Flow Measurement

Measurement of static pressure

• Wall tap / piezometer: a small hole drilled flush and normal to the wall, connected to a manometer or gauge. It reads static pressure because the streamlines at the wall are parallel to the hole and do not disturb the flow. The hole must be small and burr-free.
• Piezometer tube: a vertical open tube connected to the wall tap; the liquid height gives gauge pressure head. It is suitable only for moderate positive pressures in liquids.
• Static probe: a slender probe aligned with the flow and with small holes on its side, used inside the stream.
• Manometers and Bourdon gauges: read the pressure transmitted from the tap.

Pitot tube

A Pitot tube is a tube with an open end facing the flow, bent at right angles. The fluid at the tip is brought to rest (stagnation point), where the kinetic energy is converted into pressure.

Applying Bernoulli's equation between a point 1 upstream, where the velocity is $V$ and pressure is $p$, and the stagnation point 0, where the velocity is zero:

$$\frac{p}{\rho g} + \frac{V^2}{2g} = \frac{p_0}{\rho g}$$

So the rise of liquid in the tube above the static level equals the velocity head:

$$h = \frac{V^2}{2g} \quad \Rightarrow \quad V = \sqrt{2gh}$$

In practice a coefficient $C_v$ (0.98 to 1.0) is included: $V = C_v\sqrt{2gh}$.

Pitot-static tube

It combines the Pitot tube (stagnation pressure) and a concentric outer tube with side holes (static pressure) in one probe, and connects both to a differential manometer.

                  stagnation hole
                        |
   ---------------------+-----> flow
   ===========[inner tube]=====> p0  to manometer
   ==[outer tube, side holes]==> p   to manometer
                   o  o  static holes

$$p_0 - p = \tfrac12\rho V^2 \quad \Rightarrow \quad V = C_v\sqrt{\frac{2(p_0 - p)}{\rho}}$$

For a differential manometer with reading $h_m$ of fluid of density $\rho_m$, $p_0 - p = (\rho_m - \rho)g h_m$. Because the instrument gives point velocity, the mean velocity in a pipe is found by traversing the cross-section.

Given data

$h_w = 0.025\ \text{m}$ of water, $\rho_a = 1.2\ \text{kg/m}^3$, $\rho_w = 1000\ \text{kg/m}^3$, $C_v = 0.98$, $D = 0.4\ \text{m}$.

Pressure difference (velocity pressure)

The manometer reads the difference between stagnation and static pressures:

$$p_0 - p = \rho_w g h_w = 1000 \times 9.81 \times 0.025 = 245.25\ \text{Pa}$$

(The weight of the air column is negligible compared with water.)

Centre-line velocity

$$\begin{aligned} V_c &= C_v\sqrt{\frac{2(p_0 - p)}{\rho_a}} \\ &= 0.98\sqrt{\frac{2 \times 245.25}{1.2}} \\ &= 0.98\sqrt{408.75} = 0.98 \times 20.22 \\ &= 19.81\ \text{m/s} \end{aligned}$$

Mean velocity and discharge

$$\begin{aligned} V_{mean} &= 0.85 \times 19.81 = 16.84\ \text{m/s} \\ A &= \frac{\pi}{4}(0.4)^2 = 0.1257\ \text{m}^2 \\ Q &= A V_{mean} = 0.1257 \times 16.84 = 2.12\ \text{m}^3/\text{s} \end{aligned}$$

Answer: $V_c = 19.8\ \text{m/s}$; $Q = 2.12\ \text{m}^3/\text{s}$

(a) Venturi meter

A Venturi meter consists of a converging cone, a short throat and a diverging cone. The pressure drop between inlet (section 1) and throat (section 2) is used to find the discharge.

