Flow measurement

The figure values are only partly readable, so the following reading is assumed: the pipe diameters are $D_1 = 15$ cm and $D_2 = 9$ cm (throat); the piezometer at the inlet stands 1 cm above the pipe axis ($p_1/\gamma = 0.01$ m), and the piezometer difference between the inlet and the throat is 5 cm of water. The venturimeter is horizontal and the throat is connected by a tube to a reservoir open to the atmosphere.

(i) Velocity in the smaller diameter section (throat)

Bernoulli between 1 and 2 (horizontal, no loss) and continuity $A_1V_1 = A_2V_2$:

$$\frac{p_1 - p_2}{\gamma} = \frac{V_2^2 - V_1^2}{2g} = \frac{V_2^2}{2g}\left[1 - \left(\frac{D_2}{D_1}\right)^4\right]$$ $$\begin{aligned} \left(\frac{9}{15}\right)^4 &= 0.1296\\ V_2 &= \sqrt{\frac{2g\,(0.05)}{1 - 0.1296}} = \sqrt{\frac{0.981}{0.8704}} = 1.062\ \text{m/s} \end{aligned}$$

Flow $Q = A_2V_2 = \dfrac{\pi}{4}(0.09)^2(1.062) = 6.76\times 10^{-3}$ m³/s.

(ii) Maximum height h lifted from the reservoir

Pressure head at the throat:

$$\frac{p_2}{\gamma} = \frac{p_1}{\gamma} - 0.05 = 0.01 - 0.05 = -0.04\ \text{m (gauge)}$$

The fluid in the lifting tube is at rest, so the throat pressure just balances a column of fluid of height $h$ above the reservoir surface (open to atmosphere): $p_2 = -\gamma h$.

$$h_{max} = \left|\frac{p_2}{\gamma}\right| = 0.04\ \text{m} = 4\ \text{cm}$$

Fluid rises up the tube until it reaches the height at which the throat suction is balanced.

   inlet 1          throat 2
 ----------\      /-----------
            \____/
                |   <- tube, h = 4 cm lift
 ~~~~~~~~~~~~~~~|~~~ reservoir (open to air)

Answer: (i) $V_2 \approx 1.06$ m/s (so $Q \approx 6.8$ L/s); (ii) $h_{max} = 0.04$ m. (Method: $V_2 = \sqrt{2g\Delta h/(1-(D_2/D_1)^4)}$ and $h = -p_2/\gamma$.)

Given: orifice $b = 0.30$ m wide, $d = 0.10$ m high, bottom edge 5 cm above the bed, so the top edge is 0.15 m above the bed. Upstream depth $H_1 = 0.300$ m, downstream depth $H_2 = 0.225$ m, $C_d = 0.63$.

Both depths are above the top edge of the orifice (0.15 m), so the orifice is fully submerged (drowned): the discharge is governed by the difference of the water levels, not by the depth below the surface.

 upstream 0.300 m  |  |  downstream 0.225 m
 ~~~~~~~~~~~~~~~~~~|  |~~~~~~~~
                   |__|  orifice  (0.15 m top)
                   |  |  (0.05 m bottom)
 ------------------------ bed

Head difference:

$$H = H_1 - H_2 = 0.300 - 0.225 = 0.075\ \text{m}$$

Area of orifice $a = 0.30\times 0.10 = 0.03\ \text{m}^2$.

$$\begin{aligned} Q &= C_d\,a\sqrt{2gH}\\ &= 0.63\times 0.03\times\sqrt{2\times 9.81\times 0.075}\\ &= 0.63\times 0.03\times 1.2130\\ &= 0.02293\ \text{m}^3/\text{s} \end{aligned}$$

The velocity of approach is small (it is $0.153$ m/s if the canal is 0.5 m wide), so it is neglected. The canal width does not enter the calculation.

Answer: $Q \approx 0.0229$ m³/s (about 22.9 L/s).

Given: tank $D = 2$ m, so $A = \dfrac{\pi}{4}(2)^2 = 3.1416$ m²; orifice $d = 0.1$ m, $a = \dfrac{\pi}{4}(0.1)^2 = 0.007854$ m²; $h_0 = 3$ m; air pressure 450 kPa absolute, atmospheric 100 kPa; $C_d = 1$ (no friction); air pressure kept constant.

