Thermodynamics A/C Techniques Dep. 1 st Year Class First Term 2018-2019 Lecture 12 : Phase-Change Processes of pure substance (Examples) by: Asst. lect. Karrar Al-Mansoori 1 (Pressure of Saturated Liquid in a Tank) Example 12-1/ A rigid tank contains 50 kg of saturated liquid water at 90°C. Determine the pressure in the tank and the volume of the tank. Solution: A rigid tank contains saturated liquid water. The pressure and volume of the tank are to be determined. Analysis: The state of the saturated liquid water is shown on a T-v diagram. Since saturation conditions exist in the tank, the pressure must be the saturation pressure at 90°C: P =Psat @ 90°C = 70.183 kPa (Table A–4) The specific volume of the saturated liquid at 90°C is: v = vf @ 90°C =0.001036 m 3 /kg (Table A–4) Then the total volume of the tank becomes : V =m . v = (150 kg)(0.001036 m 3 /kg)= 0.0518 m 3 (Temperature of Saturated Vapor in a Cylinder) Example 12–2 / A piston–cylinder device contains 2 ft 3 of saturated water vapor at 50- psia pressure. Determine the temperature and the mass of the vapor inside the cylinder. Solution: A cylinder contains saturated water vapor. The temperature and the mass of vapor are to be determined. Figure (12-1): Schematic and T-V diagram for Ex 12-1
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Thermodynamics A/C Techniques Dep.
1st Year Class First Term 2018-2019
Lecture 12 : Phase-Change Processes of pure substance (Examples)
by: Asst. lect. Karrar Al-Mansoori
1
(Pressure of Saturated Liquid in a Tank)
Example 12-1/ A rigid tank contains 50 kg of saturated liquid water at 90°C.
Determine the pressure in the tank and the volume of the tank.
Solution: A rigid tank contains saturated liquid water. The pressure and volume
of the tank are to be determined.
Analysis: The state of the saturated liquid water is shown on a T-v diagram.
Since saturation conditions exist in the tank, the pressure must be the saturation
pressure at 90°C:
P =Psat @ 90°C = 70.183 kPa (Table A–4)
The specific volume of the saturated liquid at
90°C is:
v = vf @ 90°C =0.001036 m3/kg (Table A–4)
Then the total volume of the tank becomes :
V =m . v = (150 kg)(0.001036 m3/kg)= 0.0518 m3
(Temperature of Saturated Vapor in a Cylinder)
Example 12–2 / A piston–cylinder device contains 2 ft3 of saturated water vapor at 50-
psia pressure. Determine the temperature and the mass of the vapor inside the cylinder.
Solution: A cylinder contains saturated water vapor. The temperature and the
mass of vapor are to be determined.
Figure (12-1): Schematic and T-V
diagram for Ex 12-1
Thermodynamics A/C Techniques Dep.
1st Year Class First Term 2018-2019
Lecture 12 : Phase-Change Processes of pure substance (Examples)
by: Asst. lect. Karrar Al-Mansoori
2
Analysis: The state of the saturated water vapor is shown on a P-v diagram.
Since the cylinder contains saturated vapor at 50 psia, the temperature inside
must be the saturation temperature at this pressure:
T = Tsat @ 50 psia = 280.99°F (Table A–5E)
The specific volume of the saturated vapor at 50
psia is:
v = vg @ 50 psia = 8.5175 ft3/lbm (Table A–5E)
Then the mass of water vapor inside the cylinder
becomes
𝒎 = 𝑽
𝒗 =
𝟐 𝒇𝒕𝟑
𝟖.𝟓𝟏𝟕𝟓 𝒇𝒕𝟑
𝒍𝒃𝒎⁄
= 𝟎. 𝟐𝟑𝟓 𝒍𝒃𝒎
(Volume and Energy Change during Evaporation)
Example 12–3 / A mass of 200 g of saturated liquid water is completely vaporized at
a constant pressure of 100 kPa. Determine (a) the volume change and (b) the amount
of energy transferred to the water.
Solution: Saturated liquid water is vaporized at constant pressure. The volume
change and the energy transferred are to be determined.
Analysis: The process described is illustrated on a P-v diagram below. The
volume change per unit mass during a vaporization process is vfg , which is the
difference between vg and vf. Reading these values from (Table A–5) at 100 kPa
and substituting yield :
Figure (12-1): Schematic and P-v
diagram for Example 12–2.
Thermodynamics A/C Techniques Dep.
1st Year Class First Term 2018-2019
Lecture 12 : Phase-Change Processes of pure substance (Examples)
by: Asst. lect. Karrar Al-Mansoori
3
vfg = vg - vf = 1.6941 - 0.001043 = 1.6931 m3/kg.
Thus,
∆V = m .vfg = (0.2 kg) (1.6931 m3/kg) = 0.3386 m3
(b) The amount of energy needed to vaporize a unit
mass of a substance at a given pressure is the
enthalpy of vaporization at that pressure, which is:
hfg = 2257.5 kJ/kg for water at 100 kPa.
Thus, the amount of energy transferred is:
m.hfg = (0.2 kg)(2257.5 kJ/kg) = 451.5 kJ
(Pressure and Volume of a Saturated Mixture)
Example 12–4 / A rigid tank contains 10 kg of water at 90°C. If 8 kg of the water is in
the liquid form and the rest is in the vapor form, determine (a) the pressure in the tank
and (b) the volume of the tank.
Solution: A rigid tank contains saturated mixture. The pressure and the volume of
the tank are to be determined.
Analysis (a): The state of the saturated liquid–vapor mixture is shown in
(Fig. 12-4). Since the two phases coexist in equilibrium, we have a saturated
mixture, and the pressure must be the saturation pressure at the given temperature:
Figure (12-3): Schematic and P-v
diagram for Example 12–3.
Thermodynamics A/C Techniques Dep.
1st Year Class First Term 2018-2019
Lecture 12 : Phase-Change Processes of pure substance (Examples)
by: Asst. lect. Karrar Al-Mansoori
4
P = Psat @ 90°C = 70.183 kPa ( Table A– 4 )
(b) At 90°C, we have vf = 0.001036 m3/kg, and vg = 2.3593 m3/kg ( Table A–4 ). One
way of finding the volume of the tank is to determine the volume occupied by each
phase and then add them:
V = Vf + Vg = mf . vf + mg .vg
= (8 kg)(0.001036 m3/kg) + (2 kg)(2.3593 m3/kg)
= 4.73 m3
➢ Another way is to first determine the quality x, then the average specific volume v,