A village maintains a large tank with an open top,...

A village maintains a large tank with an open top, containing water for emergencies. The water can drain from the tank through a hose of diameter $6.60 \mathrm{cm} .$ The hose ends with a nozzle of diameter $2.20 \mathrm{cm} .$ A rubber stopper is inserted into the nozzle. The water level in the tank is kept 7.50 $\mathrm{m}$ above the nozzle. (a) Calculate the friction force exerted on the stopper by the nozzle. (b) The stopper is removed. What mass of water flows from the nozzle in 2.00 $\mathrm{h}$ ? (c) Calculate the gauge pressure of the flowing water in the hose just behind the nozzle.

Key Concepts

Bernoulli's Equation

Bernoulli's equation is an expression of the conservation of energy for flowing fluids, linking pressure, kinetic energy, and potential energy along a streamline. It provides insight into how the pressure in a fluid decreases as its velocity increases and is used to analyze the energy conversion processes within the tank, hose, and nozzle system. This equation underpins the calculation of pressure drops when the fluid begins to flow and is essential for understanding the gauge pressure just behind the nozzle.

Hydrostatic Pressure

Hydrostatic pressure is the pressure exerted by a fluid at rest due to the weight of the fluid above a given point. It is calculated as the product of the fluid’s density, gravitational acceleration, and the height of the fluid column. This concept is crucial when determining the gauge pressure at a certain depth or elevation below the water surface, as seen in how the water level in the tank creates pressure at the nozzle level.

Torricelli's Law

Torricelli's law relates the speed of fluid exiting an orifice under the influence of gravity to the height of the fluid above the opening. It effectively shows that the exit velocity is the square root of twice the gravitational acceleration multiplied by the height. This principle is applied when calculating the velocity of water exiting the nozzle once the stopper is removed, which in turn helps determine the volumetric flow rate and mass of water discharged over a period.

Continuity Equation

The continuity equation in fluid mechanics expresses the conservation of mass in a flowing fluid, stating that the product of cross-sectional area and velocity remains constant along a streamline. This concept is key when analyzing flow through different sections of a system—such as the hose and the nozzle—since the decrease in area at the nozzle leads to an increase in velocity, affecting the flow rate and the resulting dynamics downstream.

Pressure Forces on Surfaces

Pressure forces on surfaces arise from the product of pressure and the area over which the pressure acts. In the context of the nozzle with a stopper, the water’s static pressure results in a force being exerted on the stopper. Understanding how to calculate this force is important for determining the frictional forces at play, and it involves multiplying the gauge pressure (due to the height of the water column) by the cross-sectional area over which it acts.

Transcript

00:06 In this particular problem, the water level is a height h over the output nozzle.

00:22 Output nozzle.

00:26 So over here if i have a pressure of the atmospheric pressure, over here i have the pressure of the atmospheric pressure.

00:36 But now the nozzle is trying to come out.

00:39 So the frictional forces also in this direction and now i have to find what is the pressure inside so if i want to equate the forces then force by friction plus p -not equals p of the water p of the water which is h -row g so the fictional force comes out to be h -r -g minus atmospheric pressure, which is 7 .5 times density is 1000 times 9 .8 minus p -not is 1 .013 times 10 to the 5.

02:47 However, there is p .013 times 10 to the 5.

02:51 Not over here as well so we do not have to take this into account so this is the applied force which is the pressure difference we have to multiply it with the area and the area is pi r square so times pi times the radius is given by by 1 .1 centimeter 0 .0 1 meter square calculating this we get the frictional force to be 27 .9 newton now what happens is let's say the area over here is a2 and area over here where the that's the nozzle and that's the pipe over here a1 so you will have the pressures does not change the so at the very tippy top we do not have any height so we will be using in initial half row v1 square thus row g let us call this height to be y1 and let us call this to be y1 and now let us call from the ground the top height to be y2 and the diameter of the nozzle to be a 2 in that case half row v1 square plus row g y1 equals when you are at the top the velocity is zero so the first term goes to zero and then you have row g y2 so from here the densities cancel out and you have v1 equals you can take the density common you will have y2 minus y1 and there's a two and then the square root so now that will come out to be square root of two actually the densities cancel out as well but the g stays so the difference in the height is 7 .5 meter and g is 9 .8 all together this comes out to be 12 .125 meters per second...

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