Homework #1 Solutions
Fall 2010 - Solution
1. The zeroth law of thermodynamics states that two bodies are in thermal equilibrium if both have the same
temperature reading, even if they are not in contact.
2. A barometer is used to measure the height of a building by recording
reading at the bottom and at the top of the building. The height of the building
is to be determined.
Assumptions The variation of air density with altitude is negligible.
Properties The density of air is given to be  = 1.18 kg/m3. The density of
mercury is 13,600 kg/m3.
Analysis Atmospheric pressures at the top and at the bottom of the building are
Ptop  ( ρ g h) top

1N
 (13,600 kg/m 3 )(9.807 m/s 2 )(0.730 m) 
 1 kg  m/s 2

 97.36 kPa
Pbottom  (  g h) bottom

1N
 (13,600 kg/m 3 )(9.807 m/s 2 )(0.755 m) 
 1kg  m/s 2

 100.70 kPa
 1 kPa

 1000 N/m 2





 1 kPa

 1000 N/m 2





730 mmHg
h
755 mmHg
Taking an air column between the top and the bottom of the building and writing a force balance per unit base area,
we obtain
Wair / A  Pbottom  Ptop
( gh) air  Pbottom  Ptop

1N
(1.18 kg/m 3 )(9.807 m/s 2 )( h)
 1 kg  m/s 2

 1 kPa

 1000 N/m 2


  (100.70  97.36) kPa


It yields
h = 288.6 m
which is also the height of the building.
3. The pressure in a pressurized water tank is measured by a multi-fluid
manometer. The gage pressure of air in the tank is to be determined.
Assumptions The air pressure in the tank is uniform (i.e., its variation with
elevation is negligible due to its low density), and thus we can determine the
pressure at the air-water interface.
Properties The densities of mercury, water, and oil are given to be 13,600, 1000,
and 850 kg/m3, respectively.
Analysis Starting with the pressure at point 1 at the air-water interface, and moving
along the tube by adding (as we go down) or subtracting (as we go up) th e gh
terms until we reach point 2, and setting the result equal to Patm since the tube is
open to the atmosphere gives
P1   water gh1   oil gh2   mercury gh3  Patm
Solving for P1,
P1  Patm   water gh1   oil gh2   mercury gh3
or,
P1  Patm  g (  mercury h3   water h1   oil h2 )
Noting that P1,gage = P1 - Patm and substituting,
P1,gage  (9.81 m/s 2 )[(13,600 kg/m 3 )( 0.46 m)  (1000 kg/m 3 )( 0.2 m)

 1 kPa
1N


 (850 kg/m 3 )( 0.3 m)] 
 1 kg  m/s 2  1000 N/m 2 


 56.9 kPa
Discussion Note that jumping horizontally from one tube to the next and realizing that pressure remains the same in
the same fluid simplifies the analysis greatly.
4. The gage pressure in a pressure cooker is maintained constant at 100 kPa by a petcock.
The mass of the petcock is to be determined.
Assumptions There is no blockage of the pressure release valve.
Analysis Atmospheric pressure is acting on all surfaces of the petcock, which balances
itself out. Therefore, it can be disregarded in calculations if we use the gage pressure
as the cooker pressure. A force balance on the petcock (Fy = 0) yields
Patm
P
W = mg
W  Pgage A
(100 kPa)(4 10 6 m 2 )  1000 kg/m  s 2

g
1 kPa
9.81 m/s 2

 0.0408 kg
m
Pgage A





5. A vertical tube open to the atmosphere is connected to the vein in the arm of a person. The height that the blood
will rise in the tube is to be determined.
Assumptions 1 The density of blood is constant. 2 The gage pressure of blood is 120
mmHg.
Properties The density of blood is given to be  = 1050 kg/m3.
Analysis For a given gage pressure, the relation P  gh can be expressed for
Blood
mercury and blood as P   blood ghblood and P   mercury ghmercury . Setting these two
h
relations equal to each other we get
P   blood ghblood   mercury ghmercury
Solving for blood height and substituting gives
hblood 
 mercury
 blood
hmercury 
13,600 kg/m 3
1050 kg/m 3
(0.12 m)  1.55 m
Discussion Note that the blood can rise about one and a half meters in a tube connected to the vein. This explains
why IV tubes must be placed high to force a fluid into the vein of a patient.
6. Complete the following table for H2 O:
T, C
P, kPa
h, kJ / kg
x
Phase description
120.21
200
2045.8
0.7
Saturated mixture
140
361.53
1800
0.565
Saturated mixture
177.66
950
752.74
0.0
Saturated liquid
80
500
335.37
---
Compressed liquid
350.0
800
3162.2
---
Superheated vapor
7. A rigid container that is filled with water is cooled. The initial temperature and final pressure are to be
determined.
Analysis The initial state is superheated vapor. The temperature is determined to be
P1  25 0 psia

 T  550F
3
v 1  2.29 ft /lbm  1
H2O
250 psia
1 lbm
2.29 ft3
(Table A - 6E)
This is a constant volume cooling process (v = V /m = constant). The
final state is saturated mixture and thus the pressure is the saturation
pressure at the final temperature:
P
T2  100 F

 P  Psat @ 100F  0.9505 psia (Table A - 4E)
3
v 2  v 1  2.29 ft /lbm  2
1
2
v
8. A piston-cylinder device that is filled with water is cooled. The final pressure and volume of the water are to be
determined.
Analysis The initial specific volume is
v1 
V1
m

2.3615 ft 3
 2.3615 ft 3 /lbm
1 lbm
H2O
400°F
1 lbm
2.3615 ft3
This is a constant-pressure process. The initial state is determined to be
superheated vapor and thus the pressure is determined to be
T1  400 F

 P  P2  200 psia (Table A - 6E)
3
v 1  2.3615 ft /lbm  1
The saturation temperature at 200 psia is 381.8°F. Since the final
temperature is less than this temperature, the final state is compressed
liquid. Using the incompressible liquid approximation,
P
2
1
v 2  v f @ 100F  0.01613 ft 3 /lbm (Table A - 4E)
The final volume is then
V 2  mv 2  (1 lbm)(0.01613ft /lbm)  0.01613 ft
3
v
3