Air enters a car radiator at 50 km/hr, 25 C. The radiator is 1 m2 facial area. 2 kg/s of water enters the radiator from the engine at 10 bar, 200 C and leaves at 10 bar, 180 C. What is the temperature of the air as it leaves the radiator?
I did an energy balance for air and then for water, but I am not sure which terms can be ignored. After i obtain these eqns, i can set them equal to each other and solve for T (from
Q =m(c_p)(Change in T))?
The Attempt at a Solution
For Air: W = Q + (m)(h_in + [(v_in)^2]/2 +g*z_in - h_out - [(v_out)^2]/2 -g*z_out)
m = mass flow rate
Q = heat flow rate
W = work flow rate
h = enthalpy
v = velocity
I assume W, z_in, z_out are zero and also that m_in = m_out
So i get Q = m(h_in - h_out + [(v_in)^2]/2)
Q = m(h_out - h_in) = (0.2)(762.8 - 2827.9) = -2065.1 kW