Consider a vessel filled with saturated water vapor at 143 °C. a) Calculate the
ID: 477378 • Letter: C
Question
Consider a vessel filled with saturated water vapor at 143 °C. a) Calculate the internal pressure in the vessel. [3.94 bar] b) Assume the vessel is a horizontal cylinder with radius 2m and length 10m. What is the mass (kg) of water vapor? [267 kg] c) In rained and the temperature in the vessel dropped to 10°C. As a result, some of the vapor condensed. What is the quality of the final state? [4.42 x 10-3] d) Calculate the amount of energy removed from the system by rainwater in problem c. [1.47 x 108 J] e) What is the pressure of the vessel in the final state? [0.0123 bar] f) Calculate the net force (N) exerted on the vessel in the final state.
Explanation / Answer
1.From steam tables, 143 deg.c corresponds to a saturation pressure of 398 Kpa = 3.98 bar.
2. Specific volume of water vapor under the conditions= 468 cm3/gm=468*0.001m3/Kg=0.468m3/kg
Volume of cylinder = PI*r2*h, r= is radius and h= length, Volume= (22/7)*(2)2*10 = 125.7 m3
Hence mass of water vapor = 125.7/0.468=268 kg
3. Volume of vessel = 125.7 m3
let x= volume fraction of vapor at 10 deg.c , then 1-x= volume fraction of liquid at 10 deg.c
Specific volume of liquid = 1cm3/g =0.001m3/kg and that of vapor = 106400cm3/gm=106.4 m3/kg
0.468 =106.4*x+(1-x)*0.001, 106.4x+0.001-0.001x= 0.468
Hence x*(106.4-0.001)= 0.468-0.001, x = 0.004389
4. Enthalpy of mixture at 143 deg.c = 2736 Kj/Kg. Total enthalpy = 268*2736 KJ=733248
Enthalpy of mixture at 10 deg.c = 0.004389*2519.9+(1-0.004389)*41.99=52.87 Kj/Kg
Total enthalpy =52.87*268= 14169 Kj
Heat transferred during rain = initial enthalpy- enthalpy at 10 deg.c = 733248-14169=719079 Kj=7.2*108 Joules
5. Pressure at the final state = 1.227 Kpa= 1.227/101.3 bar=0.012 bar