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

Answer:

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

$\mathbf{ F_x =1510.63 \ lbf}$

$\mathbf{F_y =6242.17 \ lbf}$

Explanation:

From the given information;

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

$A_1v_1= A_2v_2$

$\dfrac{\pi}{4}D_1^2 *v_1 = \dfrac{\pi}{4}D_2^2 *v_2$

where;

$D_1$ = inlet diameter = 6 - In

$D_2 =$ outlet diameter = 4 -In

inlet velocity $v_1 = 5.5 \ ft/s$

outlet velocity = ???

$\dfrac{\pi}{4}*6^2 *5.5 = \dfrac{\pi}{4}*4^2 *v_2$

$36*5.5= 16 v_2$

$198 = 16 v_2$

$v_2 = \dfrac{198}{16}$

$v_2 = 12 .375 \ ft/s$  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:

$\sum F_x = \sum \dfrac{mdv}{dt}x \\ \\ P_1A_1 - P_2A_2 * \ cos 45 + F_x = (m_2)V_2* \ cos 45 - ( m_1)v$

where;

$m = \rho AV$

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

$35*\dfrac{\pi}{4}*6^2 - 32.5*\dfrac{\pi}{4}*4^2 * \ cos 45 + F_x = \dfrac{52.5}{12^3}( \dfrac{\pi}{4}*4^2*(148.5)^2 * cos 45 - \dfrac{\pi}{4}*6^2*66^2)$

$700.81 + F_x = 2211.44$

$F_x = 2211.44-700.81$

$\mathbf{ F_x =1510.63 \ lbf}$

$\sum F_y =\dfrac{m dvy}{dt}$

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

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

$F_ y = 32.5 * \dfrac{\pi}{4}*4^2* sin 45 + \dfrac{52.5}{12^3} *\dfrac{\pi}{4}*4^2*148.5^2*sin45$

$F_y = 288.79 +5953.38$

$\mathbf{F_y =6242.17 \ lbf}$

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

$\mathbf{ F_x =1510.63 \ lbf}$

$\mathbf{F_y =6242.17 \ lbf}$