Superheated water vapor at 180 psia and 500^{\circ}F is allowed

William Burnett

William Burnett

Answered question

2022-01-17

Superheated water vapor at 180 psia and 500F is allowed to cool at constant volume until the temperature drops to 250F At the final state, determine (a) the pressure, (b) the quality, and (c) the enthalpy.
Also, show the process on a Tv diagram with respect to the saturation lines. Answers: (a) 29.84 psia, (b) 0.219, (c) 426Btulbm

Answer & Explanation

Stuart Rountree

Stuart Rountree

Beginner2022-01-18Added 29 answers

The pressure is obtained from table A-4E for the given final temperature and it is:
P2=29.844ψa
To calculate the quality, we look in table A-6E for the initial general specific volume and in table A-4E for the specific volume of the constituents:
q2=ααliqαvapαliq
=(3.04330.017)ft3lbm(13.8160.017)ft3lbm
=0.22
The enthalpy is then calculated again using values from A-4E
h2=hliq+1(hvaphliq)
=218.63Btulbm+0.22(1164218.63)Btulbm
=426.61Btulbm
Chanell Sanborn

Chanell Sanborn

Beginner2022-01-19Added 41 answers

Step 1
Given:
Pressure of superheated water vapor, P1=180ψa
Initial temperature, T1=500F
Final temperature, T2=250F
Constant volume process
Step 2
a) As the initial state is superheated vapor, the specific volume at the initial pressure and temperature can be obtained from superheated water table
At P1=180ψa and T1=500F
Specific volume, v=3.0433ft3lbm
As the volume is constant throughout the process.
Final specific volume =v=3.0433ft3lbm
Therefore, the final state is defined by temperature and specific volume. We do not know which table to use to determine the other properties because we have no cule as to whether we have saturated mixture, compressed liquid or superheated vapor, which can bedetermined from the specific volume of satureted liquid (vf) and specific volume of satureted vapor (vg) from the saturated water-temperature table. from the saturated water-temperature table.
At T2=250C
vf=0.01700ft3lbm
vg=13.816ft3lbm
In our case, the given specific volume v=3.0433ft3lbm lies in between the values of vf and vg at the given temperature. Thus, we have saturated liquid-vapor mixture. Thus the final pressure of the state will be the saturation pressure at the final temperature which can be obtained from saturated water-temperature table.
Therefore, at T=250F
Final pressure, P2Psatα250F
P2=29.844ψa
Step 3
b) The quality (x) of mixture can be determined from
v=vf+(x×(vgvf))
Therefore, quality x=vvfvgvr
x=3.04330.0170013.8160.01700
x=0.219
Step 4
c) From saturated water-temperature table
at T2=250F
hf=218.63Btulbm
hfg=945.41Btulbm
Therefore, enthalphy of mixture is
h=hf+(x×hfg)
h=218.63+(0.219×945.41)
h=425.67Btulbm

alenahelenash

alenahelenash

Expert2022-01-23Added 556 answers

Step 1 a) To solve this problem we first need to determine the specific volume v1 of the superheated water at the given temperature T1=500F and pressure p1=180psia We do that using the table: v1=3.0433ft3lbm In the process the specific volume does not change v1=v2v To determine the pressure at the end of the process p2 we first need to determine the phase of the water at the end of the process. At T2=250F the specific volume of the liquid water is vf=0.01700ft3lbm1 and vapor vg=13.816ft3lbm1. Determine specific volume is between the two meaning the water is saturated mixture. The pressure p2 is then determined using the appropriate software. p2=29.8psia Step 2 b) The quality of the mixture x is calculated using the specific volumes v, vf and vg x=vvfvgvf x=3.0433ft3lbm0.01700ft3lbm13.816ft3lbm0.01700ft3lbm Step 3 c) Enthalphy of the mixture h we can calculated using the enthalpy of liquid water hf=218.63Btulbm1 and evaporation enthalphy hfg=945.41Btulbm1 at T2=250F and the calculated quality x. h=hf+x×hfg h=218.63Btulbm+0.219×945.41Btulbm h=425.7Btulbm

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