1. E6.1.1
   ​

   Explain why the viscosity of a liquid decreases while that of a gas increases with a temperature rise. The following is a table of measurements for a fluid at constant temperature.

   Determine the dynamic viscosity of the fluid.

   $\frac{du}{dy}(s^{-1})$	0.0	0.2	0.4	0.6	0.8
   $\tau(Nm^{-2})$	0.0	1.0	1.9	3.1	4.0

   using Newton’s law of viscosity

   $$\tau=\mu\frac{\partial u}{\partial y}$$

   where $\mu$ is the viscosity. So viscosity is the gradient of a graph of shear stress against the velocity gradient of the above data, or

   $$\mu=\frac{\tau}{\partial u/\partial y}$$

   Plot the data as a graph

   

   Calculate the gradient for each section of the line

   $\frac{du}{dy}(s^{-1})$	0.0	0.2	0.4	0.6	0.8
   $\tau(Nm^{-2})$	0.0	1.0	1.9	3.1	4.0
   Gradient	-	5.0	4.75	5.17	5.0

   Thus the mean gradient = viscosity = $\mu=4.98Ns/m^{2}$.

2. E6.1.2
   ​

   The density of an oil is $850kg/m^{3}$. Find its relative density and Kinematic viscosity if the dynamic viscosity is $5\times 10^{-3}kg/ms$.

   $\rho_{oil}=850kg/m^{3}$   $\rho_{water}=1000kg/m^{3}$

   $\gamma_{oil}=850/1000=0.85$

   Dynamic viscosity $=\mu=5\times 10^{-3}kg/ms$

   Kinematic viscosity $=\nu=\mu/\rho$

   $$\nu=\frac{\mu}{\rho}=\frac{5\times 10^{-3}}{1000}=5\times 10^{-3}\;m^{2}s^{-1}$$
3. E6.1.3
   ​

   The velocity distribution of a viscous liquid (dynamic viscosity $\mu=0.9Ns/m^{2}$) flowing over a fixed plate is given by $u=0.68y-y^{2}$ ($u$ is the velocity in $m/s$ and $y$ is the distance from the plate in $m$).

   What are the shear stresses at the plate surface and at $y=0.34m$?

   $$u=0.68y-y^{2}$$

   $$\frac{\partial u}{\partial y}=0.68-2y$$

   At the plate surface $y=0m$

   $$\frac{\partial u}{\partial y}=0.68$$

   Calculate the shear stress at the plate face

   $$\tau=\mu\frac{\partial u}{\partial y}=9\times 0.68=0.612N/m^{2}$$

   At $y=0.34m$,

   $$\frac{\partial u}{\partial y}=0.68-2\times 0.34=0.0$$

   As the velocity gradient is zero at $y=0.34m$ then the shear stress must also be zero.

4. E6.1.4
   ​

   $5.6m^{3}$ of oil weighs $46800\;N$. Find its mass density, $\rho$ and relative density, $\gamma$.

   Weight $46800=mg$

   Mass $m=46800/9.81=4770.6kg$

   Mass density $\rho=Mass/volume=4770.6/5.6=852kg/m^{3}$

   Relative density

   $$\gamma=\frac{\rho}{\rho_{water}}=\frac{852}{1000}=0.852$$
5. E6.1.5
   ​

   From the table of fluid properties the viscosity of water is given as $0.01008\;poises$.

   What is this value in $Ns/m^{2}$ and $Pa\;s$ units? $\mu=0.01008\;poise$

   $1\;poise=0.1\;Pa\;s=0.1Ns/m^{2}$

   $\mu=0.001008\;Pa\;s=0.001008Ns/m^{2}$

6. E6.1.6
   ​

   In a fluid the velocity measured at a distance of 75mm from the boundary is $1.125m/s$. The fluid has absolute viscosity $0.048\;Pa\;s$ and a relative density of 0.913. What is the velocity gradient and shear stress at the boundary assuming a linear velocity distribution?

   $\mu=0.048\;Pa\;s$ and $\gamma=0.913$

   $$\frac{\partial u}{\partial y}=\frac{1.125}{0.075}=15\;s^{-1}$$

   $$\tau=\mu\frac{\partial u}{\partial y}=0.048\times 15=0.720\;Pa\;s$$

Pressure and Manometers

1. E6.2.1
   ​

   What will be (a) the gauge pressure and (b) the absolute pressure of water at a depth 12m below the surface?
   $\rho_{\text{water}}=1000kg/m^{3}$, and $p_{\text{atmospheric}}=101kN/m^{2}$.
   [Ans: $117.72kN/m^{2}$, $218.72kN/m^{2}$] a)

   	$\displaystyle p_{gauge}$	$\displaystyle=\rho gh$
   		$\displaystyle=1000\times 9.81\times 12$
   		$\displaystyle=117720N/m^{2}$
   		$\displaystyle=118kN/m^{2}$

   b)

   	$\displaystyle p_{absolute}$	$\displaystyle=p_{gauge}+p_{atmospheric}$
   		$\displaystyle=(117720+101000)N/m^{2}$
   		$\displaystyle=219kN/m^{2}$

2. E6.2.2
   ​

   At what depth below the surface of oil, relative density 0.8, will produce a pressure of $120kN/m^{2}$? What depth of water is this equivalent to?
   [Ans: $15.3m$, $12.2m$]

   a)

   	$\displaystyle\rho$	$\displaystyle=\gamma\rho_{water}$
   		$\displaystyle=0.8\times kg/m^{3}$

   	$\displaystyle p$	$\displaystyle=\rho gh$
   	$\displaystyle h$	$\displaystyle=\frac{p}{\rho g}=\frac{120\times 10^{3}}{800\times 9.81}=15.29m\text{ of oil}$

   b)

   	$\displaystyle\rho$	$\displaystyle=1000kg/m^{3}$
   	$\displaystyle h$	$\displaystyle=\frac{p}{\rho g}=\frac{120\times 10^{3}}{1000\times 9.81}=12.23m\text{ of water}$

3. E6.2.3
   ​

   What would the pressure in $kN/m^{2}$ be if the equivalent head is measured as $400mm$ of (a) mercury $\gamma=13.6$ (b) water ( c) oil specific weight $7.9kN/m^{3}$ (d) a liquid of density $520kg/m^{3}$?
   [Ans: $53.4kN/m^{2}$, $3.92kN/m^{2}$, $3.16kN/m^{2}$, $2.04kN/m^{2}$]

   a)

   	$\displaystyle\rho$	$\displaystyle=\gamma\rho_{\text{water }}$
   		$\displaystyle=13.6\times 1000\mathrm{\leavevmode\nobreak\ kg}/\mathrm{m}^{3}$
   	$\displaystyle p$	$\displaystyle=\rho gh$
   		$\displaystyle=\left(13.6\times 10^{3}\right)\times 9.81\times 0.4=53366\mathrm{\leavevmode\nobreak\ N}/\mathrm{m}^{2}$

   b)

   	$\displaystyle p$	$\displaystyle=\rho gh$
   		$\displaystyle=\left(10^{3}\right)\times 9.81\times 0.4=3924\mathrm{\leavevmode\nobreak\ N}/\mathrm{m}^{2}$

   c)

   	$\displaystyle\omega$	$\displaystyle=\rho g$
   	$\displaystyle p$	$\displaystyle=\rho gh$
   		$\displaystyle=\left(7.9\times 10^{3}\right)\times 0.4=3160\mathrm{\leavevmode\nobreak\ N}/\mathrm{m}^{2}$

   d)

   	$\displaystyle p$	$\displaystyle=\rho gh$
   		$\displaystyle=520\times 9.81\times 0.4=2040\mathrm{\leavevmode\nobreak\ N}/\mathrm{m}^{2}$

4. E6.2.4
   ​

   A manometer connected to a pipe indicates a negative gauge pressure of 50 mm of mercury. What is the absolute pressure in the pipe in Newtons per square metre is the atmospheric pressure is 1 bar?
   [ $93.3\mathrm{kN}/\mathrm{m}^{2}$ ]

   	$\displaystyle p_{\text{atmosphere }}$	$\displaystyle=1\text{ bar }=1\times 10^{5}\mathrm{\leavevmode\nobreak\ N}/\mathrm{m}^{2}$
   	$\displaystyle p_{\text{absolute }}$	$\displaystyle=p_{\text{gauge }}+p_{\text{atmospheric }}$
   		$\displaystyle=\rho gh+p_{\text{atmospheric }}$
   		$\displaystyle=-13.6\times 10^{3}\times 9.81\times 0.05+10^{5}\mathrm{\leavevmode\nobreak\ N}/\mathrm{m}^{2},(\mathrm{\leavevmode\nobreak\ Pa})$
   		$\displaystyle=93.33\mathrm{kN}/\mathrm{m}^{2},(\mathrm{kPa})$

5. E6.2.5
   ​

   What height would a water barometer need to be to measure atmospheric pressure of 1 bar?
   [Ans: $>10m$]

