Hi guys,
I am doing the pressure drop calculation to re-design the inlet pipeline to the new PSVs. An enlargement (6-12’’) is used to connect the existing nozzle with the new upstream piping of the new valves. To compare the results in the calculation, I have used the method given in Shell DEP and the method described in Crane Technical paper no. 410M ( see the attachment).
According to Shell DEP 31.38.01.11 chapter 2.3.3 Table 1, the equivalent length Le of a 6-12’’ sudden enlargement is found to be around 14,3 m of 12’’ pipe. This equivalent length would cause a pressure drop of ca. 0,041bar. The other methods based on the equivalent length give the same pressure drop result.
According to Crane
The resistance to the flow due to sudden enlargement may be expressed by:
K1= (1 – β^2 )^2
β = D1/D2. D1 and D2 is the small and large internal diameter ( 6’’= 154 mm, 12’’=323,8 mm)
For a 6-12’’ enlargement, the calculated value of K1 = 0,55 and the pressure drop is around 0,03 bar.
In term of larger pipe the resistance factor of sudden and gradual enlargement can be calculated by using the equation given below.
K2 = 2.6 sin(θ / 2)(1 – β^2 )^2/β^4
θ is the conical angle of the enlargement.
Θ of the 6-12’’ enlargement is calculated to be around 42 degrees and K2= 7,89. By using this resistance factor K2, the pressure drop is ca. 0,4 bar, which is 10 times larger than the result found by using method based on the equivalent length.
Since our pipe is 12’’ and I guess we have to use K2 to calculate the pressure drop.
As you see, the pressure drop found by using method given in Crane is much higher and I would appreciate your opinion on this case. What method should I use?
Regards,
My