Air enters a heat exchanger at a rate of 5000 cubic feet per minute at a temperature of 55 °f and pressure of 14.7 psia. the air is heated by hot water flowing in the same exchanger at a rate of 11,200 pounds per hour with a decrease in temperature of 10 °f. at what temperature does the air leave the heat exchanger?

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

find the sum of the two horizontal vectors and the sum of the two vertical vectors

isoceles right angled triangle