A clever chemical engineer has devised the thermally operated elevator shown at

Question

A clever chemical engineer has devised the thermally operated elevator shown at the bottom of this page. The elevator compartment is made to rise by electrically heating the air contained in the piston and cylinder drive mechanism, and the elevator is lowered by opening a valve at the side of the cylinder, allowing the air to escape slowly once the elevator compartment returns to the lower level, a small compressor removes the remaining air in the cylinder and replaces it with air at 20 degree celcius and a pressure sufficient to support the elevator compartment. The cycle can then be repeated. There is no heat transfer between the piston, cylinder, and the gas; the weight of the piston, elevator, and elevator contents is 4000 kg; the piston has an area of 2.5 m2; and the volume in the cylinder when the elevator is at its lowest level is 25 m3. There is no friction between the piston and the cylinder, and the air in the cylinder is assumed to be an ideal gas with Cp = 30 J/mol K.

a.) What is the pressure in the cylinder throughout the process?

b.) How much heat must be added to the air during the process of raising the elevator 3 m, and what is the final temperature of the gas?

c.) What fraction of the heat added is used in doing work? What fraction is used in raising the temperature of the gas?

d.) How many moles of air must be allowed to escape in order for the elevator to return to the lowest level?

Explanation / Answer

a) Given,

The initial temperature,Ti = 200C

The area of piston, A = 2.5 m2

The volume contained in the cylinder V = 25 m3
Cp = 30 J/mol-K

mass, m= 4000 kg

a)

The weight of the piston, w = mg

= 4000 kg x 9.81m/s2

= 39240 N

Pressure inside the cylinder will always be equal to just hold theweight.
so, Pi = W/A = 39240/2.5 = 15696 N/m2

Consider that the process is reversible process so that the pressure will always remain same.

Hence, the pressure inside the cylinder throughout the process =15696 N/m2

b)

The no. of moles of air inside, n = PV/RT

= (15696*25)/(8.314*293.15)

= 161
The 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
dT = Tf - Ti = 87.945
Tf = 20+87.945 = 107.945 0C
The heat added to the system, Q = nCpdT

= 161*30*87.945
= 424774.35 J

= 424.77kJ

c)

The fraction of heat used in doing work = W/Q

=117720 / 424774.35
= 0.2771

= 27.71%
Therefore, the fraction used in raising the temperature = 1-0.2771 = 0.7229

=72.2 %

d)

Let n' is the moles allowed to escape
Let outside temperature be Ti = 200C,
By 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'

=> Tf' = PV/n'R,
Cv[(n-n')*PV/n'R + n'Ti -nTf] = mgh
Now, Cv = Cp-R = 30-8.314 = 21.686 J/mol-K

Hence [(161-n')*15696*25/(n'*8.314) + n'*293.15 - 161*381.095] =117720/21.686

=> [7598797.21/n' - 47197.49 + 293.15n' - 61356.29] = 5428.38

=> 7598797.21/n' + 293.15n' = 5428.38+47197.49+ 61356.29=113982.16

=> 25921.19/n' + n' = 388.81

=> n'2-388.81n' +25921.19 =0

By solving this we get,

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

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