Application of the Bernoulli Theorem

Application of the Bernoulli Theorem

1.0 Solving problemsApplication of the Bernoulli theorem should be rational and systematic. Suggested procedure is as follows:

(i) Draw a sketch of the system and label all cross sections of the stream under consideration.

(ii) Apply the Bernoulli equation in the direction of flow. Select a datum plan for each equation written.

(iii) Evaluate the energy upstream at section 1. The energy can be given in terms of heads (in m). Evaluate the total head at section 2. Use average velocities and gauge pressures where possible.

(iv) Add any energy contributed by mechanical device, such as pumps, and subtract any energy loss during the flow or energy extracted by mechanical devices, such as turbines.

(v) Use the above information in the Bernoullis equation.

(vi) If the velocities at the two sections are unknown, use the equation of continuity to relate them and, hence, solve the problem.

1.1 The Pitot tubeThe Pitot tube is used to measure the velocity of a stream and consists of a simple L-shaped tube facing in the oncoming flow. Referring to fig. 1, if the velocity of the stream at A is u, a particle moving from A to the mouth of the tube B will be brought to rest so that u0 at B is zero.

Fig 1(a)

Fig. 1(b)

Applying Bernoullis theorem,Total energy per unit weight at A=Total energy per unit weight at B

u2/2g + p/(g=u02/2g + p0 /(g

p0 /(g=u2/2g + p/(g

Thus p0 will be greater than p. Now,p/(g = z and p0 /(g = h + zTherefore,

u2/2g = (p0 p) /(g = hVelocity at A = u = ((2gh)When the pitot tube is used in a channel, the value of h can be determined directly, but if it to be used in a pipe, the difference between the static pressure and the pressure at the impact hole must be measured with a differential pressure gauge, as shown in fig. 1 (b).

Theoretically, the measured velocity u = ((2gh), but Pitot tubes may require calibration. The true velocity is given by u = C((2gh), where C is the coefficient of the instrument.1.2 Changes of pressure in a tapering pipe

Changes of velocity in a tapering pipe were determined using the continuity of flow equation. Change of velocity will be accompanied by a change in the kinetic energy per unit weight and consequently, by a change in pressure, modified by any change of elevation or energy loss, which can be determined by the use of Bernoullis equation.ExampleA pipe is inclined at 450 to the horizontal converges over a length l of 2 m from the diameter d1 of 200 mm diameter to a diameter d2 of 100 mm at the upper end. Oil of relative density 0.9 flows through the pipe at a mean velocity at the lower end at 2 ms-1. Find the pressure difference across the 2 m length ignoring any loss of energy, and the difference in level that would be shown on the mercury manometer, specific gravity of mercury is 13.6 and the leads to the manometer are filled with the oil.

Let A1, , p1, d1, z1 and A2, , p2, d2, z2 be the area, mean velocity, pressure, diameter and elevation at the lower and upper sections, respectively. For continuity of flow, assuming the density of the oil to be constant,

A1 = A2

So that

= (A1/A2)

A1 = ((/4) d12 and A2 = ((/4) d22Thus, = (A1/A2) = (0.2/0.1)22 = 8 ms-1Applying Bernoullis equation to the lower and upper sections assuming no energy losses,

Total energy per unit weight at section 1=Total energy per unit weight at section 2

2/2g + p1/(oilg + z1=2/2g + p2 /(oilg + z2

p1 - p2= (oil(2 - 2) + (oilg ( z2- z1)

Now,

z2- z1 = l sin 450 = 20.707 = 1.414 mAnd, since the relative density of the oil is 0.9, then (oil = 0.91000 = 900 kg/m3. Substituting in equation (1), p1 - p2 = 900 (82-22) + 9009.811.414

= 39484 Nm-2For the manometer, the pressure in each limb will be the same at the level XX;

Therefore,

p1 + (oilg z1 = p2 + (oilg(z2 - h) + (mangh

h =

(oil = 0.91000 = 900 kg/m3 and (man = 13.61000 = 13600 kg/m3 h = [0.9/(13.6-0.9)][39484/(9009.81)-1.414] m

= 0.217 m1.3 Principle of the venturimeter

Pressure difference can be used to determine the volume flow rate of flow for any particular configuration. The venturimeter uses this effect for measurement of flow in pipelines. It consists of a short converging conical tube leading to a cylindrical portion, called the throat, of smaller diameter than that of the pipeline, which is followed by a diverging section in which the diameter increases again to that of the main pipeline. The pressure difference from which the volume flow rate can be determined is measured between the entry section1 and the throat section 2, often by means of a U-tube manometer. The axis of the meter may be inclined at any angle. Assuming that there is no loss of energy, and applying Bernoullis equation to section 1and 2,

EMBED Equation.3 v22 v12 = 2g[(p1 - p2)/(g +( z1 -z2 )]

(2)For continuous flow,

A1 = A2

So that

= (A1/A2)

Substituting in equation (2),

v12[(A1/A2)-1] = 2g[(p1 - p2)/(g +( z1 -z2 )]

Volume flow rate,

Q = A1 = [A1A2/(A12 A22)1/2]((2gH)Where H = [(p1 - p2)/(g +( z1 -z2 )] if m = A1/A2Q = [A1/(m2-1)1/2]((2gH)

