Question

Steam enters an adiabatic diffuser at 200 kPa and 150°C with a velocity of 600 m/s. Determine the minimum velocity that the steam can have at the outlet when the outlet pressure is 400 kPa.

Answer 1

`v_(2)=646.99` m/s

Step-by-step explanation:

Given

At inlet :

Pressure,
`P_(1)` = 200 kPa

Temperature,
`T_(1)` = 150°C

Velocity,
`v_(1)` = 600 m/s

Now from steam table at super heated region, we know,

`h_(1)` = 2769.1 KJ/kg

Now at
`P_(2)` = 400 kPa at saturated region.

For maximum velocity at exit, steam should be at saturated vapour at exit.

Therefore, at
`P_(2)` = 400 kPa,
`h_(1)` =
`h_(g)` = 2739.8 KJ/kg

Therefore,

`h_(1)+(v_(1)^(2))/(2) = h_(2)+(v_(2)^(2))/(2)`

`2769.1* 10^(3)+(600^(2))/(2) = 2739.8* 10^(3)+(v_(2)^(2))/(2)`

`2949100 = 2739800+(v_(2)^(2))/(2)`

`v_(2)=646.99` m/s

Hence the answer is
`v_(2)=646.99` m/s