Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It only takes a minute to sign up.
Sign up to join this community
Using Bernoulli's equation, it can be shown that the water velocity depends on the height of a tank relative to the end of a hose. In reality, I find a smaller hose produces a greater velocity on exit than a larger hose. What have I overlooked to cause this disagreement?
$$v=\sqrt{2 g h-\frac{\text{losses}}{\rho}}$$
The tank sits 26m above and 210m distant from a tap at which a large hose or a small hose can be connected.
$$\pmb { \underline { \text {Tougher Case }}}(L_{Pipe} \neq 0)$$
Since the information that there is a long pipe between the Hose and the tank was unknown therefore I would use a better model i.e., Poiseuille Flow Equation.
Poiseuille Flow equation for fluid motion in pipe states that:
$$\Delta p = \frac {8L \eta \pi v_{out}}{A}$$ i.e.,
$$ v_{out} = \frac { \Delta p r^2}{8L \eta}$$
Now $\Delta p$ is the pressure difference between the point at tank to the the point at tap, therefore $\Delta p = \rho g h$.
$$ \Rightarrow v_{out} = \frac { \rho g h r^2}{8L \eta}$$
$$\pmb { \underline { \text {Crunching the Numbers}}}$$
$$r = 6 mm \Rightarrow r^2=0.36 \times 10^{-4} m^2$$
$$g =9.8 m \ s^{-2}$$
$$ L = 210 m$$
$$ \eta = 1 \times 10^{-3} Pa \ s $$
See here for the values of $\eta$
$$ \rho = 10^3 kg m^{-3}$$
Now inputting these values into the equation $$ v_{out} = \frac { \rho g h r^2}{8L \eta}$$ We get,
$$ v_{out} = 5.46 m \ s^{-1}$$
which is quite close to the true value.
Note that the value that we get here isn't quite accurate because (as Wikipedia states):
The equation does not hold close to the pipe entrance.
The equation fails in the limit of low viscosity, wide and/or short pipe. Low viscosity or a wide pipe may result in turbulent flow, making it necessary to use more complex models, such as Darcy–Weisbach equation.
$$\pmb { \underline { \text {Simpler Case }}}(L_{Pipe}=0)$$
Radius of hole is quite small in comparison to the Radius of the tank and hence it can be easily neglected without causing any deviation in the velocity of the water coming out of the hole.
What have I overlooked to cause this disagreement?
As you may see below (in the derivation of the equation) there isn't a thing in that significant enough that you have neglected. So why this discrepancy? The most probable reason is the viscous drag of the fluid and the friction applied by the wall (both of these are assumed to be insignificant to cause any deviation).
So how can you perform the experiment properly?
i.e., Don't use this
Rather use this
$$\pmb { \underline { \text {Torricelli's Law (without approximation)}}}$$
Consider a tank of radius $R$ containing a liquid of density $\rho$ with a small hole$^1$ of radius $r$ in its side at a height $h$, from the top. The air above the liquid is at pressure $P_a$. The water at the top has speed (say) $v_{top}$. The liquid flowing out through the hole has a speed $v_{hole}$ and pressure $P_{hole}$. From equation of continuity we have:
$$r^2 v_{hole} = R^2 v_{top} $$ $$ \Rightarrow v_{top} = v_{hole} \frac {r^2}{R^2} \tag 1 $$
Now applying Bernoulli equation at top and at hole we get:
$$P_a+ \frac {1}{2} \rho v_{top}^2 +\rho g (0)=P_{hole}+ \frac {1}{2} \rho v_{hole}^2 +\rho g h \tag 2$$
Substituting from equation $(1)$ into $(2)$ we get:
$$\frac {1}{2} \rho v_{hole}^2 \left (1 - {\left (\frac {r}{R}\right )}^4 \right) = \rho g h+ (P_{hole}-P_a)$$
$$ \Rightarrow \boxed {v_{hole} = \sqrt{ \frac {2 \left ( \rho g h+ (P_{hole}-P_a) \right )}{\rho \left (1 - {\left (\frac {r}{R} \right )}^4 \right)}}}$$
Now as you say $R=3.5m$ and $r=12mm$ this means that $r/R$ is about $10^{-3}$ therefore you can neglect this term $(r/R)^4$ (as it would be about $10^{-12}$) and you equation is:
$$\boxed {v_{hole} = \sqrt{ \frac {2 \left ( \rho g h+ (P_{hole}-P_a) \right )}{\rho}}}$$
i.e., $$v_{top} = \sqrt{ 2 g h + \frac {2 \left (P_{hole}-P_a) \right )}{\rho}}$$
When $P_{hole}>>P$ and $2 g h$ may be ignored, then the speed of efflux is determined by the container pressure. Such a situation occurs in rocket propulsion. On the other hand, if the tank is open to the atmosphere, then $P = P_a$ (as air has very low density) and
$$ \boxed {v_{hole} = \sqrt {2gh} }$$
Bernouilli:
$$\frac{1}{2}\rho v^2+gh\rho+p=C$$
This means:$$\frac{1}{2}\rho(v^2+2gh)+p=C\Rightarrow$$
$$v^2 +2gh + \frac{2p}{\rho}=\frac{2C}{\rho}\Rightarrow$$
$$v^2=\frac{2C}{\rho}-20h-\frac{2p}{\rho}\Rightarrow$$
$$v^2=2(\frac{C}{\rho}-10h-\frac{p}{\rho})\Rightarrow$$
$$v=\sqrt2 \sqrt{(\frac{C}{\rho}-10h-\frac{p}{\rho})}\Rightarrow$$
$$v=\sqrt2 \sqrt{\frac{1}{\rho}(C-10h\rho-p)}\Rightarrow$$
$$v=\sqrt{\frac{2}{\rho}}\sqrt{(C-10h\rho-p)}\Rightarrow$$
$$v=\frac{\sqrt{C'}}{\sqrt{p}}$$
in which
$$C'=\sqrt{\frac{2}{\rho}} \sqrt{C-10h\rho}$$
We don't know the value of $C$ (so neither that of $C'$), obviously. You can choose $h$ and $p$ to conform to the equality of Bernouilli though, and find the $v$ for these particular values. Does this answer your question somehow, is it useful? Or am I running around in circles? Please let me know!