   	$\displaystyle p_{\text{atmosphere }}$	$\displaystyle\approx 1\mathrm{bar}=1\times 10^{5}\mathrm{\leavevmode\nobreak\ N}/\mathrm{m}^{2}$
   	$\displaystyle 10^{5}$	$\displaystyle=\rho gh$
   	$\displaystyle h$	$\displaystyle=\frac{10^{5}}{1000\times 9.81}=10.19\mathrm{\leavevmode\nobreak\ m}\text{ of water }$
   	$\displaystyle h$	$\displaystyle=\frac{10^{5}}{\left(13.6\times 10^{3}\right)\times 9.81}=0.75\mathrm{\leavevmode\nobreak\ m}\text{ of mercury }$

6. E6.2.6
   ​

   An inclined manometer is required to measure an air pressure of 3mm of water to an accuracy of $\pm 3\%$. The inclined arm is $8mm$ in diameter, and the larger arm has a diameter of $24mm$. The manometric fluid has a density $740kg/m^{3},$ and the scale may be read to $\pm 0.5mm$.
   What is the angle required to ensure the desired accuracy may be achieved?
   [Ans: $7.6^{\circ}$]

   
   $$p_{1}-p_{2}=\rho_{man}gh=\rho_{man}g\left(z_{1}+z_{2}\right)$$

   Volume moved from left to right $\quad=z_{1}A_{1}=\frac{z_{2}}{\sin\theta}A_{2}=xA_{2}$

   	$\displaystyle=z_{1}$	$\displaystyle\frac{\pi D^{2}}{4}=\frac{z_{2}}{\sin\theta}\frac{\pi d^{2}}{4}=x\frac{\pi d^{2}}{4}$
   	$\displaystyle z_{1}$	$\displaystyle=\frac{z_{2}}{\sin\theta}\frac{d^{2}}{D^{2}}=x\frac{d^{2}}{D^{2}}$
   	$\displaystyle p_{1}-p_{2}$	$\displaystyle=\rho_{\operatorname{man}}gx\left(\sin\theta+\frac{d^{2}}{D^{2}}\right)$
   	$\displaystyle\rho_{\text{water }}gh$	$\displaystyle=\rho_{\operatorname{man}}gx\left(\sin\theta+\frac{d^{2}}{D^{2}}\right)$
   	$\displaystyle\rho_{\text{water }}gh$	$\displaystyle=\left(0.74\times\rho_{\text{water }}\right)gx\left(\sin\theta+\frac{0.008^{2}}{0.024^{2}}\right)$
   	$\displaystyle h$	$\displaystyle=0.74x(\sin\theta+0.1111)$

   The head being measured is $3\%$ of $3\mathrm{\leavevmode\nobreak\ mm}=0.003\times 0.03=0.00009\mathrm{\leavevmode\nobreak\ m}$
   This $3\%$ represents the smallest measurement possible on the manometer, $0.5\mathrm{\leavevmode\nobreak\ mm}=0.0005\mathrm{\leavevmode\nobreak\ m}$, giving

   	$\displaystyle 0.00009$	$\displaystyle=0.74\times 0.0005(\sin\theta+0.1111)$
   	$\displaystyle\sin\theta$	$\displaystyle=0.132$
   	$\displaystyle\theta$	$\displaystyle=7.6^{\circ}$

   [This is not the same as the answer given on the question sheet.]

7. E6.2.7
   ​

   Determine the resultant force due to the water acting on the 1m by 2m rectangular area AB shown in the figure 2 below.
   [$43560N$, $2.37m$ from O]

   

   The magnitude of the resultant force on a submerged plane is:

   	$\displaystyle\mathrm{R}$	$\displaystyle=\text{ pressure at centroid }\text{ {$\times$} }\text{ area of surface }$
   	$\displaystyle R$	$\displaystyle=\rho g\bar{z}A$
   		$\displaystyle=1000\times 9.81\times(1.22+1)\times(1\times 2)$
   		$\displaystyle=43556\mathrm{\leavevmode\nobreak\ N}/\mathrm{m}^{2}$

   This acts at right angles to the surface through the centre of pressure.

   $$Sc=\frac{I_{OO}}{A\bar{x}}=\frac{2\text{ nd moment of area about a line through }\mathrm{O}}{1\text{ st moment of area about a line through }\mathrm{O}}$$

   By the parallel axis theorem (which will be given in an exam), $I_{oo}=I_{GG}+A\bar{x}^{2}$, where $I_{GG}$ is the $2^{\text{nd }}$ moment of area about a line through the centroid and can be found in tables.

   $$Sc=\frac{I_{GG}}{A\bar{x}}+\bar{x}$$
   

   As the wall is vertical, $Sc=D$ and $\bar{x}=\bar{z}$,

   	$\displaystyle Sc$	$\displaystyle=\frac{1\times 2^{3}}{12(1\times 2)(1.22+1)}+(1.22+1)$
   		$\displaystyle=2.37\mathrm{\leavevmode\nobreak\ m}\text{ from }\mathrm{O}$

8. E6.2.8
   ​

   Determine the resultant force due to the water acting on the 1.25m by 2.0m triangular area CD shown in the figure 2 above. The apex of the triangle is at C.
   [Ans: $23.8\times 10^{3}\;N$, $2.821m$ from P]

Depth to centre of gravity is $\bar{z}=1.0+2\frac{2}{3}\cos 45=1.943m$.

Distance from P is $\bar{x}=\bar{z}/\cos 45=2.748m$
Distance from P to the centre of pressure is

Forces on submerged surfaces

1. E6.9
   ​

   Obtain an expression for the depth of the centre of pressure of a plane surface wholly submerged in a fluid and inclined at an angle to the free surface of the liquid.
   A horizontal circular pipe, $1.25m$ diameter, is closed by a butterfly disk which rotates about a horizontal axis through its centre. Determine the torque which would have to be applied to the disk spindle to keep the disk closed in a vertical position when there is a 3m head of fresh water above the axis.
   [Ans: 1176 Nm]

   

   Diagram of the forces on the disc valve, based on an imaginary water surface.
   $\bar{h}=3m$, the depth to the centroid of the disc
   $h^{\prime}=$ depth to the centre of pressure (or line of action of the force)
   Calculate the force:

   	$\displaystyle F$	$\displaystyle=\rho g\bar{h}A$
   		$\displaystyle=1000\times 9.81\times 3\times\pi\left(\frac{1.25}{2}\right)^{2}$
   		$\displaystyle=36.116\mathrm{kN}$

   Calculate the line of action of the force, $h^{\prime}$.

   	$\displaystyle h^{\prime}$	$\displaystyle=\frac{2\text{ nd moment of area about water surface }}{1\text{ st moment of area about water surface }}$
   		$\displaystyle=\frac{I_{oo}}{A\bar{h}}$

   By the parallel axis theorem $2^{\text{nd }}$ moment of area about O (in the surface) $I_{oo}=I_{GG}+A\bar{h}^{2}$ where $I_{GG}$ is the 2nd moment of area about a line through the centroid of the disc and $I_{GG}=\pi r^{4}/4$.

   	$\displaystyle h^{\prime}$	$\displaystyle=\frac{I_{GG}}{A\bar{h}}+\bar{h}$
   		$\displaystyle=\frac{\pi r^{4}}{4\left(\pi r^{2}\right)3}+3$
   		$\displaystyle=\frac{r^{2}}{12}+3=3.0326\mathrm{\leavevmode\nobreak\ m}$

   So the distance from the spindle to the line of action of the force is

   $$x=h^{\prime}-\bar{h}=3.0326-3=0.0326m$$

   And the moment required to keep the gate shut is

   $$\text{ moment }=Fx=36.116\times 0.0326=1.176\mathrm{kN}\mathrm{\leavevmode\nobreak\ m}$$
2. E6.10
   ​

   (HARDER) A dock gate is to be reinforced with three horizontal beams. If the water acts on one side only, to a depth of $6m$, find the positions of the beams measured from the water surface so that each will carry an equal load. Give the load per meter.
   [Ans: $58860N/m$, $2.31m$, $4.22m$, $5.47m$]

   

   The resultant force per unit length of gate is the area of the pressure diagram. So the total resultant force is

   $$R=\frac{1}{2}\rho gh^{2}=0.5\times 1000\times 9.81\times 6^{2}=176580N(\text{ per }m\text{ length })$$

   Alternatively the resultant force is, $\mathrm{R}=$ Pressure at centroid $\times$ Area , (take width of gate as 1 m to give force per m)

   $$R=\rho g\frac{h}{2}\times(\mathrm{h}\times 1)=176580N(\text{ per }m\text{ length })$$

