Flow Measurement

Definitions

• Energy grade line (EGL): line showing the total head at each section: $H = \dfrac{p}{\rho g} + \dfrac{V^2}{2g} + z$.
• Hydraulic grade line (HGL): line showing the piezometric head: $\dfrac{p}{\rho g} + z$. It is the level to which water would rise in a piezometer tube.

$$\text{EGL} = \text{HGL} + \frac{V^2}{2g}$$

For a pipe of uniform diameter, the EGL and HGL are parallel and the vertical gap equals the velocity head.

Sketch (lower reservoir, pump, upper reservoir)

 head
  ^                                    upper reservoir
  |                                   ___ EGL = HGL
  |                           ,------'
  |                         ,'  <- Hp added by pump
  | ____ EGL = HGL         /
  |      `--._______      |   (EGL falls with friction,
  |  lower res.     `-.___|    HGL one V^2/2g below EGL)
  +-------------------[P]---------------------> x

Features

1. At a reservoir: velocity is nearly zero and pressure is atmospheric (gauge zero), so EGL and HGL both coincide with the free surface level.
2. Along the pipe: the EGL falls continuously because of friction losses, with slope $h_f/L = fV^2/(2gD)$. The HGL falls parallel to it for constant diameter.
3. At an entrance or valve (minor loss): the EGL drops suddenly by $K\,V^2/2g$.
4. At a pump: the pump adds head $H_p$, so both EGL and HGL rise suddenly by $H_p$ across it.
5. At a turbine: the turbine extracts head $H_t$, so both lines fall suddenly by $H_t$.
6. At a change of diameter: the gap between EGL and HGL changes, as $V^2/2g$ changes.
7. At a free discharge: the HGL passes through the centre of the outlet (pressure atmospheric) and the EGL lies a velocity head above it.
8. Pressure: the pressure head at any point is the vertical distance between the pipe centre line and the HGL. If the pipe is above the HGL, pressure is below atmospheric (negative); if it falls to vapour pressure, cavitation or siphon breaking can occur.

Given data

$Q = 0.06\ \text{m}^3/\text{s}$, $D = 0.2\ \text{m}$, $f = 0.02$, $L_s = 20\ \text{m}$, $L_d = 600\ \text{m}$, static lift $z = 30\ \text{m}$, $\sum K = 3$ (minor), exit $K = 1$, $\eta = 0.70$.

Velocity and velocity head

$$\begin{aligned} A &= \frac{\pi}{4}(0.2)^2 = 0.03142\ \text{m}^2 \\ V &= \frac{Q}{A} = \frac{0.06}{0.03142} = 1.910\ \text{m/s} \\ \frac{V^2}{2g} &= \frac{1.910^2}{19.62} = 0.1859\ \text{m} \end{aligned}$$

Energy equation between the two reservoir surfaces

Both surfaces are at atmospheric pressure and velocity is nearly zero:

$$0 + 0 + H_p = z + h_f + h_{minor} + h_{exit}$$

Friction loss (total length 620 m):

$$h_f = f\frac{L}{D}\frac{V^2}{2g} = 0.02 \times \frac{620}{0.2} \times 0.1859 = 11.53\ \text{m}$$

Minor losses: $3 \times 0.1859 = 0.558\ \text{m}$

Exit loss: $1 \times 0.1859 = 0.186\ \text{m}$

Head developed by the pump

$$H_p = 30 + 11.53 + 0.558 + 0.186 = 42.27\ \text{m}$$

Power

Hydraulic (water) power:

$$P_w = \rho g Q H_p = 1000 \times 9.81 \times 0.06 \times 42.27 = 24\,880\ \text{W} = 24.9\ \text{kW}$$

Shaft power:

$$P_s = \frac{P_w}{\eta} = \frac{24.88}{0.70} = 35.5\ \text{kW}$$

Answer: $H_p = 42.3\ \text{m}$; hydraulic power $= 24.9\ \text{kW}$; shaft power $= 35.5\ \text{kW}$

Rules for pipe combinations

• Series: the same discharge flows in each pipe, and the head losses add.
• Parallel: the head loss between the two junctions is the same in each branch, and the discharges add.

Loss coefficients

The Darcy-Weisbach equation in terms of discharge:

$$h_f = \frac{8 f L}{\pi^2 g D^5}Q^2 = K Q^2$$ $$\begin{aligned} K_1 &= \frac{8 \times 0.02 \times 800}{\pi^2 \times 9.81 \times (0.3)^5} = 544.0\ \text{s}^2/\text{m}^5 \\ K_2 &= \frac{8 \times 0.025 \times 600}{\pi^2 \times 9.81 \times (0.2)^5} = 3873\ \text{s}^2/\text{m}^5 \\ K_3 &= \frac{8 \times 0.025 \times 900}{\pi^2 \times 9.81 \times (0.15)^5} = 24\,482\ \text{s}^2/\text{m}^5 \end{aligned}$$

Equivalent of the parallel pipes

For equal head loss $h_p = K_2Q_2^2 = K_3Q_3^2$, $Q = Q_2 + Q_3$:

$$\frac{1}{\sqrt{K_{eq}}} = \frac{1}{\sqrt{K_2}} + \frac{1}{\sqrt{K_3}} = 0.016068 + 0.006391 = 0.022459$$ $$K_{eq} = 1982.5\ \text{s}^2/\text{m}^5$$

Total discharge

The total head available is $120 - 90 = 30\ \text{m}$:

$$30 = (K_1 + K_{eq})Q^2 = (544.0 + 1982.5)Q^2$$ $$Q = \sqrt{\frac{30}{2526.5}} = 0.1090\ \text{m}^3/\text{s}$$

Head losses and branch flows

$$\begin{aligned} h_1 &= 544.0 \times (0.1090)^2 = 6.46\ \text{m} \\ h_p &= 30 - 6.46 = 23.54\ \text{m} \end{aligned}$$ $$Q_2 = \sqrt{\frac{23.54}{3873}} = 0.0780\ \text{m}^3/\text{s}, \qquad Q_3 = \sqrt{\frac{23.54}{24\,482}} = 0.0310\ \text{m}^3/\text{s}$$

Check: $Q_2 + Q_3 = 0.0780 + 0.0310 = 0.1090\ \text{m}^3/\text{s}$. This agrees.

Answer: $Q = 0.109\ \text{m}^3/\text{s}$; $Q_2 = 0.078\ \text{m}^3/\text{s}$; $Q_3 = 0.031\ \text{m}^3/\text{s}$