Hot water is to be used in a room heater, entering the heater at 80°C and leaving at 50°C, to maintain the air in the room at 20°C. To achieve this, the required heat transfer rate is 1.0 kW. The predicted overall heat transfer coefficient based on the outside area of the heater surface is 20 W m-2 K-1. Calculate the log mean temperature difference for heat transfer. Key in your answer with two significant figures.
The method I used to answer the question is as follows: Modelling the process as a counter current heat exchanger: water enters at 80 deg C and leaves at 50 deg C, air enters at 20 deg C and leaves at 20 deg C (no change in air temperature throughout due to heat loss to surroundings). dT1 = 80-20 = 60 deg C dT2 = 50-20 = 30 deg C dT(logmean) = (dT1-dT2)/ln(dT1/dT2) = 43.28 deg C = 43 deg C (2 s.f.) Alternatively: model the process as heat transfer from one bulk fluid to another, i.e. water to air. The bulk mean temperature of the water is Tbm = (80+50)/2 = 65 deg C Then Q = U*A*dT(logmean) = U*A*(Tbm - Tair) therefore, dT(logmean) = Tbm - Tair = 65-20 = 45 deg C
