0.4441 kg / s
Sec 4.3 : Applications of the Mass Rate Balance
Ammonia
10
Saturated vapor
state
V B  1.036 m / m in
Inlet A
Exit B
Exit C
P (bar)
14
4
4
T (°C)
28
-1.90
-1.90
v (m3/kg)
1.6714x10-3
0.3094
1.5597x10-3
m A (kg/s)
0.5
0.0559
0.4441
3
Liquid
m A  0.5 kg / s
Saturated
liquid
What can we learn from the continuity equation:
State A:
mA 
m   VA 
VA
v
VA

V
v
vA
V A  v A m A  (1.6714  10 m / kg )(0.5 kg / s )
-3
3
V B  1.036 m / m in
3
 0.0008357 m / s  0.05014 m / m in
3
State C:
mC 
3
VC
vC
V C  v C m C  (1.5597  10 m / kg )(0.4441 kg / s )
3
-3
 0.0006927 m / s  0.04156 m / m in
3
3
Sec 4.3 : Applications of the Mass Rate Balance
Ammonia
state
Saturated
vapor
Liquid
Saturated
liquid
11
Inlet A
Exit B
Exit C
P (bar)
14
4
4
T (°C)
28
-1.90
-1.90
v (m3/kg)
1.6714x10-3
0.3094
1.5597x10-3
m (kg/s)
0.5
0.0559
0.4441
0.05014
1.036
0.04156
(m3/min)
V
Determine the size of the inlet pipe so that the velocity does not exceed
VA = 20 m/s
from the continuity equation:
Area of a circular cross section:
VA  VA A
therefore:
d 
4VA
 VA
VA  VA
d
A 
d
2
4
2
4
3

4 (0 .0 5 1 4 m / m in ) 1 m in
 (20m / s)
60 s
 0 .0 0 7 3 8 m  0 .7 3 8 cm
Sec 4.3 : Applications of the Mass Rate Balance
12
Example 2: ( from Prob 4.16)
Liquid water at 70 oF enters a pump through an inlet pipe having a diameter of 6
in. The pump operates at steady state and supplies water to two exit pipes having
diameters of 3 in and 4 in. The velocity of the 3 in pipe is 1.31 ft/s. At the exit of
the 4 in pipe the velocity is 0.74 ft/s. The temperature of the water in each exit is
72 deg F. Determine
a) the mass flow rate in
lb/s in the inlet and
Exit B
each of the exit pipes.
b) the volumetric flow
rate at the inlet in
ft3/min.
Exit C
Inlet A
Identify what you know:
state
Inlet A
V [ft/s]
Exit B
Exit C
1.31
0.74
T [°F]
70
72
72
d [in]
6
3
4
dm/dt [lbm/s]
dV/dt [ft3/min]
Sec 4.3 : Applications of the Mass Rate Balance
Use continuity to find
volumetric flow rates
m   VA 
VA
V

v
v
state
Inlet A
V [ft/s]
Exit B
Exit C
1.31
0.74
T [°F]
70
72
72
d [in]
6
3
4
Volumetric Flow Rate:
V
13
Area:
 V A
A 
d
2
4
Exit B
V B  VB
 dB
2
 (1.31 ft / s )
4
 (3 in )
4
2
2
1 ft
12 in
60 s
 3.858 ft / m in
3
1 m in
Exit C
V C  VC
 dC
4
2
 (0.74 ft / s )
 (4 in )
4
2
1 ft
12 in
2
60 s
1 m in
 3.874 ft / m in
3
Sec 4.3 : Applications of the Mass Rate Balance
Next find the mass
flow rates
m 
state
v
Specific Volumes may be
found on Table A-2E:
Exit B
mB 
mB 
VB

Exit C
1.31
0.74
T [°F]
70
72
72
d [in]
6
3
4
vA = vf @ 70 oF =0.01605 ft3/lbm
vB = vC = vf @ 72 oF = 0.01606 ft3/lbm
3.858 ft / m in 1 m in
3
0.01606 ft / lb m
Exit C
Inlet A:
Exit B
3
vB
vB
Inlet A
V [ft/s]
V
VB
14
60 s
 4.004 lb m / s
3

3.874 ft / m in 1 m in
3
0.01606 ft / lb m
60 s
 4.020 lb m / s
(apply mass balance)
m A  m B  m C  4.004  4.020  8.024 lb m / s
Sec 4.3 : Applications of the Mass Rate Balance
15
Finally, the volumetric flow rate of the inlet may be found:
V A  v A m A  (0 .0 1 6 0 5 ft / lb m )(8 .0 2 4 lb m / s )
60 s
3
 7 .7 2 7 ft / m in
3
1 m in
Summary:
state
Inlet A
Exit B
Exit C
V [ft/s]
7.727
1.31
0.74
T [°F]
70
72
72
d [in]
6
3
4
v [ft3/lbm]
0.01605
0.01606
0.01606
dm/dt [lbm/s]
8.024
4.004
4.020
dV/dt [ft3/min]
7.727
3.858
3.874
16
End of slides for Lecture 12
```