Answer to Question #103873 in Molecular Physics | Thermodynamics for nicole

Question

A nozzle is a device for increasing the velocity of a steadily flowing stream. At the inlet to a certain nozzle, the enthalpy of the fluid passing is 3000 kJ/kg and the velocity is 60 m/s. At the discharge end, the enthalpy is 2762 kJ/kg. The nozzle is horizontal and there is negligible heat loss from it. Calculate a. The velocity of the fluid at the exit b. the rate of flow of fluid when the inlet area is 0.1 m^2 and the specific volume at the inlet is 0.19 m^3/kg c. The exit area of the nozzle when the specific volume at the nozzle exit is 0.5 m^3/kg

Answer 1

a. From the conservation of energy:

$$h _ 1 ​ +0.5v _ 1^ 2 ​ =h _2 ​ +0.5v _ 2^ 2 ​$$

$$3000⋅10 ^ 3 +0.5(60) ^ 2 =2762⋅10 ^ 3 +0.5v _ 2^ 2 ​$$

$$v _ 2 ​ =692.5\frac{m}{s}$$

b.

$$\frac{dm}{dt}=\frac{\frac{dV}{dt}}{\frac{dV}{dm}}$$

$$\frac{dV}{dt}=Av_1$$

$$\frac{dm}{dt}=\frac{60\cdot 0.1}{0.19}=31.6\frac{kg}{s}$$

c.

$$\frac{dm}{dt}=\frac{v_2A'}{\frac{dV}{dm}'}$$

$$31.6=\frac{692.5A'}{0.5}$$

$$A'=0.0228\ m^2$$