4An isolated rigid tank is compartmented into 2 sections \'A\" and \"B\" using a

Question

4An isolated rigid tank is compartmented into 2 sections 'A" and "B" using a thin membrane as shown in the figure. Compartment 'A" with a volume of 4.49 m2 contains water at a pressure of 100 kPa and temperature of 200*C; while compartment "B with a volume of 0.51 m2 contains saturated water with a quality of 84% at 200 kPa. The membrane is gently removed (without disturbing the system) and the water s allowed to establish a new equilibrium state in the entire tank. Determine the state of water in the final equilibrium state. What is the final eauilibrium pressure?(14 points) nem brane

Explanation / Answer

For compartment A:

P = 100 kPa

T = 200 C

V = 4.49 m3

At P = 100 kPa, Tsat = 100 C

As T > 100 C , the steam is superheated.

From the superheated steam table at T = 200 C and P = 100 kPa,

specific enthalpy HA = 2875.48 kJ/kg

specific volume v = 2.17249 m3/kg

mass of water mA = Volume / specific volume = 4.49 / 2.17249 = 2.06675 kg

For compartment B:

x = 0.84

P = 200 kPa

V = 0.51 m3

Here the steam is saturated as it has quality of 0.84.

From the saturated steam table at P = 200 kPa,

T = 120.212 C

specific enthalpy of saturated water Hl = 504.684 kJ/kg

specific enthalpy of saturated steam Hv = 2706.24 kJ/kg

specific volume of saturated water vl = 0.00106052  m3

specific volume of saturated steam vv = 0.885735  m3

specific volume v = xVv + (1-x)Vl = 0.84*0.885735 + (1-0.885735)*0.00106052 = 0.744 m3 /kg

specific enthalpy HB = xHv + (1-x)Hl = 0.84*2706.24 + (1-0.84)*504.684 = 2354 kJ/kg

mass of water mB = 0.51 / 0.744 = 0.68548 kg

Now,

final mass mf = mass of A + mass of B = 2.06675 + 0.68548 = 2.75223 kg

By enthalpy balance,

mfHf = mAHA + mBHB

2.75223*Hf = 2.06675*2875.48 + 0.68548*2354

.: Hf = 2745.6 kJ/kg

Final volume Vf = VA + VB = 4.49 + 0.51 = 5 m3

final specific volume = Vf / mf = 5 / 2.75223 = 1.8167 m3/kg

Now, the final pressure has to be between 100 and 200 kPa.

For saturated steam at 200 kPa, Hv = 2706.24 kJ/kg

As the final enthalpy > Hv at 200 kPa, the steam is superheated at the equilibrium state.

Now, look for specific enthalpy of 2745.6 kJ/kg and specific volume of 1.8167 m3/kg in the superheated steam table,

T = 135 C

P = 102 kPa

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