   This is the resultant force exerted by the gate on the water.
   The three beams should carry an equal load, so each beam carries the load $f$, where

   $$f=\frac{R}{3}=58860N$$

   If we take moments from the surface,

   		$\displaystyle DR=fd_{1}+fd_{2}+fd_{3}$
   		$\displaystyle D(3f)=f\left(d_{1}+d_{2}+d_{3}\right)$
   		$\displaystyle 12=d_{1}+d_{2}+d_{3}$

   Taking the first beam, we can draw a pressure diagram for this, (ignoring what is below),
   

   We know that the resultant force, $F=\frac{1}{2}\rho gH^{2}$, so $H=\sqrt{\frac{2F}{\rho g}}$

   $$H=\sqrt{\frac{2F}{\rho g}}=\sqrt{\frac{2\times 58860}{1000\times 9.81}}=3.46\mathrm{\leavevmode\nobreak\ m}$$

   And the force acts at $2\mathrm{H}/3$, so this is the position of the $1^{\text{st }}$ beam,

   $$\text{ position of 1st beam }=\frac{2}{3}H=2.31\mathrm{\leavevmode\nobreak\ m}$$

   Taking the second beam into consideration, we can draw the following pressure diagram,
   

   The reaction force is equal to the sum of the forces on each beam, so as before

   $$H=\sqrt{\frac{2F}{\rho g}}=\sqrt{\frac{2\times(2\times 58860)}{1000\times 9.81}}=4.9\mathrm{\leavevmode\nobreak\ m}$$

   The reaction force acts at $2\mathrm{H}/3$, so $\mathrm{H}=3.27\mathrm{\leavevmode\nobreak\ m}$. Taking moments from the surface,

   	$\displaystyle(2\times 58860)\times 3.27$	$\displaystyle=58860\times 2.31+58860\times d_{2}$
   	$\displaystyle\text{ depth to second beam }d_{2}$	$\displaystyle=4.22\mathrm{\leavevmode\nobreak\ m}$

   For the third beam, from before we have,

   	$\displaystyle 12$	$\displaystyle=d_{1}+d_{2}+d_{3}$
   	$\displaystyle\text{ depth to third beam }d_{3}$	$\displaystyle=12-2.31-4.22=5.47\mathrm{\leavevmode\nobreak\ m}$

3. E6.11
   ​

   The profile of a masonry dam is an arc of a circle, the arc having a radius of $30m$ and subtending an angle of $60^{\circ}$ at the centre of curvature, which lies in the water surface. Determine (a) the load on the dam in $N/m$ length, (b) the position of the line of action to this pressure.
   [Ans: $4.28\times 10^{6}N/m$ length at depth $19.0m$]

   

   		$\displaystyle h=30\sin 60=25.98\mathrm{\leavevmode\nobreak\ m}$
   		$\displaystyle a=30\cos 60=15.0\mathrm{\leavevmode\nobreak\ m}$

   Calculate $\mathrm{F}_{\mathrm{v}}=$ total weight of fluid above the curved surface (per m length)

   	$\displaystyle F_{v}$	$\displaystyle=\rho g(\text{ area of sector - area of triangle })$
   		$\displaystyle=1000\times 9.81\times\left[\left(\pi 30^{2}\times\frac{60}{360}\right)-\left(\frac{25.98\times 15}{2}\right)\right]$
   		$\displaystyle=2711.375\mathrm{kN}/\mathrm{m}$

   Calculate $\mathrm{F}_{\mathrm{h}}=$ force on projection of curved surface onto a vertical plane

   	$\displaystyle F_{h}$	$\displaystyle=\frac{1}{2}\rho gh^{2}$
   		$\displaystyle=0.5\times 1000\times 9.81\times 25.98^{2}=3310.681\mathrm{kN}/\mathrm{m}$

   The resultant,

   	$\displaystyle F_{R}$	$\displaystyle=\sqrt{F_{v}^{2}+F_{h}^{2}}=\sqrt{3310.681^{2}+2711.375^{2}}$
   		$\displaystyle=4279.27\mathrm{kN}/\mathrm{m}$

   acting at the angle

   	$\displaystyle\tan\theta$	$\displaystyle=\frac{F_{v}}{F_{h}}=0.819$
   	$\displaystyle\theta$	$\displaystyle=39.32^{\circ}$

   As this force act normal to the surface, it must act through the centre of radius of the dam wall. So the depth to the point where the force acts is,

   $$y=30\sin 39.31^{\circ}=19m$$
4. E6.12
   ​

   (HARDER) The arch of a bridge over a stream is in the form of a semi-circle of radius $2m$. The bridge width is $4m$. Due to a flood, the water level is now $1.25m$ above the crest of the arch. Calculate (a) the upward force on the underside of the arch, (b) the horizontal thrust on one half of the arch.
   [Ans: $263.6kN$, $176.6kN$]

   

   
   

   a) The upward force on the arch $=$ weight of (imaginary) water above the arch.

   	$\displaystyle R_{v}$	$\displaystyle=\rho g\times\text{ volume of water }$
   	volume	$\displaystyle=\left((1.25+2)\times 4-\frac{\pi 2^{2}}{2}\right)\times 4=26.867\mathrm{\leavevmode\nobreak\ m}^{3}$
   	$\displaystyle R_{v}$	$\displaystyle=1000\times 9.81\times 26.867=263.568\mathrm{kN}$

   b)

   The horizontal force on half of the arch, is equal to the force on the projection of the curved surface onto a vertical plane.
   

   	$\displaystyle F_{h}$	$\displaystyle=\text{ pressure at centroid {$\times$} area }$
   		$\displaystyle=\rho g(1.25+1)\times(2\times 4)$
   		$\displaystyle=176.58\mathrm{kN}$

5. E6.13
   ​

   The face of a dam is vertical to a depth of $7.5m$ below the water surface then slopes at $30^{\circ}$ to the vertical. If the depth of water is $17m$ what is the resultant force per metre acting on the whole face?
   [Ans: $1563.29kN$]

   

   
   

   $\mathrm{h}_{2}=17.0\mathrm{\leavevmode\nobreak\ m}$, so $\mathrm{h}_{1}=17.0-7.5=9.5.\quad\mathrm{x}=9.5/\tan 60=5.485\mathrm{\leavevmode\nobreak\ m}$.
   Vertical force $=$ weight of water above the surface,

   	$\displaystyle F_{v}$	$\displaystyle=\rho g\left(h_{2}\times x+0.5h_{1}\times x\right)$
   		$\displaystyle=9810\times(7.5\times 5.485+0.5\times 9.5\times 5.485)$
   		$\displaystyle=659.123\mathrm{kN}/\mathrm{m}$

   The horizontal force $=$ force on the projection of the surface on to a vertical plane.

   	$\displaystyle F_{h}$	$\displaystyle=\frac{1}{2}\rho gh^{2}$
   		$\displaystyle=0.5\times 1000\times 9.81\times 17^{2}$
   		$\displaystyle=1417.545\mathrm{kN}/\mathrm{m}$

   The resultant force is

   	$\displaystyle F_{R}$	$\displaystyle=\sqrt{F_{v}^{2}+F_{h}^{2}}=\sqrt{659.123^{2}+1417.545^{2}}$
   		$\displaystyle=1563.29\mathrm{kN}/\mathrm{m}$

   And acts at the angle

   	$\displaystyle\tan\theta$	$\displaystyle=\frac{F_{v}}{F_{h}}=0.465$
   	$\displaystyle\theta$	$\displaystyle=24.94^{\circ}$

6. E6.14
   ​

   A tank with vertical sides is square in plan with $3m$ long sides. The tank contains oil of relative density 0.9 to a depth of $2.0m$ which is floating on water a depth of $1.5m$. Calculate the force on the walls and the height of the centre of pressure from the bottom of the tank.
   [Ans: $165.54kN$, $1.15m$]

Consider one wall of the tank. Draw the pressure diagram:

density of oil $\rho_{\text{oil }}=0.9\rho_{\text{water }}=900\mathrm{\leavevmode\nobreak\ kg}/\mathrm{m}^{3}$.
Force per unit length, $\mathrm{F}=$ area under the graph $=$ sum of the three areas $=f_{1}+f_{2}+f_{3}$

To find the position of the resultant force F , we take moments from any point. We will take moments about the surface.

Application of the Bernoulli Equation

1. E6.3.1
   ​

   In a vertical pipe carrying water, pressure gauges are inserted at points A and B where the pipe diameters are 0.15 m and 0.075 m respectively. The point $B$ is 2.5 m below $A$ and when the flow rate down the pipe is 0.02 cumecs, the pressure at B is $14715\mathrm{\leavevmode\nobreak\ N}/\mathrm{m}^{2}$ greater than that at A .

