108000 kghr of water at 20 C is heated to 50 C by 1800kghr o

108,000 kg/hr of water at 20 C is heated to 50 C by 1800kg/hr of hot air at 200 C for district heating purposes. Both the fluids arc at atmospheric pressure. Calculate the thermal load in this energy transfer (5) If the fluids are employed in a counter flow arrangement in a suitable heat exchanger, calculate the LMTD (10). Provide an estimate for heat transfer coefficient for the water side, air side, and the overall heat transfer coefficient (cite sources). (5)

Solution

A) Heat load = Heat gained by water = Heat lost by air

Heat load = m x s x T

= 108000 x 4.2 x (50 -20) KJ / Hr

= 13608 x 10³ KJ / Hr

13608 x 10³ = M_airx C_px T_air

Or, (1800 x 60) x 1.026 x (200 - T_exit) = 13608 x 10³           [C_p at 200 C = 1.026 KJ/Kg-K]

Or, T_exit = 200 - (13608 x 10³)/(1800 x 60 x 1.026)

                = 77.19 C

B) LMTD = (dt_o - dt_i) / ln(dt_o / dt_i)

For Counter-flow heat exchangers

dt_i = T_water-in – T_air-out = 20 – 77.19 = - 57.19 ^oC

dt_o= T_water-out – T_air-in = 50 – 200 = - 150 ^oC

LMTD = [-(150-57.19) / ln (150/57.19)] = - 96.25 ^oC

C) To calculate the heat transfer co-efficients, heat transfer areas will have to be provided.