Pressure head of air above the water (gauge):

$$h_p = \frac{450 - 100}{9.81} = 35.678\ \text{m}$$

The effective head at the orifice is $h + h_p$, so the exit velocity is $V = \sqrt{2g(h + h_p)}$.

Rate equation: $-A\dfrac{dh}{dt} = a\sqrt{2g(h + h_p)}$.

$$t = \frac{A}{a\sqrt{2g}}\int_{h_2}^{h_1}\frac{dh}{\sqrt{h + h_p}} = \frac{2A}{a\sqrt{2g}}\left[\sqrt{h_1 + h_p} - \sqrt{h_2 + h_p}\right]$$ $$\frac{a\sqrt{2g}}{2A} = \frac{0.007854\times 4.4294}{2\times 3.1416} = 0.0055368\ \text{m}^{1/2}/\text{s}$$

(i) Time for half the water to discharge

The water level falls from 3 m to 1.5 m (half the volume, since the tank is cylindrical).

$$\begin{aligned} \sqrt{3 + 35.678} &= 6.2192,\qquad \sqrt{1.5 + 35.678} = 6.0974\\ t &= \frac{6.2192 - 6.0974}{0.0055368} = 22.0\ \text{s} \end{aligned}$$

(ii) Level after 10 s

$$\begin{aligned} \sqrt{h + h_p} &= 6.2192 - 0.0055368\times 10 = 6.1638\\ h + h_p &= 37.992\\ h &= 37.992 - 35.678 = 2.31\ \text{m} \end{aligned}$$

Answer: (i) $t \approx 22.0$ s; (ii) water level after 10 s $\approx 2.31$ m above the outlet.

A submerged orifice discharges below the surface of the liquid in the downstream tank. It may be fully or partially submerged.

Fully submerged orifice

Let the upstream and downstream liquid surfaces be at heights $H_1$ and $H_2$ above the centre of the orifice, $H = H_1 - H_2$. Apply Bernoulli between a point in the upstream tank (velocity negligible) and the vena contracta, where the pressure is the downstream hydrostatic pressure $\gamma H_2$:

$$H_1 + 0 = H_2 + \frac{V_c^2}{2g}\;\Rightarrow\; V_c = \sqrt{2g\,(H_1 - H_2)} = \sqrt{2gH}$$

With area of the orifice $a$ and coefficient of contraction $C_c$, $Q = C_c a V_c$. Including the velocity coefficient $C_v$: $C_d = C_cC_v$.

$$Q = C_d\,a\sqrt{2g\,(H_1 - H_2)}$$

The discharge depends only on the head difference.

Partially submerged orifice

The downstream level lies between the top and bottom edges. The flow is the sum of two parts.

Let $b$ be the width, $H_1$ and $H_2$ the heads from the upstream surface to the top and bottom edges, and $H$ the difference between the upstream and downstream levels ($H_1 < H < H_2$).

• Free part (top edge to the downstream level): the pressure outside is atmospheric, so it behaves like a free rectangular orifice with head varying from $H_1$ to $H$:

$$Q_1 = \frac{2}{3}C_d b\sqrt{2g}\left[H^{3/2} - H_1^{3/2}\right]$$

• Submerged part (downstream level to bottom edge): height $(H_2 - H)$ under the constant head difference $H$:

$$Q_2 = C_d\,b\,(H_2 - H)\sqrt{2gH}$$

Total discharge:

$$Q = \frac{2}{3}C_d b\sqrt{2g}\left[H^{3/2} - H_1^{3/2}\right] + C_d b\,(H_2 - H)\sqrt{2gH}$$

