Thermodynamics

EXAMPLE 6–3Heat Rejection by a RefrigeratorChapter 6 | 291KitchenThe food compartment of a refrigerator, shown in Fig. 6–24, is maintained at4°C by removing heat from it at a rate of 360 kJ/min. If the required powerinput to the refrigerator is 2 kW, determine (a) the coefficient of performanceof the refrigerator and (b) the rate of heat rejection to the room thathouses the refrigerator.·Q H·W net,in = 2 kWRSolution The power consumption of a refrigerator is given. The COP andthe rate of heat rejection are to be determined.Assumptions Steady operating conditions exist.·QAnalysis (a) The coefficient of performance of the refrigerator isL = 360 kJ/minCOP R Q# LW # 360 kJ>minnet,in2 kW a 1 kW60 kJ>min b 3 Foodcompartment4°CThat is, 3 kJ of heat is removed from the refrigerated space for each kJ ofwork supplied.(b) The rate at which heat is rejected to the room that houses the refrigeratoris determined from the conservation of energy relation for cyclic devices,Q # H Q # L W # 60 kJ>minnet,in 360 kJ>min 12 kW2 a1 kW b 480 kJ /minDiscussion Notice that both the energy removed from the refrigerated spaceas heat and the energy supplied to the refrigerator as electrical work eventuallyshow up in the room air and become part of the internal energy of theair. This demonstrates that energy can change from one form to another, canmove from one place to another, but is never destroyed during a process.FIGURE 6–24Schematic for Example 6–3.EXAMPLE 6–4Heating a House by a Heat PumpA heat pump is used to meet the heating requirements of a house and maintainit at 20°C. On a day when the outdoor air temperature drops to 2°C,the house is estimated to lose heat at a rate of 80,000 kJ/h. If the heatpump under these conditions has a COP of 2.5, determine (a) the powerconsumed by the heat pump and (b) the rate at which heat is absorbed fromthe cold outdoor air.Solution The COP of a heat pump is given. The power consumption andthe rate of heat absorption are to be determined.Assumptions Steady operating conditions exist.Analysis (a) The power consumed by this heat pump, shown in Fig. 6–25,is determined from the definition of the coefficient of performance to beCOP = 2.5House20°CHP·Q HQ·L = ?80,000 kJ/hHeat loss·W net,in = ?W # net,in Q# H 80,000 kJ>h 32,000 kJ/h 1or 8.9 kW2COP HP 2.5(b) The house is losing heat at a rate of 80,000 kJ/h. If the house is to bemaintained at a constant temperature of 20°C, the heat pump must deliverOutdoor air at –2°CFIGURE 6–25Schematic for Example 6–4.