# Calculating reaction force of pipe elbow

I am currently studying for the mechanical FE exam. There is a problem in the fluid dynamics section that is not making sense to me.

Problem statement:

I solved it using momentum balance and I got 50.2 kN.

The answer is supposed to be 64 kN.

The answer key shows a pressure calculation with $$P_2A_2$$ that gets added to the 50.2 kN. I'm confused why $$P_2$$ has anything to do with the reaction force.

An example I can relate to this is: if you capped the end of a garden hose and held it up with no flow but much pressure, there won't be a reaction force. The pressure is pushing equally on all sides so the net force should be zero. When the end is uncapped, then you get momentum transfer and a consequent thrust force.

Chat GPT tells me that the pressure drop through the elbow will influence the reaction force. This is what I would expect.

Pressure is force per unit of area. therefore the force on the outlet:

$$F_{discharge}=P_2*Area_{discharge}$$

This is pointing to the left and causes an additional reaction at the elbow to the right!

I think you've already worked out that the required reaction force is equal to the force exerted on the fluid in the bend by the pipe.

What you get from the momentum balance is the force exerted on the fluid in the bend by everything else in the universe.

To get from the force exerted on the fluid in the bend by everything else in the universe to the force exerted on the fluid in the bend by the pipe, you have to subtract off the force exerted on the fluid in the bend by the fluid immediately upstream and downstream of it.

The force exerted by the fluid immediately upstream only affects the vertical momentum balance. Since we're only interested in the horizontal component of the force, we can forget about that and just work out the force exerted by the fluid immediately downstream, which is the product of the pressure and cross-sectional area at the downstream end of the bend, and acts to the left. We said we'd subtract that off, or equivalently, we add on a force of the same size to the right.

(And yes, there will be reaction forces in the garden hose example. Place your hose loosely coiled on the ground (and pre-primed to be full of water), with both the tap at the upstream end and the nozzle-valve at the downstream end closed; then open the tap at the upstream end (while keeping the valve at the downstream end closed), thus introducing your "no flow but much pressure" situation. You'll see the hose start to uncoil under those reaction forces.)

• Ok, I will have to try the garden hose experiment at home! Commented Dec 31, 2023 at 11:10

Let’s take the example that you used, of a garden hose with a cap on it. Now draw the free body diagram of just the cap. In it you will have the force of the water pushing it out (F = P.A) and a force the hose exerts on the cap at their joint that holds it in place. These two balance out keeping the cap in place.

Now moving on to the free body diagram of the hose, you will again have a force one way due to the pressure and an other that comes from the reaction force of the force keeping the cap in place.

What I am trying to get at is that from the perspective of the hose or the elbow in the first question the reaction force that compensates for the force exerted due to water pressure is an external force that needs to be applied to keep the object in place.

In some other case, when the flow is turned off, the downstream pressure might also disappear. You aren't dealing with that case.

If the elbow is in a vacuum, and the outlet is at air pressure, air pressure will be forcing the pipe to the left. (Conversely, if the outlet was in a vacuum, and the elbow was exposed to air pressure, air pressure would be forcing the elbow towards the right.)

In the case where flow is forcing the pipe towards the left, the external pressure difference is still / always also forcing the pipe to the left. In the illustrated case, we aren't dealing with an air pressure difference, but we know, because it is stated, that there is 200KPa downstream pressure.

• Your answer is intuitive, but I would expect the fluid in the elbow to be at the same pressure as the downstream (neglecting minor loss). There would be no pressure differential to force the elbow to the left. I am beginning to think that the solution in my book is incorrect... Commented Jan 2 at 12:44
• There is an unequal external pressure. It's shown on the diagram as 200KPa. You aren't the first to find unbalanced external pressure non-intuitive -- look at the history of vacuum / air pressure demonstrations. Commented Jan 3 at 3:12