 upstream                downstream
 ~~~~~~~~~~|  |  H2  ~~~~~~~~ (lower level)
   H1 ---> |__|  free part
           |__|  submerged part
   H2 --->

Slope of Cipolletti weir is 1 horizontal : 4 vertical

A Cipolletti weir is a trapezoidal weir whose side slopes are chosen so that the extra discharge through the two triangular ends compensates for the reduction in discharge caused by end contractions on a rectangular weir.

Let $L$ be the crest length and $H$ the head, and $\theta$ the side slope angle with the vertical.

Rectangular weir with two end contractions (Francis):

$$Q_1 = \frac{2}{3}C_d\sqrt{2g}\,(L - 0.1\times 2H)H^{3/2}$$

Rectangular weir without end contractions:

$$Q_0 = \frac{2}{3}C_d\sqrt{2g}\,LH^{3/2}$$

so the loss due to contractions is

$$Q_0 - Q_1 = \frac{2}{3}C_d\sqrt{2g}\,(0.2H)H^{3/2}$$

Discharge through the two triangular ends (together forming one triangular notch of half-angle $\theta$):

$$Q_2 = \frac{8}{15}C_d\sqrt{2g}\tan\theta\,H^{5/2}$$

For Cipolletti, $Q_2 = Q_0 - Q_1$:

$$\frac{8}{15}\tan\theta\,H^{5/2} = \frac{2}{3}(0.2)H^{5/2}\;\Rightarrow\;\tan\theta = \frac{0.1333\times 15}{8} = 0.25$$

So $\tan\theta = \dfrac{1}{4}$: 1 horizontal to 4 vertical ($\theta = 14.04^\circ$ from the vertical).

The discharge is then $Q = \dfrac{2}{3}C_d L\sqrt{2g}\,H^{3/2}$ (with $L$ the base length), without contraction correction.

Velocity of approach

The velocity of approach $V_a = Q/A_{channel}$ adds a velocity head $h_a = V_a^2/2g$ to the head over the weir. The effective head is $H + h_a$:

$$Q = \frac{2}{3}C_dL\sqrt{2g}\left[(H + h_a)^{3/2} - h_a^{3/2}\right]$$

(for a triangular notch, $Q = \dfrac{8}{15}C_d\tan\dfrac{\theta}{2}\sqrt{2g}\left[(H + h_a)^{5/2} - h_a^{5/2}\right]$). Since $V_a$ depends on $Q$, solve by trial: first find $Q$ ignoring $h_a$, then $V_a = Q/A$, then $h_a$, then recompute $Q$ until it converges. For $h_a$ small (large channel), use the Francis form $Q = \tfrac23 C_d L\sqrt{2g}\,H^{3/2}$.

Cippoletti weir sides have slope 1:4

A Cippoletti weir is a trapezoidal weir whose side slope makes the extra flow through the triangular ends equal the loss from end contractions of a rectangular weir. For crest length $L$ and head $H$:

• Rectangular weir with 2 end contractions (Francis): $Q_1 = \frac{2}{3}C_d\sqrt{2g}\,(L - 0.2H)H^{3/2}$
• Loss due to contractions: $\Delta Q = \frac{2}{3}C_d\sqrt{2g}\,(0.2H)H^{3/2}$
• Two triangular ends (slope $\tan\theta$ from the vertical): $Q_2 = \frac{8}{15}C_d\sqrt{2g}\tan\theta\,H^{5/2}$

Equate $Q_2 = \Delta Q$:

$$\frac{8}{15}\tan\theta = \frac{2}{3}(0.2)\;\Rightarrow\;\tan\theta = 0.25 = \frac{1}{4}$$

So the side slope is 1 horizontal : 4 vertical.

Discharge formula for the trapezoidal notch (from first principles)

Base width $b = 0.1$ m, top width $0.5$ m, depth $0.3$ m. Half-width increase = $(0.5 - 0.1)/2 = 0.2$ m over 0.3 m, so the side slope is $\tan\theta = 0.2/0.3 = 0.6667$ (horizontal per unit vertical, from the vertical). This is not a Cippoletti slope (0.25).

