A piston-cylinder arrangement containing steam at 200 kPa and 200°C is cooled at

Question

A piston-cylinder arrangement containing steam at 200 kPa and 200°C is cooled at a constant volume of 0.5 m3 until it reaches 140 kPa.

(a) Calculate the amount of mass of steam in the cylinder (kg)

(b) Find the quality (x) of the steam once it reaches 140 kPa

Then the piston is allowed to float freely (maintaining a constant pressure, but changing volume), while 50% of the steam is released from the piston. Once the release is done, you determine that the remaining steam is 80% vapor by mass.

(c) Find the final TOTAL volume of steam in the cylinder (m3)

(d) Assuming RH2O is 0.4615 kJ/kg-K, Determine the TOTAL volume of the steam assuming it was an ideal gas (m3)

(e) Calculate the percent error in the specific volume calculated using the ideal gas law (from part (d)) compared the actual specific volume found from the steam tables (part (c)).

Assuming there are 3.28 ft per meter, (f) Calculate the final volume from part (c) in cubic feet. (ft3)

I have that answer I just can't figure out how to get them. Please help and if you can show all work that be great.

The answer are: a)0.463kg b)x=0.8737 c)0.229m^3 d)0.2917m^3 e)27.4% f)8.08ft^3

Explanation / Answer

Given

P1 = 200 kPa, T1 = 200 C, V1 = 0.5 m^3

P2 = 140 kPa and V2 = 0.5 m^3

The properties at state 1 are

v1 = 1.08 m^3/kg, h1 = 2870 kJ/kg

The properties at state 2 are

v2 = 1.08 m^3/kg and h2 = 2408 kJ/kg

a)

m = V1/v1 = V2/v2 = 0.463 kg

b)

The saturated properties at P2 =1 40 kPa are

vf = 0.001051 and vg = 1.237

Tsat = 109.3 C

Now

v2 = (1-x)*vf + x*vg

x = 0.8737

c)

Given

50 % of the steam is released

Therefore mass of steam in the piston-cylinder arrangement is

m3 = 0.5*0.463 = 0.2315 kg

also

x = 0.8

(since the pressure remains same, the saturated properties of steam at 140 kPa are noted in part b

v3 = (1-x)*vf + x*vg

v3 = 0.2*0.001051 + 0.8*1.237

v3 = 0.9898 m^3/kg

The volume is given by

V3 = m3*v3 = 0.229 m^3

d)

For ideal gas

V3 = mRT3/P3 =0.2315*0.4615*(273+109.3)/140

V3 = 0.2917 m^3

e)

The specific volume for the ideal gas is

v3 = V3/m3 = 0.2917/0.2315 = 1.26

From the steam tables

v3 = 0.9898 m^3/kg

Error = (1.26-0.9898)/(0.9898) = 27.3%

f)

V3 = m3*v3 = 0.229 m^3

convert m^3 into ft^3

V3 = 8.0870587 ft^3

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