Monday, 28 March 2011

POWER REQUIRED BY PUMP


Diameter of pipe = 1 ft = 0.3048m
Cross sectional area of pipe = πD2/4
                                            = π/4 (0.3048)2
                                            = 0.07297 m2
Viscosity of Methanol = 0.5478 Kg / m.sec
Density of Methanol = 787.225 Kg/m3
Mass flow rate = 0.686 Kg/sec
Volumetric flow rate = 1.4594 m3/sec
Velocity, U = Volumetric flow rate /Area
                        = 1.4594 / 0.07297
                        = 20.0 m/sec

Reynolds number = ρud/µ
                             = 787.225*0.3048*20.0/0.5478
                            = 8760.36
If      e = 0.05
      e/d = 0.002
 R/ρu2 = 0.0278
Head losses due to friction:
      
hf = -Pf/ρg = 8(R/ρu2)(l/d)(u2/2g)
   
 = 8*0.0278*(30/0.3048)(202/2*9.81)
     = 446.60m

Z = 25m
from Table 3.2, 0.8 velocity heads are lost through each 90o bend so that the loss through two bends is 1.6 velocity head.
                = 1.6*202/(2*9.81)
                    = 32.62m
Total head = 446.60 + 25 + 32.62
                = 504.62m
Mass flow rate = 3.132 Kg/sec
So,    Theoretical Power required = Mass flow rate*Total head * g

                                          = 3.132*504.62*9.81
                                          = 15504 W
 or                                      =  15.5 KW

Net Positive Suction Head (NPSH):

From an energy balance equation, the head at the suction point of the pump is:
hi = (Poh) + x – (u2/2g) – hf

the losses in the suction pipe = 1.5 m , at (u2/2g) = hf = 1.5
the net positive suction head (NPSH) is:

NPSH = hi – (Pv/ρg)
Where
 Pv is the vapor pressure of the liquid being pumped.
Po is the vessel pressure above the liquid.
NPSH = (Po/ρh) + x – (u2/2g)  –  (Pv/ρg)

Po = 760-640 = 120mmHg = 16,000 N/m2
Pv = 760-710 = 50 mmHg = 6670 N/m2
NPSH = 3.5 -1.5 + (16,000 +6670)/(787.225*9.81)
NPSH = 4.94 m

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