Take a horizontal strip at depth $h$ below the water surface (upstream head $H$ above the base). Its width is $dA$-width $= b + 2\tan\theta\,(H - h)$ and its thickness $dh$. The velocity is $\sqrt{2gh}$:

$$dQ = C_d\left[b + 2\tan\theta\,(H - h)\right]\sqrt{2gh}\;dh$$ $$\begin{aligned} Q &= C_d\sqrt{2g}\int_0^H\left[(b + 2H\tan\theta)h^{1/2} - 2\tan\theta\,h^{3/2}\right]dh\\ &= C_d\sqrt{2g}\left[\frac{2}{3}(b + 2H\tan\theta)H^{3/2} - \frac{4}{5}\tan\theta\,H^{5/2}\right] \end{aligned}$$ $$Q = \frac{2}{3}C_d\sqrt{2g}\,b\,H^{3/2} + \frac{8}{15}C_d\sqrt{2g}\tan\theta\,H^{5/2}$$

(a rectangular notch plus a triangular notch).

Discharge at H = 0.228 m

$$\begin{aligned} \frac{2}{3}C_d\sqrt{2g}\,bH^{3/2} &= \frac{2}{3}(0.6)(4.4294)(0.1)(0.228)^{1.5} = 0.01929\ \text{m}^3/\text{s}\\ \frac{8}{15}C_d\sqrt{2g}\tan\theta\,H^{5/2} &= \frac{8}{15}(0.6)(4.4294)(0.6667)(0.228)^{2.5} = 0.02346\ \text{m}^3/\text{s} \end{aligned}$$ $$Q = 0.01929 + 0.02346 = 0.04274\ \text{m}^3/\text{s}$$

Answer: $Q = \frac{2}{3}C_d\sqrt{2g}\,bH^{3/2} + \frac{8}{15}C_d\sqrt{2g}\tan\theta\,H^{5/2}$; for the given notch $Q \approx 0.0427$ m³/s (42.7 L/s).

Given (assumed from the figure): cylinder 2 m diameter, 2 m high on top of a conical frustum 4 m high tapering from 2 m diameter (top) to 1 m diameter (bottom); orifice $d = 0.10$ m at the bottom; $C_d = 0.6$. Initially full, so the head above the orifice is $H = 4 + 2 = 6$ m.

   |<--2 m-->|
   |         |  2 m
   |_________|
    \       /
     \     /  4 m (frustum)
      \___/
       o  <- orifice 10 cm (bottom dia 1 m)

Orifice area $a = \dfrac{\pi}{4}(0.1)^2 = 0.007854$ m².

$$k = C_d\,a\sqrt{2g} = 0.6\times 0.007854\times 4.4294 = 0.020873\ \text{m}^{5/2}/\text{s}$$

Rate equation: $-A(h)\,dh = k\sqrt{h}\,dt\;\Rightarrow\; t = \dfrac{1}{k}\displaystyle\int\dfrac{A(h)}{\sqrt h}dh$.

Time to lower 6 m to 4 m (cylinder, $A = \pi$ m²)

$$t_1 = \frac{A}{k}\cdot 2\left(\sqrt{6} - \sqrt{4}\right) = \frac{3.1416\times 2\times 0.4495}{0.020873} = 135.3\ \text{s}$$

Time to lower 4 m to 0 (frustum)

At height $h$ above the orifice, the diameter is $D = 1 + h/4$ (1 m at $h = 0$, 2 m at $h = 4$), so

$$A(h) = \frac{\pi}{4}\left(1 + \frac{h}{4}\right)^2 = \frac{\pi}{4}\left(1 + 0.5h + 0.0625h^2\right)$$ $$\begin{aligned} t_2 &= \frac{\pi}{4k}\int_0^4\left(h^{-1/2} + 0.5h^{1/2} + 0.0625h^{3/2}\right)dh\\ &= \frac{\pi}{4k}\left[2h^{1/2} + \frac{1}{3}h^{3/2} + 0.025h^{5/2}\right]_0^4\\ &= \frac{\pi}{4k}\left[4 + 2.667 + 0.8\right] = \frac{0.7854\times 7.467}{0.020873} = 280.9\ \text{s} \end{aligned}$$

Total time

$$t = t_1 + t_2 = 135.3 + 280.9 = 416.3\ \text{s} \approx 6.94\ \text{min}$$

Answer: time of emptying $\approx 416$ s (about 6 min 56 s).

Given: tank area $A = 3.2$ m²; two orifices each $a = 8.8$ mm² $= 8.8\times 10^{-6}$ m² at 5 m and 2 m above the bottom; $C_d = 0.62$; level falls from 8 m to 3 m.

The upper orifice (at 5 m) works only while the level is above 5 m. So the problem has two stages:

