A well-insulated rigid tank contains 3 kg of saturated liquid-vapor mixture of w

verskalksv

verskalksv

Answered question

2021-11-07

A well-insulated rigid tank contains 3 kg of saturated liquid-vapor mixture of water at 200 kPa. Initially, three-quarters of the mass is in the liquid phase. An electric resistance heater placed in the tank is now turned on and kept on until all the liquid in the tank is vaporized. Determine the entropy change of the steam during this process. Answer: 11.1 kJ/K

Answer & Explanation

SaurbHurbkj

SaurbHurbkj

Beginner2021-11-08Added 16 answers

Step 1 
Given: 
Substance: A mixture of water and vapor, in which the water is saturated
m=3kg 
p1=200kPa 
mf1=34m=2.25kg 
Process: constant volume 
x2=1 
Required:S=? 
Step 2 
Solution: 
First we need to calculate the quality in state 1.
x1=mgm=mmf1m=32.253=0.25 
Now from the steam TABLE A-5 
p1=200kPavf1=0.001061m3kg 
vg1=0.88578m3kg 
sf1=1,5302kJkgK 
sfg1=5.5968kJkgK 
Using these data, we can calculate specific entropy in state 1 from the equation:
s1=sf1+x1g=1.5302+0.255.5968=2.9294kJkgK 
We will obtain entrophy in the state 2 from the FIGURE A-10, but to do so first we have to calculate specific volume in the state 1 which is the same as the one in state 2 because this is a constant volume process. 
Equation for specific volume is: 
v1=vf1+x1vfg=0.001061+0.25(0.885780.001061)=0.22m3kg=v2 
Step 3 
Now we can use Molier diagram for water (Figure A-10) and obtain specific entrophy in the state 2. 
p2=1v2=10.22=4.55kgm3;x2=1s2=6.65kJkgK 
Finally entrophy change is calculated from: 
S=m(s2s1)=3(6.652.9294) 
S=11.16kJK 
Result 
S=11.16kJK

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