An air-cooled low-pressure-steam condenser is shown below. The tube bank is four rows deep in the direction

Question:

An air-cooled low-pressure-steam condenser is shown below.

The tube bank is four rows deep in the direction of air flow. There are 80 tubes total. The tubes have ID = 2.2 cm and OD 2.5 cm and are 9 m long. The tubes have circular fins on the outside. The tube-plus-fin area is 16 times the bare tube area (i.e., the fin area is 15 times the bare tube area neglect the tube surface covered by fins). The fin efficiency is 0.75. Air flows past the outside of the tubes. On a particular day, the air enters at 22.8?C and leaves at 45.6?C. The air flow rate is 3.4 x 105 kg/h. The steam temperature is 55?C and has a condensing coefficient of 104 W/(m2 K). The steam-side fouling coefficient is 104 W/(m2K). The tube wall conductance per unit area is 105 W/(m2K). The air-side fouling resistance is negligible. The air-side-film heat transfer coefficient is 285 W/(m2K). (Note this value has been corrected for the number of transverse tube rows.)(a)What is the log-mean temperature difference between the two streams?(b)What is the rate of heat transfer?(c)What is the rate of steam condensation?(d) Estimate the rate of steam condensation if there were no fins.GIVENThe condenser shown aboveNumber of tubes (N) = 80Number of rows (Nr) = 4 (in air flow direction)Tube diametersDi = 2.2 cm = 0.022 mDo = 2.5 cm = 0.025 mTube length (L) = 9 mAir temperatureTa,in = 22.8?CTa,out = 45.6?CAir flow rate (ma) = 3.4 x 105 kg/h = 94.4 kg/sSteam temperature = 55?C (constant)Fin area = 15 (tube area)Fin efficiency (hf) = 0.75Steam sideTransfer coefficient (hi ) = 104 W/(m2 K)Fouling coefficient (1/Ri) = 104 W/(m2 K)Tube wall conductance per unit area (1/Rk) = 105 W/(m2 K)Air side: Transfer coefficient (ho) = 285 W/(m2 K)Fouling resistance on the air side is negligibleASSUMPTIONSAir side transfer coefficient is the same with or without finsTube surface covered by the fins isnegligible