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

I

Explanation / Answer

(a)

n = PV/RT

P= 200*1000 Pa; V =0.5 m3

T = 200+273.15 = 473.15 K

R = 8.314 J/K-mol

Moles of steam = 200*1000*0.5/(8.314*473.15) = 25.4 moles

Mass of steam (M) = Moles *MW = 25.4*18/1000

= 0.46 kg

(b)

P2 = 140 kPa

T2 = T1*P2/P1 = 473.15* 140 / 200 = 331.2 K = 58 C

T2 < Tsat = 109 C. Therfore we have water in liquid and vapor phases.

Let mass of vapor = x kg

From steam table:

Specific volume of liquid (Lv) = 0.001 m3/kg

Specific volume of vapor (Vv) = 1.237 m3/kg

Lv*(M-x) + Vv*(x) = 0.5

x = 0.4 kg

Steam quality, X = x/M = 0.88

(c)

Mass of steam left after release of 50% = M/2 = 0.23 kg

Mass of vapor (Mv) = 0.8 * M/2 = 0.18 kg

Volume of vapor = Mv * Vv = 0.18 * 1.237 = 0.227 m3

(d)

Moles of vapor = M/2/MW = 0.23/18*1000 = 12.6 mol

R = 0.4615 * 18 = 8.307 J/K-mol

Volume of vapor = nRT/P

=12.6 * 8.307 * Tsat / 140000;   (Tsat = 109 C)

= 0.29 m3

(e)

% error = (0.29 - 0.227)*100/0.227 = 26 %

(f)

Volume in cubic ft, = 0.227 * 3.283

= 8 ft3

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