Thermodynamics

448 | ThermodynamicsBrick27°Cwall0°C·Qthis process since the state and thus the exergy of the wall do not changeanywhere in the wall. 3 Heat transfer through the wall is one-dimensional.Analysis We first take the wall as the system (Fig. 8–36). This is a closedsystem since no mass crosses the system boundary during the process. Wenote that heat and exergy are entering from one side of the wall and leavingfrom the other side.Applying the rate form of the exergy balance to the wall gives¡0 (steady)X # in X # out X # destroyed dX system >dt 0⎫ ⎪⎬⎪⎭⎫⎪⎬⎪⎭⎫⎪⎪⎪⎪⎬⎪⎪⎪⎪⎭20°C 5°C30 cmFIGURE 8–36Schematic for Example 8–10.Rate of net exergy transfer Rate of exergy Rate of changeby heat, work, and mass destruction in exergyQ # a 1 T 0T b Q # a 1 T 0inT b X # destroyed 0out11035 W2 a1 273 K273 Kb 11035 W2 a1 293 K 278 K b X# destroyed 0Solving, the rate of exergy destruction in the wall is determined to beX # destroyed 52.0 WNote that exergy transfer with heat at any location is (1 T 0 /T)Q at thatlocation, and the direction of exergy transfer is the same as the direction ofheat transfer.To determine the rate of total exergy destruction during this heat transferprocess, we extend the system to include the regions on both sides ofthe wall that experience a temperature change. Then one side of the systemboundary becomes room temperature while the other side, the temperatureof the outdoors. The exergy balance for this extended system(system + immediate surroundings) is the same as that given above,except the two boundary temperatures are 300 and 273 K instead of 293and 278 K, respectively. Then the rate of total exergy destruction becomesX # destroyed,total 11035 W2 a1 273 K273 Kb 11035 W2a1 300 K 273 K b 93.2 WThe difference between the two exergy destructions is 41.2 W and representsthe exergy destroyed in the air layers on both sides of the wall. Theexergy destruction in this case is entirely due to irreversible heat transferthrough a finite temperature difference.Discussion This problem was solved in Chap. 7 for entropy generation. Wecould have determined the exergy destroyed by simply multiplying theentropy generations by the environment temperature of T 0 273 K.EXAMPLE 8–11Exergy Destruction during Expansion of SteamA piston–cylinder device contains 0.05 kg of steam at 1 MPa and 300°C.Steam now expands to a final state of 200 kPa and 150°C, doing work. Heatlosses from the system to the surroundings are estimated to be 2 kJ during thisprocess. Assuming the surroundings to be at T 0 25°C and P 0 100 kPa,