Question

Water flows in a pipe of diameter 0.5 m. The dianeter of the to 1,0 m. A U-tube manometer is of the enlargement with joining ercury levels 5 mm. Determine the flow rate as well as the pressure 3 Water pipe suddenly enlarges connected to either side pipes which contain water. The difference in m head loss as a result of the enlargement.

Answer 1

Answer:

Q = 0.943$m^{3}/s$

$h_{L}$ = 0.6605 m

Explanation:

Given :

Diameter, d₁ = 0.5 m

Area, A₁ = $\frac{\pi }{4}\times 0.5^{2}$

             = 0.19625 $m^{2}$

Enlargement diameter, d₂ = 1 m

Enlargement Area, A₂ = $\frac{\pi }{4}\times 1^{2}$

             = 0.785 $m^{2}$

Manometric difference, h = 35 mm

                                          =35 X $10^{-3}$ m

From manometer , we get

$P_{1}+\rho _{w}.g.z_{1}+\rho _{m}.g.h=P_{2}+\rho _{w}.g.(z_{1}+h)$

$P_{1}-P_{2}=(\rho _{w}-\rho _{m}).g.h$

$\frac{P_{1}-P_{2}}{\rho _{w}.g}=(1-\frac{\rho _{m}}{\rho _{w}})\times h$

                                                = $(1-13.6)\times 35\times 10^{-3}$

                                                = -0.441

Now from newtons first law,

$\frac{P_{1}-P_{2}}{\rho _{w}.g}=\frac{V_{2}^{2}-V_{1}V_{2}}{g}$

-0.441 = $\frac{Q^{2}}{9.81}\times (\frac{1}{A_{2}^{2}}-\frac{1}{A_{1}A_{2}})$

-0.441 = $\frac{Q^{2}}{9.81}\times (\frac{1}{0.785^{2}}-\frac{1}{(0.19625\times 0.785)^{2}})$

Therefore. Q = 0.943 $m^{3} /s$

Now V₁ = $\frac{Q}{A_{1}}$

            = $\frac{0.943}{0.19625}$

            = 4.80 m/s

       V₂ =  $\frac{Q}{A_{2}}$

            = $\frac{0.943}{0.785}$

            = 1.20 m/s

Therefore, heat loss due to sudden enlargement is given by

  $h_{L}=\frac{(V_{1}-V_{2})^{2}}{2g}$

$h_{L}=\frac{(4.80-1.20)^{2}}{2\times 9.81}$

               = 0.6605 m