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Homework Solutions
1 Complete the following table for H 2 O
Row T(oC) P(kPa) v(m^3 /kg) Phase description 1 50 4. 2 200 Saturated vapor 3 250 400 4 110 600
Use the Tables A-4 through A-8 for water starting on page 914. For the first row data, we first try to see if the given specific volume of 4.16 m^3 /kg lies in the liquid, gas, or mixed region. We do this by looking at the values of vf and vg at the given temperature of 50oC. On page 914, we find that vf(50oC) = 0.001012 m^3 /kg and vg(50oC) = 12.026 m^3 /kg. Since the given volume of 4. m^3 /kg lies between the saturated liquid volume, vf, and the saturated vapor volume, vg, at the given temperature of 50oC, we conclude that the state is in the mixed region. Since we are in the mixed region, the pressure is the saturation pressure at the given temperature of 50oC. From the same location in Table A-4 we find Psat(50oC) = 12.352 kPa. The quality at this point can be found as follows.
3 3
3 3
kg
m
kg
m
kg
m
kg
m
v v
v v x g f
f
The point in row two is defined as saturated vapor. This means that the temperature is the saturation pressure at 200 kPa and the volume is the volume of the saturated vapor at this pressure. From Table A-5 on page 916 we find Tsat(200 kPa) = 120.21oC and vg(200 kPa) = 0.88578 m^3 /kg.
The data in row three are found in the superheat table, A-6, on page 918; the volume at this point is 0.59520 m^3 /kg. Note that we the pressures in this table are in MPa so we have to recognize that 400 kPa = 0.4 MPa.
In searching for the state in row four, we find that the temperature of 110oC is less than the saturation temperature at 600 kPa, which is found in Table A-5 on page 916 to be 158.83oC.
This can also be found if we guessed that this was a gas and tired to locate this pressure and temperature in the superheat region: Table A-6 on page 918. We see that that the minimum temperature in the section of this table for 600 kPa (stated as 0.6 MPa) is 200oC. To see if this is a liquid we can find the saturation temperature at the top of the table section, next to the 0.6 MPa value. Because the given temperature of 110oC is less than the saturation temperature of 158.83oC, the state is a compressed liquid. The simplest approach is to approximate the specific volume at this state as the volume of the saturated liquid at the given temperature of 110oC. This volume is found to be 0.001052 m^3 /kg (Table A-4, page 914).
The alternative approach would be a double interpolation in the compressed liquid table on page
- Because this table does not have a temperature of 110oC, we have to interpolate data between the table temperatures of 100oC and 120oC. Because the minimum pressure in this table is 5 MPa, we have to interpolate between the saturation pressure and the pressure of 5 MPa to get the values at the desired pressure of 600 kPa.
First interpolate to find the value at 100oC and 600 kPa.
kg
m kPa kPa kPa kPa
kg
m
kg
m
kg
m
kPa P C kPa P C
v C kPa v C v C kPa v C
o o sat sat
o f
o o f
o
3
3 3
3
- 0010509 600 101. 42 5000 101. 42
Next interpolate to find the value at 120oC and 600 kPa.
kg
m kPa kPa kPa kPa
kg
m
kg
m
kg
m
kPa P C kPa P C
v C kPa v C v C kPa v C
o o sat sat
o f
o o f
o
3
3 3
3
- 0010598 600 198. 67 5000 198. 67
Finally we can interpolate between the 100oC and 120oC at the pressure of 600 kPa.
kg
m C C C C
kg
m
kg
m
kg
m
C C
C C
v C kPa v C kPa v C kPa v C kPa
o o o o
o o o o
o o o o
3
3 3
3
- 0010553 110 100 120 100
We see that this answer is close to the value of 0.001052 m^3 /kg for the saturated liquid volume at the given temperature of 110^0 C.
The answers for this row and the previous rows are shown in the table below.
Row T(oC) P(kPa) v(m^3 /kg) Phase description 1 50 12.3 52 4.16 Mixed region with x = 0. 2 120.23 200 0.8857 8 Saturated vapor 3 250 400 0.595 20 Superheated vapor 4 110 600 0.0010 52 Compressed liquid
3 A piston-cylinder device initially contains 50L of liquid water at 25oC and 300 kPa. Heat is
added to the water at constant pressure until the entire liquid is vaporized. (a) What is the mass of the water? (b) What is the final temperature? (c) Determine the total enthalpy change. (d) Show the process on a T-v diagram with respect to saturation lines.
We find that the initial state of 25oC and 300 kPa is a compressed liquid, because the temperature of 25oC is less than the saturation temperature at 300 kPa. (See Table A-5 on page 916 that gives Tsat( 300 kPa = 133.52oC. Alternatively we could find that the pressure of 300 kPa is greater than the saturation pressure at 25 oC, which is found as 3.1598 kPa in Table A-4 on page 915.) We approximate this compressed liquid state as the saturated liquid at the given temperature of 25oC, from Table A-4. This gives an initial volume, v 1 ≈ vf(25oC) = 0.001003 m^3 /kg and an initial enthalpy, h 1 ≈ hf(25oC) = 104.89 kJ/kg.
