Fluid Mechanics Test 3

Concept:

A pitot tube is used to measure the velocity of fluid flowing through a pipe.

The velocity of flow, \(V = \sqrt {2 \times g \times h}\)

where h = pressure head difference

Also, pressure head difference (h) is given by the difference of stagnation pressure head (h_s) and static pressure head (h_t).

h = h_s – h_t

Calculation:

h_s = 6 m, h_t = 5 m

⇒ h = 6 – 5 = 1 m

The velocity of Flow = \(\sqrt {2 \times g \times h} = \sqrt {2 \times 9.81 \times 1} = 4.43m/sec\)

Concept:

In the Eulerian method, a finite region through which fluid flows in and out is used.  Here the position and velocity of fluid particles of definite mass is not tracked. But, within the region, the field variables which are continuous functions of space dimensions (x, y, z) and time (t), are defined to describe the flow. This finite region is referred as control volume or flow domain.

A control volume is a Mathematical Model used to create a physical process.  A control volume is the fixed volume in a given space through which fluid flows (liquid or gas) with constant velocity.

A control volume approach generally focuses on a particular volume and studies the fluid passing through it.

For continuous flow problems like pipe flow or external flow situations, the control volume approach is used.

Concept:

Force exerted on the jet or nozzle = Rate of change in momentum

\({\rm{Force\;}}\left( {\rm{F}} \right) = {\rm{\;}}\frac{{{\rm{Final\;Momentum}} - {\rm{Initial\;Momentum}}}}{{{\rm{Time}}}}\)

\(\therefore {\rm{F}} = \frac{{{\rm{Mass\;}} \times {\rm{Final\;Velocity}} - {\rm{Mass\;}} \times {\rm{Initial\;Velocity}}}}{{{\rm{Time}}}} = \frac{{{\rm{Mass}}}}{{{\rm{Time}}}} \times \left( {{\rm{Final\;Velocity}} - {\rm{Initial\;Velocity}}} \right)\)

Now, consider two points. Point 1 is being inside the nozzle and point 2 is being on the tip of the nozzle where the water is being discharged in the atmosphere. Applying Bernoulli’s equation we get,

\(\frac{{{{\rm{P}}_1}}}{{{\rm{\rho g}}}} + \frac{{{\rm{V}}_1^2}}{{2{\rm{g}}}} + {{\rm{z}}_1} = \frac{{{{\rm{P}}_2}}}{{{\rm{\rho g}}}} + \frac{{{\rm{V}}_2^2}}{{2{\rm{g}}}} + {{\rm{z}}_2}{\rm{\;}}\)

Now, z₁ = z₂ and P₂ = 0 as atmospheric pressure.

\(\therefore \frac{{{\rm{V}}_2^2}}{{2{\rm{g}}}} = \frac{{{{\rm{P}}_1}}}{{{\rm{\rho g}}}} + \frac{{{\rm{V}}_1^2}}{{2{\rm{g}}}}\)

∴ V₂ > V₁ as P₁ is positive pressure.

∴ Force exerted on the jet or nozzle \(,{\rm{F}} = \frac{{{\rm{Mass}}}}{{{\rm{Time}}}} \times \left( {{\rm{Final\;Velocity}} - {\rm{Initial\;Velocity}}} \right)\)

\({\rm{Force\;}}\left( {\rm{F}} \right) = \frac{{{\rm{Mass}}}}{{{\rm{Time}}}} \times \left( {{{\rm{V}}_2} - {{\rm{V}}_1}} \right)\)

∴ F > 0 as V₂ > V₁

Concept:

For venturimeter, discharge

\(Q = {C_d} \cdot \frac{{{A_1}{A_2}\sqrt {2gh} }}{{\sqrt {A_1^2 - A_2^2} }}\)

Calculation:

Given:

S₀ = 0.85, x = 20 cm = 0.2 m, D₁ = 0.25 m, D₂ = 0.1 m, C_d = 0.98

Now,

\({A_1} = \frac{\pi }{4}D_1^2 = \frac{\pi }{4}\left( {{{0.25}^2}} \right) = 0.0490\)

\({A_2} = \frac{\pi }{4}\left( {{{0.1}^2}} \right) = 0.00785\)

Now,

\({\rm{Pressure\;head\;difference\;}}\left( h \right) = x \cdot \left( {\frac{{{S_m}}}{{{S_0}}} - 1} \right)\)

\(h = 0.2\left( {\frac{{13.6}}{{0.85}} - 1} \right)\)

∴ h = 3m

Note:

The given manometer reading (x = 20 cm) is not pressure head. We have to calculate pressure head using above formula.

\({\rm{Rate\;of\;flow\;}}\left( Q \right) = \frac{{{C_d} \cdot {A_1}{A_2}\sqrt {2gh} }}{{\sqrt {A_1^2 - A_2^2} }}\)

\(Q = \frac{{\left( {0.98} \right)\left( {0.0490} \right)\left( {0.00785} \right)\sqrt {2\: \times \:9.81\: \times \:3} }}{{\sqrt {{{0.0490}^2}\: - \:{{0.00785}^2}} }}\)

Q = 0.05979 m³/s