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Chapter: Civil : Mechanics Of Fluids : Flow Through Pipes

Bernoulli's theorem for steady flow of an incompressible fluid

It states that in a steady, ideal flow of an incompressible fluid, the total energy at any point of the fluid is constant. The total energy consists of pressure energy, kinetic energy and potential energy or datum energy.

Assumptions made in the derivation   of   Bernoulli's   equation

 

(i). The fluid is ideal, i.e., Viscosity is zero.  (ii). The flow is steady

 

(iii). Te flow is incompressible.                (iv). The flow is irrotational.

 

Bernoulli's  theorem for steady flow of an incompressible fluid.

 

It states that in a steady, ideal flow of an incompressible fluid, the total energy at any point of the fluid is constant. The total energy consists of pressure energy, kinetic

 

energy and potential energy or datum energy. These energies per unit weight of the

 

fluid are:

 

Pressure Energy                    =   p   /   ?g

 

Kinetic energy                       = v2 / 2g

 

Datum Energy                      = z

 

The                                             mathematically,   Bernoulli's   theo

 

(p/w) + (v2 / 2g) + z = Constant.

Problem 1

 

Water is flowing through a pipe of 5 cm diameter under a pressure of 29.43 N/cm2

 

(gauge) and with mean velocity of 2.0 m/s. find the total head or total energy per unit weight of the water at cross -section, which is 5 cm above the datum line.

          Diameter of the pipe                5 cm = 0.5 m.      

          Pressure      ? = 29.43 N/cm2 = 29.23 N/m2        

`        velocity,      v        = 2.0 m/s.  

                                     

          datum head z        = 5 m

          total head             = Pressure head + Velocity head + Datum head  

          pressure head                 =   (p/   ?g) 4 / (2X9=(.81)) 29=30.m43X10       

          kinetic head          = (v2/ 2g) = ( 2X2/(2X9.81)) = 0.204 m    

          Total head            2       

                             =   (p/+ (v(?g))/2g)+z    

                             = 30 + 0.204 + 5 = 35.204m  

Problem 2

Water is flowing through two different pipes, to which an inverted differential manometer having an oil of sp. Gr 0.8 is connected the pressure head in the pipe A is 2 m of water, find the pressure in the pipe B for the manometer readings.

 



Problem 3

 

The diameters of a pipe at the sections 1 and 2 are 10 cm and 15 cm respectively. Find the discharge through the pipe if the velocity of water flowing through the pipe section 1 is 5 m/s. determine also the velocity at section 2.

 




Problem 4

 

A pitot -static tube is used to measure the velocity of water in a pipe. The stagnation pressure head is 6mm and static pressure head is 5m. Calculate the velocity of flow assuming the co-efficient of tube equal to 0.98.

 




Problem 5

A sub-marine moves horizontally in a sea and has its axis 15 m below the surface of water. A pitot tube properly placed just in front of the sub-marine and along its axis connected to the two limbs of a U -tube containing mercury. The difference of mercury level is found to be 170 mm. find the speed of the sub-marine knowing that the sp.gr. of mercury is 13.6 and that of sea-water is 1.026 with respect fresh

 



Problem 6

 

The water is flowing through a pipe having diameters 20 cm and 10 cm at sections 1 and 2 respectively. The rate of flow through pipe is 35 lit/sec. the section 1 is 6m above datum. If the pressure at section 2 is 4m above the datum. If the pressure at section 1 is 39.24 N/cm2, find the intensity of pressure at section 2.

 




Problem 7

In a vertical pipe conveying oil of specific gravity 0.8, two pressure gauges have been installed at A and B where the diameters are 16 cm and 8 cm respectively. A is 2 m above B. the pressure gauge readings have shown that the pressure at B is greater than at A by 0.981 N/cm2. Neglecting all losses, calculate the flow rate. If the gauges at A and B are replaced by tubes filled with the same liquid and connected to a U - tube containing mercury, calculate the difference of level of mercury in the two limbs of the U-tube.

 






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