Answer to Question #137934 in Molecular Physics | Thermodynamics for Vinaykumar

Question

At the inlet to a certain nozzle the enthalpy of fluid passing is 2800 kJ/kg and the velocity is 50 m/s and at the discharge the enthalpy is 2600 kJ/kg. The nozzle is horizontal and there is negligible heat loss from it. i. Find the velocity at exit of the nozzle. ii. If the inlet area is 900cm2 and the specific volume at inlet is 0.187 m3/kg, find the mass flow rate. iii. If the specific volume at the nozzle exit is 0.498 m3/kg, find the exit area of the nozzle. 

Answer 1

Solution

At inlet

enthalpy h₁=2800 KJ/Kg

Velocity v₁=50m/s

St discharge end

enthalpy h₂=2600 KJ/Kg

1) using energy conservation

$$h_1 ​ +0.5v _ 1^ 2 ​ =h _2 ​ +0.5v _ 2^ 2 ​$$ $$2800.10 ^ 3 +0.5(50) ^ 2 =2600⋅10 ^ 3 +0.5v _ 2^ 2 ​$$ $$v _ 2 ​ =206.15\frac{m}{s}$$

Therefore velocity at exit of nozzle is$206.15\frac{m}{s}.$

2) now mass flow rate is given by

$$\frac{dm}{dt}=\frac{\frac{dV}{dt}}{\frac{dV}{dm}}$$ $$\frac{dV}{dt}=Av_1$$ $$\frac{dm}{dt}=\frac{50\cdot(0.09)}{0.187}=24.06\frac{kg}{s}$$

3) now exit area

$$\frac{dm}{dt}=\frac{v_2A'}{\frac{dV}{dm}'}$$ $$24.06=\frac{206.15A'}{0.498}$$ $$A'=0.058\ m^2$$