4.1
A reservoir is circular in plan and the sides slope at an angle of tan^-1(1/5) to the horizontal. When the reservoir is full the diameter of the water surface is 50m. Discharge from the reservoir takes place through a pipe of diameter 0.65m, the outlet being 4m below top water level. Determine the time for the water level to fall 2m assuming the discharge to be cumecs where a is the cross sectional area of the pipe in m² and H is the head of water above the outlet in m.
[1325 seconds]

From the question: H = 4m a = p(0.65/2)² = 0.33m²

In time dt the level in the reservoir falls dh, so

Integrating give the total time for levels to fall from h₁ to h₂.

As the surface area changes with height, we must express A in terms of h.

A = pr²

But r varies with h.

It varies linearly from the surface at H = 4m, r = 25m, at a gradient of tan^-1 = 1/5.

r = x + 5h

25 = x + 5(4)

x = 5

so A = p( 5 + 5h )² = ( 25p + 25ph² + 50ph )

Substituting in the integral equation gives

From the question, h₁ = 4m h₂ = 2m, so

4.2
A rectangular swimming pool is 1m deep at one end and increases uniformly in depth to 2.6m at the other end. The pool is 8m wide and 32m long and is emptied through an orifice of area 0.224m², at the lowest point in the side of the deep end. Taking C_d for the orifice as 0.6, find, from first principles,
a) the time for the depth to fall by 1m b) the time to empty the pool completely.
[299 second, 662 seconds]

The question tell us a_o = 0.224m², C_d = 0.6

Apply Bernoulli from the tank surface to the vena contracta at the orifice:

p₁ = p₂ and u₁ = 0.

We need Q in terms of the height h measured above the orifice.

And we can write an equation for the discharge in terms of the surface height change:

Integrating give the total time for levels to fall from h₁ to h₂.

a) For the first 1m depth, A = 8 x 32 = 256, whatever the h.

So, for the first period of time:

b) now we need to find out how long it will take to empty the rest.

We need the area A, in terms of h.

So

Total time for emptying is,

T = 363 + 299 = 662 sec

4.3
A vertical cylindrical tank 2m diameter has, at the bottom, a 0.05m 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 3m to 1m 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 1.7m above the orifice.
[a) 3.314m, b) 881 seconds, c) 0.252m/min]

From the question: Q_in = 0.0095 m³/s, d_o=0.05m, C_d =0.6

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

p₁ = p₂ and u₁ = 0. .

With the datum the bottom of the cylinder, z₁ = h, z₂ = 0

We need Q in terms of the height h measured above the orifice.

For the level in the tank to remain constant:

inflow = out flow

Q_in = Q_out

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

Integrating between h₁ and h_2, to give the time to change surface level

h₁ = 3 and h₂ = 1 so

T = 881 sec

1. Q_in changed to Q_in = 0.02 m³/s

From (1) we have . The question asks for the rate of surface rise when h = 1.7m.

i.e.

The rate of increase in volume is:

As Q = Area x Velocity, the rate of rise in surface is

4.4
A horizontal boiler shell (i.e. a horizontal cylinder) 2m diameter and 10m long is half full of water. Find the time of emptying the shell through a short vertical pipe, diameter 0.08m, attached to the bottom of the shell. Take the coefficient of discharge to be 0.8.
[1370 seconds]

From the question W = 10m, D = 10m d_o = 0.08m C_d = 0.8

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

p₁ = p₂ and u₁ = 0. .

With the datum the bottom of the cylinder, z₁ = h, z₂ = 0

We need Q in terms of the height h measured above the orifice.

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

Integrating between h₁ and h_2, to give the time to change surface level

But we need A in terms of h.

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

Substitute this into the integral term,

4.5
Two cylinders standing upright contain liquid and are connected by a submerged orifice. The diameters of the cylinders are 1.75m and 1.0m and of the orifice, 0.08m. The difference in levels of the liquid is initially 1.35m. Find how long it will take for this difference to be reduced to 0.66m if the coefficient of discharge for the orifice is 0.605. (Work from first principles.)
[30.7 seconds]

by continuity,

defining, h = h₁ - h₂

Substituting this in (1) to eliminate dh₂

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

So

Integrating

4.6
A rectangular reservoir with vertical walls has a plan area of 60000m². Discharge from the reservoir take place over a rectangular weir. The flow characteristics of the weir is Q = 0.678 H^3/2 cumecs where H is the depth of water above the weir crest. The sill of the weir is 3.4m above the bottom of the reservoir. Starting with a depth of water of 4m in the reservoir and no inflow, what will be the depth of water after one hour?
[3.98m]

From the question A = 60 000 m², Q = 0.678 h ^3/2

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

Integrating between h₁ and h_2, to give the time to change surface level

From the question T = 3600 sec and h₁ = 0.6m

Total depth = 3.4 + 0.58 = 3.98m

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