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:

      

h_f = -∆P_f/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 = (P_o/ρ_h) + x – (u2/2g) – h_f

the losses in the suction pipe = 1.5 m , at (u²/2g) = h_f = 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 = (P_o/ρh) + x – (u²/2g)  –  (P_v/ρg)

P_o = 760-640 = 120mmHg = 16,000 N/m²

P_v = 760-710 = 50 mmHg = 6670 N/m²

NPSH = 3.5 -1.5 + (16,000 +6670)/(787.225*9.81)

NPSH = 4.94 m