Question

A 5 m³ tank contains superheated steam at 300°C and 22500 mmHg. The tank (and the steam inside) is cooled until the pressure inside is reduced to 2 bar. At this point, liquid water and saturated steam coexist together in the tank. How much heat was removed from the tank, what is the final temperature, and how many kilograms of steam condensed in the process?

Answer 1

Volume of super heated steam V = 5 cu.m

pressure = 22500 mmHg = 30 bar = 3000kPa

T = 300 C = 573 K

PV = nRT

number of moles in steam n = PV/RT = 3000e+3*5/(8.3*573) = 3154 moles

mass of steam = 3154 *0.018 = 56.77 kg

suppose a fraction r of this is converted to water by cooling

volume of water V_w = 3154r*18.0 e-6 cu.m ( 1 mole of water = 18g = 18 cc)

= 0.0568r

no of steam moles after cooling = 3154(1- r)

Pressure = 2bar

boiling point of water at 2 bar = 120 C = 393 K

after cooling steam temp T = 393 K

pressure P = 2 bar = 200 kPa

V_s = nRT/P = 3154(1- r) *8.3*393/200e+3 = 52.09(1-r) cu.m

V_w + V_s = 0.0568r + 52.09 (1- r) = 5

r = 0.905

90.5 % of the steam is converted to water

mass of steam converted to water = 3154*0.905*0.018 kg = 51.38 kg

specific heat of steam = 1920 J/kg/K

heat lost by steam from 300 to 120 C = 1920*56.77 *180 = 19.62 e+6 J

latent heat of steam = 2.25 J/kg

latent heat of condensation = 51.38 *2.25e+6 = 115.6 e+6 J

Total heat lost = 115.6e+6 + 19.62e+6 = 135.22 e+6 J