Problem Statement
Known
- Water flows through a pipe reducer.
- The static pressures at positions (1) and (2) are measured by the inverted U-tube manometer containing oil of specific gravity, SG, less than one.
Find
Determine the volumetric flow rate \(Q\) for a manometer reading \(h\).
Analysis
To relate \(p_1\) and \(p_2\) through the manometer:
\[ p_1-\gamma(z_2-z_1)-\gamma(h+\ell)+\mathrm{SG}\,\gamma\,h+\gamma\ell=p_2 \] \[ p_1-p_2=\gamma(z_2-z_1)+\gamma h(1-\mathrm{SG}) \qquad (1) \]Apply Bernoulli Equation along streamline from point 1 to point 2.
\[ p_1+\gamma z_1+\frac12\rho V_1^2=p_2+\gamma z_2+\frac12\rho V_2^2 \] \[ p_1-p_2=\gamma(z_2-z_1)+\frac12\rho V_2^2\left[1-\left(\frac{V_1}{V_2}\right)^2\right] \]The conservation of mass gives:
\[ \dot{m}_1=\dot{m}_2 \Rightarrow \rho Q_1=\rho Q_2 \Rightarrow Q_1=Q_2 \Rightarrow V_1A_1=V_2A_2 \] \[ V_1\frac{\pi}{4}D_1^2=V_2\frac{\pi}{4}D_2^2 \] \[ \frac{V_1}{V_2}=\left(\frac{D_2}{D_1}\right)^2 \] \[ p_1-p_2=\gamma(z_2-z_1)+\frac12\rho V_2^2\left[1-\left(\frac{D_2}{D_1}\right)^4\right] \qquad (2) \]The comparison between eq (1) and eq (2) gives
\[ \gamma h(1-\mathrm{SG})=\frac12\rho V_2^2\left[1-\left(\frac{D_2}{D_1}\right)^4\right] \] \[ V_2=\frac{Q}{A_2}=\frac{Q}{\frac{\pi}{4}D_2^2}=\frac{4Q}{\pi D_2^2} \] \[ Q=\left(\frac{\pi^2D_2^4gh(1-\mathrm{SG})}{8\left[1-\left(\frac{D_2}{D_1}\right)^4\right]}\right)^{1/2} \]Comment: The term \(z_2-z_1\) is not involved in the relationship between \(Q\) and \(h\).
Inputs & Outputs
Flow rate, \(Q\)
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Downstream speed, \(V_2\)
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Upstream speed, \(V_1\)
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