Since we know the total volume, V = 50L = 0.05 m^3 and the initial specific volume, v 1 = 0.001003 m^3 /kg, we can find the mass in the system as the ratio of these two quantities.
kg
m
m
v
V
m 3
3
1
1
m = 49.85 kg
We are told that the entire liquid is vaporized at the final state. If we assume that there is no superheat, the final state must be that of saturated vapor. Since we are also told that the process occurs at constant pressure, the final pressure, P 2 = P 1 = 300 kPa. The final temperature is the
saturation temperature at 300 kPa from Table A-5; i.e ., T 2 = Tsat(300 kPa) = 133.55C.
The total enthalpy change is the difference between the initial and final enthalpy, multiplied by the system mass. We found the initial enthalpy above; we know that the final state is a saturated vapor at 300 kPa; thus h 2 = hg(300 kPa) = 2724.9 kJ/kg, and the total enthalpy change is given by the following equation.
H = m(h 2 – h 1 ) = (49.85 kg)(2724.9 kJ/kg – 104.89 kJ/kg = 1.306x10^5 kJ/kg
The T-v diagram with the process indicated is shown above. On this diagram the process is a solid line (blue if you view this in color) and the saturation curve is the dotted line (purple in color). The temperature of the compressed liquid increases, at constant pressure, until the saturation temperature at the constant pressure of 300 kPa (133.55oC) is reached. This initial part of the process is so close to the saturation line that we cannot distinguish the two on the plot. Once the saturation temperature is reached, the additional heat vaporizes the liquid water. This part of the
T-v Diagram for Water
0.
1
10
100
1000
0.0001 0.01 1 100 10000
Specific volume (m3/kg)
Temperature (degrees Celsius)
initial point
final point
process takes place at constant temperature. (In the mixed region a constant pressure process is a constant temperature pressure process and vice versa .)
4 The pressure in an automobile tire depends on the temperature of the air in the tire. When
the air temperature is 25oC, the pressure gage reads 210 kPa. If the volume of the tire is 0.025 m^3 , determine the pressure rise in the tire when the temperature in the tire rises to 50 oC. Also, determine the amount of air that must be bled off to restore pressure to its original value at this temperature. Assume the atmospheric pressure to be 100 kPa.
For this problem we can use the ideal gas equation. The gas constant for air is found from Table A-1 on page 908 to be 0.2870 kJ/kg·. At the initial state with an atmospheric pressure of 100 kPa and a gage pressure of 210 kPa, the absolute pressure is 210 kPa + 100 kPa = 310 kPa. The absolute temperature initially is 25 + 273.15 = 298.15 K. The absolute temperature at the final state is 50 + 273.15 = 323.15 K.
In the first part of this problem, the mass is constant. We are not told anything about how the tire expands as the temperature increases. Let’s make the simple assumption that the volume remains constant.
According to the ideal gas equation, PV = mRT. If we compare two states for which mass and volume are the same we have the following result.
1
2 2 1 2
2
1
1
T
T
P P
P
T
P
T
const mR
V
Substituting the given data into the final equation gives the pressure at the temperature of 50oC.
kPa K
K
kPa T
T
P P 336
1
2 2 1
The pressure rise is simply the difference P 2 – P 1 = (336 – 310) kPa = 26 kPa.
We contemplate a third state, after we bleed off the air. In this final state the pressure equals the initial pressure (P 3 = P 1 = 310 kPa), while the final temperature remains at 50oC. (T 3 = T 2 = 50 oC.) The mass difference between these two states is computed from the ideal gas equation as follows.
K K
kgK
kJ
kPa m
R T T
PV
RT
PV
RT
PV
m m m
- 15
3
1 3
1
3
3
1
1 1 3
m = 0.0070 kg
6 Determine the specific volume of superheated water vapor at 10 MPa and 400oC using (a)
the ideal gas equations, (b) the generalized compressibility charts and (c) the steam tables. Also, determine the error involved in the first two cases.
We can solve part (c) first. The steam-table specific volume is found from Table A-6 on page 908 for the given pressure and temperature.
vsteam tables = 0.026436 m^3 /kg
To use the ideal-gas equation of state, we first obtain the gas constant for H 2 O from table A-1 on page 884: R = 0.4615 kJ/(kg•K) for water. Applying this to the pressure of 10 MPa = 10,000 kPa and a temperature of 400oC = 673.15 K gives the following specific volume.
3
idealgas
kPa m
1 kJ ( 10 , 000 kPa)
( 673. 15 K)
kg K
- 4615 kJ
P
RT
v
= 0.03106 m^3 /kg
The error in the ideal gas calculation is found by comparing the result to the specific volume obtained from the steam tables.
kg
- 026416 m
kg
026416 m kg
03106 m
Error (^3)
3 3
^ =^ 17.6%
In order to use the generalized compressibility chart, we first have to compute the reduced pressure and temperature. These are simply the ratio of the given pressure and temperature to the critical pressure and temperature, respectively. We find the critical values for water from Table A-1 on page 908: Tc = 647.1 K and Pc = 22.06 MPa. The error in the ideal gas calculation is found by comparing the result to the specific volume obtained from the steam tables.
K
K
T
T
T
MPa
MPa
P
P
P
c
r c
r
With these values of reduced pressure and temperature, we can find the compressibility factor from the chart in Figure A-15 on page 932. This gives Z = 0.84. With this value of Z, we can find the desired specific volume as follows.
kg
Zv m P
ZRT
v (^) Zchart idealgas
3 ( 0. 84 ) 0. 03106 = 0.026 m^3 /kg
As with the ideal-gas calculation, the error is found by comparing the result to the specific volume obtained from the steam tables.
kg
m
kg
m kg
m
Error (^) 3
3 3
