Question

A piston–cylinder assembly contains propane, initially at 27°C, 1 bar, and a volume of 0.2 m3. The propane undergoes a process to a final pressure of 4 bar, during which the pressure–volume relationship is pV1.1 = constant. For the propane, evaluate the work and heat transfer, each in kJ. Kinetic and potential energy effects can be ignored.

Answer 1

Work done = -19.7 KJ

Heat transferred = 17.4 KJ

Explanation:

Given-

Temperature, T = 27°C

Volume, V = 0.2 m³

Pressure, $P_{1}$= 1 bar

$v_{2}$ = 4 bar

pV¹°¹ = constant

From superheated propane table, at  $P_{1}$= 1 bar and$T_{1}$  = 27⁰C

$v_{1}$ = 0.557 m³/kg

$v_{2}$ = 473.73 KJ/kg

(a) Work = ?

We know,

V1¹°¹ = p2V2¹°¹

$V2 = (\frac{P1}{P2})^\frac{1}{1.1} * V1 \\\\V2 = \frac{1}{4}^\frac{1}{1.1} * 0.557 \\\\V2 = 0.158 m^3/kg$

At  = 4 bar and v = 0.158 m³/kg

u2 = 548.45K J/kg

To find work done in the process:

$W = \frac{P2V2 - P1V1}{1-n} \\\\W = \frac{m(P2V2 - P1V1)}{1-n} \\\\W = \frac{v}{u} * \frac{P2V2 - P1V1}{1-n}\\ \\W = \frac{0.2}{0.5571} * \frac{4 X 0.158 - 1 X 0.577}{1-1.1} X 10^5 \frac{Pa}{Bar} \frac{1KJ}{10^3Nm} \\ \\W = -19.75KJ$

(b) Heat transfer = ?

$Q = m(u2 - u1) + W\\\\Q = \frac{0.2}{0.5571} * (548.45 - 473.73) + (-19.7)\\\\Q = 17.4KJ$