Homebuilding has undergone significant advances in the last 20 years, due largel
ID: 1853289 • Letter: H
Question
Homebuilding has undergone significant advances in the last 20 years, due largely to replacement of hand tools (hammers, saws, screwdrivers) with the power equivalents. Nowadays if you watch roofers lay shingles you hear the ''thump-thump" of a pneumatic nailer instead of the "whack-whack-whack" of a framing hammer hitting a nail. A pneumatic nailer is nothing more than a piston-cylinder device. When the trigger is pulled it allows compressed air (initially at 75 F and a pressure between 70 and 120 psig) to expand; the resulting stroke drives a nail through the shingle and into the roofing plywood. Assume that the piston is 1 inch in diameter, and that the nailing stroke is 2 inches. The air is initially contained in a chamber that is 2 in long at a pressure of 90 psig and temperature of 75 F. The average force exerted by the piston on the nail is 50 lbf. The process occurs very rapidly so can be approximated as adiabatic. What is the average air pressure exerted on the piston? Report your answer in psig. How much work is done on the piston during air expansion? Report your answer in ft-lbf. What is the temperature of the air at the end of expansion? Report your answer in F.Explanation / Answer
GIVEN: T1 = 75[F] = 535 [R]
P1 = 90 [psig]
d = 1 [in]
stroke = 2 [in]
height = 2 [in]
Force = 50 [lbf]
ASSUMPTIONS:
Assume that when they say "adiabatic" they also imply it's reversible. If so, then we have an "isentropic" situation.
Because It's adiabatic, we need to assume a constant specific heat (k). The normal value of k = 1.4
k = 1.4
SOLVE: Find the Bore (surface area of cylinder):
(B = rac{pi(d^{2})}{4} = rac{pi(1^{2})}{4} = 0.785398 in^{2})
Force = Pressure x Area....or in our case
Force = Pressure x Bore
The pressure here is P2:
P2 = Force / Bore
P2 = 50 [lbf] / (0.785398[in^2]
P2 = 63.66 [psig]
Now because this is an isentropic situation, follow the formulas:
(T_2 = T_1igg( rac{p2}{p1}igg)^ rac{k-1}{k})
(T_2 = 535[R] * igg( rac{63.66}{90}igg)^ rac{1.4-1}{1.4} = 484.6 [R] = 24.6 [F])
To solve for Work we need to find the volume. Volume is Bore x stroke
V = B x S
V = 0.785398[in^2] * 2[in]
V = 1.570796 [in^3]
(W = (P_1 - P_2)V * rac{k}{k-1})
(W = ( rac{1}{12})*(90 - 63.66) * (0.785398) * rac{1.4}{1.4-1} = 12.06 [ft-lbf])
FYI the (1/12) is converting inches to feet because the final units are "ft-lbf"
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