(
3
)
A four
-
stage single acting reciprocating air compressor is designed for minimum work with perfect intercooling.
The inlet pressure and temperature is
1
.
0
1
2
9
bar and
3
2
°
C
respec

Question

(
3
)
 ﻿A four
-
stage single acting reciprocating air compressor is designed for minimum work with perfect intercooling.
The inlet pressure and temperature is 
1
.
0
1
2
9
 ﻿bar and 
3
2
°
C
 ﻿respectively, the delivery pressure is 
1
5
2
 ﻿bar.
The clearance is negligible and the polytropic index of compression for each stage is 
1
.
2
8
 
.
(
a
)
 ﻿Calculate EACH of the following per kilowatt of indicated power:
(
i
)
 ﻿the free air delivery at 
1
 ﻿bar and 
0
°
C
(
ii
)
 ﻿the isothermal efficiency of the machine;
(
b
)
 ﻿Sketch the cycle on a pressure
-
Volume diagram showing EACH of the following:
(
i
)
 ﻿the stage pressures;
(
ii
)
 ﻿the isothermal curve;
(
iii
)
 ﻿the work saved by intercooling.
Note: for air 
c
p
=
1
.
0
0
5
k
J
/
kgK and 
R
=
0
.
2
8
7
k
J
k
g
K

VIPSolutions · Accepted Answer

# Four-Stage Reciprocating Air Compressor Calculation

Given Data:
- Number of stages, $ n = 4 $
- Inlet pressure $ P_ = 1.0129 $ bar = $ 101,290 $ Pa
- Inlet temperature, $ T_1 = 32^\circ\text{C} = 305.15 $ K
- Delivery pressure, $ P_5 = 152 $ bar = $ 15,200,000 $ Pa
- Negligible clearance
- Polytropic index, $ n_p = 1.28 $
- $ c_p = 1.005 $ kJ/kgK
- $ R = .287 $ kJ/kgK

---

## (a) Calculations Per kW of Indicated Power

### (i) Free Air Delivery (FAD) at $1\,\text{bar}$ and $^\circ\text{C}$

#### Step 1: Intermediate Pressure for Each Stage

For minimum work with perfect intercooling:
$$
\text{Pressure ratio per stage} = \left(\frac{P_5}{P_1}\right)^{1/n}
$$
$$
= \left(\frac{152}{1.0129}\right)^{1/4}
= (150.11)^{.25} \approx 3.47
$$
So,
$$
P_2 = P_1 \times 3.47 \approx 3.515 \, \text{bar}
$$
$$
P_3 = P_2 \times 3.47 \approx 12.2 \, \text{bar}
$$
$$
P_4 = P_3 \times 3.47 \approx 42.35 \, \text{bar}
$$
$$
P_5 = P_4 \times 3.47 \approx 147 \, \text{bar} \ (\text{matches } 152 \, \text{bar})
$$

#### Step 2: Work Done Per kg (Perfect Intercooling)

Work per stage:
$$
W_{\text{stage}} = \frac{n_p}{n_p-1} R T_1 \left[ \left(\frac{P_2}{P_1}\right)^{\frac{n_p-1}{n_p}} - 1 \right]
$$
Total work (4 stages):
$$
W_{\text{total}} = n \times W_{\text{stage}}
$$
But with perfect intercooling, the air is cooled to $T_1$ before each stage.

$$
\frac{n_p}{n_p-1} = \frac{1.28}{1.28-1} = \frac{1.28}{.28} = 4.571
$$
$$
\left(\frac{P_2}{P_1}\right)^{\frac{n_p-1}{n_p}} = 3.47^{(1.28-1)/1.28} = 3.47^{.21875} \approx 1.31
$$
$$
W_{\text{stage}} = 4.571 \times .287 \times 305.15 \times (1.31 - 1)
$$
$$
= 4.571 \times .287 \times 305.15 \times .31
$$
$$
= (1.312 \times 305.15) \times .31
$$
$$
= 400.61 \times .31 = 124.19\,\text{kJ/kg}
$$
$$
\text{Total work for 4 stages:}
$$
$$
W_{\text{total}} = 4 \times 124.19 = 496.76\,\text{kJ/kg}
$$

