Question

A centrifugal compressor which represents an open system takes in air at a pressure of 2 bar and a temperature of 28 c at the rate of 1.4 m/s. compression takes place according to the law PV
^1.3 =C and the delivery pressure is 3.5 bar Determine (i) the input power and (ii) the heat transfer rate.

Answer 1

Answer on Question#78859 - Physics - Molecular physics - Thermodynamics

A centrifugal compressor which represents an open system takes in air at a pressure of $P_{i} = 2$ bar and a temperature of $28{}^{\circ}\mathrm{C}$ at the rate of $1.4\mathrm{m / s}$. Compression takes place according to the law $PV^{1.3} = C$ and the delivery pressure is $P_{f} = 3.5$ bar. Determine

(i) the input power and

(ii) the heat transfer rate.

Solution:

Initial temperature in Kelvin:

$$T _ {i} = 2 8 {}^ {\circ} \mathrm {C} = 3 0 1 \mathrm {K}$$

The elementary work done by compressor is given by

$$d A = P d V,$$

Since $P = C / V^{1.3}$, we obtain

$$A = C \int_ {V _ {i}} ^ {V _ {f}} \frac {d V}{V ^ {1 . 3}} = \frac {1 0}{3} C \left(V _ {i} ^ {- 0. 3} - V _ {f} ^ {- 0. 3}\right) = \frac {1 0}{3} C V _ {i} ^ {- 0. 3} \left(1 - \left(\frac {V _ {i}}{V _ {f}}\right) ^ {0. 3}\right)$$

According to the given law of compression:

$$\left(\frac {V _ {f}}{V _ {i}}\right) ^ {1. 3} = \frac {P _ {i}}{P _ {f}}$$

Using this expression we can rewrite the work in the following way:

$$A = P _ {i} V _ {i} \left(1 - \left(\frac {P _ {f}}{P _ {i}}\right) ^ {\frac {3}{1 3}}\right)$$

The change of the internal energy of the $\nu$ moles of air is given by

$$\Delta U = 2. 5 \nu R \Delta T = 2. 5 \nu R \left(T _ {f} - T _ {i}\right)$$

Using the ideal gas law $(PV = \nu RT)$ we obtain

$$\nu = \frac {P _ {i} V _ {i}}{R T _ {i}}$$

Also due to $PV^{1.3} = C$ we have

$$T _ {f} = T _ {i} \left(\frac {P _ {f}}{P _ {i}}\right) ^ {\frac {2 3}{1 3}}$$

The heat transfer is given by

$$Q = A + \Delta U$$

The work for one cubic meter of air:

$$A _ {1} = 2 \mathrm {b a r} \cdot 1 \mathrm {m} ^ {3} \left(1 - \left(\frac {3 . 5 \mathrm {b a r}}{2 \mathrm {b a r}}\right) ^ {\frac {3}{1 3}}\right) = - 1 3 7 8 5 \mathrm {J}$$

Change of internal energy of one cubic meter of air:

$$\Delta U _ {1} = 2. 5 P _ {l} V _ {l} \left(\left(\frac {P _ {f}}{P _ {i}}\right) ^ {\frac {2 3}{1 3}} - 1\right) = 2. 5 \cdot 2 \mathrm {b a r} \cdot 1 \mathrm {m} ^ {3} \left(\left(\frac {3 . 5 \mathrm {b a r}}{2 \mathrm {b a r}}\right) ^ {\frac {2 3}{1 3}} - 1\right) = 8 4 5. 7 4 \mathrm {k J}$$

Thus (for one cubic meter)

$$Q _ {1} = A _ {1} + \Delta U _ {1} = - 1 3 7 8 5 \mathrm {J} + 8 4 5. 7 4 \mathrm {k J} = 8 3 2 \mathrm {k J}$$

The input power (since the rate of pump is $1.4\mathrm{m}^3/\mathrm{s}$):

$$N = - A _ {1} \cdot 1. 4 \frac {\mathrm {m} ^ {3}}{\mathrm {s}} = 1 9. 3 \mathrm {k W}$$

The heat transfer rate (due to the cooling of compressed air):

$$\frac {d Q}{d t} = Q _ {1} \cdot 1. 4 \frac {\mathrm {m} ^ {3}}{\mathrm {s}} = 1. 1 6 5 \mathrm {M W}$$

Answer: input power: 19.3 kW, heat transfer rate: 1.165 MW.

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