Question 11.2: Oil of kinematic viscosity 10^-5 m²/s flows at a steady rate...
Oil of kinematic viscosity 10^{-5} m ^{2}/s flows at a steady rate through a cast iron pipe of 100 mm diameter and of 0.25 mm average surface roughness.
If the loss of head over a pipe length of 120 m is 5 m of the oil, what is the flow rate through the pipe?
Since the velocity is unknown, Re is unknown. Relative rougness e/D = 0.25/100 = 0.0025.
A guess of the friction factor at this relative roughness is made from Fig. 11.2 as f = 0.026
Then Eq. (11.6b) gives a first trial
h_{f}=\frac{\Delta p^{*}}{\rho g}=f \frac{L}{D}\left(V^{2} / 2 g\right) (11.6b)
5=0.026 \frac{120}{0.10} \frac{V^{2}}{2 \times 9.81}
when, V = 1.773 m/s
Then, \operatorname{Re}=\frac{1.773 \times 0.10}{10^{-5}}=1.773 \times 10^{4}
The value of Re, with \varepsilon / D as 0.0025, gives f = 0.0316 (Fig. 11.2). The second step of iteration involves a recalcualtion of V with f = 0.0316, as
5=0.0316 \times \frac{120}{0.1} \frac{V^{2}}{2 \times 9.81}
which gives V = 1.608 m/s
and \operatorname{Re}=\frac{1.608 \times 0.10}{10^{-5}}=1.608 \times 10^{4}
The value of f at this Re (Fig. 11.2) becomes 0.0318. The relative change between the two successive values of f is 0.63% which is insignificant. Hence the value of V = 1.608 m/s is accepted as the final value.
Therefore, the flow rate Q = 1.608 \times(\pi / 4) \times(0.10)^{2} = 0.013 m ^{3}/s
