CHAPTER
3 Mass
balance Consider part of a flow system, such for example as that shown in Fig. 3.3. 
This consists of a continuous pipe that changes its diameter, passing into and out of a unit of processing plant, which is represented by a tank. The processing equipment might be, for example, a pasteurizing heat exchanger. Also in the system is a pump to provide the energy to move the fluid.
In
the flow system of Fig. 3.3 we can apply the law of conservation
of mass to obtain a mass balance. Once the system is working steadily,
and if there is no accumulation of fluid in any part the system, the quantity
of fluid that goes in at section 1 must come out at section 2. If the
area of the pipe at section 1 is A_{1} , the velocity at
this section, v_{1} and the fluid density r_{1},
and if the corresponding values at section 2 are A_{2},
v_{2},
If the fluid is incompressible r_{1} = r_{2} so in this case
Equation (3.5) is known as the continuity equation for liquids and is frequently used in solving flow problems. It can also be used in many cases of gas flow in which the change in pressure is very small compared with the system pressure, such as in many airducting systems, without any serious error.
From eqn. (3.4):
where
suffixes 1, 2, 3 denote respectively raw milk, skim milk and cream. Also,
since volumes will be conserved, the total leaving volumes will equal
the total entering volume and so
This expression can be substituted for v_{2} in the mass balance equation to give:
From the known facts of the problem we have:
where r_{w} is the density of water. Substituting these values in eqn. (b) above we obtain:
Also from eqn. (a) we then have, substituting 0.23 m s^{1} for v_{3},
In addition to the mass balance, the other important quantity we must consider in the analysis of fluid flow, is the energy balance. Referring again to Fig. 3.3, we shall consider the changes in the total energy of unit mass of fluid, one kilogram, between Section 1 and Section 2. Firstly,
there are the changes in the intrinsic energy of the fluid itself which
include changes in: Fluid
maintained above the datum level can perform work in returning to the
datum level. The quantity of work it can perform is calculated from the
product of the distance moved and the force resisting movement; in this
case the force of gravity. This quantity of work is called the potential
energy of the fluid.
Fluid that is in motion can perform work in coming to rest. This is equal to the work required to bring a body from rest up to the same velocity, which can be calculated from the basic equation
where v (m s^{1}) is the final velocity of the body, a (m s^{2}) is the acceleration and s (m) is the distance the body has moved. Also work done = W = F x s, and from Newton's Second Law, for m kg of fluid
and so E_{k} = W = mas = mav^{2}/2a = mv^{2}/2 The energy of motion, or kinetic energy, for 1 kg of fluid is therefore given by E_{k} where
Fluids exert a pressure on their surroundings. If the volume of a fluid is decreased, the pressure exerts a force that must be overcome and so work must be done in compressing the fluid. Conversely, fluids under pressure can do work as the pressure is released. If the fluid is considered as being in a cylinder of crosssectional area A (m^{2}) and a piston is moved a distance L (m) by the fluid against the pressure P (Pa) the work done is PAL joules. The quantity of the fluid performing this work is ALr (kg). Therefore the pressure energy that can be obtained from one kg of fluid (that is the work that can be done by this kg of fluid) is given by E_{r} where
When a fluid moves through a pipe or through fittings, it encounters frictional resistance and energy can only come from energy contained in the fluid and so frictional losses provide a drain on the energy resources of the fluid. The actual magnitude of the losses depends upon the nature of the flow and of the system through which the flow takes place. In the system of Fig. 3.3, let the energy lost by 1 kg fluid between section 1 and section 2, due to friction, be equal to E_{ƒ} (J). If there is a machine putting energy into the fluid stream, such as a pump as in the system of Fig. 3.3, the mechanical energy added by the pump per kg of fluid must be taken into account. Let the pump energy added to 1 kg fluid be E_{c} (J). In some cases a machine may extract energy from the fluid, such as in the case of a water turbine. Heat might be added or subtracted in heating or cooling processes, in which case the mechanical equivalent of this heat would require to be included in the balance. Compressibility terms might also occur, particularly with gases, but when dealing with low pressures only they can usually be ignored. For the present let us assume that the only energy terms to be considered are E_{p}, E_{k}, E_{r}, E_{f}, E_{c}.
We are now in a position to write the energy balance for the fluid between
section 1 and section 2 of Fig. 3.3.
In the special case
where no mechanical energy is added and for a frictionless fluid,
and since this is true for any sections of the pipe the equation can also be written
Equation (3.9) is known as Bernouilli's equation. First discovered by the Swiss mathematician Bernouilli in 1738, it is one of the foundations of fluid mechanics. It is a mathematical expression, for fluid flow, of the principle of conservation of energy and it covers many situations of practical importance. Application of the equations of continuity, eqn. (3.4) or eqn. (3.5), which represent the mass balance, and eqn. (3.7) or eqn. (3.9), which represent the energy balance, are the basis for the solution of many flow problems for fluids. In fact much of the remainder of this chapter will be concerned with applying one or another aspect of these equations. The Bernouilli equation is of sufficient importance to deserve some further discussion. In the form in which it has been written in eqn. (3.9) it will be noticed that the various quantities are in terms of energies per unit mass of the fluid flowing. If the density of the fluid flowing multiplies both sides of the equation, then we have pressure terms and the equation becomes:
and the respective terms are known as the potential head pressure, the velocity pressure and the static pressure. On the other hand, if the equation is divided by the acceleration due to gravity, g, then we have an expression in terms of the head of the fluid flowing and the equation becomes:
and
the respective terms are known as the potential head, the velocity head
and the pressure head. If there is a constriction in a pipe and the static pressures are measured upstream or downstream of the constriction and in the constriction itself, then the Bernouilli equation can be used to calculate the rate of flow of the fluid in the pipe. This assumes that the flow areas of the pipe and in the constriction are known. Consider the case in which a fluid is flowing through a horizontal pipe of crosssectional area A_{1} and then it passes to a section of the pipe in which the area is reduced to A_{2}. From the continuity equation [eqn. (3.5)] assuming that the fluid is incompressible:
Since the pipe is horizontal
Substituting in eqn. (3.8)
and since r_{1} = r_{2} as it is the same fluid throughout and it is incompressible,
From eqn. (3.12), knowing P_{1}, P_{2}, A_{1}, A_{2}, r_{1}, the unknown velocity in the pipe, v_{1}, can be calculated. Another application of the Bernouilli equation is to calculate the rate of flow from a nozzle with a known pressure differential. Consider a nozzle placed in the side of a tank in which the surface of the fluid in the tank is Z ft above the centre line of the nozzle as illustrated in Fig. 3.4. Take the datum as the centre of the nozzle. The velocity of the fluid entering the nozzle is approximately zero, as the tank is large compared with the nozzle. The pressure of the fluid entering the nozzle is P_{1} and the density of the fluid r_{1}. The velocity of the fluid flowing from the nozzle is v_{2} and the pressure at the nozzle exit is 0 as the nozzle is discharging into air at the datum pressure. There is no change in potential energy as the fluid enters and leaves the nozzle at the same level. Writing the Bernouilli equation for fluid passing through the nozzle:
but P_{1} /r_{1} = gZ (where Z is the head of fluid above the nozzle) therefore
Now
substituting in eqn. (3.12)
EXAMPLE
3.7. Mass flow rate from a tank
Now area of pipe, A
Volumetric flow rate, Av
Mass flow rate, rAv
EXAMPLE
3.8. Pump horsepower Volume of flow, V
Area of pipe, A
Velocity in pipe, v
And so applying
eqn. (3.7)
Therefore total power required
and, since 1 h.p. = 7.46 x 10^{2} J s^{1},
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