Question

Determine the velocity of fluid in m/s at the throat (ie at smaller diameter of the pipe), if the flow is coming from left to right in a system as shown below Specific gravity of the flowing fluid is 1.2. Density of water is 1000 kg/m3 P1 = 735 kPa P2 = 550 kPa Flow Direction 31 mm 19 mm 215 0.225 0598 0.000

Answer 1

In the figure given below,I have drawn the diagram of the given problem in which I have taken two sections 1&2 at which values to be calculated.

2 P = 735 kPa P = 550 kPa Flout Dig T 31 mm 19 mm . (2) Specific gravity of flowing of level - 1.2. (s.al Sensity of Water, Subates = 1000 bag / m?

Pressure at section 1,P₁=735 kPa

Pressure at section 2,P₂=550 kPa

Diameter of pipe at section 1,D₁=31 mm

Diameter of pipe at section 2,D₂=19 mm

Specific gravity of the flowing fluid,(S.G.)_f =1.2

Density of water,$\small \rho _{water}$ =1000 kg/m³

So, density of the flowing fluid,$\small \rho _{f}$=1.2*$\small \rho _{water}$=1.2*1000   

S.G Pwater

=> $\small \rho _{f}$ =1200 kg/m³

Solution:

Applying continuity equation b/w section 1 & section 2:

Mass flow rate at section 1=Mass flow rate at section 2

=> $\small \rho _{f}$ *A₁*v₁ =$\small \rho _{f}$*A₂*v₂ {Since flow is incompressible}

=> A₁*v₁₌A₂*v₂

where A₁ & A₂ are the c/s area of pipe at section 1 & section 2 respecctively.

v₁ & v₂ are the velocities of fluid flow at section 1 & section 2 respecctively.

=> Du = - DV

Putting the values of D₁ and D₂ in the above eq.

we get; 31²*v₁ =19²*v₂

=> v₁ =0.37565*v₂

Now appplying Bernoulli's eq. b/w section 1 and section 2

$\small \frac{P_{1}}{\rho _{f}g}+\frac{v_{1}^2}{2g}+Z_{1}=\frac{P_{2}}{\rho _{f}g}+\frac{v_{2}^2}{2g}+Z_{2}$

g is the accelaration due to gravity =9.81 m/sec²

{since nothing is mentioned about the losses during the flow.so I have not considered loss term in the above eq.}

Since, pipe is horizontal which means Z₁ =Z₂=Z.

Above eq. becomes

$\small \frac{P_{1}}{\rho _{f}g}+\frac{v_{1}^2}{2g}+Z=\frac{P_{2}}{\rho _{f}g}+\frac{v_{2}^2}{2g}+Z$

$\small =>\frac{P_{1}}{\rho _{f}g}+\frac{v_{1}^{2}}{2g}=\frac{P_{2}}{\rho _{f}g}+\frac{v_{2}^{2}}{2g}$

$\small or\frac{v_{2}^{2}-v_{1}^{2}}{2}=\frac{P_{1}-P_{2}}{\rho _{f}}$

Putting the values in the above eq. , we get

$\small \frac{v_{2}^{2}-(0.3757v_{2})^{2}}{2}=\frac{735*10^3-550*10^3}{1200}$

$\small or\frac{0.8588v_{2}^{2}}{2}=154.167$

$\small =>v_{2}^2=359.0289$

$\small =>v_{2}=18.9481m/sec ---[Ans]$

So,velocity of fluid at throat,v₂=18.9481 m/sec [Ans]

Note:No options given below  are for this problem.You can recheck that.

***Thank you***