Air is compressed from 100 kpa and 150c to a pressure of 1000 kpa while being cooled at a rate of 20 kj/kg by circulating water through the compressor casing. the volume flow rate of the air at the inlet conditions is 140 m3/min, and the power input to the compressor is 520 kw determine: (a) the mass flow rate of the air, and (b) the temperature at the compressor exit.

a) 115.7 kg/min

b) 240 C

Explanation:

To determine the mass flow rate we need the specific volume of the air. We can use the gas state equation for that:

p * v = R(air) * T    (no mass term because we want specific volume)

v = R * T / p

The specific R for air is R = 287 J/(kg * K)

Also 150 C = 423 K

Then

v = 287 * 423 / 100e3 = 1.21 m^3/kg

Them the mass flow is

G / v = 140 / 1.21  = 115.7 kg/min

The flow rate can also be expressed as 115.7 / 60 = 1.93 kg/s

Then we find that the compressor applies a work of

L = 520 / 1.93 = 269.4 kJ/kg to the air

By enthalpy:

H2 = H1 + Q + L

H2 = cv * T2 + p2 * v2

H1 = cv * T1 + p1 * v1

Then

cv * T2 + p2 * v2 = cv * T1 + p1 * v1 + Q + L

Also

(p1 * v1) / T1 = (p2 * v2) / T2

p2 * v2 = (p1 * v1 * T2) / T1

Therefore:

cv * T2 + (p1 * v1 * T2)/T1 = cv * T1 + p1 * v1 + Q + L

T2 * (cv + (p1 * v1)/T1 = cv * T1 + p1 * v1 + Q + L

T2 = (cv * T1 + p1 * v1 + Q + L)/((cv + (p1 * v1)/T1)

The cv of air is cv = 0.72 kJ/(kg*K)

T2 = (0.72 * 423 + 100e3 * 1.21 - 20 + 269.4)/((0.72 + (100e3 * 1.21)/423) = 513 K = 240 C

answer:

an example of solubility is salt or sugar dissolving in water.

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