Water at 20^◦ C flows in a 0.7-m-diameter horizontal pipe. An orifice meter with a 0.3-m diameter is mounted in the flanged pipe just upstream of point 2 as illustrated in Figure EP 5.16 in order to measure the flowrate in the pipe. A pressure tap, p1 is installed at point 1, where the full pipe

Question

Water at [latex] 20^{\circ} C[/latex] flows in a 0.7-m-diameter horizontal pipe. An orifice meter with a 0.3-m diameter is mounted in the flanged pipe just upstream of point 2 as illustrated in Figure EP 5.16 in order to measure the flowrate in the pipe. A pressure tap, [latex] p_{1} [/latex] is installed at point 1, where the full pipe flow occurs just upstream of the orifice, and a pressure tap, [latex] p_{2} [/latex] is installed at point 2, where the vena contracta occurs just downstream of the orifice. The pressure at point 1, [latex] p_{1} [/latex] is measured to be [latex] 3× 10^{5} N/m^{2} [/latex] p_{1} [/latex] , and the pressure at point 2, [latex] p_{2} [/latex] is measured to be [latex] 2.5 × 10^{5} N/m^{2}[/latex] . The orifice discharge coefficient, [latex] C_{o} [/latex] is 0.6. (a) Determine the ideal flowrate. (b) Determine the actual flowrate.

Accepted Answer

(a) In order to determine the ideal velocity at either point 1 or point 2, the Bernoulli equation is applied between points 1 and 2. However, in order to determine the relationship between the ideal velocities at points 1 and 2, the continuity equation is applied between points 1 and 2; thus, we have two equations and two unknowns. Furthermore, in order to determine the ideal flowrate in the pipe, the continuity equation is applied at either points 1 or 2. And, finally, the diameter of flow at point 2 is assumed to be equal to the diameter of the orifice meter, because the vena contracta occurs a point 2.

[latex]D_{1}: = 0.7 m[/latex] [latex]D_{o}: = 0.3 m[/latex] [latex]D_{2}: = D_{o} = 0.3 m[/latex]

[latex]A_{1}: = \frac{\pi .D^{2}_{1}}{4} = 0.385 m^{2}[/latex] [latex]A_{2}: = \frac{\pi .D^{2}_{2}}{4} = 0.071 m^{2}[/latex]

[latex]P_{1}: = 3 imes 10^{5} \frac{N}{m^{3}} [/latex] [latex]P_{2}: = 2.5 imes 10^{5} \frac{N}{m^{2}} [/latex] [latex]Z_{1}: = 0 m[/latex] [latex]Z_{2}: = 0 m[/latex]

[latex] ho : = 998 \frac{kg}{m^{3}} [/latex] [latex]g: = 9.81 \frac{m}{sec^{2}} [/latex] [latex]\gamma : = ho .g = 9.79 imes 10^{3} \frac{kg}{m^{2}s^{2}} [/latex]

Guess value: [latex]V_{1}: = 1 \frac{m}{sec} [/latex] [latex]V_{2}: = 1 \frac{m}{sec} [/latex] [latex]Q: = 1 \frac{m^{3}}{sec} [/latex]

Given

[latex]\frac{P_{1}}{\gamma } + Z_{1} + \frac{V^{2}_{1}}{2. g} = \frac{P_{2}}{\gamma } + Z_{2} + \frac{V^{2}_{2}}{2. g}[/latex]
[latex] V_{1} .A_{1} = V_{2} .A_{2} [/latex]
[latex] Q = V_{2} .A_{2} [/latex]
[latex] \left ( \begin{matrix} V_{1} \ V_{2} \ Q \end{matrix} ight ) : = Find (V_{1}, V_{2}, Q ) [/latex]

[latex]V_{1}: = 1.87 \frac{m}{s} [/latex] [latex]V_{2}: = 10.183 \frac{m}{s} [/latex] [latex]Q: = 0.72 \frac{m^{3}}{s} [/latex]

(b) The actual flowrate, Q a is computed from Equation 5.51a [latex] Q_{a} = \frac{\sqrt{\frac{2}{ ho }\left(\Delta p - \frac{ au _{w}L}{R_{h}} ight) } }{\sqrt{\frac{2\Delta p}{ ho } } } \frac{A_{a}}{A_{i}}Q_{i} = C_{d} Q_{i} [/latex] as follows:

[latex] Q_{i}: = Q = 0. 72 \frac{m^{3}}{s} [/latex] [latex] C_{o}: = 0.6 [/latex] [latex] Q_{a}: = C_{o}. Q_{i} = 0.432 \frac{m^{3}}{s} [/latex]

Question 5.16: Water at 20^◦ C flows in a 0.7-m-diameter horizontal pipe. A...

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