Unsteady flow in pipes

A surge tank is an open (or air-cushioned) storage chamber connected to a pressure conduit (penstock or tunnel) near its downstream end, to control pressure changes caused by sudden changes of flow.

Primary purposes

1. Reduce water hammer: it reflects the pressure wave at the tank, so the pipe length subject to water hammer is only the short length between the tank and the valve or turbine, instead of the whole conduit up to the reservoir.
2. Supply and absorb flow: when the turbine load increases, the tank supplies the extra water immediately; when the load is rejected, it takes the surplus water, so that the long upstream conduit does not have to accelerate or decelerate rapidly.

Secondary purposes

• Improves speed regulation of the turbine, because the water for the changed load is available at once.
• Reduces the pressure on the long tunnel/penstock and allows lighter pipe design.
• Acts as a reservoir of water for the first moments after a demand change, and allows the conduit to be left under steady head.

Water hammer

When the velocity of flowing water in a closed pipe is changed suddenly (valve closed quickly, pump stopped), the kinetic energy of the moving column is converted into pressure energy, producing a pressure wave that travels along the pipe at a speed $c$ (celerity) and is reflected at the ends. The pressure change is $\Delta p = \rho cV_0$. This repeated pressure rise and fall is called water hammer (it produces noise and vibration like hammering).

One cycle of the wave after sudden valve closure

Consider a pipe of length $L$ with a reservoir at the upstream end and a valve at the downstream end, initial velocity $V_0$ towards the valve. The valve is closed suddenly at $t = 0$. (Friction neglected; the wave speed $c$ is constant.)

Stage 1: $0 < t < L/c$ (wave travels from the valve to the reservoir)

 reservoir |<-------------- L -------------->| valve
           |=====[ V=V0 ]=====|====[ V=0 ]===X  closed
                     <---- c <----- pressure wave +dp

The layer at the valve is stopped at once and compressed; pressure rises by $\Delta p$ and the pipe expands. The wave front moves upstream at $c$; the water behind it ($V = 0$) is at high pressure; ahead of it, the water still moves at $V_0$. At $t = L/c$ the whole pipe is at $p_0 + \Delta p$ with $V = 0$.

Stage 2: $L/c < t < 2L/c$ (wave travels from the reservoir to the valve)

 reservoir |=====[ V=-V0 ]=====|====[ V=0, +dp ]===X
                     ----> c ----->  (reflected wave, pressure back to p0)

At the reservoir the pressure cannot stay above the reservoir pressure, so the water flows back into the reservoir with $V = -V_0$ and a wave $-\Delta p$ travels back towards the valve, restoring the pressure to normal. At $t = 2L/c$ the whole pipe is at normal pressure with $V = -V_0$ (flow towards the reservoir).

Stage 3: $2L/c < t < 3L/c$ (wave travels from the valve to the reservoir)

 reservoir |=====[ V=-V0, p0 ]=====|====[ V=0, p0-dp ]===X
                     <---- c <----- negative wave

The valve prevents any further backward flow, so $V$ drops to 0 at the valve and the pressure falls by $\Delta p$ below normal (possible column separation if the pressure falls below vapour pressure). The negative wave travels towards the reservoir. At $t = 3L/c$ the pipe is at $p_0 - \Delta p$, $V = 0$.

Stage 4: $3L/c < t < 4L/c$ (wave travels from the reservoir to the valve)

 reservoir |=====[ V=+V0, p0 ]=====|====[ V=0, p0-dp ]===X
                     ----> c ----->

The low pressure in the pipe draws water from the reservoir at $V = +V_0$; the wave restores normal pressure as it moves to the valve. At $t = 4L/c$ conditions are the same as at $t = 0$ just before closure (flow $V_0$ towards the valve with pressure $p_0$), and the cycle repeats. In reality friction damps the oscillations.

The period of one cycle is $T = 4L/c$.

Case. A long pipeline of length $L$ from a reservoir (upstream) to a valve (downstream) carries water at velocity $V_0$. The valve is closed instantaneously at $t = 0$. Neglect friction. Let $\Delta H = \dfrac{cV_0}{g}$ (pressure rise $\Delta p = \rho cV_0$) and the static head be $H_0$ (shown as the zero line below).

At the valve end

The pressure wave starts at the valve at $t = 0$, so the pressure there rises at once by $+\Delta H$ and stays at $H_0 + \Delta H$ until the negative wave reflected from the reservoir returns at $t = 2L/c$. Then the pressure drops by $2\Delta H$ to $H_0 - \Delta H$ (the closed valve prevents backward flow) and remains so until $t = 4L/c$, when the positive wave from the reservoir restores it to $H_0 + \Delta H$. The variation is a square wave of period $4L/c$. Computed from the wave tracking (two cycles, $+\Delta H$ above and $-\Delta H$ below the zero line):

************            ************            *


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            ************            ************
(top = 1.0, bottom = -1.0; '-' is zero line)

At the mid-point of the pipe ($L/2$ from the valve)

• $0 < t < L/2c$: no change (the wave has not arrived yet); pressure $H_0$.
• $L/2c < t < 3L/2c$: the wave reaches the mid-point; pressure $H_0 + \Delta H$ (duration $L/c$).
• $3L/2c < t < 5L/2c$: the reflected negative wave from the reservoir cancels it; pressure $H_0$.
• $5L/2c < t < 7L/2c$: the wave reflected from the valve (negative) arrives; pressure $H_0 - \Delta H$.
• $7L/2c < t < 9L/2c$: the positive wave from the reservoir restores pressure $H_0$; then the cycle repeats from $t = 9L/2c$.

