Air (an ideal with constant specific heats, cp = 1.005 kJ/kg-K, cv = 0.718 kJ/kg

Question

Air (an ideal with constant specific heats, cp = 1.005 kJ/kg-K, cv = 0.718 kJ/kg-K, R = 0.287 kJ/kg-K) is trapped in a cylinder with a freely floating piston. The piston and cylinder are insulated from any heat transfer. Initially the air is at 100 kPa and 20 C. The diameter of the cylinder is 0.100 m and initial height of the piston (from the bottom of the cylinder to the bottom of the piston) is 0.2 m. a. Determine the mass of air in the cylinder and the mass of the piston (assume the ambient surrounding air pressure is 0 kPa absolute). Ans. 1.868 x 10-3kg. A 80.0 kg block is suspended 0.5 m above the top of the piston (0.5 m from the top of the piston to the bottom of the block). It is released. The system comes to a final equilibrium position and state. b. Assuming no change in temperature of the piston or block, determine the final air pressure and temperature, and the final position of the piston (i.e. the distance between the bottom of the cylinder and the bottom of the piston, in the final state).200 kPa, 586 K, H2 = 0.20 m. c. Considering the result of part (b), what is the change in entropy of the air during the described process? Is this process reversible? 0.497 kJ/(kg-K); not reversible.

Explanation / Answer

a) Piston Area = 3.14/4*0.1^2 = 0.00785 m^2

Initial volume V1 = Area*height = 0.00785*0.2 = 0.00157 m^3

P1*V1 = mRT1

100 *0.00157 = m*0.287*(20+273)

Mass m = 0.001867 kg = 1.867 grams

Piston weight = pressure*area = 100*10^3 *0.00785 = 785 N

Piston mass = 785/9.81 = 80 kg

b) Final pressure = Final weight/area = (80+80)*9.81/0.00785 = 200 kPa

Since, there is no net external work, dv = 0 and hence, V2 = V1. Therefore, H2 = H1 = 0.2 m

For constant volume process, P1/T1 = P2/T2

So, 100/(20+273) = 200/T2

T2 = 586 K

c) Change in entropy ds = Cp ln(T2/T1) - R ln(P2/P1)

ds = 1.005* ln(586/(20+273)) - 0.287* ln(200/100)

ds = 0.497 kJ/kg-K

Since ds > 0, process is not reversible.

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