A chemical engineer has devised a thermally operatedelevator. The elevator compa

Question

A chemical engineer has devised a thermally operatedelevator. The elevator compartment is made to rise byelectrically heating the air contained in the piston and cylinderdrive mechanism, and the elevator is lowered by opening a valve atthe side of the cylinder, allowing the air in the cylinder toescape slowly. Once the elevator compartment is back to thelower level, a small compressor forces out the air remaining in thecylinder and replaces it with air at 20 degrees C, and a pressurejust sufficient to support the elevator compartment. Thecycle can then be repeated. There is no heat transfer betweenthe piston, cylinder, and the gas; the weight of the piston,elevator, and elevator contents is 4000 kg; the piston has an areaof 2.5 m2, and the volume contained in the cylinder whenthe elevator is at its lowest level is 25 m3. There is no friction between the piston and the cylinder, and theair in the cylinder is assumed to be an ideal gas withCp = 30 J/mol-K. a. What is the pressure in the cylinder throughout theprocess? b. How much heat must be added to the air during the processof raising the elevator 3 m, and what is the final temperature ofthe gas? c. What fraction of the heat added is used in doing work, andwhat fraction is used in raising the temperature of the gas? d. how many moles of air must be allowed to escape in orderfor the elevator to return to its lowest level? e. What do you think of this engineer's ideas? Justifyyour opinion. If someone could please solve and explain this problem, itwould be really excellent!!! A chemical engineer has devised a thermally operatedelevator. The elevator compartment is made to rise byelectrically heating the air contained in the piston and cylinderdrive mechanism, and the elevator is lowered by opening a valve atthe side of the cylinder, allowing the air in the cylinder toescape slowly. Once the elevator compartment is back to thelower level, a small compressor forces out the air remaining in thecylinder and replaces it with air at 20 degrees C, and a pressurejust sufficient to support the elevator compartment. Thecycle can then be repeated. There is no heat transfer betweenthe piston, cylinder, and the gas; the weight of the piston,elevator, and elevator contents is 4000 kg; the piston has an areaof 2.5 m2, and the volume contained in the cylinder whenthe elevator is at its lowest level is 25 m3. There is no friction between the piston and the cylinder, and theair in the cylinder is assumed to be an ideal gas withCp = 30 J/mol-K. a. What is the pressure in the cylinder throughout theprocess? b. How much heat must be added to the air during the processof raising the elevator 3 m, and what is the final temperature ofthe gas? c. What fraction of the heat added is used in doing work, andwhat fraction is used in raising the temperature of the gas? d. how many moles of air must be allowed to escape in orderfor the elevator to return to its lowest level? e. What do you think of this engineer's ideas? Justifyyour opinion. If someone could please solve and explain this problem, itwould be really excellent!!!

Explanation / Answer

Given initial temperature,Ti = 200C area of piston, A = 2.5 m2 volume contained in the cylinder V = 25 m3 Cp = 30 J/mol-K mass of the piston, elevator, and elevator contents, m = 4000kg a)weight of the piston, elevator, and elevator contents, w = 4000kg*9.81m/s2 = 39240 N {g=9.81m/s2} Pressure inside the cylinder will always be equal to just hold theweight. so, Pi = W/A = 39240/2.5 = 15696 N/m2 assuming process as reversible process, pressure will always remainthesame              {to hold the weight and No jerk in the motion of the elevator} so, Pressure inside the cylinder throughout the process =15696 N/m2 .................................ans(a) b)number of moles of air inside, n = PV/RT =(15696*25)/(8.314*293.15) = 161 work done by gas to raise the elevator by 3 m = mgh = 4000kg*9.81m/s2*3m = 117720 J Now applying 1st law of thermodynamics, Q = U+W or, nCpdT = nCvdT +W or, n(Cp-Cv)dT = W but, we know that, Cp-Cv =R = 8.314 J/mol-K so, 161*8.314*dT = 117720 or, dT = Tf - Ti = 87.945 or, Tf = 20+87.945 = 107.945 0C and heat added to the system, Q = nCpdT = 161*30*87.945          = 424774.35 J = 424.77kJ  ..........................................................ans(b) c) fraction of heat used in doing work = W/Q =117720/424774.35                                                         = 0.2771 = 27.71% so fraction used in raising the temperature = 1-0.2771 = 0.7229=72.2 % ..............................ans(c) d)Let n' is the moles allowed to escape assuming outside temperature = Ti = 200C, applying conservation of energy (n-n')CvTf' + n'CvTi = nCvTf + mgh, whereTf' is the temperature of the system afterreturn to initial state Now, n' = PV/RTf' or,Tf' = PV/n'R, subtituting in the aboveequation Cv[(n-n')*PV/n'R + n'Ti -nTf] = mgh Now, Cv = Cp-R = 30-8.314 = 21.686 J/mol-K so, [(161-n')*15696*25/(n'*8.314) + n'*293.15 - 161*381.095] =117720/21.686 or, [7598797.21/n' - 47197.49 + 293.15n' - 61356.29] = 5428.38 or, 7598797.21/n' + 293.15n' = 5428.38+47197.49+ 61356.29=113982.16 or, 25921.19/n' + n' = 388.81 or, n'2-388.81n' +25921.19 =0 solving the quadratic equation gives us n' = 85.445 and 303.364 n' should be less than o equal to n (n=161) so, no of moles allowed to escape, n' = 85.44 .............................................................................ans(d) e) It should not be considered as the good idea as only 27.17% ofthe heat supplied is used in actual work. also there is work ofcompressor involved we does not come in the calculationsabove. ........................................ans(e)

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