   ******                  ******


***------******------******------******------****


               ******                  ******
(top = 1.0, bottom = -1.0; '-' is zero line)

The mid-point pressure wave has the same height ($\pm\Delta H$) but is out of phase with the valve and lasts for $L/c$ at each level, with $H_0$ in between, i.e. it has a "stepped" (rectangular pulse) shape.

Summary. At the valve the pressure alternates between $H_0 + \Delta H$ and $H_0 - \Delta H$ every $2L/c$; at the mid-point it follows the sequence $H_0 \to H_0 + \Delta H \to H_0 \to H_0 - \Delta H \to H_0$ with each stage lasting $L/c$ (except the first stage, which lasts $L/2c$). The period of both is $4L/c$.

Data. $L = 2300$ m, $D = 0.15$ m, wall thickness $t = 2.8$ mm, $Q = 2200$ l/min $= 0.03667$ m³/s, $K = 2\times10^9$ Pa, $E = 2.08\times10^{11}$ Pa. The valve is closed suddenly and completely.

Wave speed and pressure rise

Cross-section area $A = \dfrac{\pi}{4}(0.15)^2 = 0.017671$ m², so $V_0 = \dfrac{Q}{A} = 2.075$ m/s.

$$c = \sqrt{\frac{K/\rho}{1 + \dfrac{KD}{Et}}} = \sqrt{\frac{2\times10^9/1000}{1 + \dfrac{(2\times10^9)(0.15)}{(2.08\times10^{11})(0.0028)}}} = \frac{1414.2}{\sqrt{1 + 0.5151}} = 1148.9\ \text{m/s}$$ $$\Delta p = \rho cV_0 = 1000(1148.9)(2.075) = 2.384\ \text{MPa},\qquad \Delta H = \frac{\Delta p}{\rho g} = 243.0\ \text{m}$$

Time for the wave to travel the pipe: $L/c = 2.002$ s; $2L/c = 4.004$ s; period $4L/c = 8.007$ s.

(i) Pressure variation at the valve end

$\Delta p$ rises at $t = 0$ and stays at $+2.38$ MPa above the static pressure until $t = 2L/c = 4.00$ s. Then it falls to $-2.38$ MPa until $t = 4L/c = 8.01$ s, then rises again. (Square wave, period 8.01 s.)

************            ************            *


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            ************            ************
(top = 1.0, bottom = -1.0; '-' is zero line)

(ii) Pressure variation at 57.5 m from the upstream tank

The point is $d = 2300 - 57.5 = 2242.5$ m from the valve. Wave arrival times:

• $t_1 = d/c = 1.952$ s: pressure rises by $+2.38$ MPa
• $t_2 = (2L - d)/c = 2.052$ s: the wave reflected at the tank (negative) returns, pressure falls back to the static value
• $t_3 = (2L + d)/c = 5.956$ s: the negative wave reflected at the valve arrives, pressure falls to $-2.38$ MPa
• $t_4 = (4L - d)/c = 6.056$ s: the positive wave reflected at the tank arrives, pressure returns to the static value

So at this point the pressure rise is a short pulse: $+2.38$ MPa for only $0.100$ s (from 1.95 to 2.05 s), then the static value (up to 5.96 s), then $-2.38$ MPa for 0.100 s (5.96 to 6.06 s), then static until $t = 4L/c + t_1$, and the cycle repeats.

(not to scale in time; pulses are narrow)
 +2.38 |       __
       |      |  |
     0 +------+--+----------__----------> t (s)
       |    1.95 2.05    5.96  6.06      8.0
 -2.38 |                   |__|

(The pulses are only about 0.1 s wide because the point is close to the tank.)

Answer: $c = 1149$ m/s; $\Delta p = 2.38$ MPa. At the valve: $\pm2.38$ MPa square wave of half-period 4.00 s. At 57.5 m from the tank: short $\pm2.38$ MPa pulses of about 0.1 s each cycle of 8.01 s.

Data. $L = 2300$ m, $D = 0.15$ m, $t = 2.8$ mm, $Q = 2200$ l/min $= 0.03667$ m³/s, $K = 2\times10^9$ Pa, $E = 2.08\times10^{11}$ Pa. Instantaneous closure.

