Question 7.P.18: Petrol of sp. gravity 0.7 and kinematic viscosity of 0.417 ×......

In this case the determination off involves the velocity as the Reynolds number depends on velocity. The flow rate depends on velocity. A trial solution is necessary. So a value of f = 0.02 is first assumed.

Pressure difference = 0.95 bar or 0.95 × 10^{5} N/m². Converting the same to head of fluid,

0.95 × 10^{5}/700 × 9.81 = 13.834 m of petrol column.

13.834 = (fLu²/2gD) + (u²/2g)

= [(0.02 × 800 × u²)/(2 × 9.81 × 0.25)] + u²/2 × 9.81

= (3.26 + 0.051)u²

∴                         u = 2.045 m/s.

Now             Re = uD/v = 2.045 × 0.25/0.417 × 10^{–6} = 1.226 × 10^{6}

Ref. section 7.11, eqn 7.11.12,

f = 0.0032 + (0.221/Re^{0.237}) = 0.01117

or                        1/\sqrt{f}  = 1.8.  \log  Re  –  1.5186             ∴           f = 0.01122

so the value 0.02 is on the higher side. Now using the value 0.01117,

13.834 = [0.01117 × 800 × u²)/(2 × 9.81 × 0.25)] + [u²/(2 × 9.81)]

= 1.8727  u²

∴                    u = 2.7185 m/s, Re = 2.7185 × 0.25/0.417 × 10^{–6} = 1.63 × 10^{6}

f = 0.1065.

This is nearer the assumed value and further refinements can be made by repeating the procedure.

Flow rate                  = 2.7185 × π × 0.25²/4 = 0.1334\ \mathrm{m³/s} = 93.4\ \mathrm{kg/s}