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# Part 8. Typical Scenarios

### i) Natural Flow

When the flow between say two tanks at different heights is unrestricted (i.e. no flow control valves), then a situation known as “natural flow” occurs. This means that the flow of water will increase until the frictional energy losses combined with the kinetic energy exactly equal the potential energy of the water in the top tank.

Figure 4. shows a typical situation, which is normal for the flow from a spring box to a reservoir tank:

Consider the Bernoulli equation for the top and bottom tanks: $P_1 + \frac{1}{2}.\rho.v_1^2 + \rho.g.h_1 = P_2 + \frac{1}{2}.\rho.v_2^2 + \rho.g.h_2 + f_h.\rho.g$

The important points to notice here are:

1. The pressure energy terms are equal at each tank (P1 = P2) because they are both at atmospheric pressure, so they can be eliminated.
2. The velocity of the water at the top tank is zero (v1 = 0), so the kinetic energy term for the top tank can be eliminated.

The equation can thus be rewritten as below and is typical for natural flow scenarios:

It shows that the potential energy of the water in the top tank must be converted into an exactly equal amount of combined frictional and kinetic energy.

Worked Example 1 shows how to calculate the natural flow rate.

A more complicated natural flow situation occurs when pipes of varying diameters are used to carry the flow to the reservoir tank. In this situation a similar process as the above is applied to each pipe, the Bernoulli Equations derived and combined to produce an equation that contains both friction terms. Then by trial and error a solution is obtained from the friction tables. Worked Example 2 gives the solution to such a problem.

### ii) Controlled Flow: Valves Open

In many situations the flow through a system is controlled by valves (particularly at distribution points like tap stands etc.). In these situations, unlike natural flow, there is a residual pressure in the pipes just behind the control valve. This is because the flow is limited and unable to burn off all the potential energy as frictional and kinetic energy. The control valve carries out this task, burning off the residual pressure as frictional energy.

Figure 5. shows a typical reservoir tank, control valve and tap scenario:

Consider the Bernoulli Equation for this scenario: $P_1 + \frac{1}{2}.\rho.v_1^2 + \rho.g.h_1 = P_2 + \frac{1}{2}.\rho.v_2^2 + \rho.g.h_2 + f_h.\rho.g$

The important points to note are:

1. The velocity of the water at the top tank is zero (v1 = 0), so the kinetic energy term for the top tank can be eliminated.
2. The difference between the two pressure terms (P2 P1) will be the residual pressure at the valve.

The equation can thus be rewritten as below and is typical for controlled flow scenarios: $P_2 - P_1 = \rho.g.h_1 - \rho.g.h_2 - \frac{1}{2}.\rho.v_2^2 - f_h.\rho.g$

As the flow rate is controlled, we know its value. Consequently, from the Flow Equation (8) we can calculate the velocity of the water (v2). This allows us to calculate the frictional head loss (fh), from tables. Thus we now have all the variables on the right hand side of the Bernoulli Equation and can calculate the residual head (P2 – P1).

Worked Example 3 shows this process in more detail.

### iii) Controlled Flow: Valves Shut

Consider the previous example, but this time we are going to shut the control valve. There is no flow. $P_1 + \frac{1}{2}.\rho.v_1^2 + \rho.g.h_1 = P_2 + \frac{1}{2}.\rho.v_2^2 + \rho.g.h_2 + f_h.\rho.g$

The important points with regard to the Bernoulli Equation are:

1. All the velocity terms are now zero (v1 = v2 = 0), so all the kinetic energy terms disappear.
2. There is no flow so there are no frictional losses (fh = 0).

So the Bernoulli equation collapses to the following, which is typical of all Valves Shut situations: $P_2 - P_1 = \rho.g.h_1 - \rho.g.h_2$

This is the maximum pressure situation at the valve and may design the pipe specification.

Worked Example 4 shows the calculations required.

In this situation we are usually interested in the power requirement for the pump. We know the flow rate required (Q) and will have chosen the pipe size (A). However this exercise can be carried out for various pipe sizes to determine the changing pump power requirements.

