Question

Professor Modyn owns a device called a pneumatic accumulator. This device maintains constant pressure of 217.00 psia as air enters by using a piston. Initially the device is filled with air at 76.74 °F and the volume is 0.317 ft3 (State 1). Air at 191.2 °F and 217.00 psia is fed into the accumulator until the volume triples and the temperature increases to 181.2 °F (State 2). Assume air is an ideal gas with a CP = 3.5R and MW = 29. No air leaves the system.

a) Find the amount of air added. (lbmol)

b) Calculate the work interaction term. (ft*lbf)

c) Calculate the heat interaction term. (ft*lbf)

Answer 1

a) Find the amount of air added. (lbmol)

R = 10.73 (psia)(ft3)/(lbmol)(0R)

state1:
T1 = 76.74 + 459.67 = 536.41 R
P1 = 217 psia
V1 = 0.317 ft3
n1 = P1*V1/(R*T1) = 217*0.317/(10.73*536.41) = 0.012 lbmol

state2:
T2 = 181.2 + 459.67 = 640.87 R
P2 = P1 =217 psia
V2 = 3*V1 = 3*0.317 = 0.951 ft3
n2 = P2*V2/(R*T2) = 217*0.951/(10.73*640.87) = 0.03 lbmol

amount of air added = n2 - n1 = 0.03-0.012 = 0.018 lbmol

b) Calculate the work interaction term. (ft*lbf)
P = 217 psia = 217 lbf/in2
V1 = 0.317 ft3 = 0.317*12*12*12 in3 = 547.776 in3
w = -P*(V2 - V1) = -P*(3*V1 - V1) = -2*P*V1 = -2*217*547.776 = -237734.784 lbf in/12
w = -19811.232 lbf ft

c) Calculate the heat interaction term. (ft*lbf)

n1 = 0.012 lbmol
n2 = 0.03 lbmol
amount of air added = n2 - n1 = 0.03-0.012 = 0.018 lbmol
R = 1,545.348 ft lbf /(°R lb-mol)

cp = 3.5*R =3*1545.348 ft lbf /(°R lb-mol) = 4636.044 ft lbf /(°R lb-mol)

T1 = 76.74 + 459.67 = 536.41 R
T2 = 181.2 + 459.67 = 640.87 R
Air temperature, Tair = 191.2 + 459.67 = 650.87 R


q = n1*cp*(T2-T1) - (n2-n1)*cp*(Tair-T2) = cp*[n1*(T2-T1) - (n2-n1)*(Tair-T2)]
= 4636.044*[0.012*(640.87-536.41) - 0.018*(650.87-640.87)] = 4977 lbf ft