Question

Oxygen at 50 lbf/in.2, 200 F is in a piston/cylinder arrangement with a volume of 4 ft3. It is now compressed in a polytropic process with exponent, n = 1.2, to a final temperature of 400 F. Calculate the heat transfer for the process.

Answer 1

Answer:

The value of heat transfer during the process Q = - 29.49 KJ

Explanation:

Given data

$P_{1}$ = 50 $\frac{lbf}{in^{2} }$ = 344.7 k pa

$V_{1} = 4 ft^{3}$ = 0.113 $m^{3}$

$T_{1} = 200$ F = 366.4 K

$T_{2} = 400 F$ = 477.6 K

Poly tropic index n = 1.2

gas constant for oxygen = 0.26 $\frac{KJ}{kg K}$

From ideal gas equation

$P_{1}$ $V_{1}$ = m R $T_{1}$

Put all the values in above equation we get

⇒ 344.7 × 0.113 = m × 0.26 × 366.4

⇒ m = 0.408 kg

Heat transfer in poly tropic process is given by

Q = $\frac{\gamma - n}{( \gamma - 1)( n - 1)} [ {m R (T_{1} - T_{2} ) ]$

Put all the values in above formula we get

⇒ Q = $\frac{1.4 - 1.2}{( 1.4 - 1)( 1.2 - 1)} [ {m R (T_{1} - T_{2} ) ]$

⇒ Q = 2.5 × 0.408 × 0.26 × ( 366.4 - 477.6 )

⇒ Q = - 29.49 KJ

This is the value of heat transfer during the process & negative sign shows that  heat is lost during the process.