   1                2                
   |   converging   |  throat   diverging
 ==\__              _____              __/==
        \__________/     \____________/
   p1 ---|          |--- p2

Apply Bernoulli's equation between 1 and 2 (horizontal, ideal fluid):

$$\frac{p_1}{\rho g} + \frac{V_1^2}{2g} = \frac{p_2}{\rho g} + \frac{V_2^2}{2g}$$

Let $h = \dfrac{p_1 - p_2}{\rho g}$ (differential head). Continuity: $a_1V_1 = a_2V_2$, so $V_1 = V_2\,a_2/a_1$.

$$h = \frac{V_2^2}{2g}\left[1 - \left(\frac{a_2}{a_1}\right)^2\right] \quad \Rightarrow \quad V_2 = \frac{a_1}{\sqrt{a_1^2 - a_2^2}}\sqrt{2gh}$$

Theoretical discharge $Q_{th} = a_2V_2$. Including the coefficient of discharge $C_d$ (about 0.96 to 0.99) for losses:

$$Q = C_d\,\frac{a_1a_2}{\sqrt{a_1^2 - a_2^2}}\sqrt{2gh}$$

For a differential mercury manometer with reading $x$: $h = x\left(\dfrac{S_m}{S_f} - 1\right)$.

(b) Numerical

$$\begin{aligned} a_1 &= \frac{\pi}{4}(0.3)^2 = 0.07069\ \text{m}^2, \quad a_2 = \frac{\pi}{4}(0.15)^2 = 0.01767\ \text{m}^2 \\ h &= x\left(\frac{13.6}{1} - 1\right) = 0.2 \times 12.6 = 2.52\ \text{m of water} \end{aligned}$$ $$\begin{aligned} Q &= 0.98 \times \frac{0.07069 \times 0.01767}{\sqrt{0.07069^2 - 0.01767^2}} \times \sqrt{2 \times 9.81 \times 2.52} \\ &= 0.98 \times \frac{1.2492\times10^{-3}}{0.06844} \times 7.032 \\ &= 0.1258\ \text{m}^3/\text{s} \end{aligned}$$

Velocity at the inlet = $0.1258/0.07069 = 1.78\ \text{m/s}$, and at the throat = $7.12\ \text{m/s}$.

Answer: $Q = 0.126\ \text{m}^3/\text{s} = 126\ \text{L/s}$

Rectangular notch

Let $L$ be the crest length and $H$ the head over the crest. Consider a horizontal strip of thickness $dh$ at depth $h$ below the free surface.

   ~~~~~~~~~~~~~~~~~~~~~~   free surface
    |        ^
    |        | h   strip dh
    |  ______|_________
    | |   L           |  H
    |_|_______________|_ crest

Velocity through the strip: $\sqrt{2gh}$. Area of strip: $L\,dh$.

$$dQ = C_d\,L\,dh\,\sqrt{2gh}$$ $$Q = C_d\,L\sqrt{2g}\int_0^H h^{1/2}dh = \frac23\,C_d\,L\sqrt{2g}\;H^{3/2}$$

If the velocity of approach $V_a$ is significant, $H$ is replaced by $(H + h_a)$ with $h_a = V_a^2/2g$ and the equation becomes $Q = \tfrac23 C_d L\sqrt{2g}\,[(H + h_a)^{3/2} - h_a^{3/2}]$. For end contractions, Francis's formula uses an effective length $L - 0.1nH$.

Triangular (V) notch

Let the notch angle be $\theta$. The width of the strip at depth $h$ is $2(H - h)\tan(\theta/2)$.

$$dQ = C_d\,2(H - h)\tan\frac{\theta}{2}\,\sqrt{2gh}\,dh$$ $$Q = 2C_d\tan\frac{\theta}{2}\sqrt{2g}\int_0^H (H - h)h^{1/2}dh = \frac{8}{15}\,C_d\tan\frac{\theta}{2}\sqrt{2g}\;H^{5/2}$$

Advantage of V-notch

• For small discharges, the head $H$ is large enough to be measured accurately, since $Q \propto H^{5/2}$.
• A single notch covers a wide range of flows.
• No end contractions, and no ventilation problem.

Numerical

Rectangular weir:

$$Q = \frac23 \times 0.62 \times 4.429 \times 1.5 \times (0.25)^{1.5} = \frac23 \times 0.62 \times 4.429 \times 1.5 \times 0.125 = 0.343\ \text{m}^3/\text{s}$$

90° V-notch ($\tan 45^\circ = 1$):

$$Q = \frac{8}{15} \times 0.6 \times 1 \times 4.429 \times (0.15)^{2.5} = 0.01235\ \text{m}^3/\text{s} = 12.35\ \text{L/s}$$

Answer: rectangular weir $Q = 0.343\ \text{m}^3/\text{s}$; V-notch $Q = 12.35\ \text{L/s}$