When gas expands in a cylinder with radius r , the pressure P at any given time

Question

When gas expands in a cylinder with radius r, the pressure P at any given time is a function of the volume V: P = P(V). The force exerted by the gas on the piston (see the figure) is the product of the pressure and the area: F = ?r2P. The work done by the gas when the volume expands from volume V1 to volume V2 is.

In a steam engine the pressure and volume of steam satisfy the equation PV1.4 = k, where k is a constant. (This is true for adiabatic expansion, that is, expansion in which there is no heat transfer between the cylinder and its surroundings.) Calculate the work done by the engine during a cycle when the steam starts at a pressure of 180 lb/in2 and a volume of 500 in3 and expands to a volume of 800 in3. (Round your answer to two decimal places.)

Explanation / Answer

We have given Pressure P=180 lb/in2 =180*144 lb/ft2 since 1lb/in2=144 lb/ft2

P=180*144 lb/ft2=25920 lb/ft2

Given Volume V1=500 in3 and V2= 800 in3

V1=500 in3 =(500/1728)ft3 since 1 inch =1/12 ft so 1inch^3=1/1728 ft^3

V2=800 in3=(800/1728)ft3

We have to find P(V), remember since PV^(1.4) = k, we can solve for k by using the pressure of the steam and the initial volume.

PV^(1.4) = k

k= PV^(1.4) since rewriting the equation

P=k*V^(-1.4)

k=25920 lb/ft2* ((500/1728)ft3 )^(1.4)=4567.007 lb/ft since intial volume V=V1=(500/1728)ft3

k=4567.007 lb/ft

so P(V)=k*V^(-1.4)=(4567.007)*V^(-1.4)

So work is given by W=integration of from (V1=(500/1728) to V2=(800/1728))(PdV)

W=integration of from (V1=(500/1728) to V2=(800/1728))((4567.007)*V^(-1.4)dV)

W=[(4567.007)*[(V^(-0.4))/(-0.4))]] from (V1=(500/1728) to V2=(800/1728))

W=[(4567.007)*[((800/1728)^(-0.4))/(-0.4))]]-[(4567.007)*[((500/1728)^(-0.4))/(-0.4))]]=3213.49 ft-lb

W=3213.49 ft-lb

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.