Answer:
Explanation:
To start with the mass at the entrance of the channel
[tex]m = \rho V[/tex]
where
[tex]\rho[/tex] = density of water = 62.37 lb/ft³
V = velocity of the flow = 100 ft³/s
m = 62.37 lb/ft³ × 100 ft³/s
m = 6237 lb/s
m = 62.37 × 10² lb/s
Since the channels are identical, the mass flow rate are as well equal in both sections i.e
[tex]m_1 = m_2[/tex]
Also; the flow rate is noted to be at a steady state, frictionless and the fluid is deemed to be incompressible
Therefore using the law of conservation of mass flow:
m = [tex]m_1 +m_2[/tex]
m = [tex]2m_1[/tex]
[tex]m_1 = \frac{1}{2}m[/tex]
Applying the momentum equilibrium for steady one-dimensional flow:
[tex]\sum F ^ {\to} = \sum_{out} \beta m V^ {\to} - \sum_{in} \beta m V^ {\to}[/tex]
Considering the momentum equation along the x-axis is:
[tex]F_{Rx} = m_1V_1cos \theta _1 + m_2V_2 cos \theta_2 -mV[/tex]
where;
m = mass flow rate before hitting the splliter
V = velocity of flow before hitting the splliter
[tex]m_1 =[/tex]mass flow rate channel one
[tex]V_1[/tex] = velocity flow rate channel one
[tex]\theta _1 =[/tex] angle by the spit channel one to the horizontal
[tex]m_2[/tex] = mass flow rate channel two
[tex]V_2[/tex] = velocity flow rate channel two
[tex]\theta _2[/tex] = angle by the spit channel two to the horizontal
From the above recent equation:
[tex]F_{Rx} = \frac{m}{2} V_1cos \theta _1 + \frac{m}{2}V_2 cos \theta_2 -mV[/tex]
[tex]F_{Rx} = \frac{1}{2} m V(cos \theta _1 +cos \theta_2) -mV[/tex]
[tex]F_{Rx} = m V(\frac{cos \theta _1 +cos \theta_2}{2} -1)[/tex]
Replacing our values :
[tex]F_{Rx} = 62.37*10^2*18(\frac{cos 45^0 +cos 315^0}{2} -1)[/tex]
[tex]F_{Rx} = 62.37*10^2*18({0.707-1)[/tex]
[tex]F_{Rx} = -32,893.938 \ lb[/tex]
Thus, the force needed in the x-direction to keep the splliter position is 32,893.938 lb (i.e in the opposite direction of the water jet)
Again:
[tex]\sum F ^ {\to} = \sum_{out} \beta m V^ {\to} - \sum_{in} \beta m V^ {\to}[/tex]
Considering the momentum equation along the z-axis is:
[tex]F_{Rx} = m_1V_1sin \theta _1 + m_2V_2 sin \theta_2 -0[/tex]
[tex]F_{Rx} = \frac{m}{2} V_1sin \theta _1 + \frac{m}{2}V_2 sin \theta_2[/tex]
[tex]F_{Rx} = \frac{m}{2} V_1(sin \theta _1 +sin \theta_2})[/tex]
Replacing our values :
[tex]F_{Rx} = \frac{62.37*10^2}{2} *18(sin 45^0 +sin 315^0})[/tex]
[tex]F_{Rx} = 0[/tex]
Thus, there is no force needed in the z-direction. Since the forces are equal and opposite in direction to each other.
