Question

Oil with a density of 850kg/m3850 kg/m^3 and kinematic viscos­ity of 0.00062m2/s0.00062 m^2/s is being discharged by a 8-mm-diameter, 40-m-long horizontal pipe from a storage tank open to the atmosphere. The height of the liquid level above the center of the pipe is 4 m. Disregarding the minor losses, determine the flow rate of oil through the pipe.

Solution

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Step 1

1 of 4

Given:\textbf{Given:}

ρ=850kgm3\rho = 850 \dfrac{kg}{m^3}

ν=62×105m2s\nu = 62 \times 10^{-5} \dfrac{m^2}{s}

D=0.008D = 0.008 mm

L=40L = 40 mm

h=4h = 4 mm

Approach:\textbf{Approach:}

We have steady and incompressible flow. The entrance effects are negligible, so the flow is fully developed. The entrance and exit loses are alos negligible.

First step is to calculate pressure at the bottom of the tank:

P1,gage=ρgh=8509.814=33.354kPa\begin{align*} P_{1,gage} &= \rho g h = 850 \cdot 9.81 \cdot 4\\ &= 33.354 kPa \end{align*}

Disregarding inlet and outlet losses, the pressure drop across the pipe is:

ΔP=P1P2=P1Patm=P1,gage=33.354Pa\begin{align*} \Delta P &= P_1 - P_2 = P_1 - P_{atm} = P_{1,gage}\\ &=\boxed{33.354 Pa}\\ \end{align*}

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