Bloggers: Continuity Equation & Bernoulli Equation

There are basically two equations needed to design gravity flow water systems:

The Continuity Equation
The Bernoulli Equation

With these two relations and an understanding of frictional effects most systems can be designed and/or analysed.

The following are the basic equations which can be used to derive the Continuity and Bernoulli Equations:

(a) Velocity = Distance Moved/Time

$v=s/t$

Metres/Second (m/s)(1)

Example: If I run 100 metres in 10 seconds (in a straight line), I have a velocity of 100/10 = 10 metres per second (m/s).

(b) Acceleration = Velocity Change/Time

$A=\left ( V_2-V_1 \right )\setminus t$

Metres/Second2 (m/s2)(2)

Example: If at the start of the 100m race I have a velocity of zero (I am in the blocks) and after 2 seconds I have a velocity of 10 m/s then my acceleration is (10-0)/2 = 5 metres per second per second (a = (v2 – v1)/t ). This means I increase my velocity by 5 metres/second every second that passes. This is my average acceleration.

(c) Force = Mass.Acceleration

$F=M.a$

Kilogramme Metres/Second2 (Kg.m/s2) or Newtons (N)(3)

Example: If I throw a bucket of water at you, you will feel a force when the water hits you because the water is decelerating. Suppose the water had a mass of 10 Kg , was travelling at 5m/s before it hit you and took 1 second to slow down. The deceleration of the water is 5/1 = 5 metres per second per second and so the force applied to you is 5 x 10 = 50 Newtons (F = M.a).

(d) Pressure = Force/Area

$P=F/A$

Newtons/Meter 2 (N/m2) or Pascals (Pa)(4)

Example: I wear my training shoes and step on your foot. If I then step on your foot wearing my stilettos, you will feel the difference in pressure. This is because the force (my weight) is the same, but the area over which it is applied (the heel of the stiletto) is smaller, so the pressure is higher. If I have a mass of 70 Kg, then my force due to gravity is 70 x 9.81 = 687 Newtons. If the area of my training shoe is 0.0075 m2 (75 cm2) then the pressure due to it on your foot is 687/0.0075 = 91600 N/m2 or Pa. If the area of my stiletto heel is 0.0001 m2 (1cm2) then the pressure due to it on your foot is 687/0.0001 = 6870000 N/m2 or Pa. This is why it hurts !

(e) Density = Mass/Volume

$\rho=M/V$

Kilogrammes/Meter3 (Kg/m3)(5)

Example: A bucket of water and a bucket of air do not have the same weight. This is because, although the volume of each substance is the same (a bucket full), the water is about 1000 times denser than air. Density (ρ, pronounced ‘ro’) is a measure of how closely the atoms or molecules of a substance are packed together. The density of water is 1000 Kg/m3. Which means that if I have 10 Kg of water it will occupy a volume of 10/1000 = 0.01 m3 (10 litres).

(f) Energy = Force.Distance Moved

$E=F.S$

Newton Metres (N.m) or Joules (J)(6)

Example: If I pull a bucket of water on a rope to the top floor of my house then I am expending energy (I get tired). This is because I am applying a force (through my arms) to lift the bucket through a distance h (the height of my house) against the force of gravity. If the water has a mass of 10 Kg and knowing that the gravitational acceleration on earth is 9.81 m/s/s, then the force required to lift the water is 10 x 9.81 = 981 Newtons. If the height of the house is 10 metres, then the energy required to lift the bucket is 981 x 10 = 9810 Joules (E = F.S).

(g) Power = Energy/Time

$W=E/ t$

Newton Metres/Second (N.m/s) or Watts (W)(7)

Example: If I lift the bucket from the previous example to the top of the house in 10 seconds then the power required to carry out the task is 9810/10 = 981 Watts (W = E/t).

Consider an incompressible fluid (water is almost incompressible) flowing along a pipe, as in Figure 1.

An incompressible fluid flowing along a pipe.

Its volume (V) is given by:

$V=A.L$

Therefore the volume passing per second (the volumetric flow rate Q) is given by:

$Q = V/t = A.L/t$

But we can write velocity as distance moved/time (see Equation (1)), so we can replace L/t by v:

$Q = A.v$

(9)

This is the FLOW EQUATION.

Now consider pipes of different areas A1 and A2 as shown in Figure 2.

The volumetric flow rate (Q) must be the same for both pipes, because we cannot gain or lose any fluid.

Therefore from Equation (8) above:

$Q = A_1.v_1 = A_2.v_2$

(10)

This is the CONTINUITY EQUATION and it is true for any number of changes in pipe diameter for a single pipe arrangement (a single flow path).

The rule for multiple flow paths for incompressible fluids is:

$\text{Sum of the Flows in} = \text{Sum of the Flows out}$

This is written mathematically as:

$\sum Q_{in}=\sum Q_{out}$ (10)

Consider the pipe system shown below (in section) in Figure 3:

In this case the flow in is given by:

$\sum Q_{in}=A_1v_1$

And the flow out is given by:

$\sum Q_{out}=A_2v_2+A_3v_3$

So from Equation X we can write:

$A_1v_1 = A_2v_2 + A_3v_3$ (11)

This is true for any number of flows in and out.

Anywhere in a perfect system (i.e. there are no frictional effects), for an incompressible fluid there are three types of energy existing:

Pressure Energy. Example: If you blow up the tyre of a car with a pump you are turning your physical energy of working the pump into pressure energy in the tyre.
Kinetic Energy. Example: This is the energy contained in a moving fluid. If a wave hits you at the beach, you feel the kinetic energy contained within it.
Potential Energy. Example: Gravity is trying to pull water to the lowest point on the earth’s surface. So when water is at a high point it contains energy, which can “potentially” allow it to flow down.

