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SIZIF [17.4K]
3 years ago
5

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
Engineering
1 answer:
ohaa [14]3 years ago
6 0

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}

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Read more about Carburizing Time at; brainly.com/question/13031810

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