   Assuming the losses in the pipe between A and B can be expressed as $k\frac{v^{2}}{2g}$ where $v$ is the velocity at A , find the value of $k$.
   If the gauges at A and B are replaced by tubes filled with water and connected to a U -tube containing mercury of relative density 13.6, give a sketch showing how the levels in the two limbs of the U-tube differ and calculate the value of this difference in metres.
   $[k=0.319,0.0794\mathrm{\leavevmode\nobreak\ m}]$
   

   Part i)

   		$\displaystyle d_{A}=0.15m\quad d_{B}=0.075m\quad Q=0.02\mathrm{\leavevmode\nobreak\ m}^{3}/\mathrm{s}$
   		$\displaystyle p_{B}-p_{A}=14715\mathrm{\leavevmode\nobreak\ N}/\mathrm{m}^{2}$
   		$\displaystyle h_{f}=\frac{kv^{2}}{2g}$

   Taking the datum at B , the Bernoulli equation becomes:

   		$\displaystyle\frac{p_{A}}{\rho g}+\frac{u_{A}^{2}}{2g}+z_{A}=\frac{p_{B}}{\rho g}+\frac{u_{B}^{2}}{2g}+z_{B}+k\frac{u_{A}^{2}}{2g}$
   		$\displaystyle z_{A}=2.5\quad\quad z_{B}=0$

   By continuity: $\mathrm{Q}=\mathrm{u}_{\mathrm{A}}\mathrm{A}_{\mathrm{A}}=\mathrm{u}_{\mathrm{B}}\mathrm{A}_{\mathrm{B}}$

   		$\displaystyle u_{A}=0.02/\left(\pi 0.075^{2}\right)=1.132\mathrm{\leavevmode\nobreak\ m}/\mathrm{s}$
   		$\displaystyle u_{B}=0.02/\left(\pi 0.0375^{2}\right)=4.527\mathrm{\leavevmode\nobreak\ m}/\mathrm{s}$

   giving

   	$\displaystyle\frac{p_{B}-p_{A}}{1000g}-z_{A}+\frac{u_{B}^{2}-u_{A}^{2}}{2g}$	$\displaystyle=-k\frac{u_{A}^{2}}{2g}$
   	$\displaystyle 1.5-2.5+1.045-0.065$	$\displaystyle=-0.065k$
   	$\displaystyle k$	$\displaystyle=0.319$

   Part ii)

   	$\displaystyle p_{xxL}$	$\displaystyle=\rho_{w}gz_{B}+p_{B}$
   	$\displaystyle p_{xxR}$	$\displaystyle=\rho_{m}gR_{p}+\rho_{w}gz_{A}-\rho_{w}gR_{p}+p_{A}$
   	$\displaystyle p_{xxL}$	$\displaystyle=p_{xxR}$
   	$\displaystyle\rho_{w}gz_{B}+p_{B}$	$\displaystyle=\rho_{m}gR_{p}+\rho_{w}gz_{A}-\rho_{w}gR_{p}+p_{A}$
   	$\displaystyle p_{B}-p_{A}$	$\displaystyle=\rho_{w}g\left(z_{A}-z_{B}\right)+gR_{P}\left(\rho_{m}-\rho_{w}\right)$
   	$\displaystyle 14715$	$\displaystyle=1000\times 9.81\times 2.5+9.81R_{p}(13600-1000)$
   	$\displaystyle R_{p}$	$\displaystyle=-0.079m$

2. E6.3.2
   ​

   A Venturimeter with an entrance diameter of 0.3 m and a throat diameter of 0.2 m is used to measure the volume of gas flowing through a pipe. The discharge coefficient of the meter is 0.96 .
   Assuming the specific weight of the gas to be constant at $19.62\mathrm{\leavevmode\nobreak\ N}/\mathrm{m}^{3}$, calculate the volume flowing when the pressure difference between the entrance and the throat is measured as 0.06 m on a water U-tube manometer.
   [ $0.816\mathrm{\leavevmode\nobreak\ m}^{3}/\mathrm{s}$ ]
   

   What we know from the question:

   	$\displaystyle\rho_{g}g$	$\displaystyle=19.62\mathrm{\leavevmode\nobreak\ N}/\mathrm{m}^{2}$
   	$\displaystyle C_{d}$	$\displaystyle=0.96$
   	$\displaystyle d_{1}$	$\displaystyle=0.3\mathrm{\leavevmode\nobreak\ m}$
   	$\displaystyle d_{2}$	$\displaystyle=0.2\mathrm{\leavevmode\nobreak\ m}$

   Calculate Q.

   $$u_{1}=Q/0.0707$$
   $$u_{2}=Q/0.0314$$

   For the manometer:

   	$\displaystyle p_{1}+\rho_{g}gz$	$\displaystyle=p_{2}+\rho_{g}g\left(z_{2}-R_{p}\right)+\rho_{w}gR_{p}$
   	$\displaystyle p_{1}-p_{2}$	$\displaystyle=19.62\left(z_{2}-z_{1}\right)+587.423$		(1)

   For the Venturimeter

   	$\displaystyle\frac{p_{1}}{\rho_{g}g}+\frac{u_{1}^{2}}{2g}+z_{1}$	$\displaystyle=\frac{p_{2}}{\rho_{g}g}+\frac{u_{2}^{2}}{2g}+z_{2}$
   	$\displaystyle p_{1}-p_{2}$	$\displaystyle=19.62\left(z_{2}-z_{1}\right)+0.803u_{2}^{2}$		(2)

   Combining (1) and (2)

   		$\displaystyle 0.803u_{2}^{2}=587.423$
   		$\displaystyle u_{2_{\text{ideal }}}=27.047\mathrm{\leavevmode\nobreak\ m}/\mathrm{s}$
   		$\displaystyle Q_{\text{ideal }}=27.047\times\pi\left(\frac{0.2}{2}\right)^{2}=0.85\mathrm{\leavevmode\nobreak\ m}^{3}/\mathrm{s}$
   		$\displaystyle Q=C_{d}Q_{\text{idea }}=0.96\times 0.85=0.816\mathrm{\leavevmode\nobreak\ m}^{3}/\mathrm{s}$

3. E6.3.3
   ​

   A Venturimeter is used for measuring flow of water along a pipe. The diameter of the Venturi throat is two fifths the diameter of the pipe. The inlet and throat are connected by water filled tubes to a mercury U-tube manometer. The velocity of flow along the pipe is found to be $2.5\sqrt{H}\mathrm{\leavevmode\nobreak\ m}/\mathrm{s}$, where $H$ is the manometer reading in metres of mercury. Determine the loss of head between inlet and throat of the Venturi when $H$ is 0.49 m . (Relative density of mercury is 13.6).
   [0.23m of water]
   

   For the manometer:

   	$\displaystyle p_{1}+\rho_{w}gz_{1}$	$\displaystyle=p_{2}+\rho_{w}g\left(z_{2}-H\right)+\rho_{m}gH$
   	$\displaystyle p_{1}-p_{2}$	$\displaystyle=\rho_{w}gz_{2}-\rho_{w}gH+\rho_{m}gH-\rho_{w}gz_{1}$		(1)

   For the Venturimeter

   	$\displaystyle\frac{p_{1}}{\rho_{w}g}+\frac{u_{1}^{2}}{2g}+z_{1}$	$\displaystyle=\frac{p_{2}}{\rho_{w}g}+\frac{u_{2}^{2}}{2g}+z_{2}+\text{ Losses }$
   	$\displaystyle p_{1}-p_{2}$	$\displaystyle=\frac{\rho_{w}u_{2}^{2}}{2}+\rho_{w}gz_{2}-\frac{\rho_{w}u_{1}^{2}}{2}-\rho_{w}gz_{1}+L\rho_{w}g$		(2)

   Combining (1) and (2)

   	$\displaystyle\frac{p_{1}}{\rho_{w}g}+\frac{u_{1}^{2}}{2g}+z_{1}$	$\displaystyle=\frac{p_{2}}{\rho_{w}g}+\frac{u_{2}^{2}}{2g}+z_{2}+\text{ Losses }$
   	$\displaystyle L\rho_{w}g$	$\displaystyle=Hg\left(\rho_{m}-\rho_{w}\right)-\frac{\rho_{w}}{2}\left(u_{2}^{2}-u_{1}^{2}\right)$		(3)

   but at 1 . From the question

   	$\displaystyle u_{1}$	$\displaystyle=2.5\sqrt{H}=1.75\mathrm{\leavevmode\nobreak\ m}/\mathrm{s}$
   	$\displaystyle u_{1}A_{1}$	$\displaystyle=u_{2}A_{2}$
   	$\displaystyle 1.75\times\pi\frac{d^{2}}{4}$	$\displaystyle=u_{2}\pi\left(\frac{2d}{10}\right)^{2}$
   	$\displaystyle u_{2}$	$\displaystyle=10.937\mathrm{\leavevmode\nobreak\ m}/\mathrm{s}$

   Substitute in (3)

   	Losses	$\displaystyle=L=\frac{0.49\times 9.81(13600-1000)-(1000/2)\left(10.937^{2}-1.75^{2}\right)}{9.81\times 1000}$
   		$\displaystyle=0.233\mathrm{\leavevmode\nobreak\ m}$

