A simple ideal Brayton cycle with air as the working fluid has a pressure ratio of
1
0
.
Air enters the compressor at a temperature of
5
4
0
R
,
and furthermore, air enters the turbine at a te

Question

A simple ideal Brayton cycle with air as the working fluid has a pressure ratio of 
1
0
.
 ﻿Air enters the compressor at a temperature of 
5
4
0
 ﻿R
,
 ﻿and furthermore, air enters the turbine at a temperature of 
2
1
0
0
 ﻿R
.
 ﻿Taking into account the variation of specific heats with temperature 
(
exact approach
)
,
 ﻿do the following: a
)
 ﻿Draw the diagramTs of the Brayton cycle in question b
)
 ﻿Calculate the air temperature at the compressor outlet. c
)
 ﻿Calculate the reverse work ratio.
r
b
u
 
-
 ﻿d
)
 ﻿Calculate the thermal efficiency of the cycle. give complete answer with answer of all subparts

VIPSolutions · Accepted Answer

# Brayton Cycle with Variable Specific Heats

Given:
- Pressure ratio, $ r_p = 10 $
- Compressor inlet temperature, $ T_1 = 540~R $
- Turbine inlet temperature, $ T_3 = 210~R $
- Working fluid: Air (ideal gas, variable $ c_p $)

---

## a) T-s Diagram of the Brayton Cycle

The ideal Brayton cycle has four processes:
1. **1-2:** Isentropic compression (compressor)
2. **2-3:** Constant-pressure heat addition
3. **3-4:** Isentropic expansion (turbine)
4. **4-1:** Constant-pressure heat rejection

```mermaid
%%{init: {"themeVariables": {"fontSize": "16px"}}}%%
graph TD
    A[T1, s1] -- Isentropic compression --> B[T2, s1]
    B -- Heat addition @ p=const --> C[T3, s2]
    C -- Isentropic expansion --> D[T4, s2]
    D -- Heat rejection @ p=const --> A
```
*Alt text: T-s diagram with ideal Brayton cycle showing points 1 to 4 as described above.*

---

## b) Compressor Outlet Temperature ($T_2$), Considering Variable Specific Heats

For isentropic compression of an ideal gas with variable $ c_p $:

$$
\frac{s_2 - s_1}{R} =  = \int_{T_1}^{T_2} \frac{c_p(T)}{RT} dT - \ln\left(\frac{P_2}{P_1}\right)
$$

$$
\int_{T_1}^{T_2} \frac{c_p(T)}{RT} dT = \ln(r_p)
$$

Let:
- $ P_2/P_1 = r_p = 10 $
- $ T_1 = 540~R $

### Steps:
1. Define $ \phi(T) = \int_{T_}^T \frac{c_p(T)}{R T} dT $, with $ T_ $ as a reference (but difference is what matters).
2. Using air tables or NASA polynomials (for exact, as requested) for $ c_p $ as a function of $ T $.

But commonly, tables provide the function $ s^(T) $, the absolute entropy at 1 atm:

$$
s_2^ - s_1^ = R \ln\left(\frac{P_1}{P_2}\right)
$$

$$
s_2^ = s_1^ + R \ln\left(\frac{P_1}{P_2}\right)
$$

But $ s_2^ = s^(T_2) $, $ s_1^ = s^(T_1) $.

Given $ R = .06855~\mathrm{Btu}/(\mathrm{lb}_\mathrm{m}\cdot\mathrm{R}) $, and using US air tables:

$$
s^(T_2) = s^(T_1) + R \ln\left(\frac{P_1}{P_2}\right)
$$

$$
s^(T_2) = s^(T_1) + R \ln\left(\frac{1}{10}\right) = s^(T_1) - R \ln(10)
$$

#### From air tables (US units, Btu/lbm·R):

- At $ T_1 = 540~R $: $ s^ \approx 1.032~\mathrm{Btu}/(\mathrm{lb}_\mathrm{m}\cdot\mathrm{R}) $
- $ R = .06855~\mathrm{Btu}/(\mathrm{lb}_\mathrm{m}\cdot\mathrm{R}) $
- $ \ln(10) = 2.3026 $

So:

$$
s^(T_2) = 1.032 - (.06855 \times 2.3026) = 1.032 - .1579 = .874~\mathrm{Btu}/(\mathrm{lb}_\mathrm{m}\cdot\mathrm{R})
$$

Now, look up the temperature $ T_2 $ at which $ s^(T_2) = .874~\mathrm{Btu}/(\mathrm{lb}_\mathrm{m}\cdot\mathrm{R}) $.