Velocity and wave speed

$$V_0 = \frac{Q}{\pi D^2/4} = \frac{0.03667}{0.017671} = 2.075\ \text{m/s}$$ $$c = \sqrt{\frac{K/\rho}{1 + \dfrac{KD}{Et}}} = \frac{\sqrt{2\times10^{6}}}{\sqrt{1 + \dfrac{2\times10^9(0.15)}{2.08\times10^{11}(0.0028)}}} = \frac{1414.2}{\sqrt{1 + 0.5151}} = 1148.9\ \text{m/s}$$

Check closure type: $2L/c = 4.00$ s, and the valve closes instantaneously, so the closure is rapid (full Joukowsky rise).

Water hammer pressure

$$\Delta p = \rho cV_0 = 1000(1148.9)(2.075) = 2.384\times10^6\ \text{Pa} = 2.38\ \text{MPa}\quad(\Delta H = 243.0\ \text{m of water})$$

Stress developed

Hoop (circumferential) stress due to the pressure rise, thin-walled pipe:

$$\sigma = \frac{\Delta p\,D}{2t} = \frac{(2.384\times10^6)(0.15)}{2(0.0028)} = 63.9\ \text{MPa}$$

(The longitudinal stress is about half of this, $\Delta p D/4t$.)

Pressure-time variation at 1725 m from the reservoir

The point is $d = 2300 - 1725 = 575$ m from the valve ($x = 1725$ m from the reservoir). Wave arrival times ($c = 1148.9$ m/s):

• $t_1 = d/c = 0.500$ s: wave from the valve arrives; pressure rises by $+2.38$ MPa
• $t_2 = (2L - d)/c = 3.503$ s: wave reflected from the reservoir arrives (negative); pressure falls back to the static value
• $t_3 = (2L + d)/c = 4.504$ s: negative wave reflected from the valve arrives; pressure falls to $-2.38$ MPa
• $t_4 = (4L - d)/c = 7.507$ s: positive wave reflected from the reservoir arrives; pressure returns to the static value
• The cycle repeats with period $4L/c = 8.007$ s.

   ******************


***------------------******------------------****


                           ******************
(top = 1.0, bottom = -1.0; '-' is zero line)

(Pressure change relative to static pressure: +1 = $+\Delta p$, $-1$ = $-\Delta p$; time from 0 to $4L/c = 8.0$ s.)

Answer: $\Delta p = 2.38$ MPa (243 m of water); hoop stress $\sigma = 63.9$ MPa; the pressure at 1725 m from the reservoir is a stepped wave with the changes at $t$ = 0.50, 3.50, 4.50 and 7.51 s.

Data. $V = 4$ m/s, $L = 4500$ m, $c = 1500$ m/s, sudden valve closure at the downstream end.

Maximum pressure rise (Joukowsky)

Sudden closure means $t_c \le 2L/c = 6.0$ s (rapid closure), so the full rise occurs:

$$\Delta H = \frac{cV}{g} = \frac{1500(4)}{9.81} = 611.6\ \text{m of water}$$ $$\Delta p = \rho cV = 1000(1500)(4) = 6.0\times10^{6}\ \text{Pa} = 6.0\ \text{MPa}$$

Pressure-time variation at 400 m upstream of the valve

Distance from the valve $d = 400$ m. Wave times:

• $t_1 = \dfrac{d}{c} = \dfrac{400}{1500} = 0.267$ s: the pressure wave arrives, pressure rises by $+611.6$ m ($+6.0$ MPa)
• $t_2 = \dfrac{2L - d}{c} = 5.733$ s: the negative wave reflected from the reservoir arrives, pressure returns to normal
• $t_3 = \dfrac{2L + d}{c} = 6.267$ s: the negative wave reflected from the valve arrives; pressure drops by $-611.6$ m
• $t_4 = \dfrac{4L - d}{c} = 11.733$ s: the positive wave reflected from the reservoir arrives; pressure returns to normal
• The cycle repeats every $4L/c = 12.0$ s.

  *********************


**---------------------***---------------------**


                          *********************
(top = 1.0, bottom = -1.0; '-' is zero line)

(+1 = $+\Delta H$, $-1$ = $-\Delta H$ relative to the normal pressure; time from 0 to $4L/c = 12$ s.)

Answer: Maximum pressure rise $= 612$ m of water ($6.0$ MPa). At 400 m from the valve the pressure is $+612$ m for 0.267 s to 5.733 s (and falls by the same amount from 6.267 s to 11.733 s); see the table, stepped wave of period 12 s.

Data. $H = 10$ m, $D = 0.15$ m, $L = 400$ m, $f = 0.025$, minor losses $= 15\dfrac{V^2}{2g}$. The valve is opened suddenly, so the water column (assumed rigid, incompressible) accelerates until the head is used up by losses.

