# Bernoulli appliance to garden hose

1. Jan 8, 2010

### Ferex

I know the topic has been dealt in the past, I read it but still couldn't extrapolate a satisfying answer.
I would like to know how Berboulli equation applies to a hose without restrictions in which fluid (water) exits at atmospheric pressure, the typical garden situation without any nozzle.
I have, let's suppose 3 bars at the initial section (1) and 0 bars (atmospheric pressure) at the discharge section (2). We consider z1 = z2, p1 = 3 bars, p2 = 0, the question is: does v1 = v2?

The second question is: if I put my finger at the exit section causing a little reduction in the section (but still allowing water flow) what happens to pressure??

Many thanks, Ferex

2. Jan 8, 2010

### mugaliens

Bernoulli's equation would not be applicable in your case unless you had two hoses of differing diameters, at which point the smaller diameter hose would have a faster flow and less pressure, regardless of which hose is first. It's really only applicable to the difference between the pressures immediately before and after the junction between the hoses.

Navier-Stokes, however, covers how pressure varies as a function of flow rate, but for incompressible flows of Newtonian fluids, such as water, it simiplifies to a linear solution involving pressure gradient, viscosity, and other body forces such as smoothness of the pipe or hose.

For a rigid hose, v1 is nearly equal to v2, the very slight difference being due to the very slight compressibility of water. For all practical garden hose purposes, however, they're equal, as water is considered to be incompressible for pressures encountered in a garden hose! Even at the 4 km depth in the ocean, at approximately 6,430 psi, there's only a 1.8% decrease in volume.

Pressure will vary linearly along the hose. If you restrict the exit such that P2>0, then P1 will increase as well, but not by the same amount, as your velocity will be reduced, and pressure differential per unit of length will be reduced as well.

Last edited: Jan 8, 2010
3. Jan 8, 2010

### minger

What your calculating is a delta velocity. The difference in pressure causes net force which drives the fluid. If the diameter stays constant, then the velocity also must due to conservation of mass (for incompressible flow).

However, when given a pressure difference, you must consider where that pressure is measured. That pressure is going to be at a location where the velocity is zero. Think of the water tower or some other plenum where the velocity of the fluid can be essentially negligible.

As you open the spigot, suddenly there is a delta P across the line, and the net force drives the fluid. So, to answer, no v1 does not equal v2 because they are taken at location of differing pressures (assuming z1=z2 of course).

4. Jan 9, 2010

### nvn

Ferex: In scenario 1, the hose outlet is perfectly straight, with no expansion nor contraction. Therefore, the exit head loss, he, is he = 0. Section 1 is immediately after the hose inlet, section 2 is immediately before the hose outlet, and section 3 is immediately after the hose outlet. The pressure at section 1 is given in post 1, p1 = 300 000 Pa.

p2 = p3 = 0, v1 = v2 = v3 = v, z1 = z2 = z3. For water at T = 15 C, density is rho = 998.7 kg/m^3, and absolute viscosity is mu = 1.14e-3 Pa*s. Let's assume the hose absolute roughness is e = 0.010 mm, the hose diameter is D = 15 mm, and the hose length is L = 5 m. The relative roughness is therefore e/D = 0.0007. Friction head loss is hf = 0.5*f*L*(v^2)/D. Initially assume fully turbulent flow. Therefore, going into the Moody chart, we see f = ~0.0180 for e/D = 0.0007. The Bernoulli equation, written between sections 1 and 2, is therefore as follows.

p1/rho + 0.5*v1^2 = p2/rho + 0.5*v2^2 + hf + he​

You could also divide through the above equation by g, if you wish, but I did not. Therefore, (p1 - p2)/rho + 0.5(v1^2 - v2^2) = hf + 0. This gives, p1/rho + 0 = 0.5*f*L*(v^2)/D. Solving for v gives, v = [2*p1*D/(rho*f*L)]^0.5. Therefore, we currently have v = {2(300 000 Pa)(0.015 m)/[(998.7 kg/m^3)(0.0180)(5 m)]}^0.5 = 10.01 m/s.

Now compute the Reynolds number. Re = rho*v*D/mu = 998.7*10.01*0.015/1.14e-3 = 1.32e5. Going back into the Moody chart for e/D = 0.0007, we see f = 0.0191, which does not match our previous value. Therefore, recompute v = [2*300 000*0.015/(998.7*0.0191*5)]^0.5 = 9.714 m/s.

Now Reynolds number becomes Re = 1.28e5. Going back into the Moody chart, we see f = 0.0192, which reasonably matches our previous value. Therefore, v = [2*300 000*0.015/(998.7*0.0192*5)]^0.5 = 9.689 m/s. The flow rate is Q = v*A = 0.001 712 (m^3)/s = 1.71 L/s.

In scenario 2, where the hose exit is contracted by your finger, v1 = v2 ≠ v3, p1 ≠ p2 ≠ p3, p3 = 0. Let v = v1 = v2. Let's say D3 = 10.61 mm; therefore, A3/A2 = 0.50. Compute the hose exit head loss, he, which will now be nonzero due to the outlet contraction. he = 0.5*kc*v3^2, which would perhaps be he = 0.5*0.2*v3^2 (?). We know v3 = v2*A2/A3 = 2*v; therefore, he = 0.4*v^2. Now write the Bernoulli equation between sections 1 and 3, and solve for v, which gives v = {(p1/rho)/[(0.5*f*L/D) + 1.9]}^0.5. Solve for v iteratively, as we did before. I ended up with f = 0.0213, and v = 7.424 m/s. Now write the Bernoulli equation between sections 1 and 2, and solve for p2. Simplifying gives, p2 = p1 - 0.5*rho*f*(L/D)*v^2 = 300 000 - 0.5*998.7*0.0213*(5/0.015)*7.424^2 = 104 590 Pa = 104.6 kPa.

5. Jan 10, 2010

### Ferex

Oh yes! great work! I couldn't be more satisfied, thanks for your work nvn, and thanks to others who answered me. Now everything is clear!!

6. Jan 11, 2010