#### Transcript 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
4
][ ] [ ]
=
net m
input
+ net m
dmCV
i  m
e
m
dt
output
CV = control volume
i = inlet, e = exit
mi
General form for multiple inlets/exits
dmCV
i  m
e
 m
dt
dmCV
dt
me
Sec 4.1.2: Mass Flow Rate
5
Continuity Principle:
mass flow is steady and continuous
m   V   (Vn tA)
where ρ = density
Vn = normal velocity component
A = cross sectional area
V = volume
Mass flow rate:
dm
 m   Vn A
dt
m    Vn dA
A
Mass Flux
  Vn
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).
AV
m   V   AV 
v
where:
m = mass flow rate
ρ = density
V = velocity
A = area
V = volumetric flow rate
v = specific volume
General form for multiple inlets/exits
dmCV
Ai Vi
Ae Ve


dt
vi
ve
dmCV
0
dt
and

Ai Vi
Ae Ve

vi
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
VB  1.036m3 / min
state
Liquid
mA  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… mA  0.5 kg / s
State B: can be found from volumetric flow rate
VB
1.036 m3 / min 1min
mB   B VB 

 0.0559 kg / s
3
vB 0.0016714 m / kg 60 s
State C: can be found from mass rate balance
mA  mB  mC
mC  mA  mB  0.5  0.0559  0.4441 kg / s
Sec 4.3 : Applications of the Mass Rate Balance
Ammonia
10
Saturated vapor
state
VB  1.036m / min
3
Liquid
mA  0.5 kg / s
Saturated
liquid
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
mA (kg/s)
0.5
0.0559
0.4441
What can we learn from the continuity equation:
State A:
mA 
VA
vA
m   VA 
VA  v A mA  (1.6714  10-3 m3 / kg )(0.5kg / s )
 0.0008357 m / s  0.05014 m / min
3
State C:
VA V

v
v
3
VB  1.036m3 / min
VC
vC
VC  vC mC  (1.5597  10-3 m3 / kg )(0.4441kg / s )
mC 
 0.0006927 m3 / s  0.04156 m3 / min
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
V (m3/min)
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
4 VA

 VA
VA  VA
d2
A
d2
4
4
4(0.0514m3 / min) 1min
 0.00738m  0.738cm
 (20m / s )
60 s
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
VA V
m   VA 

v
v
state
13
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:
Area:
V V A
A 
d
2
4
Exit B
VB  VB
 dB2
4
 (1.31 ft / s )
 (3in) 2 1 ft
4
2
60 s
 3.858 ft 3 / min
12in 1min
Exit C
VC  VC
 dC 2
4
 (0.74 ft / s )
 (4in) 2 1 ft
4
2
60 s
 3.874 ft 3 / min
12in 1min
Sec 4.3 : Applications of the Mass Rate Balance
Next find the mass
flow rates
state
Specific Volumes may be
found on Table A-2E:
Exit C
Inlet A:
Inlet A
V [ft/s]
V
m
v
Exit B
14
Exit B
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
VB
3.858 ft 3 / min 1min
mB 

 4.004 lbm / s
3
vB 0.01606 ft / lbm 60 s
VB
3.874 ft 3 / min 1min
mB 

 4.020 lbm / s
3
vB 0.01606 ft / lbm 60 s
(apply mass balance)
mA  mB  mC  4.004  4.020  8.024 lbm / s
Sec 4.3 : Applications of the Mass Rate Balance
15
Finally, the volumetric flow rate of the inlet may be found:
60 s
VA  v A mA  (0.01605 ft / lbm )(8.024lbm / s )
 7.727 ft 3 / min
1min
3
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
```