Equation of motion

Applying Newton's law to the water column (mass $\rho AL$) with driving force $\rho gAH$ and resisting force from losses:

$$\rho AL\frac{dV}{dt} = \rho gAH - \rho gA\,K\frac{V^2}{2g}, \qquad K = f\frac{L}{D} + 15 = 66.67 + 15 = 81.67$$ $$\frac{dV}{dt} = \frac{g}{L}\left(H - K\frac{V^2}{2g}\right)$$

Ultimate (steady) velocity

At the final steady state, $dV/dt = 0$:

$$V_f = \sqrt{\frac{2gH}{K}} = \sqrt{\frac{2(9.81)(10)}{81.67}} = 1.550\ \text{m/s}\quad(Q_f = 27.39\ \text{l/s})$$

Solution

Writing $\dfrac{dV}{dt} = \dfrac{gH}{L}\left(1 - \dfrac{V^2}{V_f^2}\right)$ and integrating from $V = 0$ at $t = 0$:

$$\int_0^V\frac{dV}{1 - V^2/V_f^2} = \frac{gH}{L}t \;\Rightarrow\; V_f\tanh^{-1}\frac{V}{V_f} = \frac{gH}{L}t$$ $$V = V_f\tanh\left(\frac{gH}{LV_f}\,t\right) = 1.550\tanh(0.15823\,t)$$

where $\dfrac{gH}{LV_f} = \dfrac{9.81(10)}{400(1.550)} = 0.15823\ \text{s}^{-1}$.

(a) Time for 95% of ultimate flow

$$\tanh(0.15823\,t) = 0.95 \;\Rightarrow\; 0.15823\,t = \tanh^{-1}(0.95) = 1.8318$$ $$t = \frac{1.8318}{0.15823} = 11.58\ \text{s}$$

(b) Flow after 10 s

$$V = 1.550\tanh(0.15823\times10) = 1.550\tanh(1.5823) = 1.550(0.9190) = 1.424\ \text{m/s}$$ $$Q = AV = 0.017671(1.424) = 25.17\ \text{l/s}\quad(91.9\%\text{ of the ultimate flow})$$

Answer: (a) $t_{95} \approx 11.6$ s; (b) $Q \approx 25.2$ l/s at 10 s (the ultimate flow is 27.4 l/s).

Data. $D = 1.20$ m, $Q = 1.40$ m³/s, static head 300 m ($p_0 = \rho gH = 2.943$ MPa), $E = 210$ kN/mm² $= 210\times10^9$ Pa, safe stress $\sigma_s = 0.1$ kN/mm² $= 100$ MPa, $K = 2.10$ kN/mm² $= 2.1\times10^9$ Pa. Sudden closure, so the full Joukowsky rise applies.

$$V_0 = \frac{Q}{\pi D^2/4} = \frac{1.4}{1.1310} = 1.238\ \text{m/s}$$

Water hammer pressure

The wave speed depends on the wall thickness $t$ (unknown):

$$c = \frac{\sqrt{K/\rho}}{\sqrt{1 + \dfrac{KD}{Et}}}, \qquad \Delta p = \rho cV_0$$

First estimate (rigid pipe, $t \to \infty$): $c = \sqrt{K/\rho} = 1449$ m/s, so $\Delta p \approx 1.79$ MPa. This is the upper limit.

Thickness

The pipe must carry the maximum pressure $p_0 + \Delta p$ at the safe hoop stress:

$$\sigma_s = \frac{(p_0 + \Delta p)D}{2t} \;\Rightarrow\; t = \frac{(p_0 + \Delta p)D}{2\sigma_s}$$

Since $\Delta p$ depends on $t$, iterate. Start with $t$ = 10 mm and repeat: each new $t$ gives a new $c$ and $\Delta p$. The substitution converges to

$$t = 26.6\ \text{mm},\qquad c = 1202.9\ \text{m/s},\qquad \Delta p = \rho cV_0 = 1.489\ \text{MPa}$$ $$p_0 + \Delta p = 2.943 + 1.489 = 4.432\ \text{MPa}, \qquad t = \frac{4.432\times10^6(1.2)}{2(100\times10^6)} = 26.6\ \text{mm}$$

To the nearest (larger) millimetre: $t = 27$ mm.

Check with $t = 27$ mm: $c = 1205.8$ m/s, $\Delta p = 1.493$ MPa ($152.1$ m of water), maximum stress $= \dfrac{(2.943 + 1.493)\times10^6(1.2)}{2(0.027)} = 98.6$ MPa $\le 100$ MPa.

Answer: Water hammer pressure rise $\approx 1.49$ MPa ($\approx 152$ m of water; $1.79$ MPa if the pipe were rigid); minimum wall thickness $\approx 27$ mm.

Data. Static lift $H_s = 12$ m. Pipe 1: $L_1 = 30$ m, $D_1 = 0.225$ m, $f = 0.028$. Pipe 2: $L_2 = 100$ m, $D_2 = 0.15$ m, $f = 0.032$. Steady flow $Q_0 = 115$ l/s. After the pump stops instantaneously the water column continues to move for a short time, decelerating under the static head and friction. Treat the water as incompressible and the pipes as rigid (rigid-column theory).

