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When calculating the installation height of a pump, is it necessary to convert the net positive suction head into the net positive suction head at the density of the liquid being transported? See the question below

2011-06-26View Original

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23. The operating pressure at the top of the distillation column is 0.08 MPa(G), the total pressure drop across the columns is 0.03 MPa, and the density of the fluid in the bottom of the column at the operating temperature is 640 kg/m3. The height of the tower bottom (from the tangent at the bottom of the tower to the first tray) is 6 m. Under normal operation, the liquid level at the tower bottom is at least 15%, and the height of the centerline of the centrifugal pump at the bottom of the tower above the ground is 0.5 m. As shown in the figure below (the length in the figure is in mm). Assume that the total pressure loss on the suction side of the pump is 0.015 Mpa, the required net positive suction head is 3 m, and the safety margin is taken as 0.5 m. Then, the minimum height (m) of the tower skirt base (H) should be no less than which of the following values? http://bbs.hcbbs.com/attachment.php?aid=NzI4NDU3fDliZDViOWM2fDEzMDkxMDE5MDF8MGVlNnpLUmJGWnpwNHZMS1NKbHI0c3A2K2R2ZENwWFVPenNYM1JmRjNoNUV6U2c%3D&noupdate=yes Download (15.07 KB) 2010-12-13 20:47 (A) (B) (C)3.4 (D) Solution: Hg=-(H-0.5+6×0.15) =0-0.015×106/(640×9.81)-(3+0.5)=5.5 (m) Is it Hg=-(H-0.5+6×0.15) =0-0.015×106/(640×9.81)-(3+0.5)*1000/640=7.8 (m)?
Reply #22011-06-26
The net positive suction head and the suction vacuum seem to be values determined under standard clean water conditions. When transporting fluids with different densities, it is necessary to adjust these values of net positive suction head and vacuum. I would appreciate advice from those who are more experienced in this area; in some examples I’ve seen, adjustments are made sometimes while at other times they aren’t, and I’m not sure why
Reply #32011-06-27
When the flow rate is constant and the fluid flow is in the resistance-square region, the net positive suction head is only related to the structure and size of the pump, and not to the properties of the liquid or factors such as the local atmospheric pressure; therefore, there is no need to adjust the liquid density. Generally speaking, the allowable suction vacuum degree of a pump is related to the properties and temperature of the liquid being transported, as well as the local atmospheric pressure, and adjustments need to be made accordingly ; Generally, the net positive suction head does not need to be corrected.
Reply #42011-06-27
OP, what is the value of 106 in the formula? I don’t understand.
Reply #52011-06-27
10 to the power of 6. . . . . . . . . .
Reply #62011-06-27
Reply to 3# kaminocmyhc: Thank you for the answer
Reply #72011-07-04
Why does it not need to consider the operating pressure at the top of the tower or the total pressure drop across the tray?
Reply #82011-07-19
This post was last edited by xue2010 on 2011-7-19 at 13:24. The type of material inside the tower and its temperature are not specified, nor is the saturated vapor pressure of the material known – and this is a key factor. The operating pressure and pressure drop within the tower also need to be taken into account. The final calculation formula should be: Pump installation height = (Pressure inside the tower – pressure drop + atmospheric pressure – steam pressure – pressure losses) / density * g + 3.5. Base installation height of the tower = Pump installation height – 0.4
Reply #92011-07-22
Reply to 8# xue2010: The pressure above the liquid in the bottom of the tower (i.e., the bottom-of-tower pressure) is the saturated vapor pressure of the liquid at that temperature; therefore, the difference between the two values is zero.
Reply #102011-07-25
Reply to 9# janelle: Yes, thinking about it later, that’s indeed the case. The net positive suction head must not need to be corrected, right?
Reply #112011-07-27
Hg = -hf/pg – (Hs + 0.5)/p = 0 – 0.015×106/(640×9.81) – (3 + 0.5)×1000/640 = -7.86 m; therefore, the height of the skirt is 7.86 + 0.5 = 8.36 m

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