Consider the Bernoulli Equation for this situation (Equation (16) from Section 7), between the two tanks : $P1 + \frac{1}{2}.\rho.v_1^2 + \rho.g.h_1 + P_p = P_2 + \frac{1}{2}.\rho.v_2^2 + \rho.g.h_2 + f_h.\rho.g$

The important points to note are:

1. The velocity of the water at the bottom tank is zero (v1 = 0), so the kinetic energy term for the bottom tank can be eliminated.
2. The pressure requirement for the pump is the term Pp .
3. The pressure energy terms are equal at each tank (P1 = P2) because they are both at atmospheric pressure, so they can be eliminated.

So we can rewrite the Bernoulli Equation as below: $\rho.g.h_1 + P_p = \frac{1}{2}.\rho.v_2^2 + \rho.g.h_2 + f_h.\rho.g$

This can be rearranged to give the pump pressure requirement: $P_p = \frac{1}{2}.\rho.v_22^ + \rho.g.h_2 - \rho.g.h_1 + f_h.\rho.g$

We know Q, therefore from the Flow Equation we can calculate v2. With this term we can calculate the frictional head loss fh. So we can calculate the Pump Pressure requirement Pp.

From Section 7 Equation (20): $P_p = W_{in}.\gamma/Q$

Knowing the flow rate Q and the pump efficiency term γ we can calculate Win as below: $W_{in} = P_p.Q/\gamma$

This is the power requirement in Watts that needs to be supplied to the pump.

If the efficiency term is not known, then assume it is equal to 1 (in practice it is always less than 1). The equation will then give the required power to be supplied by the pump (Wout).

As an aside, the power supplied to the pump (Win ) will be equal to the electrical power supplied. Electrical power (We) is given by the following Equation: $W_e = I.V$

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Where I is the current (in Amps) and V is the voltage (in Volts). Typically the voltage is known (power cables in Mexico are approximately 110V, car batteries supply 12V) so it is possible to calculate the current requirement. This is useful for specification of controllers, cables and fuses.

Worked Example 5 gives a typical calculation of pump requirements.

### v) General Equation for Distribution Systems

In most gravity flow water systems we have to deliver water from a reservoir tank to tap stands at various locations in a community. Figure 7 shows a distribution system where all the tap supply pipes connect to a junction with the reservoir tank supply line (point A).

If we assume the flow requirement at each tap is known (q) and that there are n taps, then by the Continuity Equation (11): $q_A=q_1+q_2+q_3+\cdots +q_n=n.q$

Therefore we can say that at the junction: $n.q = A_t .v_A$ $v_A = n.q /A_t$

Now consider the Bernoulli Equation from the reservoir tank to the junction at point A: $P_t+\frac{1}{2}.\rho .v_t^2+\rho.g.h_t=P_A+\frac{1}{2}.\rho .v_A^2+\rho .g.h_A+f_{t-A}.\rho .g$
1. The pressure energy term at the reservoir tank will be considered to be zero (Pt = 0) because it is at atmospheric pressure.
2. The velocity of the water at the reservoir tank is zero (vt = 0), so the kinetic energy term can be eliminated.

The Bernoulli Equation becomes (after rearranging for PA ): $P_A = \rho.g.h_t - \rho.g.h_A - \frac{1}{2}.\rho.v_A^2 - f_{t-A}.\rho.g$

The Bernoulli Equation from the junction A to any typical tap is: $P_A + \frac{1}{2}.\rho.v_n^2 + \rho.g.h_A = P_n + \frac{1}{2}.\rho.v_n^2 + \rho.g.h_n + f_{A-n}.\rho.g$

Simplifying this we get: $P_n = P_A + \rho.g.h_A - \rho.g.h_n - f_{A-n}.\rho.g$

Substitute the equation for PA into this equation and simplify: $P_n = \rho.g.h_t - \rho.g.h_n - f_{t-A}.\rho.g - f_{A-n}.\rho.g - \frac{1}{2}.\rho.v_A^2$

Dividing by ρ.g to convert into head: $P_n/\rho.g = h_t -h_n - f_{t-A} - f_{A-n} - \frac{1}{2}.v_A^2/g$

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This is the General Equation for any tap in the distribution system. It allows us to calculate two things:

1. If we know the tap and the reservoir tank supply pipe sizes we can calculate the residual head at the tap (Pn/ρ.g).
2. Alternatively if we know the desirable residual head at the taps we can design the tap supply pipe size.

Worked Example 6 shows this process in more detail.