Constants

• Areas: $A_1 = 0.03976$ m², $A_2 = 0.01767$ m².
• Inertia length: $\sum\dfrac{L}{A} = \dfrac{30}{0.03976} + \dfrac{100}{0.01767} = 6413.4\ \text{m}^{-1}$
• Friction coefficients, $h_f = rQ^2$ with $r = \dfrac{8fL}{\pi^2gD^5}$: $r_1 = 120.4$, $r_2 = 3481.9$, so $r = r_1 + r_2 = 3602.3$ s²/m⁵ (check: steady friction head at 115 l/s is $r Q_0^2 = 47.64$ m).

Equation of motion

For the whole water column (mass $\rho\sum AL$-type inertia) after pump failure the forces are the static head $H_s$ (pushing the water back) and friction (opposing the forward motion):

$$\frac{\sum (L/A)}{g}\frac{dQ}{dt} = -\left(H_s + rQ^2\right) \;\Rightarrow\; \frac{dQ}{dt} = -\frac{g}{I}\left(H_s + rQ^2\right)$$

Separate and integrate from $Q_0$:

$$\int_{Q_0}^{Q}\frac{dQ}{H_s + rQ^2} = -\frac{g}{I}t \;\Rightarrow\; \frac{1}{\sqrt{H_sr}}\left[\tan^{-1}\left(Q\sqrt{\frac{r}{H_s}}\right) - \tan^{-1}\left(Q_0\sqrt{\frac{r}{H_s}}\right)\right] = -\frac{g}{I}t$$ $$Q = \sqrt{\frac{H_s}{r}}\tan\left[\tan^{-1}\left(Q_0\sqrt{\frac{r}{H_s}}\right) - \frac{g\sqrt{H_sr}}{I}\,t\right]$$

Numbers

$$\sqrt{\frac{H_s}{r}} = \sqrt{\frac{12}{3602.3}} = 0.05772,\qquad Q_0\sqrt{\frac{r}{H_s}} = 0.115\sqrt{\frac{3602.3}{12}} = 1.9925,\qquad \tan^{-1}(1.9925) = 1.1056\ \text{rad}$$ $$\frac{g\sqrt{H_sr}}{I} = \frac{9.81\sqrt{12(3602.3)}}{6413.4} = 0.31803\ \text{s}^{-1}$$

At $t = 2$ s:

$$Q = 0.05772\tan\left(1.1056 - 0.31803(2)\right) = 0.05772\tan(0.4696) = 29.29\times10^{-3}\ \text{m}^3/\text{s}$$

(The flow stops completely after $t = 1.1056/0.31803 = 3.48$ s.)

Answer: The flow rate 2 s after the power failure is about $29.3$ l/s (still forward), decreasing from 115 l/s.

Data. $D = 0.15$ m, $L = 90$ m, static lift $H_s = 8$ m, $f = 0.028$, steady $Q_0 = 0.05$ m³/s. The pump stops instantaneously; minor losses and elasticity are neglected, so the water column decelerates as a rigid body under the static head (back-pressure) and friction.

Constants

$$A = \frac{\pi}{4}(0.15)^2 = 0.01767\ \text{m}^2,\qquad I = \frac{L}{A} = \frac{90}{0.01767} = 5093.0\ \text{m}^{-1}$$ $$r = \frac{8fL}{\pi^2gD^5} = \frac{8(0.028)(90)}{\pi^2(9.81)(0.15)^5} = 2742.0\ \text{s}^2/\text{m}^5 \quad(\text{steady friction head } rQ_0^2 = 6.85\ \text{m})$$

Equation of motion

$$\frac{L}{gA}\frac{dQ}{dt} = -\left(H_s + rQ^2\right) \;\Rightarrow\; dt = -\frac{I}{g}\frac{dQ}{H_s + rQ^2}$$

The flow into the tank stops when $Q = 0$. Integrating from $Q_0$ to 0:

$$t = \frac{I}{g}\int_0^{Q_0}\frac{dQ}{H_s + rQ^2} = \frac{I}{g\sqrt{H_sr}}\tan^{-1}\left(Q_0\sqrt{\frac{r}{H_s}}\right)$$ $$Q_0\sqrt{\frac{r}{H_s}} = 0.05\sqrt{\frac{2742.0}{8}} = 0.9257,\qquad \tan^{-1}(0.9257) = 0.7468\ \text{rad}$$ $$t = \frac{5093.0}{9.81\sqrt{8(2742.0)}}(0.7468) = \frac{0.7468}{0.28528} = 2.62\ \text{s}$$

Answer: The flow into the tank continues for about $2.62$ s (about 2.6 s) after the power failure.

Data. Normal net head at the valve $= 300$ m with $Q = 2.5$ m³/s; $D = 1$ m, $L = 1.6$ km, $f = 0.02$; valve closed gradually over $T = 8$ s with retardation $a \propto t^{5/4}$. The pipe is treated as rigid and the water as incompressible (rigid column theory); minor losses neglected.

