Applications of First Law
•
•
•
•
•
Nozzles and Diffusers
Turbines
Compressors, Pumps, Fans
Throttles
Heat Exchangers
– Separating Wall (Closed Feedwater)
ES305: Thermodynamics
Heat Exchanger
1
Hot
fluid
2
.
Q
3
Cold
fluid
4
What shall we choose as control volume?
Hot fluid? Cold fluid? Both fluids?
ES305: Thermodynamics
Example 1
Dry air enters an air conditioning system at 30oC and
0.11MPa at a volume flow rate of 1.20m3/s. The air is
cooled by exchanging heat with a stream of refrigerant22 which enters the heat exchanger at -10oC and a quality
of 20%. The refrigerant leaves as a saturated vapor.
(Assume constant pressure for both streams.) 22kJ/s of
heat are removed from the air.
Determine:
(a) the flow rate of refrigerant-22, in kg/s, and
(b) the exit temperature of the air, in oC.
ES305: Thermodynamics
Example 1
air: p = 0.11Mpa
.
T = 30oC, V = 1.2m3/s
1
p = 0.11Mpa
T = ?? oC
Hot
fluid
2
.
Q = 22 kJ/s
3
Cold
fluid
R-22: T = -10oC
.
x = 0.20, m = ?? kg/s
4
T = -10oC
saturated vapor
ES305: Thermodynamics
Example 1
ES305: Thermodynamics
Example 1
Equations:


V  mv
pv  RT


(Ideal gas air only)

Q  W sh  m  h  ke  pe 
Solution:
Choose a control volume around the
R-22 only.
kJ
h3  hf  x3 hfg  33.54  0.2  246.15  33.54  76.06
kg


kJ
h4  hg @Tf  10 C  246.15
kg
o
ES305: Thermodynamics
Example 1


0
0

Q 3  4  W sh,3  4  m R  22  h4  h3  ke  pe 
kJ

22

Q3  4
s
mR  22 

kJ
kJ
h4  h3
246.15
 76.06
kg
kg
(a)

mR  22
kg
 0.13
s
ES305: Thermodynamics
0
Example 1
Choose a control volume around the air only.
kJ
0.286
303K
RT1
m3
kgK
v1 

 0.79
p1
110kPa
kg

mair

3
m
1.2
V1
kg
s


 1.52
3
v1
s
m
0.79
kg


0

Q12  W sh,12  mair  cp,air 
0
0
T2  T1   ke  pe 
ES305: Thermodynamics
Example 1

T2  T1 
Q1 2

mair cp,air
(b)
kJ 

 22 s 


 30o C 
kg  
kJ 

 1.52 s   1.005 kg K 



T2  15.6 C
o
ES305: Thermodynamics
Example 2
Liquid water at 5bar and 140oC flows through the inside
20 tubes of a shell-and-tube heat exchanger at a total rate
of 240kg/min, and it leaves at 4.8bar and 60oC. The
water is cooled by passing air initially at 110kPa and
25oC at an inlet volume rate of 1000m3/min through the
shell of the exchanger. The exit pressure of the air is
105kPa. Determine:
(a) the exit air temperature, in oC,
(b) the inlet area for airflow, in m2, if V = 25m/s, and
(c) the inlet water velocity, in m/s, if each tube has a
diameter of 2.0cm.
ES305: Thermodynamics
Example 2
.
Water
m=240kg/min
Hot
fluid
1
p = 5bar
T = 140oC
D = 2.0cm
N = 20
.
QQ = ??
4
p = 1.05bar
2
p = 4.8 bar
T = 60oC
.
Cold
fluid
3
air
V=1000m3/min
p = 1.10bar
T = 25oC
V = 25m/s
Find: T4 [oC], A3 [m2], V1 [m/s]
ES305: Thermodynamics
Example 2
Equations:


V  mv
pv  RT


(Ideal gas, air only)

Q  W sh  m  h  ke  pe 
Solution:


