Question

Air flowing at 1000 ft3 /min, 66°F, and 1 atm enters a heat exchange unit and exits at 76°F and 1 atm. Saturated steam at 12 psia is used to heat the air and leaves the unit operation as saturated water at 12 psia. The system is not adiabatic and heat is lost to the surroundings at -3 Btu/lbm steam condensed. Calculate the mass flow rate of the steam.

Answer 1

Volumetric flow rate = 1000 ft³/min = 1000 * 0.0283168 m³/min = 28.3168 m³/min
Mass flow rate of air = density * volumetreic flow rate
you need to know the average temperature first so,
inlet temperature, T_i= 66^oF = 18.89^oC
outlet temperature, T_f = 76^oF = 24.44^oC
Average temperature = (18.89+24.44) / 2 = 21.66^oC
density of air at 21.66^oC = (29/22.4) [ 273.15 / (21.66 + 273.15) ] =1.1995 Kg/m³
Mass flow rate of air = 1.1995 Kg/m³ * 28.3168 m³/min = 33.9666 Kg/min
Heat duty required for air, q= mC_p ΔT
q = 33.9666*10³ gm/min * 1J/gm^oC * (24.44-18.89)^oC = 188514.63 J/min
let the flow rate of the steam be m₁ gm/min
Heat lost to the surroundings = 3 btu/lb_m = 3*2.326 J/gm = 6.978 J/gm
Heat losses for the steam = 6.978m₁ J/min
heat duty required by the steam =(188514.63 + 6.978m₁) J/min
Look for enthalpy of the saturated steam (h_v) and liquid (h_l) at 12psia or 82.7371 Kpa
pressure in the steam tables,
h_v = 2667.832 J/gm ; h_l = 392.6538 J/gm
change in enthalpy, ΔH = h_v - h_l = 2667.832 - 392.6538 = 2275.1782 J/gm
now heat balance equation will be,
ΔH * m₁ = (188514.63 + 6.978m₁)
2275.1782*m₁ = (188514.63 + 6.978m₁)
2268.2002 m₁ = 188514.63
m₁ = 83.1119 gm/min = 83.1119*10^-3 Kg/min = mass flow rate of the steam