Answer:
Explanation:
To start with the mass at the entrance of the channel
![m = \rho V](https://tex.z-dn.net/?f=m%20%3D%20%5Crho%20V)
where
= 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
![m_1 = m_2](https://tex.z-dn.net/?f=m_1%20%3D%20m_2)
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 = ![m_1 +m_2](https://tex.z-dn.net/?f=m_1%20%2Bm_2)
m = ![2m_1](https://tex.z-dn.net/?f=2m_1)
![m_1 = \frac{1}{2}m](https://tex.z-dn.net/?f=m_1%20%3D%20%5Cfrac%7B1%7D%7B2%7Dm)
Applying the momentum equilibrium for steady one-dimensional flow:
Considering the momentum equation along the x-axis is:
![F_{Rx} = m_1V_1cos \theta _1 + m_2V_2 cos \theta_2 -mV](https://tex.z-dn.net/?f=F_%7BRx%7D%20%3D%20m_1V_1cos%20%5Ctheta%20_1%20%2B%20m_2V_2%20cos%20%5Ctheta_2%20-mV)
where;
m = mass flow rate before hitting the splliter
V = velocity of flow before hitting the splliter
mass flow rate channel one
= velocity flow rate channel one
angle by the spit channel one to the horizontal
= mass flow rate channel two
= velocity flow rate channel two
= angle by the spit channel two to the horizontal
From the above recent equation:
![F_{Rx} = \frac{m}{2} V_1cos \theta _1 + \frac{m}{2}V_2 cos \theta_2 -mV](https://tex.z-dn.net/?f=F_%7BRx%7D%20%3D%20%5Cfrac%7Bm%7D%7B2%7D%20V_1cos%20%5Ctheta%20_1%20%2B%20%5Cfrac%7Bm%7D%7B2%7DV_2%20cos%20%5Ctheta_2%20-mV)
![F_{Rx} = \frac{1}{2} m V(cos \theta _1 +cos \theta_2) -mV](https://tex.z-dn.net/?f=F_%7BRx%7D%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20m%20V%28cos%20%5Ctheta%20_1%20%2Bcos%20%5Ctheta_2%29%20-mV)
![F_{Rx} = m V(\frac{cos \theta _1 +cos \theta_2}{2} -1)](https://tex.z-dn.net/?f=F_%7BRx%7D%20%3D%20%20m%20V%28%5Cfrac%7Bcos%20%5Ctheta%20_1%20%2Bcos%20%5Ctheta_2%7D%7B2%7D%20-1%29)
Replacing our values :
![F_{Rx} = 62.37*10^2*18(\frac{cos 45^0 +cos 315^0}{2} -1)](https://tex.z-dn.net/?f=F_%7BRx%7D%20%3D%20%2062.37%2A10%5E2%2A18%28%5Cfrac%7Bcos%2045%5E0%20%2Bcos%20315%5E0%7D%7B2%7D%20-1%29)
![F_{Rx} = 62.37*10^2*18({0.707-1)](https://tex.z-dn.net/?f=F_%7BRx%7D%20%3D%20%2062.37%2A10%5E2%2A18%28%7B0.707-1%29)
![F_{Rx} = -32,893.938 \ lb](https://tex.z-dn.net/?f=F_%7BRx%7D%20%3D%20%20-32%2C893.938%20%5C%20lb)
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:
Considering the momentum equation along the z-axis is:
![F_{Rx} = m_1V_1sin \theta _1 + m_2V_2 sin \theta_2 -0](https://tex.z-dn.net/?f=F_%7BRx%7D%20%3D%20m_1V_1sin%20%5Ctheta%20_1%20%2B%20m_2V_2%20sin%20%5Ctheta_2%20-0)
![F_{Rx} = \frac{m}{2} V_1sin \theta _1 + \frac{m}{2}V_2 sin \theta_2](https://tex.z-dn.net/?f=F_%7BRx%7D%20%3D%20%5Cfrac%7Bm%7D%7B2%7D%20V_1sin%20%5Ctheta%20_1%20%2B%20%5Cfrac%7Bm%7D%7B2%7DV_2%20sin%20%5Ctheta_2)
![F_{Rx} = \frac{m}{2} V_1(sin \theta _1 +sin \theta_2})](https://tex.z-dn.net/?f=F_%7BRx%7D%20%3D%20%20%5Cfrac%7Bm%7D%7B2%7D%20V_1%28sin%20%5Ctheta%20_1%20%2Bsin%20%5Ctheta_2%7D%29)
Replacing our values :
![F_{Rx} = \frac{62.37*10^2}{2} *18(sin 45^0 +sin 315^0})](https://tex.z-dn.net/?f=F_%7BRx%7D%20%3D%20%20%5Cfrac%7B62.37%2A10%5E2%7D%7B2%7D%20%2A18%28sin%2045%5E0%20%2Bsin%20315%5E0%7D%29)
![F_{Rx} = 0](https://tex.z-dn.net/?f=F_%7BRx%7D%20%3D%20%200)
Thus, there is no force needed in the z-direction. Since the forces are equal and opposite in direction to each other.