It is given that 50 kg/sec of air at 288.2k is iesntropically compressed from 1 to 12 atm. Assuming a calorically perfect gas, d

Question

3 years ago

13

etermines the exit temperature and the compressor's input power.

1 answer:

denis23 [38] · Answer 1 · 2023-01-07T05:12:58+03:00

The exit temperature is 586.18K and compressor input power is 14973.53kW

Data;

<h3>Exit Temperature </h3>

The exit temperature of the gas can be calculated isentropically as

$\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{y-1}{y}\\ y = 1.4\\ C_p= 1.005 Kj/kg.K\\$

Let's substitute the values into the formula

$\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{y-1}{y} \\\frac{T_2}{288.2} = (\frac{12}{1})^\frac{1.4-1}{1.4} \\ T_2 = 586.18K$

The exit temperature is 586.18K

<h3>The Compressor input power</h3>

The compressor input power is calculated as

$P= mC_p(T_2-T_1)\\P = 50*1.005*(586.18-288.2)\\P= 14973.53kW$

The compressor input power is 14973.53kW

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