Answer:
the restraining forces in the x and y-directions are :
[tex]\mathbf{ F_x =1510.63 \ lbf}[/tex]
[tex]\mathbf{F_y =6242.17 \ lbf}[/tex]
Explanation:
From the given information;
Let us first calculate the outlet velocity by using continuity equation.
[tex]A_1v_1= A_2v_2[/tex]
[tex]\dfrac{\pi}{4}D_1^2 *v_1 = \dfrac{\pi}{4}D_2^2 *v_2[/tex]
where;
[tex]D_1[/tex] = inlet diameter = 6 - In
[tex]D_2 =[/tex] outlet diameter = 4 -In
inlet velocity [tex]v_1 = 5.5 \ ft/s[/tex]
outlet velocity = ???
[tex]\dfrac{\pi}{4}*6^2 *5.5 = \dfrac{\pi}{4}*4^2 *v_2[/tex]
[tex]36*5.5= 16 v_2[/tex]
[tex]198 = 16 v_2[/tex]
[tex]v_2 = \dfrac{198}{16}[/tex]
[tex]v_2 = 12 .375 \ ft/s[/tex] to In; we have
Since 1 ft = 12 inches; thus
12.375 ft/s = (12.375 × 12 ) inches = 148.5 In/s
[tex]v_1 =[/tex] 5.5 ft/s = 66 In/s
By using the linear momentum in x-direction for the volume ; we have the relation:
[tex]\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[/tex]
where;
[tex]m = \rho AV[/tex]
[tex]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[/tex]
[tex]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)[/tex]
[tex]700.81 + F_x = 2211.44[/tex]
[tex]F_x = 2211.44-700.81[/tex]
[tex]\mathbf{ F_x =1510.63 \ lbf}[/tex]
[tex]\sum F_y =\dfrac{m dvy}{dt}[/tex]
[tex]- P_2A_2 * \ Sin \ 45 + F_y = (\rho A_2 V_2 )V_2* \ sin 45 - 0[/tex]
[tex]F_y = (\rho A_2 V_2 )V_2* \ sin 45 + P_2A_2 * \ Sin \ 45 +[/tex]
[tex]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[/tex]
[tex]F_y = 288.79 +5953.38[/tex]
[tex]\mathbf{F_y =6242.17 \ lbf}[/tex]
Therefore; the restraining forces in the x and y-directions are :
[tex]\mathbf{ F_x =1510.63 \ lbf}[/tex]
[tex]\mathbf{F_y =6242.17 \ lbf}[/tex]
