Question

You must install a centrigual pump to transfer a volatile liquid from a removal tank .If the liquid has a vapor pressure of 30 psia , the pressures in the tank is 30 psia , the level in the tank is 40

-ft above the pump inlet and the required pump NPSH is 20-ft ,what is the maximum length of the pump section side of the pump without the possibility of cavitation ? The line is 2-in schedule-70 commercial steel , the flowrate is 75 (gallon/min ) , and the fluid properties are (density :45 Ibm/ft^3 )and ( dynamic viscosity :5 cp ) .Use Colebrook equation if necessary.

D:2.067 in , A:0.0233 ft^2

Answer 1

The NPSH available at the pump suction (NPSHA) in head form is given by the below equation

NPSHA (ft) = pressures in the tank (ft) + level in the tank above the pump inlet (ft) - line pressure drop (ft) - vapor pressure of fluid(ft)....... eqn (1)

Since pressures in the tank = vapor pressure of fluid = 30 psia, (1) of NPSHA can be modified to

NPSHA (ft) = level in the tank above the pump inlet (ft) - line pressure drop (ft) ....... eqn (2)

Now as per data given

level in the tank above the pump inlet (ft) = 40 ft

line pressure drop (ft) = $\Delta$H ft

Hence from (2)

NPSHA (ft) = 40 ft - $\Delta$H (ft) .............eqn (3)

Definition of NPSHR

NPSHR is the NPSH required by pump vendor based on the pump model, impeller diameter and pumping flow rates.

It is the minimum NPSH that should be available at the pump inlet so that the pump will NOT cavitate.

Hence for any centrifugal pump to prevent any cavitation, the minimum requirement is

NPSHR = NPSHA ..... eqn (4)

As per data NPSHR = 20 ft

Based on eqn (3) and eqn (4)

20 (ft) = 40 ft - $\Delta$H (ft)

Hence $\Delta$H (ft) = 20 ft

It implies that the maximum length of the pump suction pipe will be limited to a $\Delta$H of 20 ft to avoid any pump cavitation.

Line pressure drop calculation

The line pressure drop can be evaluated by darchys equation (head form) as stated below

$\Delta$​​​​​​​H (ft) = 4* f* L* V² / (2*g*D) ......... eqn (5)

where

f = friction factor of the pipe (no unit).

L = maximum length of the pump suction pipe (ft)

V = velocity in the pipe (ft/s)

g = 32.2 ft/s²

D = Pipe internal diameter = 2.067 in = 2.067 in / 12 in/ft = 0.173 ft

a) Estimate velocity in pipe

Now velocity in pipe V (ft/s) = Q (ft³/s) / A (ft²) where Q (ft³/s) is volumetric flow rate through the pump suction pipe

By data we have Q (USGPM) = 75 USGPM & A = 0.0233 ft²

Hence Q (ft³/s) = 75 USGPM * 0.0022 ft³/s / USGPM = 0.165 ft³/s

Substituting we have

V (ft/s) = 0.165 ft³/s / 0.0233 ft^{2 =} 7.08 ft/s

b) Estimation of friction factor

The friction factor can be estimated from Moodys chart and is a replicate of colebrook equation. Reynolds number and $\epsilon$ / D are the parameters required for determining the friction factor through Moodys chart. Moody chart will be available in any standard literature.

Assumption: For a commercial steel pipe $\epsilon$ is assumed to be 0.00015 ft. Hence $\epsilon$ = 0.00015 ft

$\epsilon$ / D = 0.00015 ft / 0.173 ft = 8.67 * 10^-4

Reynolds number is defined as N_re = D * V * Rho / $\mu$ where

D (m) = Internal diameter of the pipe = 0.173 ft * 0.3048 m / ft= 0.053 m

V (m/s) = Velocity of fluid flow through the pipe = 7.08 ft/s * 0.3048 m / ft= 2.16 m/s

Rho (kg/m³) = fluid density = 45 lb/ft³ * (0.454/0.3048^3) kg/m³ / lb/ft³ = 721.5 kg/m³

$\mu$ = absolute viscocity in Pa-s = 5 cP* 10^-3 Pa-s/cP = 5 * 10^-3 Pa-s

Substituting we have

N_re = D * V * Rho / $\mu$ = 0.053 m * 2.16 m/s * 721.5 kg/m³ / 5 * 10^-3 Pa-s = 16520

From Moodys chart based on $\epsilon$ / D = 8.67 * 10^-4 & N_re = 16520 ; f = 0.007

Substituting the above values in eqn (5) we have

$\Delta$​​​​​​​H (ft) = 4* f* L* V² / (2*g*D)

20 ft = 4* 0.007 * L ft * 7.08^{^2} ft²/s² / (2*32.2 ft/s²*0.173 ft)

Hence maximum length = L ft = 158 ft