Question

Exercise 10 Consider a 2-m-high electric hot water heater that has a diameter of 40 cm and maintains the hot water at 55°C. The tank is located in a small room whose average temperature is 27°C, and the heat transfer coefficients on the inner and outer surfaces of the heater are 50 and 12 W/m2.°C, respectively. The tank is placed in another 46-cm-diameter sheet metal tank of negligible thickness, and the space between the two tanks is filled with foam insulation (k=0.04 W/m - °C). The thermal resistances of the water tank and the outer thin sheet metal shell are very small and can be neglected. The price of electricity is 0.305 dhs/kWh, and the home owner pays 1070 dhs a year for water heating. a) Draw the network of the thermal resistances. b) Evaluate each thermal resistance. 3 cm 40 cm c) Calculate the rate of the heat loss through the wall of the tank. d) Determine the fraction of the hot water energy cost of 27°C this household that is due to the heat loss from the Tv = 55°C tank. Foam insulation 2 m Water heater

Answer 1

7,- 40/2 = 200m -0.2m Y - 46/2 - 23cm 0.23 m L2m 200m admosphere Network diagram. R. Tw: 55°c Teate R R₂ Resistance due to not watertat inner surface) Resistance Ri= due to cylindrical shaped = Resistance al foam insulation. due to outside croom) atmosphere R3- Resistance P - Home Re- en Bu Hale r.) 2 TTLK where ho= Heat transfer coefficient of water (at inner surface) w=50 W/m²oC ha= Heat transfer coefficient at oudersurfeee. ha = 12 W moc A; - Surface area of inner surface - 2TL de surface area of outer surface - Tr, L conductivity of foam = 0.04 W/moc => P,-0.0079521 ºc/w Rz = 0.02893245°Clw R = 0.2780 °C/N

Total Resistance R=R,+ R₂ + R2 R=0.3148 °C/W 6 = Rate of heat loss through the wall of trank - ģR TwoTa 55-27 11 88.34 W = 0.0889 kW 0.3148 price of electricity = 0.305 dhs/khlh Cost due to heat loss from tank per year = 0.305 dhs/kWh 0.0889 kW. x 365 days * 24h = 237.5? des cost due to heat loss from tank out of Fraction of total cost. cost due to heat loss cost due to total) water treating 237.52 = 0.22 1070 days round Assumption: neater is operated continuously 365 the clock (24 hrs)

Thus we can calculate individual resistances and heat transfer rate and fraction of cost due to heatloss.