Assignment I answers

Assignment I answers - Tutorial I Chemistry for Engineers...

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Tutorial I Chemistry for Engineers Problem 1. A rigid tank contains a hot fluid that is cooled while being stirred by a paddle wheel. Initially the internal energy of the fluid is 800 KJ. During the cooling process, the fluid losses 500 KJ of heat, and the paddle wheel does 100 KJ of work on the fluid. Determine the final internal energy of the fluid. Neglect the energy stored in the paddle wheel. Solution: We take the contents of the tank as the system. This is a closed system since no mass crosses the boundary during the process. We observe that the volume of a rigid tank is constant, and the thus there is no boundary work and V 2 = V 1 . Also, heat is lost from the system and shaft work is done on the system. Assumptions: The tank is stationary and thus the kinetic and potential energy changes are zero. ∆KE = ∆PE = 0. Therefore, ∆E = ∆U and internal energy is the only form of the system’s energy that may change during this process. Applying the energy balance on the system gives E in – E out = ∆E system Wpw,in – Q out = ∆U = U 2 – U 1 100KJ – 500KJ = U 2 – 800 KJ U 2 = 400KJ The final internal energy of the system is 400KJ.
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Problem 2. A heat engine works on the Carnot cycle between temperature 900 ° C & 200 ° C. If the engine receives heat at the higher temperature at the rate of 60 kW, calculate the power of the engine. Solution T H = 900 + 273 = 1173 k T L = 200 + 273 = 473 k 597 . 0 1173 473 1173 = - = - = H L H th T T T η Also, H th H H th Q W Q W Q W η η = = = . . . . . W = 0.597 x 60 = 35.82 kW
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Problem 3. An engineer claims to have developed an engine which develops 3.4 kW while consuming 0.44 Kg of fuel of calorific value of calorific value of 41870 kJ / kg in one hour. The maximum and minimum temperatures recorded in the cycle are 1400 ° C & 350 ° C respectively is the claim of the engineer genuine. Solution: Temperature of source, T H = 1400 ° C = 1673 K Temperature of sink, T L = 350 ° C = 673 K We know that the thermal efficiency of the CARNOT cycle is the maximum between the specified temperature limits and is given as.
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