Initial velocity and reservoir head

$$V_0 = \frac{Q}{\pi D^2/4} = 3.183\ \text{m/s},\qquad h_{f0} = f\frac{L}{D}\frac{V_0^2}{2g} = 0.02(1600)\frac{3.183^2}{19.62} = 16.53\ \text{m}$$

The head at the upstream end of the pipe (reservoir level) is the valve head plus the friction loss: $H_R = 300 + 16.53 = 316.53$ m.

Velocity during closure

$a = kt^{5/4}$ and the velocity falls to zero at $t = T$: $\displaystyle V_0 = \int_0^T kt^{5/4}\,dt = \frac{4}{9}kT^{9/4}$, so $k = \dfrac{9V_0}{4T^{9/4}}$. Then

$$V(t) = V_0 - \int_0^t a\,dt = V_0\left[1 - \left(\frac{t}{T}\right)^{9/4}\right]$$

At $t = 6$ s: $\left(\dfrac{6}{8}\right)^{2.25} = 0.5235$, so

$$V = 3.183(1 - 0.5235) = 1.517\ \text{m/s}$$

Head at the valve at $t = 6$ s

Retardation $a = \dfrac{dV}{dt}$ (magnitude) $= \dfrac{9}{4}V_0\dfrac{t^{5/4}}{T^{9/4}} = 2.25(3.183)\dfrac{6^{1.25}}{8^{2.25}} = 0.6248\ \text{m/s}^2$.

Equation of motion of the water column (deceleration raises the head at the valve):

$$H_{valve} = H_R - f\frac{L}{D}\frac{V^2}{2g} + \frac{L}{g}a$$ $$H_{valve} = 316.53 - 3.75 + \frac{1600(0.6248)}{9.81} = 316.53 - 3.75 + 101.91 = 414.7\ \text{m}$$

Answer: At $t = 6$ s, the velocity in the pipe is $1.52$ m/s and the head at the valve is about $415$ m (an increase of about $102$ m, minus the lower friction loss, above the normal 300 m).

Data. $D = 0.3$ m, $t = 8$ mm, $L = 1500$ m, $Q = 200$ l/s, $K = 2200$ MPa $= 2.2\times10^9$ Pa, $E = 80\times10^9$ Pa for cast iron.

Velocity and wave speed

$$V = \frac{Q}{\pi D^2/4} = \frac{0.2}{0.070686} = 2.829\ \text{m/s}$$ $$c = \frac{\sqrt{K/\rho}}{\sqrt{1 + \dfrac{KD}{Et}}} = \frac{\sqrt{2.2\times10^6}}{\sqrt{1 + \dfrac{(2.2\times10^9)(0.3)}{(80\times10^9)(0.008)}}} = \frac{1483.2}{\sqrt{1 + 1.0312}} = 1040.7\ \text{m/s}$$

(a) Maximum time of closure for rapid closure

Closure is rapid if $t_c \le \dfrac{2L}{c}$ (the time for the pressure wave to travel to the reservoir and back):

$$t_c \le \frac{2(1500)}{1040.7} = 2.883\ \text{s}$$

(b) Peak water hammer pressure (rapid closure)

$$\Delta p = \rho cV = 1000(1040.7)(2.829) = 2.945\ \text{MPa}\quad(\Delta H = \frac{cV}{g} = 300.2\ \text{m of water})$$

(c) Length of pipe subjected to peak pressure for $t_c = 2.0$ s

Since $2.0\ \text{s} < 2L/c = 2.88$ s, the closure is rapid, but the closing takes a finite time, so only part of the pipe sees the full peak pressure. At a point $d$ from the valve, the last pressure increment of the closure arrives at $t = t_c + d/c$, while the negative wave reflected from the reservoir arrives at $t = (2L - d)/c$. The full peak $\rho cV$ is reached only if the last increment arrives before the reflection:

$$t_c + \frac{d}{c} \le \frac{2L - d}{c} \;\Rightarrow\; d \le L - \frac{c\,t_c}{2}$$ $$L_{peak} = 1500 - \frac{1040.7(2.0)}{2} = 459.3\ \text{m}$$

This is the length of pipe measured from the valve towards the reservoir that experiences the peak pressure; the remaining $c\,t_c/2 = 1040.7$ m next to the reservoir does not.

Answer: (a) $t_c \le 2.88$ s; (b) $\Delta p = 2.94$ MPa; (c) about $459$ m of the pipe adjacent to the valve experiences the peak pressure.

Data. $L = 500$ m, $V_0 = 2$ m/s, static head 300 m, instantaneous valve closure, increase of circumferential stress $\Delta\sigma \le 20$ MPa, $K = 2$ GPa, $E = 200$ GPa. Let $x = D/t$.

Relations

Wave speed: $c = \dfrac{\sqrt{K/\rho}}{\sqrt{1 + \dfrac{K}{E}\dfrac{D}{t}}} = \dfrac{1414.2}{\sqrt{1 + 0.01x}}$ (since $K/E = 0.01$).

Pressure rise: $\Delta p = \rho cV_0 = 1000(2)c = 2000c$.

