Question

The water stream strikes the inclined surface of the cart. Determine the power produced by the stream if, due to rolling friction, the cart moves to the right with a constant velocity of 2 m/s. The discharge from the 50-mm-diameter nozzle is 0.04 m3/s. One-fourth of the discharge flows down the incline, and three-fourths flows up the incline. Assume steady flow, all within the vertical plane.

Answer 1

Answer

given,

constant speed of cart on right side = 2 m/s

diameter of nozzle = 50 mm = 0.05 m

discharge flow through nozzle = 0.04 m³

One-fourth of the discharge flows down the incline

three-fourths flows up the incline

Power = ?

Normal force i.e. Fn acting on the cart

$F_n = \rho A (v - u)^2 sin \theta$

v is the velocity of jet

Q = A V

$v = \dfrac{0.04}{\dfrac{\pi}{4}d^2}$

$v= \dfrac{0.04}{\dfrac{\pi}{4}\times 0.05^2}$

v = 20.37 m/s

u be the speed of cart assuming it to be u = 2 m/s

angle angle of inclination be 60°

now,

$F_n = 1000 \times \dfrac{\pi}{4}\times 0.05^2\times (20.37 - 2)^2 sin 60^0$

F n = 2295 N

now force along x direction

$F_x = F_n sin 60^0$

$F_x = 2295 \times sin 60^0$

$F_x = 1987.52\ N$

Power of the cart

P = F x v

P = 1987.52 x 20.37

P = 40485 watt

P = 40.5 kW

Answer 2

The power produced by the stream is; 40.5 kW

What is the power produced?

We are given;

Constant speed of cart on right side; u = 2 m/s

Diameter of nozzle; d = 50 mm = 0.05 m

Discharge flow through nozzle; Q = 0.04 m³/s

Formula for the normal force acting on the cart is;

Fₙ = ρA(v - u)²sin θ

Let us find v the speed of the jet from;

Q = Av

v = Q/A

v = 0.04/(π * 0.05²/4)

v = 20.37 m/s

From online research about this question, the angle of inclination is 60°. Thus;

Fₙ = ρA(v - u)²sin θ

Fₙ = 1000 * (π * 0.05²/4) * (20.37 - 2)sin 60

Fₙ = 2295 N

Force along the x-direction will be;

Fₓ = Fₙ * sin 60

Fₓ = 2295 * sin 60

Fₓ = 1987.52 N

Power of the Cart is;

P = Fₓ * v

P = 1987.52 * 20.37

P = 40.5 kW

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