4. E6.3.4
   ​

   Water is discharging from a tank through a convergent-divergent mouthpiece. The exit from the tank is rounded so that losses there may be neglected and the minimum diameter is 0.05 m .
   If the head in the tank above the centre-line of the mouthpiece is 1.83 m . a) What is the discharge?
   b) What must be the diameter at the exit if the absolute pressure at the minimum area is to be 2.44 m of water? c) What would the discharge be if the divergent part of the mouthpiece were removed. (Assume atmospheric pressure is 10 m of water).
   $\left[0.0752\mathrm{\leavevmode\nobreak\ m},0.0266\mathrm{\leavevmode\nobreak\ m}^{3}/\mathrm{s},0.0118\mathrm{\leavevmode\nobreak\ m}^{3}/\mathrm{s}\right]$
   

   From the question:

   	$\displaystyle d_{2}$	$\displaystyle=0.05\mathrm{\leavevmode\nobreak\ m}$
   	minimum pressure	$\displaystyle=\frac{p_{2}}{\rho g}=2.44\mathrm{\leavevmode\nobreak\ m}$
   	$\displaystyle\frac{p_{1}}{\rho g}$	$\displaystyle=10\mathrm{\leavevmode\nobreak\ m}=\frac{p_{3}}{\rho g}$

   Apply Bernoulli:

   $$\frac{p_{1}}{\rho g}+\frac{u_{1}^{2}}{2g}+z_{1}=\frac{p_{2}}{\rho g}+\frac{u_{2}^{2}}{2g}+z_{2}=\frac{p_{3}}{\rho g}+\frac{u_{3}^{2}}{2g}+z_{3}$$

   If we take the datum through the orifice:

   $$z_{1}=1.83m\quad z_{2}=z_{3}=0\quad u_{1}=\text{ negligible }$$

   Between 1 and 2

   	$\displaystyle 10+1.83$	$\displaystyle=2.44+\frac{u_{2}^{2}}{2g}$
   	$\displaystyle u_{2}$	$\displaystyle=13.57\mathrm{\leavevmode\nobreak\ m}/\mathrm{s}$
   	$\displaystyle Q$	$\displaystyle=u_{2}A_{2}=13.57\times\pi\left(\frac{0.05}{2}\right)^{2}=0.02665\mathrm{\leavevmode\nobreak\ m}^{3}/\mathrm{s}$

   Between 1 and $3p_{1}=p_{3}$

   	$\displaystyle 1.83$	$\displaystyle=\frac{u_{3}^{2}}{2g}$
   	$\displaystyle u_{3}$	$\displaystyle=5.99\mathrm{\leavevmode\nobreak\ m}/\mathrm{s}$
   	$\displaystyle Q$	$\displaystyle=u_{3}A_{3}$
   	$\displaystyle 0.02665$	$\displaystyle=5.99\times\pi\frac{d_{3}^{2}}{4}$
   	$\displaystyle d_{3}$	$\displaystyle=0.0752\mathrm{\leavevmode\nobreak\ m}$

   If the mouth piece has been removed, $p_{1}=p_{2}$

   	$\displaystyle\frac{p_{1}}{\rho g}+z_{1}$	$\displaystyle=\frac{p_{2}}{\rho g}+\frac{u_{2}^{2}}{2g}$
   	$\displaystyle u_{2}$	$\displaystyle=\sqrt{2gz_{1}}=5.99\mathrm{\leavevmode\nobreak\ m}/\mathrm{s}$
   	$\displaystyle Q$	$\displaystyle=5.99\pi\frac{0.05^{2}}{4}=0.0118\mathrm{\leavevmode\nobreak\ m}^{3}/\mathrm{s}$

5. E6.3.5
   ​

   A closed tank has an orifice 0.025 m diameter in one of its vertical sides. The tank contains oil to a depth of 0.61 m above the centre of the orifice and the pressure in the air space above the oil is maintained at $13780\mathrm{\leavevmode\nobreak\ N}/\mathrm{m}^{2}$ above atmospheric. Determine the discharge from the orifice.
   (Coefficient of discharge of the orifice is 0.61 , relative density of oil is 0.9 ).
   [ $0.00195\mathrm{\leavevmode\nobreak\ m}^{3}/\mathrm{s}$ ]
   

   From the question

   	$\displaystyle\sigma$	$\displaystyle=0.9=\frac{\rho_{o}}{\rho_{w}}$
   	$\displaystyle\rho_{o}$	$\displaystyle=900$
   	$\displaystyle C_{d}$	$\displaystyle=0.61$

   Apply Bernoulli,

   $$\frac{p_{1}}{\rho g}+\frac{u_{1}^{2}}{2g}+z_{1}=\frac{p_{2}}{\rho g}+\frac{u_{2}^{2}}{2g}+z_{2}$$

   Take atmospheric pressure as 0 ,

   	$\displaystyle\frac{13780}{\rho_{o}g}+0.61$	$\displaystyle=\frac{u_{2}^{2}}{2g}$
   	$\displaystyle u_{2}$	$\displaystyle=6.53\mathrm{\leavevmode\nobreak\ m}/\mathrm{s}$
   	$\displaystyle Q$	$\displaystyle=0.61\times 6.53\times\pi\left(\frac{0.025}{2}\right)^{2}=0.00195\mathrm{\leavevmode\nobreak\ m}^{3}/\mathrm{s}$

6. E6.3.6
   ​

   The discharge coefficient of a Venturimeter was found to be constant for rates of flow exceeding a certain value. Show that for this condition the loss of head due to friction in the convergent parts of the meter can be expressed as $KQ^{2}m$ where $K$ is a constant and $Q$ is the rate of flow in cumecs.
   Obtain the value of $K$ if the inlet and throat diameter of the Venturimeter are 0.102 m and 0.05 m respectively and the discharge coefficient is 0.96 .
   [ $K=1060$ ]

7. E6.3.7
   ​

   A Venturimeter is to fitted in a horizontal pipe of 0.15 m diameter to measure a flow of water which may be anything up to $240\mathrm{\leavevmode\nobreak\ m}^{3}/$ hour. The pressure head at the inlet for this flow is 18 m above atmospheric and the pressure head at the throat must not be lower than 7 m below atmospheric. Between the inlet and the throat there is an estimated frictional loss of $10\%$ of the difference in pressure head between these points. Calculate the minimum allowable diameter for the throat.
   [0.063m]
   

   From the question:

   $$\begin{array}[]{ll}d_{1}=0.15\mathrm{\leavevmode\nobreak\ m}&Q=240\mathrm{\leavevmode\nobreak\ m}^{3}/\mathrm{hr}=0.667\mathrm{\leavevmode\nobreak\ m}^{3}/\mathrm{s}\\ u_{1}=Q/A=3.77\mathrm{\leavevmode\nobreak\ m}/\mathrm{s}&\\ \frac{p_{1}}{\rho g}=18\mathrm{\leavevmode\nobreak\ m}&\frac{p_{2}}{\rho g}=-7\mathrm{\leavevmode\nobreak\ m}\end{array}$$

   Friction loss, from the question:

   $$h_{f}=0.1\frac{\left(p_{1}-p_{2}\right)}{\rho g}$$

   Apply Bernoulli:

   	$\displaystyle\frac{p_{1}}{\rho g}+\frac{u_{1}^{2}}{2g}+$	$\displaystyle=\frac{p_{2}}{\rho g}+\frac{u_{2}^{2}}{2g}+h_{f}$
   	$\displaystyle\frac{p_{1}}{\rho g}-\frac{p_{2}}{\rho g}+\frac{u_{1}^{2}}{2g}-h_{f}$	$\displaystyle=\frac{u_{2}^{2}}{2g}$
   	$\displaystyle 25-\frac{3.77^{2}}{2g}-2.5$	$\displaystyle=\frac{u_{2}^{2}}{2g}$
   	$\displaystyle u_{2}$	$\displaystyle=21.346\mathrm{\leavevmode\nobreak\ m}/\mathrm{s}$
   	$\displaystyle Q$	$\displaystyle=u_{2}A_{2}$
   	$\displaystyle 0.0667$	$\displaystyle=21.346\times\pi\frac{d_{2}^{2}}{4}$
   	$\displaystyle d_{2}$	$\displaystyle=0.063\mathrm{\leavevmode\nobreak\ m}$

8. E6.3.8
   ​

   A Venturimeter of throat diameter 0.076 m is fitted in a 0.152 m diameter vertical pipe in which liquid of relative density 0.8 flows downwards. Pressure gauges are fitted to the inlet and to the throat sections. The throat being 0.914 m below the inlet. Taking the coefficient of the meter as 0.97 find the discharge a) when the pressure gauges read the same b)when the inlet gauge reads $15170\mathrm{\leavevmode\nobreak\ N}/\mathrm{m}^{2}$ higher than the throat gauge.
   $\left[0.0192\mathrm{\leavevmode\nobreak\ m}^{3}/\mathrm{s},0.034\mathrm{\leavevmode\nobreak\ m}^{3}/\mathrm{s}\right]$
   