• Stage 1: level 8 m to 5 m, both orifices discharging.
• Stage 2: level 5 m to 3 m, only the lower orifice (at 2 m) discharging.

$$k = C_d\,a\sqrt{2g} = 0.62\times 8.8\times 10^{-6}\times 4.4294 = 2.4167\times 10^{-5}\ \text{m}^{5/2}/\text{s}$$

Head over the upper orifice $= h - 5$; over the lower orifice $= h - 2$ ($h$ = level above tank bottom).

Stage 1 (8 m to 5 m)

$$-A\,dh = k\left[\sqrt{h - 5} + \sqrt{h - 2}\right]dt$$

Rationalise: $\dfrac{1}{\sqrt{h-5} + \sqrt{h-2}} = \dfrac{\sqrt{h-2} - \sqrt{h-5}}{3}$.

$$\begin{aligned} t_1 &= \frac{A}{k}\int_5^8\frac{\sqrt{h-2} - \sqrt{h-5}}{3}dh = \frac{A}{k}\cdot\frac{2}{9}\left[(h-2)^{3/2} - (h-5)^{3/2}\right]_5^8\\ &= \frac{3.2}{2.4167\times 10^{-5}}\times\frac{2}{9}\,[\ldots] \end{aligned}$$

Evaluating the bracket: at $h = 8$: $6^{3/2} - 3^{3/2} = 14.697 - 5.196 = 9.501$; at $h = 5$: $3^{3/2} - 0 = 5.196$. Difference $= 4.305$.

$$t_1 = \frac{3.2}{2.4167\times 10^{-5}}\times\frac{2}{9}\times 4.305 = 126\,663\ \text{s}$$

Stage 2 (5 m to 3 m, lower orifice only)

$$t_2 = \frac{2A}{k}\left[\sqrt{5-2} - \sqrt{3-2}\right] = \frac{2\times 3.2}{2.4167\times 10^{-5}}\times(1.7321 - 1) = 193\,864\ \text{s}$$

Total

$$t = t_1 + t_2 = 126\,663 + 193\,864 = 320\,527\ \text{s} \approx 89.0\ \text{h}$$

Answer: $t \approx 3.2\times 10^{5}$ s (about 89 hours) for the data as given. The very long time results from the tiny orifice areas (8.8 mm²). If the areas were 8.8 cm², all times would be 100 times smaller (about 3205 s, i.e. 53 min).

Given: $Q = 12\ \text{L/s} = 0.012\ \text{m}^3/\text{s}$.

Coefficient of discharge, triangular notch

Included angle $\theta = 45^\circ$, so $\tan(\theta/2) = \tan 22.5^\circ = 0.4142$; $H = 0.21$ m.

$$Q = \frac{8}{15}C_d\tan\frac{\theta}{2}\sqrt{2g}\,H^{5/2}$$ $$\begin{aligned} H^{5/2} &= 0.21^{2.5} = 0.020208\\ 0.012 &= \frac{8}{15}(C_d)(0.4142)(4.4294)(0.020208) = 0.019774\,C_d\\ C_d &= 0.607 \end{aligned}$$

Coefficient of discharge, rectangular notch

$L = 0.30$ m, $H = 0.078$ m.

$$Q = \frac{2}{3}C_dL\sqrt{2g}\,H^{3/2}$$ $$\begin{aligned} H^{3/2} &= 0.078^{1.5} = 0.021784\\ 0.012 &= \frac{2}{3}(C_d)(0.3)(4.4294)(0.021784) = 0.019299\,C_d\\ C_d &= 0.622 \end{aligned}$$

Error in head for 2 % error in discharge

Differentiating $Q \propto H^n$: $\dfrac{dQ}{Q} = n\dfrac{dH}{H}$.

• Triangular notch ($n = 5/2$): $\dfrac{dH}{H} = \dfrac{0.02}{2.5} = 0.008$, so $dH = 0.008\times 210 = 1.68$ mm.
• Rectangular notch ($n = 3/2$): $\dfrac{dH}{H} = \dfrac{0.02}{1.5} = 0.01333$, so $dH = 0.01333\times 78 = 1.04$ mm.

Answer: $C_d = 0.607$ (triangular) and $0.622$ (rectangular); a 2 % error in discharge corresponds to an error in head of about 1.68 mm (triangular) and 1.04 mm (rectangular), i.e. 0.8 % and 1.33 % of the head.

Given: CO₂ density $\rho = 3$ kg/m³, $p_1 = 170$ kPa, manometer fluid S.G. = 0.827 ($\rho_m = 827$ kg/m³), manometer reading $h = 8$ cm $= 0.08$ m, $d_1 = 10$ cm, $d_2 = 6$ cm. The gas is treated as incompressible (small pressure change), horizontal meter, no losses.

