# What is the aperture in the hydraulic analogy of Ohm's law

Note: My question is duplicate of Why doesn't water come out of tap/faucet at high pressure when I turn it on?. None of the answers given there explains how the continuity equation fits properly. That's why I am asking this question.

I do not understand the Hydraulic analogy explained on wikipedea.

I asked my Network analysis teacher for an explanation .
He told me that cross-section area of a pipe represents its resistance and flow of water represents current passing through that pipe. Like in a tap when we change the area of aperture the flow rate of water changes, this is analogous to change in current in a wire due to the change cross-section area of the wire.

There is a contradiction in my understanding.

Contradiction: When we decrease the area of the mouth of the tap by our thumb the amount of water flowing out remains same but if we decrease the area of aperture of the tap by turning the knob the amount of water flowing out decreases.why?
Is it due to the change in type of flow i.e the flow changes from laminar to turbulent or choked?
I found a flaw by applying equation of continuity as:

Suppose two parallel resisters are connected to voltage source as shown:

(source: tutorvista.com)
Let's name the three wires as pipe-0(having the battery) ,pipe-1(having resister $$R_1$$) and pipe 2(having resister $$R_2$$).

All three pipes are of same length. Resistances of different wires are equivalent to the respective aperture radius of different pipes.

Our analogy is to interchange the terms current $$i=dq/dt$$ (amount of charge crossed in a unit time) with flow rate of water i.e $$i \equiv dm/dt$$ (amount of mass of water crossed in unit time).

Let's remove the wire having $$R_2$$.

The current $$I$$ in pipe 0 will decrease as the resistance of circuit increase.
removing the resistance is analogous to change the area of pipe-2 to $$0$$. On the other hand removing pipe 2 will not change water flow $$dm/dt$$ in pipe-0 because of Equation of Continuity.

I wrote my understanding about the situation to tell the community that my question is not Home-work like In essence my question is
1.How the tap works? And how can we apply the equation of continuity to the water flow when we turn the knob and when we cover the tap with thumb?
2. Where I'm getting wrong with my understanding of the Hydraulic analogy.

• please explain why removing pipe 1 and pipe 2 are different scenarios – gregsan Dec 20 '13 at 10:28
• removing pipe-1 and pipe-2 are different because $R_1$ and $R_2$ are different. – user31782 Dec 20 '13 at 10:56
• suggest editing your question to include such information. there is also no reason why pipe 2 wont decrease flow unless pipe 2 is bigger than pipe 0. – gregsan Dec 20 '13 at 11:00
• @gregsan whatever the cross-section of pipe-2 is it doesn't matter.if you remove pipe2 the flow in pipe0 will remain same. – user31782 Dec 20 '13 at 11:14
• @Anupam No problem. You fixed the errors, so I changed my down vote to an up vote. Also, I was able to down vote because I used to have more than 125 reputation. – user41086 Mar 15 '14 at 17:07

How the tap work? And how we can apply equation of continuity to the water flow when we turn the knob and when we cover the tap with thumb

The tap works by changing the minimum cross-sectional area of the flow. For a given pressure difference (upstream pressure minus downstream pressure) flow rate is a function of minimum cross-sectional area. Using your thumb would do the same thing. You can stop the flow with your thumb if you are strong relative to the force of the flow.

http://people.uncw.edu/lugo/MCP/DIFF_EQ/deproj/orifice/orifice.htm

Where am getting wrong with my understanding of the Hydraulic analogy.

Probably you are misunderstanding the Equation of Continuity. The Equation of Continuity only means that the mass flow rate in equals the mass flow rate out. It does not mean that the flow in and the flow out never change. Flow rate in and flow rate out can change simultanteously. Your statement "On the other hand removing pipe 2 will not change water flow [rate]" is incorrect. Removing pipe 2 will make a big difference in total flow if it is large in cross section compare to pipe 1. It will make a small difference in total flow if it is small in cross section compare to pipe 1. Your statement "When we decrease the area of the mouth of tap by our thumb the amount of water flowing out remains same" is also incorrect. Instead, the flow rate approaches zero as you make the cross-sectional area of the unblocked portion of the mouth small.

• When we decrease the area of the mouth of tap by covering it with our thumb(or something else) the speed of water flowing out increases but if we decrease the area of the aperture of the tap by turning the knob the speed of water flowing out does not increase, why? Is there any difference between these two mechanisms? – user31782 Mar 15 '14 at 7:15
• When the area is decreased the speed at the restricted aperature increases in both situations, meaning at the point where your thumb is and at the point where the valve is. However, in the case of the tap, the area inceases again by the point of the outlet. You cannot see the point of restriction, where the speed is faster in the tap, because the tap is not transparent. When you can actually see the water come out of the tap it has slowed again because downstream of the valve restriction the area is large again. – DavePhD Mar 15 '14 at 18:52
• I understand now. Thanks for your answer. The link you have given is not detailed. They are presuming the general differential equation $\dfrac{dR}{dD}=aR(b-R)$. Would you give me some other(detailed) reference which uses SI units. – user31782 Mar 16 '14 at 4:02
• The relationship of pressure drop, flow rate, pipe length and pipe diameter is the Hagen–Poiseuille equation. Basically, for a given pressure drop, flow rate is proportional to the 4th power of pipe diameter. It is specifically the Hagen–Poiseuille equation that is the analogy to Ohm's law. en.wikipedia.org/wiki/Hagen%E2%80%93Poiseuille_equation – DavePhD Mar 16 '14 at 12:47

When you decrease the aperture in the tap by turning the tap, the water velocity does increase at the aperture (B). The problem is that the aperture is in the valve stem, where you cant observe this increase in velocity. You only observe the water trickling out the spout slowly (C) because the cross section has already increased by then.

If you put your thumb directly at C then you do in fact have direct observation of increase in velocity in relation to decrease in aperture, as per the continuity law.

• '-1' if $v$ increases at B then $.A.\rho v\Delta t$ i.e amount of water coming out per second will remain same? – user31782 Dec 20 '13 at 10:57
• the mass flow through A B and C are all the same. – gregsan Dec 20 '13 at 10:59
• then why the bucket fills slowly? what is the purpose of tap then? moreover tap is not like you drew : see this – user31782 Dec 20 '13 at 11:02
• continuity equation doesnt take into account the pressure required at A to produce a constant flow rate even as the aperture is decreased. thats where flow factor or flow coefficient comes in. in water systems your water pressure from the provider is fixed, or in electrical systems your battery voltage is fixed. so while flow rate through A B C is the same, the actual value depends on the aperture size. – gregsan Dec 20 '13 at 11:11
• Huh?? if i turn the tap 99% the flow rate becomes 0.01%. means at B flow rate depends upon open cross-section area. this conradicts Equation of continuety. – user31782 Dec 20 '13 at 11:22