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TER 201E THERMODYNAMICS
Problem Hour - 1
Problems and Solutions
Subject: Properties of Pure Substances
( Property Tables )
Prof. Dr. Ayşegül ABUŞOĞLU
2019 - 2020 Spring Term
Problems
1. Complete this table for H 2 O
2. Complete this table for H 2 O
3. Complete this table for refrigerant-134a
4. 1.8-m3 rigid tank contains steam at 220°C. One third of the volume is in the liquid phase and the rest
is in the vapor form. Determine ( a ) the pressure of the steam, ( b ) the quality of the saturated mixture,
and ( c ) the density of the mixture.
Solution: Properties At 220C v f = 0.001190 m
3 /kg and v g = 0.08609 m 3 /kg (Table A-4).
Analysis ( a ) Two phases coexist in equilibrium, thus we have a saturated liquid-vapor mixture. The pressure of the steam is the
saturation pressure at the given temperature. Then the pressure in the tank must be the saturation pressure at the specified temperature,
P T sat@220C 2320 kPa
( b ) The total mass and the quality are determined as
- 2 13. 94 518. 1 kg
1 3.94kg 0.08609m/kg
2/3 (1.8m)
504 .2kg 0.001190m/kg
1/3 (1.8m)
3
3
3
3
t
g
t f g
g
g g
f
f f
m
m x
m m m
m
m
v
V
v
V
( c ) The density is determined from
3 287.8kg/ m
( ) 0. 001190 ( 0. 0269 )( 0. 08609 ) 0. 003474 m/kg 3
v
v v v v
f x g f
5. A person cooks a meal in a 30-cm-diameter pot that is covered with a well-fitting lid and lets the food
cool to the room temperature of 20°C. The total mass of the food and the pot is 8 kg. Now the person tries
to open the pan by lifting the lid up. Assuming no air has leaked into the pan during cooling, determine if
the lid will open or the pan will move up together with the lid.
Solution: Assumptions 1 The local atmospheric pressure is 1 atm = 101.325 kPa. 2 The weight of the lid is small and thus its effect
on the boiling pressure and temperature is negligible. 3 No air has leaked into the pan during cooling.
Properties The saturation pressure of water at 20C is 2.3392 kPa (Table A-4).
Analysis Noting that the weight of the lid is negligible, the reaction force F on the lid after cooling at the pan-lid interface can be
determined from a force balance on the lid in the vertical direction to be
PA +F = PatmA
or,
= 6997 mPa= (since 1 Pa= 1 N/m )
( 101 , 325 2339. 2 )Pa 4
( 0. 3 m)
2 2
2
2
6997 N
F APatm P D Patm P
The weight of the pan and its contents is
( 8 kg)(9.81m/s)= 78.5N
2 W mg
which is much less than the reaction force of 6997 N at the pan-lid interface. Therefore, the pan will move up together with the lid
when the person attempts to open the pan by lifting the lid up. In fact, it looks like the lid will not open even if the mass of the pan and
its contents is several hundred kg.
6. Water in a 5-cm-deep pan is observed to boil at 98°C. At what temperature will the water in a 40-cm-
deep pan boil? Assume both pans are full of water.
Solution: Assumptions Both pans are full of water.
Properties The density of liquid water is approximately = 1000 kg/m 3 .
Analysis The pressure at the bottom of the 5-cm pan is the saturation pressure corresponding to
the boiling temperature of 98C:
94.39kPa sat@98 C P P (Table A-4)
The pressure difference between the bottoms of two pans is
3.43kPa 1000 kg/ms
1 kPa (1000 kg/m)(9.807m/s)(0.35m) 2
3 2
P gh
Then the pressure at the bottom of the 40-cm deep pan is
P = 94.39 + 3.43 = 97.82 kPa
Then the boiling temperature becomes
T boiling T [email protected] 99.0 C (Table A-5)
Steam
1.8 m
3
220 C
P
Patm = 1 atm mmHg
2.3392 kPa
40 cm
5 cm
( c ) The quality at the final state is specified to be x 2 = 0.5. The specific volumes at the
initial and the final states are
0.25799m/kg 300 C
1.0 MPa 3 1 1
1
^ v T
P
(Table A-6)
0 .09775m/kg
1.0MPa
3
2 2 2
2
f x fg x
P
v v v
Thus,
3 Δ ( )(0.8kg)(0.0977 5 0.25799)m/kg 0.128 2 m
3
V m v 2 v 1
11. A rigid tank initially contains 1.4-kg saturated liquid water at 200°C. At this state, 25 percent of the
volume is occupied by water and the rest by air. Now heat is supplied to the water until the tank contains
saturated vapor only. Determine ( a ) the volume of the tank, ( b ) the final temperature and pressure, and
( c ) the internal energy change of the water.
Solution: Properties The saturated liquid properties of water at 200C are: v f = 0.001157 m
3 /kg and uf = 850.46 kJ/kg (Table A-4).
Analysis ( a ) The tank initially contains saturated liquid water and air. The volume occupied by water is
3 3
V 1 m v 1 ( 1. 4 kg)( 0. 001157 m/kg) 0. 001619 m
which is the 25 percent of total volume. Then, the total volume is determined from
3 ( 0. 001619 ) 0.006476 m
- 25
V
( b ) Properties after the heat addition process are
- 004626 m /kg 1.4kg
0.006476 m 3
3
2 ^ m
V
v
- 5 kJ/kg
- 004626 m /kg
2
2
2
2
3 2
u
P
T
x
21,367kPa
371.3C
v (Table A-4 or A-5)
( c ) The total internal energy change is determined from
U m ( u 2 u 1 )( 1. 4 kg)(2201.5-850.46)kJ/kg 1892 kJ
12. A piston–cylinder device initially contains steam at 3.5 MPa, superheated by 5°C. Now, steam loses
heat to the surroundings and the piston moves down hitting a set of stops at which point the cylinder
contains saturated liquid water. The cooling continues until the cylinder contains water at 200°C.
Determine ( a ) the initial temperature, ( b ) the enthalpy change per unit mass of the steam by the time the
piston first hits the stops, and ( c ) the final pressure and the quality (if mixture).
Solution: ( a ) The saturation temperature of steam at 3.5 MPa is
T [email protected] MPa = 242.6C (Table A-5)
Then, the initial temperature becomes
T 1 = 242.6+5 = 247.6 C
Also, 2821. 1 kJ/kg
6 C
5 MPa 1 1
1
h T
P
(Table A-6)
( b ) The properties of steam when the piston first hits the stops are
001235 m/kg
7 kJ/kg
0
- 5 MPa 3 2
2
2
2 1
v
h
x
P P
(Table A-5)
T
v
2
1
Then, the enthalpy change of steam becomes
h h 2 h 1 1049. 7 2821.1 -1771kJ/kg
( c ) At the final state
1555 kPa
3
3
3
3 3 2
200 C
- 001235 m/kg
x
P
T
v v
(Table A-4)
The cylinder contains saturated liquid-vapor mixture with a small mass of vapor at the final state.
