Answer:
75 °F
Explanation:
Air has a specific heat at constant pressure of:
Cpa = 0.24 BTU/(lbm*F)
The specific heat of water is:
Cpw = 1 BTU/(lbm*F)
The first law of thermodynamics:
Q = L + ΔU
The heat exchanger is running at a steady state, so ΔU = 0. Also does not perform or consume any work L = 0.
Then:
Q = 0.
We split the heat into the heat transferred by the air and the heat trnasferred by the water:
Qa + Qw = 0
Qa = -Qw
The heat exchanged by the air is
Qa = Ga * Cpa * (tfin - ti)
And the heat exchanged by the water is:
Qw = Gw * Cpw * Δt
Replacing:
Ga * Cpa * (tfin - ti) = -Gw * Cpw * Δt
tfin - ti = (-Gw * Cpw * Δt) / (Ga * Cpa)
tfin = (-Gw * Cpw * Δt) / (Ga * Cpa) + ti
The G terms are mass flows, however we have volume flow of air.
With the gas state equation we calculate the mass:
p * V = m * R * T
m = (p * V) / (R * T)
55 °F = 515 °R
The gas constant for air is R = 53.35 (ft*lb)/(lbm* °R)
14.7 psi = 2117 lb/ft^2
m = (2117 * 5000) / (53.35 * 515) = 385 lbm
The mass flow is that much amount per minute
The mass flow of water is
11200 lbm/h = 186.7 lbm/min
Then:
tfin = (-186.7 * 1 * (-10)) / (385 * 0.24) + 55 = 75 °F
