How Does Nozzle Elevation Affect Water Flow and Pressure in a Hose System?

[asheik234]

Homework Statement

Consider a hose that carries water (density = 1000 kg/m3) leads to a nozzle that is elevated h = 1.3 meters above the ground. The nozzle has a diameter d, and the hose has a diameter D = 5d. Water flows through the hose with a speed vH = 0.6 m/s.

(a) What is the speed of the water as it leaves the nozzle?


(b) What is the gauge pressure of the water in the hose at ground level?

Homework Equations

P+pgh+(1/2)pV^2 = P+pgh+(1/2)pV^2
Area * V = Area * V

The Attempt at a Solution

I'm confused on the part where the hose is elevated upward, I know the velocity decreases, but I don't know by how much.

 

[gneill]

Is water compressible or incompressible? What does that imply for the flow rate at any point along the (constant diameter) hose?

 

[asheik234]

It is compressible, does that mean it stays the same throughout?

 

[gneill]

It is compressible, does that mean it stays the same throughout?

Are you guessing? If water is compressible, why can't we store a cubic meter of water in a one liter container?

For all practical purposes, water is NOT compressible. So for a filled, constant diameter pipe, the flow rate must be the same everywhere -- otherwise, wouldn't water "pile up" behind the slow regions and "thin out" in the speedy ones?

 

[Nickie]

I would approach this problem by first identifying the relevant equations and variables, and then using them to solve for the unknown quantities. In this case, we can use the equations for Bernoulli's principle and continuity equation to solve for the speed of the water as it leaves the nozzle and the gauge pressure of the water in the hose at ground level.

(a) Using Bernoulli's principle, we can set the pressure at the nozzle (P1) equal to the pressure at ground level (P2) and solve for the speed of the water leaving the nozzle (v2). This gives us the equation P1 + pgh1 + (1/2)pv1^2 = P2 + pgh2 + (1/2)pv2^2. Since the water is flowing horizontally, we can set the heights (h1 and h2) equal to each other, and since the nozzle is elevated, we know that the pressure at ground level is atmospheric pressure (P2 = Patm). We can also assume that the water is incompressible, so the density (p) remains constant. This gives us the equation P1 + (1/2)pv1^2 = Patm + (1/2)pv2^2. We can then use the continuity equation, which states that the volume flow rate (Q) is constant, to relate the velocities (v1 and v2) to the areas (A1 and A2) of the nozzle and hose. This gives us the equation A1v1 = A2v2. Solving for v2, we get v2 = (A1/A2)v1. Plugging this into our first equation, we get P1 + (1/2)pv1^2 = Patm + (1/2)(A1/A2)^2pv1^2. Simplifying and solving for v1, we get v1 = √((2(Patm - P1))/p(1-(A1/A2)^2)). Plugging in the given values, we get v1 = 0.44 m/s.

(b) To find the gauge pressure of the water at ground level, we can use the equation P2 = Patm + pgh2 + (1/2)pv2^2. Plugging in the given values, we get P2 = 101300 Pa + (1000 kg