In practice some losses will occur between section 1 and 2. The value of Q given by the above equations is a theoretical value which will be slightly greater than the actual value. The coefficient of discharge Cd is therefore, introduced:Actual discharge, Qactual = Cd QtheoreticalThe value of H can be found from the reading of the U-tube gauge. Assuming that the connections to the gauge are filled with the fluid flowing in the pipeline, which has density (, and that the density of the manometric liquid in the bottom of the U-tube is (man, then, since pressures at level XX must be the same in both limbs,px = p1 + ((g z1-z) = p2 + (g(z2 - z- h) + (mangh

Rearranging,

H = [(p1 - p2)/(g +( z1 -z2 )] = h ()

Therefore,

Q = [A1/(m2-1)1/2]

(3)Problem

Kerosene (SG = 0.85) flows through the Venturimeter shown below a with flowrates between 0.005 and 0.050 m3/s. Determine the range in pressure difference, p1 - p2, needed to measure these flowrates.

Answer:

Use Bernoullis equation.

EMBED Equation.3 Required: p1 - p2

z1 =z2

v1 and v2 can be determined from flowrates.

The pressure difference for the smallest flowrate is 1.16 kPa

Likewise, the pressure difference for the largest flowrate is 116 kPa1.4 Pipe orifices

The venturimeter operates by changing the cross section of the flow. A similar effect can be achieved by inserting an orifice plate which has an opening in it smaller than the internal diameter of the pipeline. The orifice plate produces a constriction of the flow.

The arrangement is cheap compared to a venturimeter, but there are substantial energy losses. The theoretical discharge can be calculated, but the actual discharge may be as little as two-thirds of this value. A coefficient must therefore be introduced. A typical value for a sharp-edged orifice is 0.65.Theory of small orifices discharging to the atmosphere:

An orifice in the side or base of a tank is an opening, usually circular through which fluid is discharged in the form of a jet, usually into the atmosphere. The volume of fluid discharged will depend upon the head of the fluid above the level of the orifice and it can, therefore, be use as a means of flow measurement. The term small orifice is applied to an orifice which has a diameter, or vertical dimension, which is small compared with the head producing the flow from point to point across the orifice.

Application of the Bernoullis equation between points (1) and (2) gives:

EMBED Equation.3 The reservoir is open to the atmosphere, therefore, p1= 0 (gauge pressure) and p2 is also equal to 0 (because it leaves as a free jet). The reservoir is large, therefore v1 is taken as 0 compared to v2. z2 = 0 and z1 = h.

EMBED Equation.3 v2 = ((2gh)

This is a statement of Toricellis Theorem, that is the velocity of the issuing jet is proportional to the square root of the head producing the flow.

Discharge, Q = Area velocity

= A((2gh)

In practice actual discharge is less than the theoretical discharge given above, which must therefore be modified by introducing a coefficient of discharge Cd, so that

Actual discharge, Qactual = Cd Qtheoretical

= Cd A((2gh)

There are two reasons for the difference between the theoretical and actual discharges. First, the velocity of the jet is less than that given by ((2gh) because there is loss of energy between section 1 and section 2.

Actual velocity at section 2 = Cv v = Cv((2gh)

Where Cv is the coefficient of velocity which has to b determined experimentally and is of the order of 0.97.

Because of the vena contracta at section 2, the actual area of the jet at section 2 is equal to CcA, where Cc is the coefficient of contraction. For a sharp-edged orifice of t form shown, it is of the order of 0.64.

Actual discharge = Actual area at B Actual velocity at B

= CcA Cv((2gh)

= CcCv A ((2gh)

Therefore Cd = Cc CvThe values of the coefficient of discharge, the coefficient of velocity and the coefficient of contraction are determined experimentally and values are available for standard configurations in BS specifications.Generally coefficients can be obtained as follows:

Coefficient of discharge = Actual discharge / Theoretical discharge

Coefficient of contraction = Area of jet at the vena contracta/Area of the orifice

Coefficient of velocity = Velocity at th vena contracta/ Theoretical velocity

Example

A jet of water discharges horizontally into the atmosphere from an orifce in the vertical side of a large open-topped tank. Drive an expression for the actual velocity v of a jet at the vena contracta if the it falls a distance y vertially in a horizontal distance x, measured from the vena contracta. If the head of water above the orifice is H, calculate the coefficient of velocity.

If the orifice has an area of 650 mm2 and the jet falls a distance y of 0.5 m in a horizontal distance x of 1.5 m from the vena contracta, calculate the values of the coefficients of velocity, discharge and contraction, given that the volume rate of flow is 0.117 m3 and the head H above the orifice is 1.2 m.

Let t be the time taken for a particle of fluid to travel from the vena contracta A to point B. Then,x = vt and y = gt2 then v = x/t and t = ((2y/g)

Eliminating t,

Velocity at the vena contracta, v = ((gx2/2y)

This is the actual velocity of the jet at the vena contracta.

Theoretical velocity = ((2gH)

Coefficient of velocity = Actual velocity/ Theoretical velocity

= v/((2gH)

= v/((x2/4yH)Putting x = 1.5 m, y = 0.5 m, H= 1.2 m area A= 65010-6 m2,

Coefficient of velocity, Cv = 0.968.

Coefficient of discharge, Cd = Qactual/[A ((2gh)]

= (0.117/60)[65010-6((29.811.2)

= 0.618Coefficient of contraction, Cc = Cd /Cv = 0.618/0.968 = 0.639z

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