Air flows steadily at the rate of 1 kg/s through an air compressor at 10 m/s, 100 kPa and 1 m³/kg and leaving at 6 m/s, 800 kPa and 0.2 m³/kg. The internal energy of the air leaving is 100 kJ/kg greater than the air entering. Cooling water in the compressor jacket absorbs heat from air at the rate of 100 kW. Find the rate of work done.

Concept:

Steady Flow Energy Equation (S.F.E.E.)

\(m\left( {{h_1} + \;\frac{{C_1^2}}{2} + g{z_1}} \right) + Q = m\;\left( {{h_2} + \frac{{C_2^2}}{2} + g{z_2}} \right) + W\)

where m = mass flow rate in kg/s, h₁ = enthalpy at inlet, h₂ = enthalpy at outlet, C₁ = velocity at inlet in m/s, C₂ = velocity at outlet in m/s, z₁ = potential head at inlet in m, z₂ = potential head at outlet in m, Q = heat flow in kW, W = work done in kW.

Calculation:

Given:

m = 1 kg/s, C₁ = 10 m/s, p₁ = 100 kPa, ν₁ = 1 m³/kg, C₂ = 6 m/s, p₂ = 800 kPa, ν₂ = 0.2 m³/kg, Q (Absorbed by cooling water from air) = -100 kW, ΔU = 100 kJ/kg

\(m\left( {{h_1} + \;\frac{{C_1^2}}{2} + g{z_1}} \right) + Q = m\;\left( {{h_2} + \frac{{C_2^2}}{2} + g{z_2}} \right) + W\)

From the above formula we have,

\(Δ K.E~=~\frac{C_2^2~-~C_1^2}{2}~=~\frac{36~-~100}{2}~=~-~32~J/kg ~=~-~0.032~kJ/kg\)

ΔP.E = 0

Δh = Δ (U + pv)

∴ Δpv = (800 × 10³ × 0.2) - (100 × 10³ × 1) = 60 kJ/kg

Δh = 100 + 60 = 160 kJ/kg 

From SFEE we have,

Q - W = m(Δh + ΔK.E + ΔP.E)

- 100 - W = 1(160 - 0.032)

W = - (159.968 + 100)

W = - 259.968 ≈ - 260 kW

W = 260 kW

- ve sign is because work is done on the compressor.