Choose a control volume around the
water only.
0 
0
0
Qwater  W sh  mwater
 1h2  ke  pe 
ES305: Thermodynamics
Example 2


Qwater  mwater

 1h2   mwater  h2  h1 
Both states 1 and 2 are compressed liquid, therefore,
use properties of saturated liquid water at T1 and T2
ES305: Thermodynamics
Example 2
ES305: Thermodynamics
Example 2

Qwater

Qwater
kg  
kJ
kJ 

  240
 589.13
 251.13


min  
kg
kg 

kJ
 81,120
min
ES305: Thermodynamics
Example 2
from 1st law on overall heat exchanger


Qair  Qwater
kJ
 81,120
min
kJ
0.286
298K
3
RT3
m
kgK
v3 

 0.778
p3
110kPa
kg

mair

m3 

 1000

min 
V3 
kg


 1290
3
v3 
min
m 
 0.778

kg 

ES305: Thermodynamics
Example 2


Q air  mair

 3 h4   mair c p,air T4  T3 
Choose cp,air at 300 K = 1.005 kJ/kg K

T4  T3 
Qair

mair cp,air
(a)
kJ 

 81,120 min 


 25o C 
kg  
kJ 

 1290 min   1.005 kg K 



T4  87.57 C
o
ES305: Thermodynamics
Example 2
3


kg
m



 1290 min   0.778 kg 
mair v 3 


A3 

V3
 m  60s 
 25 s  1min 



(b)
A3  0.667m 2
ES305: Thermodynamics
Example 2


m1 v1 4 mwater v1
V1 

2
A1
N D
N D 2
A1 
4
3


kg
m


3
 4   240  1.04 x10

min  
kg 

V1 
2  60s 
 20   0.02m  

1min


(c)
m
V1  0.66
s
ES305: Thermodynamics
Applications of First Law
•
•
•
•
•
Nozzles and Diffusers
Turbines
Compressors, Pumps, Fans
Throttles
Heat Exchangers
– Separating Wall (Closed Feedwater)
– Mixing (Open Feedwater)
ES305: Thermodynamics
Open Feedwater Heater
1
3
2
.
.
.
m

m

m
.
.
.
m h m h m h
1
1
1
2
2
2
ES305: Thermodynamics
3
3
3
Example
An open feedwater heater operates at 7bar. Compressed
liquid water at 35oC enters at one section. The water
exits the heater as saturated liquid. Determine the
temperature of the steam entering another section if the
ratio of the mass flow rate of compressed liquid to
superheated vapor is 4.52:1.
ES305: Thermodynamics
Example
ES305: Thermodynamics
Example
ES305: Thermodynamics
Example
An open feedwater heater operates at 7bars. Compressed
liquid water at 35oC enters at one section. The water exits
the heater as saturated liquid. Determine the temperature
of the steam entering another section if the ratio of the
mass flow rate of compressed liquid to superheated vapor
is 4.52:1.
1
saturated
T = 35oC
3 liquid
h=146.7kJ/kg
p = 7 bar
h=697.2kJ/kg
2

T = ??oC
m1

ES305: Thermodynamics
m2
 4.52
Example
Equations:
Conservation of Energy
0 0


0 0
0 0
Q  W sh   m  h  ke  pe    m  h  ke  pe 

out
in


m3 h3   m1 h1  m2 h2   0





Conservation of Mass



m3   m1  m2   0



ES305: Thermodynamics
Example


m3
m1

m2


1
m2
 

m 3 h3  m1 h1
0


h
2




m2
 m2


Combining:
 

  
m1
m1 



h2    1 h3   h1




 m2 
 m2

ES305: Thermodynamics
Example
 

  
m1
m1 



h2    1 h3   h1




 m2 
 m2



kJ 
kJ 
kJ
h2   4.52  1  697.2
   4.52   146.7
  3185.5
kg 
kg 
kg


ES305: Thermodynamics
Example
ES305: Thermodynamics
Example
kJ
h2  3185.5 , P2  7bar
kg
from superheated steam tables: T2  360C
ES305: Thermodynamics