Question

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.
1. 260 kW
2. 480 kW
3. 320 kW
4. 540 kW

Answer 1

Correct Answer - Option 1 : 260 kW

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.