From air tables (interpolating):

- At $ 900~R $, $ s^ \approx .878 $
- At $ 100~R $, $ s^ \approx .929 $

So, $ T_2 \approx 900~R $.

**Final Answer:**

- **Compressor outlet temperature, $ T_2 \approx 900~R $.**

---

## c) Reverse Work Ratio ($r_{bu}$)

**Back work ratio** is the ratio of compressor work to turbine work:

$$
r_{bu} = \frac{w_{comp}}{w_{turb}}
$$

Where:

- $ w_{comp} = h_2 - h_1 $
- $ w_{turb} = h_3 - h_4 $

We need to find $ h_1, h_2, h_3, h_4 $:

- $ h_1 $ at $ 540~R $
- $ h_2 $ at $ 900~R $
- $ h_3 $ at $ 210~R $
- $ h_4 $: Find using isentropic expansion from $ T_3 $ to $ T_4 $ with $ r_p = 10 $

### Find $ T_4 $:

Isentropic expansion, so:

$$
s^(T_4) = s^(T_3) + R \ln\left(\frac{P_3}{P_4}\right)
$$
But $ P_3/P_4 = 1/10 $:

$$
s^(T_4) = s^(T_3) + R \ln(1/10) = s^(T_3) - R \ln(10)
$$

- At $ T_3 = 210~R $, $ s^ \approx 1.556~\mathrm{Btu}/(\mathrm{lb}_\mathrm{m}\cdot\mathrm{R}) $:

$$
s^(T_4) = 1.556 - .1579 = 1.398~\mathrm{Btu}/(\mathrm{lb}_\mathrm{m}\cdot\mathrm{R})
$$

- From tables, at $ T_4 $:
    - At $ 170~R $, $ s^ \approx 1.375 $
    - At $ 180~R $, $ s^ \approx 1.419 $

Interpolate:

$$
T_4 \approx 170 + (1.398 - 1.375) \times \frac{100}{1.419 - 1.375}
$$

$$
= 170 + .023 \times \frac{100}{.044} = 170 + .023 \times 2272.7 \approx 170 + 52 \approx 1752~R
$$

### Find Enthalpy values (from tables):

- At $ 540~R $, $ h_1 \approx 94~\mathrm{Btu}/\mathrm{lb}_\mathrm{m} $
- At $ 900~R $, $ h_2 \approx 147~\mathrm{Btu}/\mathrm{lb}_\mathrm{m} $
- At $ 210~R $, $ h_3 \approx 319~\mathrm{Btu}/\mathrm{lb}_\mathrm{m} $
- At $ 1752~R $, interpolate between:
    - $ 170~R $: $ h \approx 270~\mathrm{Btu}/\mathrm{lb}_\mathrm{m} $
    - $ 180~R $: $ h \approx 285~\mathrm{Btu}/\mathrm{lb}_\mathrm{m} $
    - Difference: $ 15~\mathrm{Btu}/\mathrm{lb}_\mathrm{m} $ over $ 100~R $
    - Per $ 1~R $: $ .15~\mathrm{Btu}/\mathrm{lb}_\mathrm{m} $
    - $ 1752 - 170 = 52~R $
    - $ 52 \times .15 = 7.8 $
    - $ h_4 \approx 270 + 7.8 = 277.8~\mathrm{Btu}/\mathrm{lb}_\mathrm{m} $

#### Calculate Work:

$$
w_{comp} = h_2 - h_1 = 147 - 94 = 53~\mathrm{Btu}/\mathrm{lb}_\mathrm{m}
$$
$$
w_{turb} = h_3 - h_4 = 319 - 277.8 = 41.2~\mathrm{Btu}/\mathrm{lb}_\mathrm{m}
$$

But this gives $ w_{comp} > w_{turb} $, which can't occur. Let's check $ h_3 $ and $ h_4 $ again.

Actually, at $ 210~R $, $ h_3 \approx 399~\mathrm{Btu}/\mathrm{lb}_\mathrm{m} $ (using better tables).
At $ 1752~R $, \( h_4 \ap

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