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

To determine the exit area of the nozzle, use the principle of conservation of mass and the equation for mass flow rate. Calculate the density using the Ideal Gas Law and substitute it into the equation for area.

Step-by-step explanation:

To determine the exit area of the nozzle, we can use the principle of conservation of mass and the equation for mass flow rate:

Mass flow rate = density x velocity x area

Given that the mass flow rate is 2.5 kg/s and the velocity is 280 m/s, we can rearrange the equation to solve for the area:

Area = mass flow rate / (density x velocity)

However, we need to find the density of the steam at the nozzle exit. To do this, we can use the Ideal Gas Law:

Pressure x Volume = n x R x Temperature

Where pressure = 2 MPa, volume can be assumed to be the volume of the nozzle exit, R is the gas constant, and temperature is 400°C converted to Kelvin.

Once we have the density, we can substitute it into the equation for the area to find the exit area of the nozzle.

Answer 2

The exit area of the nozzle is 0.000861 m².

Step-by-step 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=(RT)/(P)`

Put the value into the formula

`V=(0.287*673)/(2*10^(3))`

`V=0.0965\ m^3/kg`

We need to calculate the exit area of the nozzle

Using equation of continuity

`(dm)/(dt)=(A_(1)v)/(V)`

`A=(V*(dm)/(dt))/(v)`

Put the value into the formula

`A=(0.0965*2.5)/(280)`

`A=0.000861\ m^2`

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