#### Step 3: Mass Flow Rate per kW Indicated Power

$$
\text{Per kW:} \quad \dot{m} = \frac{1\,\text{kW}}{W_{\text{total}}\,\text{kJ/kg}} = \frac{1\,\text{kJ/s}}{496.76\,\text{kJ/kg}} = .002014\,\text{kg/s}
$$

#### Step 4: Free Air Delivery at $1\,\text{bar}$, $^\circ\text{C}$

Standard conditions: $P_ = 1\,\text{bar}$, $T_ = 273.15\,\text{K}$
$$
\text{FAD} = \dot{m} \cdot \frac{T_1}{T_} \cdot \frac{P_}{P_1}
$$

But, using the ideal gas law, the volumetric flow at standard conditions:
$$
V_{} = \dot{m} \cdot \frac{R T_}{P_}
$$
But since $\dot{m}$ is at intake conditions, we need to convert it to "free air" basis:

$$
\text{FAD (m}^3/\text{s at 1 bar, }^\circ\text{C}) = \dot{m} \cdot \frac{R T_}{P_}
$$
$$
= .002014\,\text{kg/s} \times \frac{.287 \times 273.15}{100000}
$$
$$
= .002014 \times \frac{78.41505}{100000}
$$
$$
= .002014 \times .0007842
$$
$$
= .000001579\,\text{m}^3/\text{s}
$$

But this is too small. Let's check again; the usual conversion is:
$$
\text{FAD (m}^3/\text{s at 1 bar, }^\circ\text{C}) = \frac{\dot{m} \cdot R \cdot T_}{P_}
$$
But $R$ here is in kJ/kgK, so $P_$ should be in kPa:
$$
P_ = 100\,\text{kPa}
$$
$$
\text{FAD} = .002014 \times \frac{.287 \times 273.15}{100}
$$
$$
= .002014 \times \frac{78.41505}{100}
$$
$$
= .002014 \times .78415 = .001579\,\text{m}^3/\text{s}
$$

Or, per hour:
$$
.001579 \times 360 = 5.684\,\text{m}^3/\text{h}
$$

**Final Answer for (i):**

- **Free air delivery per kW indicated power at 1 bar, $^\circ\text{C}$:**  
  $\boxed{.00158\,\text{m}^3/\text{s}}$ (rounded to 3 decimals)

---

### (ii) Isothermal Efficiency of the Machine

Isothermal work:
$$
W_{\text{iso}} = R T_1 \ln \left(\frac{P_5}{P_1}\right)
$$
$$
= .287 \times 305.15 \times \ln\left(\frac{152}{1.0129}\right)
$$
$$
= 87.37 \times \ln(150.11) = 87.37 \times 5.012 = 437.96\,\text{kJ/kg}
$$

Isothermal efficiency:
$$
\eta_{\text{iso}} = \frac{W_{\text{iso}}}{W_{\text{total}}} = \frac{437.96}{496.76} = .882 = \boxed{88.2\%}
$$

---

## (b) p-V Diagram Sketch

### (i) Stage Pressures

The diagram should show:
- Four compression curves, each from one pressure to the next, with intermediate isothermal intercooling.
- Pressures: 1.0129 bar, 3.515 bar, 12.2 bar, 42.35 bar, 152 bar.

### (ii) Isothermal Curve

- The isothermal compression curve is a smooth, downward sloping curve from initial to final pressure.
- It lies below the actual polytropic compression curve (area under curve = minimum work).

### (iii) Work Saved by Intercooling

- The area between the single-stage polytropic curve and the multi-stage-with-intercooling curve represents the work saved.

#### Example Diagram

```plaintext
P
│            Polytropic (single stage)
│           /
│    ____--/------------------- (Isothermal)
│   /    /
│  /    /         (Multistage polytropic with intercooling)
│_/____/___________________ V
   ↑    ↑    ↑     ↑
  P1   P2   P3    P4   P5
```
- **Isothermal curve:** Lower dashed line.
- **Stage pressures:** Marked at each stage.
- **Work saved:** Shaded area between polytropic and multistage curves.

---

## **Summary Table**

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