Homework 3 Thermodynamics - ME100A
1. Water at 120 °C with a quality of 25% has its temperature raised 20 °C in a constant
volume process. What is the new quality and pressure? Sketch this process in p - v
and T - v diagrams.
2. Determine whether water at each of the following states is a compressed liquid, a
superheated vapor, or a mixture of saturated liquid and vapor. (show all states on
p – v and T – v diagrams.)
a. p = 10 MPa, v = 0.003 m3/kg
b. p = 1 MPa, T = 190 °C
c. T = 200 °C, v = 0.1 m3/kg
d. p = 10 kPa, T = 10 °C
3. (OPTIONAL) Saturated liquid water at 60°C is put under pressure to decrease the
volume by 1% keeping the temperature constant. To what pressure should it be
compressed?
(Use thermodynamic diagrams to show the path of this process)
4. The following table lists temperatures and specific volumes of water vapor at two
pressures:
Data encountered in solving problems often do not fall exactly on the grid of values
provided by property tables, and linear interpolation between adjacent table
entries becomes necessary. Using the data provided here, estimate:
a. The specific volume at T = 240 °C, p = 1.25 MPa, in m3/kg
b. The temperature at p = 1.5 MPa, v = 0.1555 m3/kg, in °C
c. The specific volume at T = 220 °C, p = 1.4 MPa, in m3/kg
Homework 3 Thermodynamics - ME100A
5. As shown in the figure, steam at 80 bar, 440 °C, enters a turbine operating at steady
state with a volumetric flow rate of 236 m3/min. Twenty percent of the entering
mass flow exits through a diameter of 0.25 m at 60 bar, 400 °C. The rest exits
through a diameter of 1.5 m with a pressure of 0.7 bar and a quality of 90%.
Determine the velocity at each exit duct, in m/s.
6. (OPTIONAL) As shown in the figure, air enters a pipe at 25 °C, 100 kPa with a
volumetric flow rate of 23 m3/h. On the outer pipe surface is an electrical resistor
covered with insulation. With a voltage of 120 V, the resistor draws a current of 4
amps. Assuming the ideal gas model with cp = 1.005 kJ/kg.K for air and ignoring
kinetic and potential energy effects, determine (a) the mass flow rate of the air, in
kg/h, and (b) the temperature of the air at the exit, in °C.
7. (OPTIONAL) Simplify the following units and convert them to SI.
' *+ ,
a. ! = 3.0
( ) -./)
' 0 12 )
b. ! = 3.0
() /
/ 345
c. ! = 3.0
12 ) 6/)
' 12
d. ! = 3.0
*+ ) /,
T =
120° x =
25 % constant volume New x is pressure
T
=
140° V
VI
=
Stat I State 2
P T constant
v =
~constan
2 p
=
1 985 bar
.
P =
3 613b ar
.
T =
120°C T =
140°
I 25 %
v v
2238
x
=
v -
Vf v -
1 8603x10'3 v =
25
= -
.
=
=
-
vg-Uf ·
8919 -
1 , 063x103
·
2238 -
1 0797
.
x104
43 85 %
=
· S08a -
1 ,
0797x103
.
c)
v Vf+Vg + V-Vf
v
g
=
10 MPA 003
=
p
=
W =
Om
Mixture Uq(u> uf
~-
ug= Uf
100 am ·
0 18037 . 0037 . OOl
b) P 1 MPA T 1902 T, T2
= =
P= IMP
superheat T =17 9 o
°
vapor
-
a) T = 200 =1
millg v = . 007
vf= 001Vg=
. .
1274
Uq(u> uf ·
12747 0077 001
d)
,
:
lPa= . Ol bar
p 1644a 16
=
T =
Mixture
bor=1 Temp = 45 8 .
Means
18CL45 compress liquid water
P T
·;
b
d
is B
ci c
&
~ W
M
cl 2480--1627 x -
-1627 m
B X 20S3
·
V
n =
- = -
-1 25-1
-
1 -1
kg
,
.
b) & 1 S MPa 280 440 260E
s ** T=
-
.
=
·
1627-- 1483
c) 2
=
Y
Finch 2 z0 in coch
· 1483-1325 u -- 1325
-
240 =
200 220-200
v=
for
1404m3/kg 230°c
·
- 175 -
-
2060 ~-2060 v= 21675
=
240-200 220-100
· 1484--21675 U --21675
1 5
.
-
I
=
14-1
=
v =
19567
mig
↑
-
60
Area .
Velocity
& 80bar Superheat vapor
-
Volume
440 03742h
144rg
v
3246
= =
-
.
i in =
AV
& 60 bor superheat vapor -
-
T =
400 h =
x +
pv
v = -
04734h =
3177 2 .
&. 7 bar x =
9
,
.
n =
376 78:
+ , 90(2660 -
176 70
.
= 2 .
12 mYkg
mu
2)
see ( 2064s
=
-2 = 22 .
it
e 236
O
L son)
As 8() 101 MIS
=
. =
23
