Calculating Pressure Drop Across Valve in Horizontal Pipe

[pray4mojo]

Hi,

20m straight length horizontal pipe (60 mm internal diameter), steam flows at rate 0.5 kg/s, at midpoint there is a valve in the line. inlet pressure to this section of line is 5 bar (gauge), pressure at end of line is 3 bar (gauge). mass quality of steam is 0.8 and 2 phase flow occurs in the line.

I have calculated the mass flow of liquid and vapor in the line, the frictional pressure drop (N/m2) for the horizontal tube (without valve) using lockart martinelli correlation and also predicted the flow pattern at entry to the valve.

However,the part I am having a problem with is..When calculating the pressure drop across the valve.After calculating the homogeneous parameters and inserting them into the pressure drop equation which is: -dp/dz= [ρ*u *du/dz] + [2*f*ρ*u^2/d] + ρ*g*sinθ

First term in the equation i took equal to zero (no change in velocity assumed), but for the last term (ρ*g*sinθ), do i take theta(θ) as equal to zero as it is a horizontal pipe?

My solution is attached.
Thanks for any help!

 

[nbryan5]

Yes, you would take theta (θ) as equal to zero as the pipe is horizontal. The last term (ρ*g*sinθ) in the equation is the gravitational pressure drop which is only applicable when the pipe is not horizontal, i.e. when it has a vertical or inclined section.