(a) Pressure at the throat

The U-tube manometer measures the pressure difference between sections 1 and 2:

$$p_1 - p_2 = (\rho_m - \rho)\,g\,h = (827 - 3)(9.81)(0.08) = 646.7\ \text{Pa}$$ $$p_2 = 170\,000 - 646.7 = 169\,353\ \text{Pa} \approx 169.35\ \text{kPa}$$

(b) Gas rate

$$\beta = \frac{d_2}{d_1} = 0.6,\qquad \beta^4 = 0.1296$$

Bernoulli with continuity: $p_1 - p_2 = \dfrac{\rho}{2}V_2^2\left(1 - \beta^4\right)$.

$$V_2 = \sqrt{\frac{2(646.7)}{3\times 0.8704}} = \sqrt{495.3} = 22.26\ \text{m/s}$$ $$\begin{aligned} A_2 &= \frac{\pi}{4}(0.06)^2 = 2.827\times 10^{-3}\ \text{m}^2\\ Q &= A_2V_2 = 0.06293\ \text{m}^3/\text{s} = 226.5\ \text{m}^3/\text{h} \end{aligned}$$

Answer: (a) $p_2 \approx 169.35$ kPa; (b) $Q \approx 0.0629$ m³/s $\approx 226.5$ m³/h.

Cippoletti notch

A Cippoletti notch (weir) is a trapezoidal notch whose sides have a slope of 1 horizontal to 4 vertical. The slope is chosen so that the extra discharge through the triangular sides offsets the discharge lost due to end contractions on a rectangular weir. Its discharge is $Q = \tfrac23 C_d L\sqrt{2g}\,H^{3/2}$, with no end-contraction correction.

Tank with orifice and inflow

Given: orifice $d = 40$ mm, $a = \dfrac{\pi}{4}(0.04)^2 = 1.2566\times 10^{-3}$ m², $C_d = 0.62$. Let the inflow be $Q_{in}$ and the tank area $A$. The level is rising, so inflow exceeds outflow:

$$Q_{in} - Q_{out} = A\frac{dh}{dt},\qquad Q_{out} = C_d\,a\sqrt{2gh}$$

Outflows:

$$\begin{aligned} Q_{out1} &= 0.62(1.2566\times 10^{-3})\sqrt{2\times 9.81\times 0.68} = 2.8458\times 10^{-3}\ \text{m}^3/\text{s}\\ Q_{out2} &= 0.62(1.2566\times 10^{-3})\sqrt{2\times 9.81\times 1.24} = 3.8429\times 10^{-3}\ \text{m}^3/\text{s} \end{aligned}$$

Equations:

$$\begin{aligned} Q_{in} - 2.8458\times 10^{-3} &= A(0.0014)\\ Q_{in} - 3.8429\times 10^{-3} &= A(0.00062) \end{aligned}$$

Subtracting:

$$(3.8429 - 2.8458)\times 10^{-3} = A(0.0014 - 0.00062)\;\Rightarrow\; A = \frac{0.9971\times 10^{-3}}{0.00078} = 1.278\ \text{m}^2$$ $$Q_{in} = 2.8458\times 10^{-3} + 1.278(0.0014) = 4.636\times 10^{-3}\ \text{m}^3/\text{s}$$

Check with the second equation: $3.8429\times 10^{-3} + 1.278(0.00062) = 4.636\times 10^{-3}$. Correct.

Answer: rate of inflow $= 4.64\times 10^{-3}$ m³/s (4.64 L/s); cross-sectional area of the tank $A = 1.28$ m².

Given: $d_1 = 0.15$ m, $Q = 240$ m³/h $= 0.06667$ m³/s, $p_1/\gamma = 18$ m (gauge), minimum throat pressure head $p_2/\gamma = -7$ m (gauge). Friction loss $= 10\%$ of the pressure head difference.

Smaller throat gives higher velocity and lower pressure, so the minimum throat diameter corresponds to $p_2/\gamma = -7$ m.

$$\begin{aligned} \Delta\left(\frac{p}{\gamma}\right) &= 18 - (-7) = 25\ \text{m}\\ h_L &= 0.1\times 25 = 2.5\ \text{m} \end{aligned}$$

Inlet velocity:

$$A_1 = \frac{\pi}{4}(0.15)^2 = 0.017671\ \text{m}^2,\qquad V_1 = \frac{0.06667}{0.017671} = 3.7726\ \text{m/s},\qquad \frac{V_1^2}{2g} = 0.7254\ \text{m}$$

Bernoulli (horizontal pipe):

