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Ra eb (tear Supla w e 200 (| tg 4000 .07 Q tos Cm dl - Euro to Seus Tempo techive 885 k Cia A E) =29 Er TheimA «idem Ze ef. nox Volue , - dd er — ISO 2-7 x (essas Q=9:7 = v ss - 0-540 weste dae ty ss cr tecto Ay police) n Eooxted da = oo” = Q3 548 NK) Oy = az x0% tou = &- Be Seg =-w + S% =-8«f 4 2305431] OM -3 ar x10u ra Q.09 Um as bom as | = | o friuenta Carmel emu Tu d o | 7 = T, ' 5 . | . K as : o at ich has , . . tor b . , & = 4 po ) alra 300 285 = |- ER d. 15 . 300 o > É Q.10 Qok Now caloulate the entropy change of reservoir. From the definition ofentropy, AS = e. AB) The negative sign is because the heat is flowing out of reservoir. Here. 7 is the temperature ofihe reservoir, Q is the energy obtained from the reservoir and is calculated frem the relation given below: O=-me AT Here se is the mass ofthe water. Substitute the corresponding values in above equation and getthe value of QD as, P=-me AT E) =- ER ara *(373,15K-273.15K) = ARA] The heat of the resenvoíris 4]84k) This is the energy requires to heat 1 kg of water so the units of () will be kuikg. Substitute the values of () in equation (2). [a] o 373.15K HH =-112] Now caleulate tha AS AS = AS ao + AS, Therefore, AS 5 |O.184 sam ke-K Q.15 Stepiois & (a) Let the two streams be stream-A and stream-B. Inlet temperature of stream-A (Tu) is 70ºC. Outlet temperature of stream-A (T,o) is 190ºC. Inlet temperature of stream-B (Tu) is 320º€. Assuming the temperature drop of siream-B is same as the temperature drop of siream-A. Calculate the ouílet temperature of stream-B as follows: Ta —Tu = Ta — Ta 190ºC—70ºC = 320º€ — Ta, Te = 320ºC-120ºC Bt = 200º€ Hence the outlet temperature of stream-B (7) 5 200ºC. Stepáors 4 (D) Total entropy change is defined as the sum of change in entropy of each stream. Write the expression for the total entropy change as follows: AS=A8, + AS, Here, ASis the total entropy change. substitute 8.73 J/molK for AS, and -6.57J/molK for AS, in the above equation to calculate the total entropy change. AS=AS, + AS, = 8.73 J/molK — 6.57 ]/molK = 2.16 J/molk Therefore, the required entropy change is 2.16 J/molK|. Q.17 Given: Tr = 4º€ [e] = 360 kI/min x d min é em 60s 1kJ's 0,=6kW Fa =2KW The refrigeration system is as follows: Surroundings | ANS Refrigerated space The coefficient of performance: The heat rejected is: D,=0+F =6+2 =8kWx bis dJ kJ/s Imin 1kW O, = 480 kJ/min b) The rate of heat transfer to the kitchen is calculated from UA nar E E -Q, O, = Õ, + ra O, =(60+50) kJ/min Ô, =110 kJ/min -. The heat transfer to the kitchen is 10, =110 kJmin Q.20 SEPpIO IT + The cooling effect and the COP of a refrigerator are given. The power input to the refrigerator is to be determined. Assumptions The refrigerator operates steadily. Analysis Rearranging the definition of the refrigerator coefficient of periormance and applying the result to this reirigerator gives [8] R Fa " - LO WM/ 1h y COP; FER EE, Step4or4 4 The power consumed by the refrigerator is expressed as W, me * Oy -0, = 22,000 - 15,000 = 7000 kJ/h Therefore the COP is cor, - Le HW, me! 15000 7000 =2.14 «. The COP of the refrigerator, |COP, = 2.14 Q.24 Stepioi3 4 (a) The power required to refrigeratoris 1,5 kW The rate of refrigeration is 4 kW The expression for coefficient of periormance is as follows: cor-€ W — 4kW “15kW = 2.667 Thus, the coefficient of performance of the reirigeration is 2.667 (b) The amount of heat rejected from condenser is equal to sum of the heat taken and the the work down by the refrigerator. 0, =0.+W =4kW+1,5 kW =5,5 kW 
