3.17 Each of the following exercises requires evaluating the quality of a two-ph

Question

3.17 Each of the following exercises requires evaluating the quality of a two-phase liquid-vapor mixture:

The quality of a two-phase liquid-vapor mixture of H2O at 40oC with a specific volume of 10m3/kg is (i) 0. (ii) 0.486 (iii) 0.512 (iv) 1

The quality of a two-phase liquid-vapor mixture of propane at 20bar with a specific internal energy of 300kJ./kg is (i) 0.166. (ii) 0.214 (iii) 0.575 (iv) 0.627

The quality of a two-phase liquid-vapor mixture of Refrigerant 134a at 90 lbf/in2 with a specific enthalpy of 90Btu/lb is (i) 0.387, (ii) 0.718, (iii) 0.806 (iv) 0.854

Explanation / Answer

From table A-16The data on properties is obtained from Engineering by Moran and Shapiro

1. Specific volume given= 10m3/kg

let the volume be 1m3

mass = (1/10) m3/kg* 1= 0.1

let x be the liquid fraction, then the vapor 0.1-x

From steam tables ( Table A2), the specific volume of liquid at 40 de.gc = 1.0078X10-3m3/kg =1000 kg/m3 and that of vapor = 19.523 m3/kg =1/19.523 kg/m3=0.05122 m3/kg

let x be the vapor fraction and 1-x becomes liquid fraction

0.1=1000*x+ *(1-x)* 0.05122

0.1 =1000x+ 0.05122-0.5122x

0.1-0.05122 =1000x-0.5122x

x =0.488 1-x=0.512 ,III is the correct answer

2.  Internal energy = liquid fraction * internal energy of liquid + Vapor fraction * its internal energy ( at 20bar)

20 bar corresponds to a saturation temperature of 56 deg.c

, Table-A16 gives Specific internal energy of liquid = 247.4 Kj/Kg and that of vapor =477.8 Kj/Kg

300 = 247.7*x+ 477.8*(1-x)

(477.8-244.7)x= 477.8-300, x =0.763 1-x = 1-0.763= 0.237 close to this answer is 0.214 ( hence 2 is correct answer)

3. From table A-11

90 lbf/in2 corresponds to 6.2 bar, liquid enthalpy =79.48 Kj/Kg and that of vapor =259.19 Kj/Kg

mixture enthalpy =90 but/lb =209.34 Kj/Kg

209.34 = x*79.48+(1-x) *259.19

x=0.277

1-x =1-0.277= 0.723 ( II is very close and hence correct answer is ii)

4. From table A-14 , -20 deg.F , specific volume of liquid = 1.5542 m3/kg and that of vapor 0.3289 m3/kg

specific volume 11 ft3/lb= 11*0.0624m3/kg= 0.6862 m3/kg

x* 1.5542+(1-x)*0.0.3289 = 0.6862,

x=0.555 ( Close answer is 0.5337)

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