Question

In an isentropic flow through nozzle, air flows at the rate of 600Kg/hr. At inlet to the nozzle, pressure is 2Mpa and temperature is 127°C. The exit pressure is 0.5Mpa. If initial air velocity is 300m/sec. Determine 

(i) Exit velocity of air, and 

(ii) Inlet and exit area of the nozzle.

Answer 1

Rate of flow of air m_f = 600Kg/hr 

Pressure at inlet P₁ = 2MPa 

Temperature at inlet T₁ = 127 + 273 = 400K 

Pressure at exit P₂ = 0.5MPa 

The velocity at inlet V₁ = 300m/sec 

Let the velocity at exit = V₂

 And the inlet and exit areas be A₁ and A₂ 

Applying SFEE between section 1 – 1 & section 2 – 2 

Q – W_S = m_f [(h₂ – h₁) +1/2( V₂² –V₁²) + g(Z₂ –Z₁)] 

Q = W_S = 0 and Z₁ = Z₂

from equation 1

V₂ = [2 x 1.005 x 10³ (400 – 269.18) + (300)² ]^1/2

V₂ = 594 m/sec .

Since P₁ν₁ = RT₁

ν₁ = 8.314 x 400/ 29 × 2000 = 0.05733 m³/kg

Also m_f .ν₁ = A₁ν₁

A₁ = 600 × 0.05733/3600 × 300 = 31.85mm²

P₂ν₂ = RT₂

ν₂ = 8.314 × 269.18/ 29 × 500 = 0.1543 m³/kg

Now m_f .ν₂ = A₂ν

A₂ = 600 × 0.1543/3600 × 594 = 43.29mm²