Calculate the thermal efficiency of the cycle.

Question

A work cycle with an ideal gas (M = 29 kg / kmol, κ = 1.4) as working medium

consists of the following reversible state changes:

a-b: isothermal compression, b-c: isentropic expansion,

c-d: isothermal compression, d-e: isobaric heat supply,

e-a: isochoric heat removal

In state a the pressure is 1 bar, the temperature 20 ° C and

the volume 2 liters.

In state c is

pressure 1 bar. The compression ratio is 3.

The volume in state b is 2/3 · Va.

Calculate the thermal efficiency of the cycle.

Answer 1

For this analysis to work, several assumptions need to be made and are as follows:

1. Given a molar mass of 29 kgl/kmol and a heat ratio (γ) 1.4, it is assumed the working fluid is air.
2. Assuming the working fluid is air, therefore the specific heats and individual gas constant are:
3. cv = 718 J/kg
4. cp = 1005 J/kg
5. R = 286 J/kg-K
6. The compression ratio throughout the cycle is 3, unless otherwise stated.

This analysis begins with a state property analysis followed by a first and second laws of thermodynamics analysis.

State Properties Analysis

State A

We are given:

1. P_a = 100kPa
2. T_a = 293 K
3. V_a = 0.002 m³
4. Process: Isothermal compression

And need to the mass for later analysis:

m = (PV)/RT) = (100,000*0.002)/(286*293) = 0.002387 kg

State B

1. P_b = 150 kPa
2. T_b= 293 K
3. V_b = V_a(2/3) = (0.002)*(2/3) = 0.0013 m³
4. Process: isentropic expansion

To find P_b, we use Boyle Gas Law = (PV)₁ = (PV)₂ = (100,000*0.002)₁ = (Pb*0.0013)₂, ∴ P_b = 150kPa

State C

We are given P_c and CR, but need to find T_c.

1. P_c = 100 kPa
2. T_c = 260.9 K
3. CR = 3

Using isentropic relations of an ideal gas to find Tc:

(T_c/T_b) = (P_c/P_b)^(γ-1)/γ = (T_c/293) = (1/1.5)^(1.4-1)/1.4 ∴ T_c = 260.9 K

State D

This is an isothermal process, Boyle's Gas Law is used to find P_d.

1. P_d = P_c (V_d/V_c) = P_c (CR) = 100,000 * 3 = 300 kPa
2. T_d = T_c = 260.9K

State E

This is an isobaric process and Charles's Gas Law is used to find T_e.

T_e/V_e = T_d/V_d → T_e = T_d*CR = 260.9*3 = 782.7K

First Law of Thermodynamics Analysis

Process A → B (Isothermal Compression)

^ΔU = 0

W_{(isothermal)(a→b)} = mRT* ln(V_a/V_b) = (PV)_a*ln(V_a/V_b) = 100,000*0.002*ln(-1.5 )= -81.1 J

Q = ^ΔU + W = 0 - 81.1 = -81.1

Process B → C (Isentropic Expansion)

^ΔU = 55.4

W_{(isentropic) (b→c)} = m*^Δh = m*cp ^ΔT = 0.002387*1005*(269.9 - 293) = -55.4 J

Q = ^ΔU + W = 55.4 - 55.4 = 0 J

Process C → D (Isothermal Compression)

^ΔU = 0

W_{(isothermal) (c→d)} = mRT* ln(V_c/V_d) = 0.002387*286*260.9*ln(1/3) = -195.7 J

Q = ^ΔU + W = 0 -195.7 = -195.7 J

Process D → E (Isobaric Heat Addition)

Q = m*^Δh = m*cp ^ΔT = 0.002387*1005*(782.7-260.9) = 1251.8 J

Process E → A (Isochloric Heat Removal)

^ΔU = m*^Δu = m*cv ^ΔT = 0.002387*718 *(293-782.7) = -839.3 J

W_{(isochloric) (e→a)} = 0

Q = ^ΔU + W = -839.3+0 = -839.3 J

Second Law of Thermodynamics Analysis

ΣQ_(out) = 81.1 + 0 + 195.7 + 839.3 = 1116.1 J

ΣQ_(in) = 1251.8 J

η = (ΣQ_(in) -ΣQ_(out))/ΣQ_(in) = (1251.8-1116.1)/1251.8 = 0.1084 = 10.84%

Lets compare our efficiency to Carnot's efficiency.

η_(Carnot) = 1-T_c/T_h = 1 - 260.9/782.7 = 0.667 = 66.7%

From this it can be concluded this thermal cycle is well below the Carnot efficiency. Implying the cycle is 1) realistic because the cycle efficiency is below the Carnot efficiency and 2) has room to improve.