   From the question:

   $$\begin{array}[]{ll}d_{1}=0.152m&A_{1}=0.01814m\\ d_{2}=0.076m&A_{2}=0.00454m\\ \rho=800\mathrm{\leavevmode\nobreak\ kg}/\mathrm{m}^{3}&\\ C_{d}=0.97&\end{array}$$

   Apply Bernoulli:

   $$\frac{p_{1}}{\rho g}+\frac{u_{1}^{2}}{2g}+z_{1}=\frac{p_{2}}{\rho g}+\frac{u_{2}^{2}}{2g}+z_{2}$$

   a) $p_{1}=p_{2}$

   $$\frac{u_{1}^{2}}{2g}+z_{1}=\frac{u_{2}^{2}}{2g}+z_{2}$$

   By continuity:

   $$\begin{gathered}Q=u_{1}A_{1}=u_{2}A_{2}\\ u_{2}=u_{1}\frac{A_{1}}{A_{2}}=u_{1}4\\ \frac{u_{1}^{2}}{2g}+0.914=\frac{16u_{1}^{2}}{2g}\\ u_{1}=\sqrt{\frac{0.914\times 2\times 9.81}{15}}=1.0934\mathrm{\leavevmode\nobreak\ m}/\mathrm{s}\\ Q=C_{d}A_{1}u_{1}\\ Q=0.96\times 0.01814\times 1.0934=0.019\mathrm{\leavevmode\nobreak\ m}^{3}/\mathrm{s}\end{gathered}$$

   b)

   		$\displaystyle p_{1}-p_{2}=15170$
   	$\displaystyle\frac{p_{1}-p_{2}}{\rho g}=$	$\displaystyle\frac{u_{2}^{2}-u_{1}^{2}}{2g}-0.914$
   	$\displaystyle\frac{15170}{\rho g}=$	$\displaystyle\frac{Q^{2}\left(220.43^{2}-55.11^{2}\right)}{2g}-0.914$
   	$\displaystyle 55.8577=$	$\displaystyle Q^{2}\left(220.43^{2}-55.11^{2}\right)$
   	$\displaystyle Q=$	$\displaystyle 0.035\mathrm{\leavevmode\nobreak\ m}^{3}/\mathrm{s}$

9. E6.3.9
   ​

   A vertical cylindrical tank 2 m diameter has, at the bottom, a 0.05 m diameter sharp edged orifice for which the discharge coefficient is 0.6 .
   a) If water enters the tank at a constant rate of 0.0095 cumecs find the depth of water above the orifice when the level in the tank becomes stable.
   b) Find the time for the level to fall from 3 m to 1 m above the orifice when the inflow is turned off.
   c) If water now runs into the tank at 0.02 cumecs, the orifice remaining open, find the rate of rise in water level when the level has reached a depth of 1.7 m above the orifice.
   [a) 3.314 m , b) 881 seconds, c) $0.252\mathrm{\leavevmode\nobreak\ m}/\mathrm{min}$ ]
   

   

   From the question: $Q_{in}=0.0095\mathrm{\leavevmode\nobreak\ m}^{3}/\mathrm{s},d_{o}=0.05m,C_{d}=0.6$
   Apply Bernoulli from the water surface (1) to the orifice (2),

   $$\frac{p_{1}}{\rho g}+\frac{u_{1}^{2}}{2g}+z_{1}=\frac{p_{2}}{\rho g}+\frac{u_{2}^{2}}{2g}+z_{2}$$

   $p_{1}=p_{2}$ and $u_{1}=0.u_{2}=\sqrt{2gh}$.
   With the datum the bottom of the cylinder, $z_{1}=h,z_{2}=0$
   We need $Q$ in terms of the height $h$ measured above the orifice.

   	$\displaystyle Q_{\text{out }}$	$\displaystyle=C_{d}a_{o}u_{2}=C_{d}a_{o}\sqrt{2gh}$
   		$\displaystyle=0.6\pi\left(\frac{0.05}{2}\right)^{2}\sqrt{2\times 9.81}\sqrt{h}$
   		$\displaystyle=0.00522\sqrt{h}$		(1)

   For the level in the tank to remain constant:

   		$\displaystyle\text{ inflow }=\text{ out flow }$
   		$\displaystyle Q_{\text{in }}=Q_{\text{out }}$
   		$\displaystyle\begin{aligned} 0.0095&=0.00522\sqrt{h}\\ h&=3.314\mathrm{\leavevmode\nobreak\ m}\end{aligned}$

   (b) Write the equation for the discharge in terms of the surface height change:

   	$\displaystyle Q\delta t$	$\displaystyle=-A\delta h$
   	$\displaystyle\delta t$	$\displaystyle=-\frac{A}{Q}\delta h$

   Integrating between $h_{1}$ and $h_{2}$, to give the time to change surface level

   	$\displaystyle T$	$\displaystyle=-\int_{h_{1}}^{h_{2}}\frac{A}{Q}dh$
   		$\displaystyle=-601.8\int_{h_{1}}^{h_{2}}h^{-1/2}dh$
   		$\displaystyle=-1203.6\left[h^{1/2}\right]_{h_{1}}^{h_{2}}$
   		$\displaystyle=-1203.6\left[h_{2}^{1/2}-h_{1}^{1/2}\right]$

   $h_{1}=3$ and $h_{2}=1$ so

   $$T=881\mathrm{sec}$$

   c) $Q_{\text{in }}$ changed to $Q_{\text{in }}=0.02\mathrm{\leavevmode\nobreak\ m}^{3}/\mathrm{s}$

   From (1) we have $Q_{\text{out }}=0.00522\sqrt{h}$. The question asks for the rate of surface rise when $h=1.7\mathrm{\leavevmode\nobreak\ m}$.
   i.e.

   $$Q_{out}=0.00522\sqrt{1.7}=0.0068\mathrm{\leavevmode\nobreak\ m}^{3}/\mathrm{s}$$

   The rate of increase in volume is:

   $$Q=Q_{\text{in }}-Q_{\text{out }}=0.02-0.0068=0.0132\mathrm{\leavevmode\nobreak\ m}^{3}/\mathrm{s}$$

   As $Q=$ Area $\times$ Velocity, the rate of rise in surface is

   		$\displaystyle Q=Au$
   		$\displaystyle u=\frac{Q}{A}=\frac{0.0132}{\left(\frac{\pi 2^{2}}{4}\right)}=0.0042\mathrm{\leavevmode\nobreak\ m}/\mathrm{s}=0.252\mathrm{\leavevmode\nobreak\ m}/\mathrm{min}$

10. E6.3.10
    ​

    A horizontal boiler shell (i.e. a horizontal cylinder) 2 m diameter and 10 m long is half full of water. Find the time of emptying the shell through a short vertical pipe, diameter 0.08 m , attached to the bottom of the shell. Take the coefficient of discharge to be 0.8 .
    [1370 seconds]

    $$d=2\mathrm{\leavevmode\nobreak\ m}$$
    

    From the question $W=10m,D=10md_{o}=0.08mC_{d}=0.8$

    Apply Bernoulli from the water surface (1) to the orifice (2),

    $$\frac{p_{1}}{\rho g}+\frac{u_{1}^{2}}{2g}+z_{1}=\frac{p_{2}}{\rho g}+\frac{u_{2}^{2}}{2g}+z_{2}$$

    $p_{1}=p_{2}$ and $u_{1}=0.u_{2}=\sqrt{2gh}$.
    With the datum the bottom of the cylinder, $z_{1}=h,z_{2}=0$
    We need $Q$ in terms of the height $h$ measured above the orifice.

