Answer to Question #87615 in Mechanics | Relativity for Unknown277793

Question

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Q1. Sections 1 and 2 are at the beginning and end of the bend of the 200 mm diameter pipe in which the quantity of flow is 0.28 m3/s. The angle of deflection of the water is 40. Calculate the force that the liquid exerts on the bend if the pressure in the pipe is 50 kPa. Assume no loss of pressure round the bend.

Q2. A 50 mm diameter stream of water strikes a 0.5 m by 0.5 m square plate which is at an angle of 30 degrees with the stream’s direction. The velocity of the water in the stream is 20 m/s and the jet strikes the door at its centre of gravity. If the plate is hinged at A, by neglecting the friction and self-weight of the plate, determine the normal force P applied at the upper edge of the plate to keep the plate in equilibrium. You may assume that the velocities of the jet remain unchanged after split.

Answer 1

A1.

$$F=ma=m\cdot \Delta v/\Delta t,$$

$$m=\rho V=\rho Q\Delta t,$$

$$\Delta v=v_2-v_1=v_1\text{cos}40^\circ-v_1=v_1(\text{cos}40^\circ-1),$$

$$v_1=Q/A=4Q/(\pi d^2),$$

finally,

$$F=\rho Q\Delta t\cdot \frac{4Q}{\pi d^2}(\text{cos}40^\circ-1)\frac{1}{\Delta t}=$$

$$=\frac{4\rho Q^2(\text{cos}40^\circ-1)}{\pi d^2}=582\text{ N}.$$

A2. This problem can be solved likewise. First, calculate the force acting on the centre of the door. To do that, transform the expression for $F$ we obtained above considering that $Q=v_1\pi d^2/4$:

$$F=\frac{4\rho (v_1\pi d^2/4)^2(\text{cos}60^\circ-1)}{\pi d^2}=$$

$$=\frac{\rho v_1^2\pi d^2(\text{cos}60^\circ-1)}{4}=391.9\text{ N}.$$

Now simply apply equations for moment of force: our force $F$ calculated above acts on a centre of plate 50 cm in length:

Assume that the moments are considered in relation to the lower edge.

$$0=0.25F-0.5P\text{sin}30^\circ,$$

$$P=\frac{0.25F}{0.5\text{sin}30^\circ}=391.9\text{ N}.$$