Kerosene of sp.gravity 0.85 flows through a verturimeter as shown in Fig. 2.22 with flow rate of 0.05 m^3/s. Determine the difference of pressure between inlet and throat of the venturimeter. The diameter of the pipe is D1 = 0.1 m and diameter of the throat D2 = 0.06 m.

Question

Kerosene of sp.gravity 0.85 flows through a verturimeter as shown in Fig. 2.22 with flow rate of [latex]0.05 {m^{3}}/{s}[/latex]. Determine the difference of pressure between inlet and throat of the venturimeter. The diameter of the pipe is [latex]D_{1} = 0.1 m[/latex] and diameter of the throat [latex]D_{2} = 0.06 m[/latex].

Accepted Answer

The flow is considered inviscid steady and incompressible and the relation between
flow rate and pressure difference is given by eq. 2.24

[latex]Q = a_{2} imes \sqrt{\frac{2\left(P_{1} - P_{2} ight) }{ ho \left(1 - \left({a_{2}}/{a_{1}} ight)^{2} ight) } }[/latex]

[latex] herefore P_{1}-P_{2} =\frac{Q^{2} ho \left[1-\left\lgroup\frac{a_{2}}{a_{1}} ight group ^{2} ight] }{2a^{2}_{2}}[/latex]

The density of kerosene [latex] ho = specitic gravity imes density of water[/latex]

[latex] = 0.85 imes 1000 = 850 {kg}/{m^{3}}[/latex]

Substituting proper

[latex] herefore P_{1}-P_{2} =\frac{\left(0.05 ight)^{2}\left(850 ight)\left[1 - \left\lgroup\frac{0.06}{0.1} ight group^{4} ight] }{2\left[\left\lgroup\frac{\pi }{4} ight group\left(0.06 ight)^{2} ight]^{2} } = 1.16 imes 10^{5} {N}/{m^{2}}[/latex]

[latex] herefore P_{1}-P_{2} =116 kPa[/latex]

Question 2.11: Kerosene of sp.gravity 0.85 flows through a verturimeter as ...

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