Question

Steam at 0.6 MPa and 300oC steadily enters a nozzle whose inlet area is 0.07 m2. The velocity of the steam at the inlet is 5 m/s. Steam leaves the nozzle at 250oC and 0.2 MPa. Heat losses from the nozzle are estimated to be 20 kW. Find (a) the velocity at the nozzle exit, and (b) the volume flow rate at the nozzle exit

Answer 1

The velocity of the steam at nozzle exit
`V_(2)` = 383.74
`(m)/(s)`

The volume flow rate at nozzle exit Q = 0.958
`(m^(3) )/(sec)`

Step-by-step explanation:

Pressure at inlet
`P_(1)` = 0.6 M pa = 600 k pa

Velocity at inlet
`V_(1)` = 5
`(m)/(s)`

Temperature at inlet
`T_(1)` = 300 °c = 573 K

Temperature at outlet
`T_(2)` = 250 °c = 523 K

Gas constant for steam R = 0.462
`(KJ)/(kg K)`

Pressure at outlet
`P_(2)` = 0.2 M pa = 200 k pa

Inlet area
`A_(1)` = 0.07
`m^(2)`

Heat loss Q = 20 KW

(a). Apply steady flow energy equation for the nozzle,

`h_(1) + (V_(1) ^(2) )/(2000) - Q = h_(2) + (V_(2) ^(2) )/(2000)`

⇒ 1.8723 ( 573 - 523 ) +
`(5^(2) )/(2000)` - 20 =
`(V_(2) ^(2) )/(2000)`

⇒
`(V_(2) ^(2) )/(2000)` = 93.615 + 0.0125 - 20

⇒
`V_(2) ^(2)` = 147255

⇒
`V_(2)` = 383.74
`(m)/(s)`

This is the velocity of the steam at nozzle exit.

(b). We know that mass flow rate through the nozzle is constant at inlet and outlet and given by

m =
`\rho_(1) A_(1) V_(1)` =
`\rho_(2) A_(2) V_(2)` --------- ( 1 )

`\rho_(1) = (P_(1) )/( R T_(1) )`

Put all the values in above equation we get,

`\rho_(1) = (600)/(0.462) (1)/(573)`

`\rho_(1)` = 2.266
`(kg)/(m^(3) )`

Similarly

`\rho_(2) = (P_(2) )/( R T_(2) )`

`\rho_(2)` =
`(200)/(0.462)` ×
`(1)/(523)`

`\rho_(2)` = 0.8277
`(kg)/(m^(3) )`

Now put all the values in equation (1),

⇒ 2.266 × 0.07 × 5 = 0.8277 ×
`A_(2)` ×
`V_(2)`

⇒
`A_(2)` ×
`V_(2)` = 0.958
`(m^(3) )/(sec)`

⇒ Q =
`A_(2)` ×
`V_(2)` = 0.958
`(m^(3) )/(sec)`

This is the volume flow rate at nozzle exit.