Thermodynamics

Chapter 8 | 453Thus,Integrating, we getdW net,in W net,in 21dQ HCOP HP a 1 T 0T b mc v dTa 1 T 0T b mc v dT mc v,avg 1T 2 T 1 2 T 0 mc v,avg ln T 2T 1 120.6 19.62 Btu 1.0 BtuThe first term on the right-hand side of the final expression above is recognizedas U and the second term as the exergy destroyed, whose valueswere determined earlier. By substituting those values, the total work input tothe heat pump is determined to be 1.0 Btu, proving our claim. Notice thatthe system is still supplied with 20.6 Btu of energy; all we did in the lattercase is replace the 19.6 Btu of valuable work by an equal amount of “useless”energy captured from the surroundings.Discussion It is also worth mentioning that the exergy of the system as aresult of 20.6 Btu of paddle-wheel work done on it has increased by 1.0 Btuonly, that is, by the amount of the reversible work. In other words, if thesystem were returned to its initial state, it would produce, at most, 1.0 Btuof work.EXAMPLE 8–13Dropping a Hot Iron Block into WaterA 5-kg block initially at 350°C is quenched in an insulated tank that contains100 kg of water at 30°C (Fig. 8–40). Assuming the water that vaporizesduring the process condenses back in the tank and the surroundings areat 20°C and 100 kPa, determine (a) the final equilibrium temperature,(b) the exergy of the combined system at the initial and the final states, and(c) the wasted work potential during this process.Solution A hot iron block is quenched in an insulated tank by water. Thefinal equilibrium temperature, the initial and final exergies, and the wastedwork potential are to be determined.Assumptions 1 Both water and the iron block are incompressible substances.2 Constant specific heats at room temperature can be used for both the waterand the iron. 3 The system is stationary and thus the kinetic and potentialenergy changes are zero, KE PE 0. 4 There are no electrical, shaft, orother forms of work involved. 5 The system is well-insulated and thus there isno heat transfer.Analysis We take the entire contents of the tank, water iron block, as thesystem. This is a closed system since no mass crosses the system boundaryduring the process. We note that the volume of a rigid tank is constant, andthus there is no boundary work.WATERT i = 30°C100 kgHeatIRONT i = 350°C5 kgFIGURE 8–40Schematic for Example 8–13.T 0 = 20°CP 0 = 100 kPa