Test: Fluid Dynamics Level - 3 - Civil Engineering (CE) MCQ

Weight of body

= 0.3 x(32.12 - 32.12cos42.5)

= 0.3 x 32.12 (1-cos42.5)

Weight of body = 2.53 N

For air, specific weight, w = ρg = 1.2 × 9.81 = 11.77 N/m³.

Applying Bernoulli's equation between point 1 and point 2 Point 1 is the far away point from where the fluid particle starts moving.

∴P₁ = 0, V₁ = 0

Solution gives: V₂ = 63.93 m/s

With such low values of V₂ air flow is nearly incompressible

Air flow entering the bell mouth

=1.28 m³/s

along the normal direction [i.e. normal to streamline] which is actually −ve of radial direction

Also as V = V(r), speed is constant along a streamline, therefore tangential acceleration (a_s) in the flow is zero.

a_s = 0

Thus by Euler’s equation along the tangential direction

i.e. pressure along a streamline is constant, it can vary only across streamlines.

Thus from equation ②

Let us consider a point a at the inlet of section pipe where the fluid is actually moving. Applying Bernoulli’s equation between a and 2

Where h represents any kind of losses in the flow. In this equation, ?? cannot be predicted as the fluid is moving, hence let us take point 1 on the free surface where the fluid particles start moving.

If we neglect friction loss

h = 0

V₂ < 8.457="" />

Thus maximum possible discharge is

= 66.4 Lt/sec

Let us consider 1 and 2 as the inlet and outlet of the pump. Applying Bernoulli’s equation between ① and ②

Considering negligible flow losses in the pump. hp represents the head supplied to the fluid by the pump.

As nothing is mentioned about relative elevation of the inlet an outlet, let us consider

By continuity equation

Thus to calculate p2 − p1 (i.e. the pressure difference across the pump, value of hp is required).

Therefore let us consider a, i.e. the top surface of oil in the reservoir and b which is just outside the exit of discharge point.

For steady state conditions any particle moves from a to b passing through the suction pipe, pump and the discharge pipe. ∴ Applying Bernoulli’s equation between a and b

hL represents head losses in the piping system. lies on the free surface of the liquid, which is exposed to the atmosphere, and from the same point, fluid particles start moving. ∴ P_a = 0

V_a = 0

Also b is a point in a free liquid jet coming out from the discharge pipe.

Also y_b - y_a = 15 m

Hence,

P₂ - P₁ = 36387 Pa

or 363.807kPa

Considering the center line of flow passing along the x-axis.

Since the nozzle discharges into atmosphere P₂ = 0 and for horizontal arrangement y₁ = y₂

= −394.85 + 1978.2

= 1583.36 N

The water exerts a force of 1583.35 N on the nozzle and it acts in a direction, opposite to direction of flow. An equal and opposite force will be exerted by the nozzle on the fluid mass.

The pressure intensity is also the same at the two sections.

P₁ = P₂ = 400 kPa = 400 × 103 N⁄m²

Based on forces on pipe bends Force along the x-axis (on the pipe): F_x = dynamic force + static force

Force along the y -axis (on the pipe):

F_y = dynamic force + static force

Magnitude of resultant force on the bend is

And the direction of resultant force with x-axis is

An equal and opposite force will be required to hold the duct in position.

Note: That any particle in a free liquid jet experiences gravitational forces only and thus it can be solved as a projectile.

(ii) Amount of water falling on the window equals the amount delivered by the nozzle

Force along the y-axis:

Fy = dynamic force + static force

The magnitude of resultant force on the bend is

= 6358.87 N

Apply Bernoulli’s equation between a point (1) on the water surface and a point (2) downstream from the aperture.

Velocity V1 on the water surface in the reservoir is practically zero because the cross sectional area of the tank is much greater than that of the aperture.

Pressure is atmospheric both at the free water surface and at the center line of the jet

i.e., velocity of efflux from the aperture is equal to the velocity of the free fall from the surface of the reservoir. This is known as Torricelli’s Theorem.

Now: h = 5 m

(ii)

Apply Bernoulli’s equation between point 2 and 3; the point 3 refers to the position of maximum elevation of the jet.

(iii) Pressure is atmospheric both at points 2 and 3, i.e., P₂ = P₃ = P_a

Thus if there are no losses, the water jet would reach the initial level of water in the tank and this is the height to which the water may be sprayed.