Answer to Question #87451 in Mechanics | Relativity for Unknown277793

Question

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

Please help me

Answer 1

According to Newton's second law, the force is

$$F=ma=m\cdot\frac{\Delta v}{\Delta t},$$

where change in speed is

$$\Delta v=v_2-v_1=v_0\text{cos}40^\circ -v_0=v_0(\text{cos}40^\circ-1).$$

The flow of water $Q$ during time $\Delta t$ gives mass

$$m=\rho Q\Delta t.$$

Substitution of the mass and the change in speed to the rightmost part of the first equation gives

$$F=\rho Q\Delta t\cdot\frac{v_0(\text{cos}40^\circ-1)}{\Delta t}=$$

$$=\rho Qv_0(\text{cos}40^\circ-1)=998\cdot0.28(0.766-1)=65.39\text{ N}.$$