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Question: Water flows out of a big tank along with a horizontal tube AB of length 1 and radius r and bent at r...

Water flows out of a big tank along with a horizontal tube AB of length 1 and radius r and bent at right angles at the other end as shown in the figure. The rate of flow is Q m3/s{m^3}/s. Calculate the moment of the force exerted by the water on the tube about the end A.

Explanation

Solution

The rate of flow of a fluid is how fast the fluid is flowing from a particular area or volume at a given time. Here, if we are considering the area, we have to consider the velocity of the fluid and if we are taking the rate of flow in terms of velocity then we have to consider time with it.

Complete step by step solution:
Here the rate of flow of water is going through a cross-sectional area A and velocity V so, the formula for the rate of flow would be:
Q=AvQ = Av; …(A = Area; V = Velocity)
The area A of cylinder is given by:
A=πr2A = \pi {r^2};
Put the above relation in the equation Q=AVQ = AV:
Q=(πr2)vQ = \left( {\pi {r^2}} \right)v;
Write the above equation in terms of V (Velocity):
⇒Qπr2=v\Rightarrow \dfrac{Q}{{\pi {r^2}}} = v;
Now, applying the force formula:
F=maF = ma;
m = mass;
a = acceleration;
Now, we know that the acceleration is change in velocity w.r.t time.
F=m(vt)F = m\left( {\dfrac{v}{t}} \right);
⇒F=v(mt)\Rightarrow F = v\left( {\dfrac{m}{t}} \right);
Here, the mass flow rate which is the mass of the liquid upon the time taken for the liquid to flow would be equal to the fluid’s density, area and velocity.
(mt)=ρAv\left( {\dfrac{m}{t}} \right) = \rho Av;
Put this relation in the formula for forceF=v(mt)F = v\left( {\dfrac{m}{t}} \right) we have:
F=v(ρAv)F = v\left( {\rho Av} \right);
Where:
F = Force;
ρ\rho = Density;
A = Area;
v = velocity;
⇒F=ρ(Av)v\Rightarrow F = \rho \left( {Av} \right)v;
We have found out that Q=AVQ = AVand Qπr2=V\dfrac{Q}{{\pi {r^2}}} = V;
⇒F=ρ×Q×Qπr2;\Rightarrow F = \rho \times Q \times \dfrac{Q}{{\pi {r^2}}};
Also, the length that is protruding out of the cylinder can’t be neglected:
F=ρQ2lπr2;F = \dfrac{{\rho {Q^2}l}}{{\pi {r^2}}};

Therefore, The moment of the force exerted by the water on the tube about the end A is F=ρQ2lπr2.F = \dfrac{{\rho {Q^2}l}}{{\pi {r^2}}}.

Note: Hear, first find out the rate of flow of fluid, then apply the formula for force and calculate the mass flow rate and apply the value of mass flow rate in the relation with force and since the length of the tank cannot be ignored, multiply the length with the rest of the formula.