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2 years ago

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An air standard cycle with constant specific heats is executed in a closed system with 0.003 kg of air and consists of the follo

wing three processes:
1-2 - V= Constant heat addition from 95 kPa and 17 Deg Celcius to 380 kPa
2-3 - Isentropic expansion to 95 kPa
3-1 - P= Constant heat rejection to initial state

a) Show cycle on P-v and T-s diagrams
b) Calculate net work per cycle in kJ
c) Determine thermal efficiency

1 answer:

Vsevolod [243]2 years ago

7 0

Answer:

a) Please see attached copy below

b) 0.39KJ

c) 20.9‰

Explanation:

The three process of an air-standard cycle are described.

Assumptions

1. The air-standard assumptions are applicable.

2. Kinetic and potential energy negligible.

3. Air in an ideal gas with a constant specific heats.

Properties:

The properties of air are gotten from the steam table.

b) T₁=290K ⇒ u₁=206.91 kj/kg, h₁=290.16 kj/kg.

P₂V₂/T₂=P₁V₁/T₁⇒ T₂=P₂T₁/P₁ = 380/95(290K)= 1160K

T₃=T₂(P₃/P₂)⁽k₋1⁾/k =(1160K)(95/380)⁽⁰°⁴/₁.₄⁾ =780.6K

Qin=m(u₂₋u₁)=mCv(T₂-T₁)

=0.003kg×(0.718kj/kg.k)(1160-290)K= 1.87KJ

Qout=m(h₃₋h₁)=mCp(T₃₋T₁)

=0.003KG×(1.005kj/kg.k(780.6-290)K= 1.48KJ

Wnet, out= Qin-Qout = (1.87-1.48)KJ =0.39KJ

c)ηth= Wnet/W₍in₎ =0.39KJ/1.87KJ = 20.9‰

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