$$\frac{p_1}{\gamma} + \frac{V_1^2}{2g} = \frac{p_2}{\gamma} + \frac{V_2^2}{2g} + h_L$$ $$\begin{aligned} \frac{V_2^2}{2g} &= 18 + 0.7254 - (-7) - 2.5 = 23.2254\ \text{m}\\ V_2 &= \sqrt{2\times 9.81\times 23.2254} = 21.347\ \text{m/s} \end{aligned}$$

Throat area and diameter:

$$A_2 = \frac{Q}{V_2} = \frac{0.06667}{21.347} = 3.123\times 10^{-3}\ \text{m}^2$$ $$d_2 = \sqrt{\frac{4A_2}{\pi}} = 0.0631\ \text{m}$$

Answer: minimum throat diameter $\approx 63$ mm (0.0631 m). The throat velocity is 21.35 m/s and the loss is 2.5 m.

Given: throat area $a_t = 0.00025$ m², outlet area $a_e = 0.001$ m², venturi vertical, discharging to atmosphere.

The figure values (2.0 m, 1.9 m, 1.0 m) are not fully readable, so assume: the throat is $z_t = 1.0$ m above the outlet plane, and the water surface in the suction tank is $H_s = 1.9$ m below the throat.

        inlet
        ||
        ||        <- 
      throat ---- suction pipe (to tank)
        /\          1.9 m lift
       /  \
      outlet (atmosphere) ... 1.0 m below throat

Condition for flow up the suction pipe: the throat pressure must fall at least to $p_t = -\gamma H_s$ (gauge). The minimum discharge is when $p_t/\gamma = -1.9$ m.

Continuity: $V_t = Q/a_t$, $V_e = Q/a_e = V_t/4$.

Bernoulli, throat to outlet (exit pressure atmospheric):

$$\frac{p_t}{\gamma} + \frac{V_t^2}{2g} + z_t = 0 + \frac{V_e^2}{2g} + 0$$ $$\begin{aligned} -1.9 + \frac{V_t^2}{2g} + 1.0 &= \frac{V_t^2}{16\times 2g}\\ \frac{V_t^2}{2g}\left(1 - \frac{1}{16}\right) &= 0.9\\ \frac{V_t^2}{2g} &= 0.96\ \text{m}\\ V_t &= \sqrt{2\times 9.81\times 0.96} = 4.340\ \text{m/s} \end{aligned}$$ $$Q_{min} = a_tV_t = 0.00025\times 4.340 = 1.085\times 10^{-3}\ \text{m}^3/\text{s}$$

Answer: minimum discharge $\approx 1.085\times 10^{-3}$ m³/s (about 1.09 L/s) for the assumed dimensions. General result: $Q_{min} = a_t\sqrt{2g\,(H_s - z_t)\big/\left(1 - (a_t/a_e)^2\right)}$.

Given: $Q = 20$ L/s $= 0.02$ m³/s; error in head $dH = 1$ mm $= 0.001$ m. The coefficients are not given, so take $C_d = 0.62$ for the rectangular notch and $C_d = 0.60$ for the right-angled triangular notch.

Percentage error formula: since $Q \propto H^n$,

$$\frac{dQ}{Q} = n\frac{dH}{H}$$

Rectangular notch ($L = 0.30$ m, $n = 3/2$)

$$H = \left[\frac{Q}{\frac{2}{3}C_dL\sqrt{2g}}\right]^{2/3} = \left[\frac{0.02}{\frac{2}{3}(0.62)(0.3)(4.4294)}\right]^{2/3} = 0.1099\ \text{m}$$ $$\frac{dQ}{Q} = 1.5\times\frac{0.001}{0.1099} = 0.01365 = 1.37\ \%$$

Right-angled triangular notch ($\theta = 90^\circ$, $\tan 45^\circ = 1$, $n = 5/2$)

$$H = \left[\frac{Q}{\frac{8}{15}C_d\sqrt{2g}}\right]^{2/5} = \left[\frac{0.02}{\frac{8}{15}(0.6)(4.4294)}\right]^{0.4} = 0.1819\ \text{m}$$ $$\frac{dQ}{Q} = 2.5\times\frac{0.001}{0.1819} = 0.01374 = 1.37\ \%$$

Answer: the error in discharge is about 1.37 % for the rectangular notch and about 1.37 % for the triangular notch. Both give nearly the same error for 1 mm error in head; the result depends only slightly on the assumed $C_d$.