Calculation of friction losses, power, developed head and available Net Positive Suction Head of a pump

2. In the spreadsheet "Calculations" a. Insert the parameter values in the cells C21-C40 (yellow cells)

- If there are not enlargements or contractions in the system, set enlargement and/or contraction coefficients in the cells C27 and C28 equal to 0 b. Turn the "SWITCH" ON (set the value in the cell G1 (red cell) equal to 1 and press ENTER) c. Iterate (by pressing F9) until the value of " f " in the cell H17 is constant d. To recalculate for different parameter values turn the "SWITCH" OFF (set the value in the cell G1 equal to 0 and press ENTER). Repeat steps 2a to 2c

Equations that are being used in the calculations

Calculation of the Fanning friction factor " f " In laminar flow, by using the equation:

In turbulent flow, by using the Colebrook equation:

Calculation of the friction losses In straight pipes

In sudden enlargement

In sudden contraction

Available Net Positive Suction Head

1. On the Tools menu, click Options, then click the Calculation tab and tick on the Iteration

- If the pump is below the liquid level in the suction tank, z1 (cell C38) is positive. If the pump is above the liquid level in the suction tank, z1 is negative

1

√ f=−4 log10( ε /D3 .7

+1 .255Re√ f )

f=16Re

hs=ΔPρ

=4 fLDv2

2

NPSH a=P−pvρg

+z1−ΣF sg

hc=0 . 55(1−A2

A1)2 v2

2

2α

he=(1− A1

A2)2 v1

2

2α

- If there are not enlargements or contractions in the system, set enlargement and/or contraction coefficients in the cells C27 and C28 equal to 0 b. Turn the "SWITCH" ON (set the value in the cell G1 (red cell) equal to 1 and press ENTER)

d. To recalculate for different parameter values turn the "SWITCH" OFF (set the value in the cell G1 equal to 0 and press ENTER). Repeat steps 2a to 2c

tab and tick on the Iteration box. On Maximum iterations box write 1000.

(cell C38) is positive. If the pump is above the liquid level in the suction tank, z1 is negative

Calculation of friction losses,required power, developed head and available NPSH of a pump

Input dataFlow rate Q = 20Inside diameter Di = 0.042 mPipe length L = 12 mSuction pipe length Ls= 2 m

Density den = 1050Viscosity visc= 0.002 Pas

Enlargement-loss coefficient Ke = 0

Contraction-loss coefficient Kc = 0.55

Valves-loss equivalent length Le/D = 300

Fittings-loss equvalent length Le/D = 32

Number of valves Nv = 1

Nf = 3

Rafness of the pipe raf = 0 m

Velocity at point 1 0 m/s

Pressure at point 1 101325 Pa

Pressure at point 2 101325 Pa

Vapor pressure in the suction line 12349 Pa

Distance of point 1 from reference level -2

Distance of point 2 from reference level 8Efficiency of the pump n= 70%

m3/h

kg/m3

Number of fittings (90o elbows)

v1=

P1=

P2=

pv =

z1*=

z2=

* If the pump is above the liquid level in the suction tank, z1 is negative

Ws

z1

, P2v2

z2

x 2

Level of reference, P1v1

x1

SWITCH= 1

Derived dataMass flow rate, = 5.8333 kg/s

Cross sectional area A = 0.00139

Mean velocity at point 2 4.010 m/sReynolds number Re = 88419Relative rafness rr = 0

Corection factor a= 1

ResultsFriction factor

f = 0.0047

14.6 14.7

Friction losses Enlargement losses Fe = 0.00 J/kg Contraction losses Fc = 4.42 J/kg Valve-equivalent length LeV = 12.60 m Fitting-equivalent length LeF = 4.03 m

Straight tube losses Fs = 103.26 J/kgTotal friction losses Ft = 107.68 J/kg

Energy and Power added to the fluid by the pump

213.8 J/kg

1247.3 W

Required Power (brake power)

1781.8 W

Developed Head H = H = 21.8 m

Available NPSH Friction losses in the suction line

-Contraction losses Fcs = 4.42 J/kg

-Fitting-equivalent length LeFs = 1.34 m -Straight tube losses Fss = 12.06 J/kg Total friction losses in the suction line Fts = 17.83 J/kg

NPSHa = 4.8 m

m2

vm2 =

-ws =

-Ws =

-Wsr = -Wsr =

m