    	$\displaystyle Q_{\text{out }}$	$\displaystyle=C_{d}a_{o}u_{2}=C_{d}a_{o}\sqrt{2gh}$
    		$\displaystyle=0.8\pi\left(\frac{0.08}{2}\right)^{2}\sqrt{2\times 9.81}\sqrt{h}$
    		$\displaystyle=0.0178\sqrt{h}$

    Write the equation for the discharge in terms of the surface height change:

    	$\displaystyle Q\delta t$	$\displaystyle=-A\delta h$
    	$\displaystyle\delta t$	$\displaystyle=-\frac{A}{Q}\delta h$

    Integrating between $h_{1}$ and $h_{2}$, to give the time to change surface level

    $$T=-\int_{h_{1}}^{h_{2}}\frac{A}{Q}dh$$

    But we need $A$ in terms of $h$
    

    

    Surface area $A=10L$, so need $L$ in terms of $h$

    	$\displaystyle 1^{2}$	$\displaystyle=a^{2}+\left(\frac{L}{2}\right)^{2}$
    	$\displaystyle a$	$\displaystyle=(1-h)$
    	$\displaystyle 1^{2}$	$\displaystyle=(1-h)^{2}+\left(\frac{L}{2}\right)^{2}$
    	$\displaystyle L$	$\displaystyle=2\sqrt{\left(2h-h^{2}\right)}$
    	$\displaystyle A$	$\displaystyle=20\sqrt{2h-h^{2}}$

    Substitute this into the integral term,

    	$\displaystyle T$	$\displaystyle=-\int_{h_{1}}^{h_{2}}\frac{20\sqrt{2h-h^{2}}}{0.1078\sqrt{h}}dh$
    		$\displaystyle=-1123.6\int_{h_{1}}^{h_{2}}\frac{\sqrt{2h-h^{2}}}{\sqrt{h}}dh$
    		$\displaystyle=-1123.6\int_{h_{1}}^{h_{2}}\sqrt{\frac{2h-h^{2}}{h}}dh$
    		$\displaystyle=-1123.6\int_{h_{1}}^{h_{2}}\sqrt{2-h}dh$
    		$\displaystyle=-1123.6\left(-\frac{2}{3}\right)\left[(2-h)^{3/2}\right]_{h_{1}}^{h_{2}}$
    		$\displaystyle=749.07[2.828-1]=1369.6\mathrm{sec}$

11. E6.3.11
    ​

    Two cylinders standing upright contain liquid and are connected by a submerged orifice. The diameters of the cylinders are 1.75 m and 1.0 m and of the orifice, 0.08 m . The difference in levels of the liquid is initially 1.35 m . Find how long it will take for this difference to be reduced to 0.66 m if the coefficient of discharge for the orifice is 0.605 . (Work from first principles.)
    [30.7 seconds]
    

    
    $$\mathrm{d}_{\mathrm{o}}=0.108\mathrm{\leavevmode\nobreak\ m}$$

    		$\displaystyle A_{1}=\pi\left(\frac{1.75}{2}\right)^{2}=2.4m^{2}\quad A_{2}=\pi\left(\frac{1}{2}\right)^{2}=0.785m^{2}$
    		$\displaystyle d_{o}=0.08m,\quad a_{o}=\pi\left(\frac{0.08}{2}\right)^{2}=0.00503m^{2}\quad C_{d}=0.605$

    by continuity,

    $$-A_{1}\delta h_{1}=A_{2}\delta h_{2}=Q\delta t$$ (1)

    defining, $h=h_{1}-h_{2}$

    $$-\delta h=-\delta h_{1}+\delta h_{2}$$

    Substituting this in (1) to eliminate $\delta h_{2}$

    	$\displaystyle-A_{1}\delta h_{1}$	$\displaystyle=A_{2}\left(\delta h_{1}-\delta h\right)=A_{2}\delta h_{1}-A_{2}\delta h$
    	$\displaystyle\delta h_{1}$	$\displaystyle=\frac{A_{2}\delta h}{A_{1}+A_{2}}$
    	$\displaystyle-A_{1}\frac{A_{2}\delta h}{A_{1}+A_{2}}$	$\displaystyle=Q\delta t$		(2)

    From the Bernoulli equation we can derive this expression for discharge through the submerged orifice:

    $$Q=C_{d}a_{o}\sqrt{2gh}$$

    So

    		$\displaystyle-A_{1}\frac{A_{2}\delta h}{A_{1}+A_{2}}=C_{d}a_{o}\sqrt{2gh}\delta t$
    		$\displaystyle\delta t=-\frac{A_{1}A_{2}}{\left(A_{1}+A_{2}\right)C_{d}a_{o}\sqrt{2g}}\frac{1}{\sqrt{h}}\delta h$

    Integrating

    	$\displaystyle T$	$\displaystyle=-\frac{A_{1}A_{2}}{\left(A_{1}+A_{2}\right)C_{d}a_{o}\sqrt{2g}}\int_{h_{1}}^{h_{2}}\frac{1}{\sqrt{h}}dh$
    		$\displaystyle=-\frac{2A_{1}A_{2}}{\left(A_{1}+A_{2}\right)C_{d}a_{o}\sqrt{2g}}\left(\sqrt{h_{2}}-\sqrt{h_{1}}\right)$
    		$\displaystyle=\frac{-2\times 2.4\times 0.785}{(2.4+0.785)\times 0.605\times 0.00503\sqrt{2\times 9.81}}(0.8124-1.1619)$
    		$\displaystyle=30.7\mathrm{sec}$

12. E6.3.12
    ​

    A rectangular reservoir with vertical walls has a plan area of $60000\mathrm{\leavevmode\nobreak\ m}^{2}$. Discharge from the reservoir take place over a rectangular weir. The flow characteristics of the weir is $Q=0.678H^{3/2}$ cumecs where $H$ is the depth of water above the weir crest. The sill of the weir is 3.4 m above the bottom of the reservoir. Starting with a depth of water of 4 m in the reservoir and no inflow, what will be the depth of water after one hour?
    $[3.98\mathrm{\leavevmode\nobreak\ m}]$
    

    

    From the question $A=60000m^{2},Q=0.678h^{3/2}$
    Write the equation for the discharge in terms of the surface height change:

    	$\displaystyle Q\delta t$	$\displaystyle=-A\delta h$
    	$\displaystyle\delta t$	$\displaystyle=-\frac{A}{Q}\delta h$

    Integrating between $h_{1}$ and $h_{2}$, to give the time to change surface level

    	$\displaystyle T$	$\displaystyle=-\int_{h_{1}}^{h_{2}}\frac{A}{Q}dh$
    		$\displaystyle=-\frac{60000}{0.678}\int_{h_{1}}^{h_{2}}\frac{1}{h^{3/2}}dh$
    		$\displaystyle=2\times 88495.58\left[h^{-1/2}\right]_{h_{1}}^{h_{2}}$

    From the question $T=3600\mathrm{sec}$ and $h_{l}=0.6\mathrm{\leavevmode\nobreak\ m}$

    		$\displaystyle 3600=176991.15\left[h_{2}^{-1/2}-0.6^{-1/2}\right]$
    		$\displaystyle h_{2}=0.5815m$

    Total depth $=3.4+0.58=3.98\mathrm{\leavevmode\nobreak\ m}$

    Notches and weirs

13. E6.3.13
    ​

    Deduce an expression for the discharge of water over a right-angled sharp edged V-notch, given that the coefficient of discharge is 0.61 .
    A rectangular tank 16 m by 6 m has the same notch in one of its short vertical sides. Determine the time taken for the head, measured from the bottom of the notch, to fall from 15 cm to 7.5 cm .
    [1399 seconds]
    

    

    From your notes you can derive:

    $$Q=\frac{8}{15}C_{d}\tan\frac{\theta}{2}\sqrt{2g}H^{5/2}$$

    For this weir the equation simplifies to

    $$Q=1.44H^{5/2}$$

    Write the equation for the discharge in terms of the surface height change:

    	$\displaystyle Q\delta t$	$\displaystyle=-A\delta h$
    	$\displaystyle\delta t$	$\displaystyle=-\frac{A}{Q}\delta h$

    Integrating between $h_{1}$ and $h_{2}$, to give the time to change surface level

    	$\displaystyle T$	$\displaystyle=-\int_{h_{1}}^{h_{2}}\frac{A}{Q}dh$
    		$\displaystyle=-\frac{16\times 6}{1.44}\int_{h_{1}}^{h_{2}}\frac{1}{h^{5/2}}dh$
    		$\displaystyle=\frac{2}{3}\times 66.67\left[h^{-3/2}\right]_{h_{1}}^{h_{2}}$

    $h_{1}=0.15m,h_{2}=0.075m$

    	$\displaystyle T$	$\displaystyle=44.44\left[0.075^{-3/2}-0.15^{-3/2}\right]$
    		$\displaystyle=1399\mathrm{sec}$

14. E6.3.14
    ​

    Derive an expression for the discharge over a sharp crested rectangular weir. A sharp edged weir is to be constructed across a stream in which the normal flow is 200 litres/sec. If the maximum flow likely to occur in the stream is 5 times the normal flow then determine the length of weir necessary to limit the rise in water level to 38.4 cm above that for normal flow. $C_{d}=0.61$.
    $[1.24\mathrm{\leavevmode\nobreak\ m}]$
    

    

    From your notes you can derive:

    $$Q=\frac{2}{3}C_{d}b\sqrt{2g}h^{3/2}$$

    From the question:

    $$\begin{array}[]{ll}Q_{1}=0.2\mathrm{\leavevmode\nobreak\ m}^{3}/\mathrm{s},&h_{1}=x\\ Q_{2}=1.0\mathrm{\leavevmode\nobreak\ m}^{3}/\mathrm{s},&h_{2}=x+0.384\end{array}$$

    where $x$ is the height above the weir at normal flow.
    So we have two situations:

