Question

A 45o reducing elbow can be found in many piping systems. Crude oil (r = 52.5 lbm/ft3) flows into the elbow with a velocity of 5.5 ft/s and is deflected through an angle of 45o. The inlet diameter is 6-in, and the outlet diameter is 4-in. The inlet and outlet gage pressures are 35 psig and 32.5 psig, respectively. If the elbow is located in a horizontal plane, determine the restraining forces in the x and y-directions in lb

Answer 1

the restraining forces in the x and y-directions are :

`\mathbf{ F_x =1510.63 \ lbf}`

`\mathbf{F_y =6242.17 \ lbf}`

Step-by-step explanation:

From the given information;

Let us first calculate the outlet velocity by using continuity equation.

`A_1v_1= A_2v_2`

`(\pi)/(4)D_1^2 *v_1 = (\pi)/(4)D_2^2 *v_2`

where;

`D_1` = inlet diameter = 6 - In

`D_2 =` outlet diameter = 4 -In

inlet velocity
`v_1 = 5.5 \ ft/s`

outlet velocity = ???

`(\pi)/(4)*6^2 *5.5 = (\pi)/(4)*4^2 *v_2`

`36*5.5= 16 v_2`

`198 = 16 v_2`

`v_2 = (198)/(16)`

`v_2 = 12 .375 \ ft/s` to In; we have

Since 1 ft = 12 inches; thus

12.375 ft/s = (12.375 × 12 ) inches = 148.5 In/s

`v_1 =` 5.5 ft/s = 66 In/s

By using the linear momentum in x-direction for the volume ; we have the relation:

`\sum F_x = \sum (mdv)/(dt)x \\ \\ P_1A_1 - P_2A_2 * \ cos 45 + F_x = (m_2)V_2* \ cos 45 - ( m_1)v`

where;

`m = \rho AV`

`P_1A_1 - P_2A_2 * \ cos 45 + F_x = (\rho A_2 V_2)V_2* \ cos 45 - ( \rho A_1 V_1)v`

`35*(\pi)/(4)*6^2 - 32.5*(\pi)/(4)*4^2 * \ cos 45 + F_x = (52.5)/(12^3)( (\pi)/(4)*4^2*(148.5)^2 * cos 45 - (\pi)/(4)*6^2*66^2)`

`700.81 + F_x = 2211.44`

`F_x = 2211.44-700.81`

`\mathbf{ F_x =1510.63 \ lbf}`

`\sum F_y =(m dvy)/(dt)`

`- P_2A_2 * \ Sin \ 45 + F_y = (\rho A_2 V_2 )V_2* \ sin 45 - 0`

`F_y = (\rho A_2 V_2 )V_2* \ sin 45 + P_2A_2 * \ Sin \ 45 +`

`F_ y = 32.5 * (\pi)/(4)*4^2* sin 45 + (52.5)/(12^3) *(\pi)/(4)*4^2*148.5^2*sin45`

`F_y = 288.79 +5953.38`

`\mathbf{F_y =6242.17 \ lbf}`

Therefore; the restraining forces in the x and y-directions are :

`\mathbf{ F_x =1510.63 \ lbf}`

`\mathbf{F_y =6242.17 \ lbf}`