Flow Measurement Devices Questions and Answers

Answer: B

A venturimeter works on Bernoulli's principle. In the converging section the velocity rises and the pressure falls, and the drop in pressure between the inlet and throat, measured by a manometer, gives the discharge. Newton's third law, Pascal's principle and Archimedes' principle are not used for measuring the flow in this way.

Answer: A

If the stagnation pressure indicated by a pitot tube equals the static pressure, then the dynamic pressure \(\frac{1}{2}\rho v^{2}\) is zero, so the velocity of the fluid is zero at that location. The stagnation pressure is the sum of the static pressure and the dynamic pressure, and the dynamic pressure is zero only when the velocity is zero. A maximum or an average velocity would give a positive dynamic pressure, and equal readings are not an error in the tube.

Answer: B

The flow velocity is 1 m/s.

The Strouhal number relates shedding frequency \(f\), body diameter \(d\) and velocity \(V\):

\[ \begin{aligned} St &= \frac{fd}{V} \\ V &= \frac{fd}{St} = \frac{10 \times 0.020}{0.2} \\ &= 1\ \text{m/s} \end{aligned} \]

The other options differ from it by factors of ten, the usual result of a unit slip with the diameter or the frequency. Keeping \(d\) in metres and \(f\) in hertz gives 1 m/s.

Answer: B

The velocity in the 0.1 m section is 1.0 m/s.

The constriction acts as a venturi meter. Continuity links the two velocities: the diameter halves, so the area falls to a quarter and \(V_{2} = 4V_{1}\). Bernoulli’s equation between the two sections then gives

\[ \begin{aligned} \Delta p &= \tfrac{1}{2}\rho(V_{2}^{2} - V_{1}^{2}) = \tfrac{1}{2}\rho(16 - 1)V_{1}^{2} \\ 7500 &= 500 \times 15 V_{1}^{2} \\ V_{1} &= 1.0\ \text{m/s} \end{aligned} \]

The throat velocity is 4.0 m/s. A value of 0.5 or 1.5 would give pressure drops of 1.9 kPa or 16.9 kPa, and 2.0 m/s would need 30 kPa.

Answer: D

The match is P - 1; Q - 3.

An orifice meter is just a thin plate with a hole clamped between flanges, so it is cheap to make and install. But the jet contracts sharply (the vena contracta) and then expands abruptly, so much of the pressure drop is never recovered: high head loss, low cost.

A venturi meter has a smooth converging section, a throat and a long gradual diffuser that recovers most of the pressure, so its permanent head loss is low; but it is long and precisely machined, so it is expensive. Options pairing the orifice with high cost or the venturi with low cost or high loss reverse these features.

Answer: B

The pressure difference is about 1.76 mm Hg.

A Pitot-static tube reads the dynamic pressure, the difference between stagnation and static pressure:

\[ \begin{aligned} \Delta p &= \tfrac{1}{2}\rho_{air}V^{2} = \tfrac{1}{2}(1.2)(20)^{2} = 240\ \text{Pa} \\ h &= \frac{\Delta p}{\rho_{Hg}g} = \frac{240}{13600 \times 10} \\ &= 1.76 \times 10^{-3}\ \text{m} = 1.76\ \text{mm} \end{aligned} \]

1760 is the same column expressed in micrometres, a units slip. 0.57 is the reciprocal of 1.76, and \(0.57 \times 10^{-5}\) compounds that with a wrong power of ten.

Answer: A

The centre-line velocity is 2 m/s.

The pitot tube reads the stagnation pressure and the static tap the static pressure. In the manometer the difference in piezometric head shows up as the 20 mm deflection of the manometric fluid; the 300 mm vertical spacing between the two openings is accounted for by the fluid columns in the connecting lines and does not appear in the result. So

\[ \begin{aligned} \frac{1}{2}\rho V^{2} &= (\rho_{m} - \rho) g h \\ &= (11000 - 1000)(10)(0.02) = 2000\ \text{Pa} \\ V &= \sqrt{\frac{2 \times 2000}{1000}} = 2\ \text{m/s} \end{aligned} \]

Using the full manometric density 11000 instead of the difference gives about 2.1 m/s, and 3, 5 and 7 m/s do not follow from a 20 mm reading with these densities.

Answer: B

The correct match is I - R; II - Q; III - P.

• Pitot probe (R): a single L-shaped tube whose open end faces the oncoming flow. The fluid is brought to rest at the opening, so the gauge reads the stagnation (total) pressure.
• Pitot-static probe (Q): two concentric tubes. The inner one is open at the nose and senses stagnation pressure; the outer one has holes in its side wall, parallel to the flow, which sense static pressure. Their difference gives the dynamic pressure and hence the velocity.
• Piezometer (P): a tapping flush with the pipe wall, which disturbs nothing and reads the static pressure only.

The options pairing P with either probe give a wall tapping the job of a probe that has to sit in the stream, and those calling Q a simple Pitot probe ignore its second, static-pressure tube.

Answer: B

The correct relation is \(C_d = C_c \times C_v\).

Actual discharge is the actual area of the jet at the vena contracta times the actual velocity there. The area is \(C_c\) times the orifice area and the velocity \(C_v\) times the theoretical velocity, so

\[ \begin{aligned} Q_{act} &= (C_c a)(C_v \sqrt{2gH}) \\ C_d &= \frac{Q_{act}}{a\sqrt{2gH}} = C_c C_v \end{aligned} \]

The coefficients multiply, so a sum as in the first option is wrong, and \(C_c = C_v/C_d\) inverts the relation (it should be \(C_c = C_d/C_v\)). The relation offered for \(C_r\) is not a standard result.

Answer: B

The coefficient of discharge is 0.50.

The coefficient of discharge is the actual discharge divided by the discharge the area and velocity would give. Convert 800 lit/s to 0.8 m³/s first.

\[ \begin{aligned} C_d &= \frac{Q}{aV} = \frac{0.8}{0.8 \times 2} \\ &= 0.50 \end{aligned} \]

0.68, 0.71 and 0.56 are values near the usual range of \(C_d\) for orifices (about 0.6 to 0.65), but they do not follow from these data.