At any point in a perfect system the sum of these “bits” of energy in different forms (Pressure, Kinetic and Potential) must equal the sum of these “bits” at any other point in the system.

This is because energy cannot be “created” or “destroyed”, it can only change its form. This is what theBERNOULLI EQUATION expresses.

Appendix 1. gives the derivation of the Kinetic Energy in terms of a pressure and Appendix 2. gives the derivation of the Potential Energy in terms of a pressure.

These different forms of energy are expressed mathematically (as pressures) in the Bernoulli Equation (for a perfect system) shown below:

$P_1 + \frac{1}{2}.\rho .v_1^2 + \rho .g.h_1 = P_2 + \frac{1}{2}.\rho .v_2^2 + \rho .g.h_2$

(12)

The terms on the left hand side of the Equation are as follows:

P1 is the pressure energy at point 1 (expressed as a pressure). [Units are N/m2 or Pa]
ρ is the density of the fluid.[Units are Kg/m3]
v1 is the velocity of the fluid at point 1. [Units are m/s]
g is the acceleration due to earth’s gravity (9.81 m/s/s).[Units are m/s/s]
h1 is the height (from a given datum) of the fluid at point 1.[Units are m]

The terms are similar on the right hand side of the Equation, but for point 2.

The left hand side of the equation represents all the “bits” of energy (expressed as pressures) at a point 1 in a perfect system and the right hand side all the “bits” of energy (expressed as pressures) at another point 2.

Of course in the real world, systems are not perfect. Energy is lost from a real system as friction. This is the sound and heat generated by the fluid as it flows through the pipes. These forms of energy are lost by the fluid rubbing against the walls of the pipe, rubbing against itself and by turbulence in the flow. The amount of frictional loss is affected by the following parameters:

The length of the pipe. The longer the pipes the greater the frictional losses.
The roughness of the pipe walls. The smoother the surface of the pipe the smaller the frictional losses.
The diameter of the pipe. The smaller the diameter of the pipe the greater the frictional losses.
The velocity of the fluid. The greater the velocity of the fluid the greater the frictional losses.
The type of flow of the fluid. Turbulent flow creates greater friction losses than laminar flow.
Changes in the shape or section of the pipe. Fittings, valves, bends etc. all increase frictional losses.

Frictional losses are non-linear. This means that if you double one of the above parameters then the frictional losses can triple or quadruple in size. Also if there is no flow there are no frictional losses. Appendix 3 gives a fuller explanation of the relationship between the above variables.

Friction is included as a term in the Bernoulli Equation (usually on the right hand side) because it represents a fraction of the overall total energy at a point in the system, although it has been lost in between those two points. Equation (13) is the Bernoulli Equation including the friction term as a pressure (fp).

$P_1 + \frac{1}{2}.\rho .v_1^2 + \rho.g.h_1 = P_2 + \frac{1}{2}.\rho .v_2^2 + \rho.g.h_2 + f_p$

(13)

In practice frictional head loss (fh) is calculated from tables that require pipe type, diameter, length and flow rate. Frictional head loss can be converted into frictional loss as a pressure by the following relation:

$f_p = f_h.\rho.g$

(14)

So Equation (13) can be rewritten as:

$P_1 + \frac{1}{2}.\rho.v_1^2 + \rho.g.h_1 = P_2 + \frac{1}{2}.\rho.v_2^2 + \rho.g.h_2 + f_h.\rho.g$

(15)

Essentially a pump adds energy to a system and a turbine takes it away. Therefore typically in the Bernoulli Equation the pump pressure (Pp) is added to the left-hand side of the equation and the turbine pressure (Pt) is added to the right. So for a system containing a pump and a turbine the Bernoulli equation would look something like this:

$P_1 + \frac{1}{2}.\rho.v_1^2 + \rho.g.h_1 + P_p = P_2 + \frac{1}{2}.\rho.v_2^2 + \rho.g.h_2 + f_h.\rho.g + P_t$

(16)

If you know the power (Wout) that is delivered by a pump, then it is possible to calculate the pressure it represents (Pp).

From Equations (6) and (7):

$W_{out} = F.S/t$

From Equation (1):

$v = S/t$

So we can write:
look something like this:

$W_{out} = F.v$

(17)

Now from Equation (4):

$P = F/A$

so:

$F = P.A$

Combining this with Equation (17):

$W_{out} = P.A.v$

From Equation (8):

$Q = A.v$

So we can write:

$W_{out} = P.Q$

Thus the energy the pump provides as a pressure (Pp) is given by:

$P_p = W_{out}/Q$

(18)

Typically pumps have an efficiency (γ)which is the ratio of the power out (Wout ) to the power in (Win ). It represents the losses in the pump due to friction and electrical efficiency. It is usually expressed as a percentage and will always be less than 100%. It should be applied in the equation as a fraction. We can write therefore:
look something like this:

$W_{out} = W_{in}.\gamma$

(19)

Combining Equations (18) and (19) gives:
look something like this:

$P_p = W_{in}.\gamma /Q$

(20)

This can be substituted into the Bernoulli Equation (16) and allows the determination of the pump power requirement or alternatively the flow rate in a system for a given pump power.

Analysis of turbines will not be dealt with in detail here but is very similar to that of pumps (but in reverse).

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Wednesday, March 02, 2016

Continuity Equation & Bernoulli Equation