One can also say, without considering losses:
$$v=\sqrt{2gh}\Rightarrow$$
$$v^2=2gh\Rightarrow$$
Now consider the kinetic energy
$$E=\frac{1}{2} m v^2$$
so (putting in the expression for $v^2$):
$$E=mgh$$
This was obvious from the very start. The kinetic energy with which the water leaves the hose is linear dependent on the variable h. $m$, The mass of water and $g$ are fixed. I don't see a problem here, even not with "losses" involved. So the smaller the diameter of the hose the faster the water will come out. Isn't that clear to you?
Why did you need Bernouilli (poor dead old man!!) at all? Just for the losses?
You have to take into account the viscosity of the fluid (Reynold's number) as well as $p$, It is these two which eventually determine the velocity of the fluid coming out of the hose. There is also feedback involved so I think it's a tough calculation. In the case of water (low viscosity) it's only $p$ that counts and the energy losses will be low in comparison to $\sqrt{19,6h}$. so...
No need for Bernoulli. And in the case of water, it's a very simple (and good) approximation. This finally answers your question: you overlooked the very low viscosity of water. As simple as that.
My suggestions:
First, check the Reynolds numbers of the two flows.
Second, have you considered the flow around the entrance to the hose? How are the two different diameter hoses connected differently?
Third, try two different diameter glass tubes inserted through a bung so that the entrance geometry is identical for both? Check the Reynolds numbers around the entrances of the tubes.
Let me say at the outset, that fluid mechanics is not my forte. However, it is my understanding the viscous energy losses depend on flow rate. The greater the flow rate, the greater the energy losses. The lower the flow rate, the lower the energy losses.
If that is the case, all other things being equal, when the diameter of the hose is decreased the flow rate decreases. This results in less viscous energy loss which, in turn, would increase the pressure behind the hose opening and the velocity of the flow.
Bottom line: If a narrower hose and lower flow rate reduces the magnitude of the losses in your equation, that would account for the increased velocity based on your equation.
Hope this helps.
The flow through the narrower pipe+hose must be experiencing a much lower friction factor than that through the wider pipe+hose. For more information see: https://www.slideshare.net/ChiragBhanagale/flow-through-pipes
It comes down to the restrictions on the flow rate caused by friction from the pipe.
An important factor to note (that I initially overlooked) is that there is a pipe coming down from the tank to a tap at which either a large or a small hose may be connected. For a larger hose, a greater flow rate will be produced. A greater flow rate will mean a greater velocity in the pipe, and with a greater velocity comes greater losses per litre of water. Therefore the water coming out the hose will have less energy per litre in a larger hose but will have a greater flow and therefore more energy.
Here is some working that brought me to this conclusion.
Using reasonable assumptions, Bernoulli's equation becomes:
$$v=\sqrt{2gh-\frac{\text{losses}}{\rho}}$$
If losses were insignificant this then becomes:
$$v=\sqrt{2gh}$$
Therefore in my case with 26m of head:
$$v=\sqrt{2\times 9.8 \times 26}$$ $$=22.6 \text{ m/s}$$
Experimental results show that for a hose with a diameter of 18mm:
$$v_{18\text{mm}}= 1.59 \text{ m/s}$$
and a hose with a diameter of 13mm:
$$v_{13\text{mm}}= 1.87 \text{ m/s}$$
Theses figures are significantly lower than the theoretical results. Therefore losses must be significant.
A good way to look at this (I think) is to view the velocity as the amount of energy per litre that has been conserved. If there were no losses then the velocity would be $22.6$ m/s. A very small flow rate (a tiny hose or hole) would produce minimal losses in the pipe and the velocity should more closely approximate $22.6$ m/s.