    	$\displaystyle 0.2=\frac{2}{3}C_{d}b\sqrt{2g}x^{3/2}=1.801bx^{3/2}$		(1)
    	$\displaystyle 1.0=\frac{2}{3}C_{d}b\sqrt{2g}(x+0.384)^{3/2}=1.801b(x+0.384)^{3/2}$		(2)

    From (1) we get an expression for $b$ in terms of $x$

    $$b=0.111x^{-3/2}$$

    Substituting this in (2) gives,

    	$\displaystyle 1.0$	$\displaystyle=1.801\times 0.111\left(\frac{x+0.384}{x}\right)^{3/2}$
    	$\displaystyle 5^{2/3}$	$\displaystyle=\left(\frac{x+0.384}{x}\right)$
    	$\displaystyle x$	$\displaystyle=0.1996\mathrm{\leavevmode\nobreak\ m}$

    So the weir breadth is

    	$\displaystyle b$	$\displaystyle=0.111(0.1996)^{-3/2}$
    		$\displaystyle=1.24\mathrm{\leavevmode\nobreak\ m}$

15. E6.3.15
    ​

    Show that the rate of flow across a triangular notch is given by $Q=C_{d}KH^{5/2}$ cumecs, where $C_{d}$ is an experimental coefficient, $K$ depends on the angle of the notch, and $H$ is the height of the undisturbed water level above the bottom of the notch in metres. State the reasons for the introduction of the coefficient.
    Water from a tank having a surface area of $10\mathrm{\leavevmode\nobreak\ m}^{2}$ flows over a $90^{\circ}$ notch. It is found that the time taken to lower the level from 8 cm to 7 cm above the bottom of the notch is 43.5 seconds . Determine the coefficient $C_{d}$ assuming that it remains constant during his period.
    [0.635]
    The proof for $Q=\frac{8}{15}C_{d}\tan\frac{\theta}{2}\sqrt{2g}H^{5/2}=C_{d}KH^{5/2}$ is in the notes.
    From the question:

    $$A=10\mathrm{\leavevmode\nobreak\ m}^{2}\quad\theta=90^{\circ}\quad h_{1}=0.08\mathrm{\leavevmode\nobreak\ m}\quad h_{2}=0.07\mathrm{\leavevmode\nobreak\ m}\quad T=43.5\mathrm{sec}$$

    So

    $$Q=2.36C_{d}h^{5/2}$$

    Write the equation for the discharge in terms of the surface height change:

    	$\displaystyle Q\delta t$	$\displaystyle=-A\delta h$
    	$\displaystyle\delta t$	$\displaystyle=-\frac{A}{Q}\delta h$

    Integrating between $h_{1}$ and $h_{2}$, to give the time to change surface level

    	$\displaystyle T$	$\displaystyle=-\int_{h_{1}}^{h_{2}}\frac{A}{Q}dh$
    		$\displaystyle=-\frac{10}{2.36C_{d}}\int_{h_{1}}^{h_{2}}\frac{1}{h^{5/2}}dh$
    		$\displaystyle=\frac{2}{3}\times\frac{4.23}{C_{d}}\left[h^{-3/2}\right]_{0.07}^{0.08}$
    	$\displaystyle 43.5$	$\displaystyle=\frac{2.82}{C_{d}}\left[0.07^{-3/2}-0.08^{-3/2}\right]$
    	$\displaystyle C_{d}$	$\displaystyle=0.635$

16. E6.3.16
    ​

    A reservoir with vertical sides has a plan area of $56000\mathrm{\leavevmode\nobreak\ m}^{2}$. Discharge from the reservoir takes place over a rectangular weir, the flow characteristic of which is $Q=1.77BH^{3/2}\mathrm{\leavevmode\nobreak\ m}^{3}/\mathrm{s}$. At times of maximum rainfall, water flows into the reservoir at the rate of $9\mathrm{\leavevmode\nobreak\ m}^{3}/\mathrm{s}$. Find a) the length of weir required to discharge this quantity if head must not exceed 0.6 m ; b) the time necessary for the head to drop from 60 cm to 30 cm if the inflow suddenly stops.
    [10.94m, 3093seconds]
    From the question:

    $$A=56000\mathrm{\leavevmode\nobreak\ m}^{2}\quad Q=1.77\mathrm{BH}^{3/2}\quad Q_{\max}=9\mathrm{\leavevmode\nobreak\ m}^{3}/\mathrm{s}$$

    a) Find B for $\mathrm{H}=0.6$

    		$\displaystyle 9=1.77B0.6^{3/2}$
    		$\displaystyle B=10.94m$

    b) Write the equation for the discharge in terms of the surface height change:

    	$\displaystyle Q\delta t$	$\displaystyle=-A\delta h$
    	$\displaystyle\delta t$	$\displaystyle=-\frac{A}{Q}\delta h$

    Integrating between $h_{1}$ and $h_{2}$, to give the time to change surface level

    	$\displaystyle T$	$\displaystyle=-\int_{h_{1}}^{h_{2}}\frac{A}{Q}dh$
    		$\displaystyle=-\frac{56000}{1.77B}\int_{h_{1}}^{h_{2}}\frac{1}{h^{3/2}}dh$
    		$\displaystyle=\frac{2\times 56000}{1.77B}\left[h^{-1/2}\right]_{0.6}^{0.3}$
    		$\displaystyle=5784\left[0.3^{-1/2}-0.6^{-1/2}\right]$
    	$\displaystyle T$	$\displaystyle=3093\mathrm{sec}$

17. E6.3.17
    ​

    Develop a formula for the discharge over a $90^{\circ}\mathrm{V}$-notch weir in terms of head above the bottom of the V. A channel conveys 300 litres $/\mathrm{sec}$ of water. At the outlet end there is a $90^{\circ}\mathrm{V}$-notch weir for which the coefficient of discharge is 0.58 . At what distance above the bottom of the channel should the weir be placed in order to make the depth in the channel 1.30 m ? With the weir in this position what is the depth of water in the channel when the flow is 200 litres/sec?
    $[0.755\mathrm{\leavevmode\nobreak\ m},1.218\mathrm{\leavevmode\nobreak\ m}]$
    Derive this formula from the notes: $Q=\frac{8}{15}C_{d}\tan\frac{\theta}{2}\sqrt{2g}H^{5/2}$
    From the question:

    $$\theta=90^{\circ}\quad C_{d}0.58\quad Q=0.3\mathrm{\leavevmode\nobreak\ m}^{3}/\mathrm{s},\quad\text{ depth of water, }Z=0.3\mathrm{\leavevmode\nobreak\ m}$$

    giving the weir equation:

    $$Q=1.37H^{5/2}$$

    a) As $H$ is the height above the bottom of the V , the depth of water $=Z=D+H$, where $D$ is the height of the bottom of the V from the base of the channel. So

    		$\displaystyle Q=1.37(Z-D)^{5/2}$
    		$\displaystyle 0.3=1.37(1.3-D)^{5/2}$
    		$\displaystyle D=0.755m$

    b) Find $Z$ when $Q=0.2\mathrm{\leavevmode\nobreak\ m}^{3}/\mathrm{s}$

    		$\displaystyle 0.2=1.37(Z-0.755)^{5/2}$
    		$\displaystyle Z=1.218m$

18. E6.3.18
    ​

    Show that the quantity of water flowing across a triangular V-notch of angle $2\theta$ is
    $Q=C_{d}\frac{8}{15}\tan\theta\sqrt{2g}H^{5/2}$. Find the flow if the measured head above the bottom of the V is 38 cm , when $\theta=45^{\circ}$ and $C_{d}=0.6$. If the flow is wanted within an accuracy of $2\%$, what are the limiting values of the head.
    $\left[0.126\mathrm{\leavevmode\nobreak\ m}^{3}/\mathrm{s},0.377\mathrm{\leavevmode\nobreak\ m},0.383\mathrm{\leavevmode\nobreak\ m}\right]$
    Proof of the v -notch weir equation is in the notes.
    From the question:

    $$H=0.38m\theta=45^{\circ}\quad C_{d}=0.6$$

    The weir equation becomes:

    	$\displaystyle Q$	$\displaystyle=1.417H^{5/2}$
    		$\displaystyle=1.417(0.38)^{5/2}$
    		$\displaystyle=0.126\mathrm{\leavevmode\nobreak\ m}^{3}/\mathrm{s}$

    $$\begin{gathered}Q+2\%=0.129\mathrm{\leavevmode\nobreak\ m}^{3}/\mathrm{s}\\ 0.129=1.417H^{5/2}\\ H=0.383\mathrm{\leavevmode\nobreak\ m}\end{gathered}$$

    	$\displaystyle Q-2\%$	$\displaystyle=0.124\mathrm{\leavevmode\nobreak\ m}^{3}/\mathrm{s}$
    	$\displaystyle 0.124$	$\displaystyle=1.417\mathrm{H}^{5/2}$
    	$\displaystyle H$	$\displaystyle=0.377\mathrm{\leavevmode\nobreak\ m}$