Question

Steam is accelerated by a nozzle steadily from zero velocity to a velocity of 280 m/s at a rate of 2.5 kg/s. If the temperature and pressure of the steam at the nozzle exit are 400°C and 2 MPa, determine the exit area of the nozzle. Solve using appropriate software.

Answer 1

Answer:

The exit area of the nozzle is 0.000861 m².

Explanation:

Given that,

Velocity = 280 m/s

Rate = 2.5 kg/s

Pressure = 2 MPa

Temperature = 400°C

We need to calculate the volume

Using equation of ideal gas

$PV=RT$

$V=\dfrac{RT}{P}$

Put the value into the formula

$V=\dfrac{0.287\times673}{2\times10^{3}}$

$V=0.0965\ m^3/kg$

We need to calculate the exit area of the nozzle

Using equation of continuity

$\dfrac{dm}{dt}=\dfrac{A_{1}v}{V}$

$A=\dfrac{V\times\dfrac{dm}{dt}}{v}$

Put the value into the formula

$A=\dfrac{0.0965\times2.5}{280}$

$A=0.000861\ m^2$

Hence, The exit area of the nozzle is 0.000861 m².