Question

A reciprocating air compressor takes in 2m³/min of air at 0.11MPa and 20°C which it delivers at 1.5MPa and 111°C to an after cooler where the air is cooled at constant pressure to 25°C. The power absorbed by the compressor is 4.15KW. Determine the heat transfer in 

(a) Compressor and 

(b) cooler. CP for air is 1.005KJ/Kg-K.

Answer 1

ν₁ = 2m³/min = 1/30 m³/sec

P₁ = 0.11MPa = 0.11 x 10⁶ N/m²

T₁ = 20°C

P₂ = 1.5 x 10⁶ N/m²

T₂ = 111°C

T₃ = –25°C

W = 4.15KW

C_P = 1.005KJ/kgk

Q_1-2 = ? and Q_2-3 = ?

From SFEE

Q – W_S = m_f [(h₂ – h₁) + 1/2( V₂ ² – V₁ ²) + g(Z₂ – Z₁)]

There is no data about velocity and elevation so ignoring KE and PE

Q_1-2 – W_1–2 = m[cp(T₂ – T₁)] ...(i) 

Now P₁ν₁ = mRT₁

m = (0.11 × 10⁶ x 1/30)/(287 × 293) = 0.0436Kg/sec; R= 8314/29 = 287 For Air

From equation (i) 

Q_1-2 – 4.15 × 10³ = 0.0436[1.005 × 10³ (111 – 20)]

Q_1-2 = 8.137KJ/sec

For process 2 – 3; W_2-3 = 0

Q_2-3 – W_2-3 = m[cp(T₂ – T₁)]

Q_2-3 – 0 = 0.0436[1.005 x 10³ (– 111 + 25)]

Q_2-3 = – 3.768KJ/sec