Momentum and forces in fluid flow
Forces due to fluids in motion
This chapter will study the forces generated by fluids in motion. For example, a power washer produces a high velocity jet of water which produces a force large enough to remove debris. Hurricane strength winds can produce forces large enough to destroy a home. The nozzle of a firehose requires considerable strength to control.
The continuity equation was a consequence of conservation of mass. Bernoulli’s equation was a consequence of conservation of energy. The forces described above are a consequence of conservation of momentum. We don’t have a special name for these equations and will just refer to them as the momentum equations for a fluid.
Newton’s second law states that rate of change of momentum of a body is directly proportional to the force applied. For steady flows this results to the following three force equations, one for each direction:
\[F_x= \rho Q \Delta v_x \nonumber\\ F_y= \rho Q \Delta v_y \nonumber\\ F_z= \rho Q \Delta v_z\]In the above expression $F_x$ is the net force acting on the fluid in the x-direction. This needs to be stressed: $F_x$ is the sum of all forces (in the x-direction) acting on the fluid. These forces may include reaction forces exerted from solid surfaces or from the fluid pressure itself (more on this later). It is not the force which the fluid imparts on an object. The next exercise will belabor this point as it is often a point of confusion.
The remaining quantities in the above force equation are the density, $\rho$ and the volume flow rate, $Q$. The term, $\Delta v_x$, represents the change in the fluid velocity in the x-direction. It is the fluid’s final velocity in the x-direction minus the fluids initial velocity in the x-direction.
A hose sprays a jet of water directly at a wall at a rate of 2 liters per second. If the velocity of the stream is 8 m/s how much force is exerted on the wall by the water?
In this problem I will assume that the water does not reflect from the wall but simply comes to rest. More realistically the fluid will be deflected from the wall in a direction along the wall. In either case the problem is solved by focusing on the change in momentum in one dimension, which we will call the $x$-direction.
The momentum equation in the x-direction is:
\[F_x= \rho Q \Delta v_x\]The fluid changes the x-component of its velocity when it hits the wall. It goes from 8 m/s before hitting the wall to 0 m/s (in the x-direction) after contact. The change in velocity (final minus initial) is therefore:
\[\Delta v_x = v_{fx}-v_{ix}=0-8~\m/s = -8~\m/s.\]The momentum equation in the x-direction is:
\[F_x= \rho Q \Delta v_x = \left(1000~\kg/\m^3\right) \left( 2~\L/s\right)\left(-8~\m/s\right)=-16~\N\]The minus sign tells us that the force is in the minus x-direction, i.e. to the left. Remember, the force we solved for, $F_x$, is the force acting on the fluid. In this problem the wall exerts a force of 16 N in the minus x-direction on the fluid. Newton’s third law then tells us that the fluid exerts a force of 16 N in the positive x-direction on the wall.
Compute the horizontal and vertical forces on the block shown below. The fluid is water with a velocity of 25 fps and a stream diameter of 1.75 inches.
In this example we will need to use the momentum equations in both the $x$ and $y$ directions. Before we get started let’s find the volume flow rate:
\[Q=vA=25~\frac{\ft}{\s}\times \frac{\pi}{4}\left(\frac{1.75~\inch}{12~\inch/\ft}\right)^2=0.418~\ft^3/\s\]Now let’s start with the x-direction:
\[F_x= \rho Q \Delta v_x\]The initial x-component of the velocity is $v_{ix}=25$ fps. The final component is $v_{fx} = -25\cos(30)=-21.65$ fps. Note the minus sign since the final x-component of the velocity is to the left.
The change in velocity (final minus initial) is therefore:
\[\Delta v_x = v_{fx}-v_{ix}=-21.65~\fps-25~\fps = -46.65~\fps.\]The momentum equation in the x-direction is:
\[F_x= \rho Q \Delta v_x = \left(1.94~\slug/\ft^3\right) \left(0.418~\ft^3/\s\right)\left(-46.65~\fps \right)=-37.83~\lb\]The minus sign tells us that the force on the fluid is to the left. The force on the block is therefor 37.38 lb to the right.
Now the y-direction:
\[F_y= \rho Q \Delta v_y\]The initial y-component of the velocity is zero. The final component is $v_{fy} = +25\sin(30) = 12.5~\fps$. Note the plus sign since the final y-component of the velocity is up. The change in velocity (final minus initial) is:
\[\Delta v_y = v_{fy}-v_{iy}=12.5~\fps - 0~\fps = 12.5~\fps.\]The force on the fluid in the y-direction is:
\[F_y= \rho Q \Delta v_y = \left(1.94~\slug/\ft^3\right) \left(0.418~\ft^3/\s\right)\left(12.5~\fps \right)=10.14~\lb\]So the block experiences a force of $10.14~\lb$ downward. The figure below shows both the x and y force on the block as well as the resultant force.
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