Question

U4.4. An engine is created in which an ideal gas follows the cyclic Platm) process shown in the PV diagram to the right. a) How much work is done by the system in one cycle 1231 b) If the cycle repeats 20 per second, what is the power of this engine? c) What is the temperature at state 3? d) What is the maximum efficiency of an engine with this temperature change? e) How much heat energy enters the system during the isothermal (1-22)? Given: during the constant pressure process (2-23) 636 J of heat leaves the system. during the constant volume process (3>1) 455 J of heat 1.4- T 76°C 2 3.0 V(L) enters the system. Find the actual efficiency of this engine. answer: U4.4 d) e: 4396 e) e ways 8% where w: total work, and Qs is the total heat into system, Q-2+ QM

Answer 1

GIVEN:

Initial pressure, P₁ = 1.4atm = 1.4 x 1.01325 = 1.41855 bar = 1.41855 x 10⁵N/m²

Pressure, P₂ = P₃ = 0.8atm = 0.8106 x 10⁵N/m²

Temperature, T₁ = T₂ = 76^oC

Volume, V₁ = V₃ = 3.0 litre = 3 x 10^-3m³

Heat energy, Q_2-3 = 636J

Heat energy, Q_3-1 = 455J

TO FIND:

1. Work done in the system in 1 cycle, W

2.Power of the engine

3.Temperature at state 3, T₃

4. Maximum efficiency

5.Heat energy during process 1-2, Q_1-2

SOLUTION:

Process 1-2 Isothermal or constant temperature process

Process 2-3 Constant pressure process

process 3-1 constant volume process

Workdone in process 1-2,

$W_{1-2}=pVln\left ( \frac{V_{2}}{V_{1}} \right )$

To find V₂

For constant temperature process,

$p_{1}V_{1}=p_{2}V_{2}$

$V_{2}=\frac{p_{1}V_{1}}{p_{2}}$

$V_{2}=\frac{1.41855\times 10^{5}\times 3\times 10^{-3}}{0.8106\times 10^{5}}$

$V_{2}=5.25\times 10^{-3}m^{3}=5.25l$

$W_{1-2}=1.41855\times 10^{5}\times 3\times 10^{-3}\times ln\left ( \frac{5.25}{3} \right )$

$W_{1-2}=238.15J$

Workdone during process 2-3

Process 2-3 is constant pressure process

Workdone,

$W_{2-3}=p_{3}V_{3}-p_{2}V_{2}$

$W_{2-3}=0.8106\times 10^{5}\times 3\times 10^{-3}-0.8106\times 10^{5}\times 5.25\times 10^{-3}$

$W_{2-3}=-182.385J$

Workdone during process 3-1 ( Constant volume process)

W_3-1 = 0

QUESTION 1

Total workdone, W = W_1-2 + W_2-3 + W_3-1

W = 238.15 - 182.385 + 0

W = 55.765J

QUESTION 2

Power developed

P = Workdone x mass flow rate of gas per cycle x 20cycles/per second

mass flow rate,

$m=\frac{V_{s}}{V_{2}}$

$m=\frac{3}{5.25}=0.571kg/cycle$

Power developed,

$P=55.765\times 0.571\times 20=637.3W$

QUESTION 3

Temperature at state 3

Process 2-3 is constant pressure process

PVT relation is

$\frac{V_{2}}{T_{2}}=\frac{V_{3}}{T_{3}}$

${T_{3}}=\frac{V_{3}}{V_{2}}\times T_{2}$

${T_{3}}=\frac{3}{5.25}\times \left ( 76+273 \right )$

${T_{3}}=199.4K$

QUESTION 4

Efficiency of engine,

$\eta _{e}=\left ( 1-\frac{T_{3}}{T_{2}} \right )\times 100$

$\eta _{e}=\left ( 1-\frac{199.4}{76+273} \right )\times 100=43$

QUESTION 5

Total heat supplied, Q_s = Q_1-2 + Q_3-1

Q_1-2 Process 1-2 is constant temperature process

Hence Workdone, W_1-2 = Q_1-2 since Internal energy U=0

$\therefore W_{1-2}=Q_{1-2}$

$Q_{1-2}=238.15J$

$\therefore Q_{s}=238.15+455=693.15J$

$Heat supplied=\frac{Heat supplied }{Total Heat supplied}=\frac{55.765}{693.15}=0.080=8$