Hoop stress increase (thin-walled pipe): $\Delta\sigma = \dfrac{\Delta p\,D}{2t} = \dfrac{\Delta p\,x}{2}$.

Solve for $D/t$

$$\Delta\sigma = \frac{x}{2}\cdot\frac{2000(1414.2)}{\sqrt{1 + 0.01x}} = 20\times10^6 \;\Rightarrow\; \frac{x}{\sqrt{1 + 0.01x}} = \frac{40\times10^6}{2000(1414.2)} = 14.142$$

Square both sides: $x^2 = 200(1 + 0.01x) = 200 + 2x$, so

$$x^2 - 2x - 200 = 0 \;\Rightarrow\; x = 1 + \sqrt{201} = 15.18$$ $$\boxed{\frac{D}{t} \approx 15.2}$$

Maximum time for rapid closure

Wave speed with this thickness:

$$c = \frac{1414.2}{\sqrt{1 + 0.01(15.18)}} = 1317.7\ \text{m/s}$$

(then $\Delta p = 1000(1317.7)(2) = 2.64$ MPa, and $\Delta\sigma = 20.0$ MPa, as required; the maximum pressure is $2.943 + 2.64 = 5.58$ MPa.)

Closure is rapid if its time $t_c \le \dfrac{2L}{c}$:

$$t_c \le \frac{2(500)}{1317.7} = 0.759\ \text{s}$$

Answer: $D/t \approx 15.2$ (wall thickness is about 1/15 of the diameter); closure is rapid for $t_c \le 0.76$ s.

Water hammer phenomenon

When the velocity of water flowing in a pipe is changed quickly (valve closed or opened suddenly, pump or turbine started or stopped), the kinetic energy of the water is converted into pressure energy and a pressure wave of high intensity travels through the pipe at the speed of sound in the water-pipe system ($c$). The wave is reflected at the reservoir and at the valve and may cause pressure rise or fall of many times the normal pressure, with noise like hammering. This is water hammer. It can burst pipes (pressure rise) or collapse them and cause column separation (pressure fall).

For instantaneous closure, the pressure rise is (Joukowsky)

$$\Delta p = \rho c V_0,\qquad \Delta H = \frac{cV_0}{g},\qquad c = \sqrt{\frac{K/\rho}{1 + \dfrac{KD}{Et}}}$$

Euler (momentum) equation for unsteady pipe flow

Consider a short element of fluid of length $dx$ in a pipe of cross-section $A$, perimeter $P$, inclined at an angle $\theta$ to the horizontal, $x$ measured along the pipe. Forces along the pipe:

• Pressure: $pA - \left(p + \dfrac{\partial p}{\partial x}dx\right)A = -\dfrac{\partial p}{\partial x}A\,dx$
• Weight component: $-\rho gA\,dx\sin\theta$
• Wall shear: $-\tau_0P\,dx$

Newton's second law, with mass $\rho A\,dx$ and acceleration $\dfrac{DV}{Dt} = \dfrac{\partial V}{\partial t} + V\dfrac{\partial V}{\partial x}$:

$$\rho A\,dx\left(\frac{\partial V}{\partial t} + V\frac{\partial V}{\partial x}\right) = -\frac{\partial p}{\partial x}A\,dx - \rho gA\sin\theta\,dx - \tau_0P\,dx$$

Divide by $\rho A\,dx$. With the piezometric head $H = \dfrac{p}{\gamma} + z$ ($\dfrac{\partial z}{\partial x} = \sin\theta$), the pressure and weight terms combine to $-g\dfrac{\partial H}{\partial x}$. The wall shear $\tau_0 = \dfrac{f\rho V|V|}{8}$ and $\dfrac{P}{A} = \dfrac{4}{D}$ give $\dfrac{\tau_0P}{\rho A} = \dfrac{fV|V|}{2D}$:

$$\boxed{\frac{\partial V}{\partial t} + V\frac{\partial V}{\partial x} + g\frac{\partial H}{\partial x} + \frac{fV|V|}{2D} = 0}$$

This is Euler's (momentum) equation for unsteady flow in a pipe. In water hammer problems the convective term $V\,\partial V/\partial x$ is small compared with $\partial V/\partial t$ and is usually dropped.

Continuity equation for unsteady flow (elastic water and pipe)

Consider a pipe element of length $dx$. Mass inflow minus outflow equals the rate of change of mass in the element:

$$\frac{\partial(\rho AV)}{\partial x} + \frac{\partial(\rho A)}{\partial t} = 0 \quad(\text{with the convective term: } \frac{D(\rho A)}{Dt} + \rho A\frac{\partial V}{\partial x} = 0)$$

Hence $\dfrac{1}{\rho A}\dfrac{D(\rho A)}{Dt} + \dfrac{\partial V}{\partial x} = 0$. A change of pressure $dp$ changes the density and the area:

• Water compressibility (bulk modulus $K$): $\dfrac{d\rho}{\rho} = \dfrac{dp}{K}$
• Pipe elasticity (thin wall, hoop stress $\dfrac{pD}{2t}$, strain $\dfrac{dD}{D} = \dfrac{D\,dp}{2tE}$): $\dfrac{dA}{A} = 2\dfrac{dD}{D} = \dfrac{D\,dp}{tE}$

