Thermodynamics

452 | Thermodynamics(b) The reversible work, which represents the minimum work input W rev,in inthis case, can be determined from the exergy balance by setting the exergydestruction equal to zero,X in X out X destroyed→ 0 (reversible) ¢X systemNet exergy transfer Exergy Changeby heat, work, and mass destruction in exergyW rev,in X 2 X 1since KE PE 0 and V 2 V 1 . Noting that T 0 (S 2 S 1 ) T 0 S system 19.6 Btu, the reversible work becomes 1.0 Btu⎫ ⎪⎬⎪⎭⎫⎪⎬⎪⎭⎫⎪⎪⎪⎬⎪⎪⎪⎭⎫⎪⎬⎪⎭⎫⎪⎬⎪⎭ 1E 2 E 1 2 P 0 1V 2 V 1 2 Q0 T 0 1S 2 S 1 2 1U 2 U 1 2 T 0 1S 2 S 1 2W rev,in mc v 1T 2 T 1 2 T 0 1S 2 S 1 2 12 lbm210.172 Btu>lbm # °F21130 702°F 19.6 Btu 120.6 19.62 BtuTherefore, a work input of just 1.0 Btu would be sufficient to accomplishthis process (raise the temperature of air in the tank from 70 to 130°F) if allthe irreversibilities were eliminated.Discussion The solution is complete at this point. However, to gain somephysical insight, we will set the stage for a discussion. First, let us determinethe actual work (the paddle-wheel work W pw ) done during this process.Applying the energy balance on the system,E in E out ¢E systemReversibleheat pumpAIR70°F → 130°FAmbient air70°F20.6 Btu19.6 BtuW net,in = 1 BtuFIGURE 8–39The same effect on the system can beaccomplished by a reversible heatpump that consumes only 1 Btu ofwork.Net energy transferby heat, work, and massChange in internal, kinetic,potential, etc., energiesW pw,in ¢U 20.6 Btu 3from part 1b24since the system is adiabatic (Q 0) and involves no moving boundaries(W b 0).To put the information into perspective, 20.6 Btu of work is consumedduring the process, 19.6 Btu of exergy is destroyed, and the reversible workinput for the process is 1.0 Btu. What does all this mean? It simply meansthat we could have created the same effect on the closed system (raising itstemperature to 130°F at constant volume) by consuming 1.0 Btu of workonly instead of 20.6 Btu, and thus saving 19.6 Btu of work from going towaste. This would have been accomplished by a reversible heat pump.To prove what we have just said, consider a Carnot heat pump that absorbsheat from the surroundings at T 0 530 R and transfers it to the air in therigid tank until the air temperature T rises from 530 to 590 R, as shown inFig. 8–39. The system involves no direct work interactions in this case, andthe heat supplied to the system can be expressed in differential form asdQ H dU mc v dTThe coefficient of performance of a reversible heat pump is given byCOP HP dQ H 1dW net,in 1 T 0 >T