Steam at 100 kPa and 150 degree C enters to an isentropic di

Steam, at 100 kPa and 150 degree C enters to an isentropic diffuser with a velocity of 500 m/s. If the exist state is 300 kPa, calculate the exist velocity of steam.

Solution

P1= 100 kPa, T1= 150^oC, V1 = 500 m/s, P2 = 300 kPa, V2=?

From Steam Tables at state 1, h1= 2776.6 kJ/kg, s1= 7.6148 kJ/kgK

The process is isentropic diffusion, so the s1=s2.

From steam tables corresponding to 300 kPa and s2=s1=7.6148, the T2= 275^o C and h2= 3019 kJ/kg.

V₁²/2 +h₁ = V²₂/2 + h₂ ==> V₂^2//2 = V₁²/2 + h₁ - h₂ where h is in Joules

But h₁ - h₂ = -242.4 kJ/kg = -242400 J/kg and V₁²/2 =125000 only which gives negative value for V₂² which is not possible.

There must be some error in the question.