So $\dfrac{d(\rho A)}{\rho A} = dp\left(\dfrac{1}{K} + \dfrac{D}{tE}\right) = \dfrac{dp}{\rho c^2}$, which defines the wave speed

$$c^2 = \frac{K/\rho}{1 + \dfrac{KD}{tE}}$$

Therefore $\dfrac{1}{\rho c^2}\dfrac{Dp}{Dt} + \dfrac{\partial V}{\partial x} = 0$. Using $p = \gamma(H - z)$, $\dfrac{Dp}{Dt} = \gamma\left(\dfrac{\partial H}{\partial t} + V\dfrac{\partial H}{\partial x} - V\sin\theta\right)$:

$$\boxed{\frac{\partial H}{\partial t} + V\frac{\partial H}{\partial x} - V\sin\theta + \frac{c^2}{g}\frac{\partial V}{\partial x} = 0}$$

The terms $V\,\partial H/\partial x$ and $V\sin\theta$ are small, so the usual form is $\dfrac{\partial H}{\partial t} + \dfrac{c^2}{g}\dfrac{\partial V}{\partial x} = 0$.

Together with the momentum equation, these are the two equations that govern water hammer. They combine to the wave equation $\dfrac{\partial^2H}{\partial t^2} = c^2\dfrac{\partial^2H}{\partial x^2}$ (friction neglected), whose solutions are waves travelling at $\pm c$, and for a sudden change of velocity $\Delta V$ they give $\Delta H = \mp\dfrac{c}{g}\Delta V$.

Water hammer phenomenon

When the velocity of flow in a closed conduit is changed rapidly, the moving water column is suddenly slowed or accelerated. Its kinetic energy is converted into pressure (strain) energy, and a pressure wave travels along the pipe at wave speed $c = \sqrt{\dfrac{K/\rho}{1 + KD/(Et)}}$, reflecting at the reservoir and at the valve. The resulting rapid pressure rise and fall, with hammering noise, is water hammer. For instantaneous closure, $\Delta p = \rho cV_0$.

Causes

1. Sudden closing or opening of valves or gates (the commonest cause).
2. Starting or stopping of pumps, or power failure to the pump.
3. Sudden change of load on a turbine (governor action closing the guide vanes or nozzle).
4. Rupture of a pipe, or a sudden change in the demand at the end of a pipe.
5. Vibration of the machine or flow control equipment, and air trapped in the pipe.
6. Failure of a pump check valve, or a change in the reservoir head.

Momentum equation for unsteady flow through a pipe

Consider a short element of fluid of length $dx$ in a pipe of cross-section $A$, perimeter $P$, inclined at an angle $\theta$ to the horizontal, $x$ measured along the pipe. Forces along the pipe:

• Pressure: $pA - \left(p + \dfrac{\partial p}{\partial x}dx\right)A = -\dfrac{\partial p}{\partial x}A\,dx$
• Weight component: $-\rho gA\,dx\sin\theta$
• Wall shear: $-\tau_0P\,dx$

Newton's second law, with mass $\rho A\,dx$ and acceleration $\dfrac{DV}{Dt} = \dfrac{\partial V}{\partial t} + V\dfrac{\partial V}{\partial x}$:

$$\rho A\,dx\left(\frac{\partial V}{\partial t} + V\frac{\partial V}{\partial x}\right) = -\frac{\partial p}{\partial x}A\,dx - \rho gA\sin\theta\,dx - \tau_0P\,dx$$

Divide by $\rho A\,dx$. With the piezometric head $H = \dfrac{p}{\gamma} + z$ ($\dfrac{\partial z}{\partial x} = \sin\theta$), the pressure and weight terms combine to $-g\dfrac{\partial H}{\partial x}$. The wall shear $\tau_0 = \dfrac{f\rho V|V|}{8}$ and $\dfrac{P}{A} = \dfrac{4}{D}$ give $\dfrac{\tau_0P}{\rho A} = \dfrac{fV|V|}{2D}$:

$$\boxed{\frac{\partial V}{\partial t} + V\frac{\partial V}{\partial x} + g\frac{\partial H}{\partial x} + \frac{fV|V|}{2D} = 0}$$

This is Euler's (momentum) equation for unsteady flow in a pipe. In water hammer problems the convective term $V\,\partial V/\partial x$ is small compared with $\partial V/\partial t$ and is usually dropped.

Use in water hammer. With $V\,\partial V/\partial x$ and friction neglected, the equation reduces to $\dfrac{\partial V}{\partial t} + g\dfrac{\partial H}{\partial x} = 0$. Together with the continuity equation $\dfrac{\partial H}{\partial t} + \dfrac{c^2}{g}\dfrac{\partial V}{\partial x} = 0$, it gives the wave solution $\Delta H = \mp\dfrac{c}{g}\Delta V$.