Basic Mechanical Engineering (3110006)

Question-4c

BE | Semester-1 Summer-2019 | 04-06-2023

Q4) (c)

Given Data:

p_{1} = 1 bar

T_{1} = 15 ° C = 288 K

r = 8

Q = 1000 \frac{kJ}{kg}

m = 1 kg

γ = 1.4

C_{v} = 0.718 \frac{kJ}{kg} K

Determine:

T_{3} = ?

η_{air} = ?

W = ?

Solution:

Now,

\frac{T_{2}}{T_{1}} = {(r)}^{γ - 1}

T_{2} = 288 \times {(8)}^{1.4 - 1}

T_{2} = 661.650 K

We have,

Q_{s} = m C_{v} (T_{3} - T_{2})

1000 = 1 \times 0.718 \times (T_{3} - 661.650)

T_{3} = 2054.407 K

Now,

η = 1 - \frac{1}{r^{γ - 1}}

η = 1 - \frac{1}{{(8)}^{1.4 - 1}}

η = 0.5646 = 56.46 %