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 nends 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.

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 nends 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

Continuity Equation

The continuity equation, a statement of the conservation of mass for an incompressible fluid, asserts that the volumetric flow rate must remain constant along a streamline. It is expressed as A?v? = A?v?, where A is the cross-sectional area and v is the velocity of the fluid. This concept is key to relating changes in the flow area (such as from a hose to a nozzle) to changes in the flow speed.

Torricelli's Law

This is a specific application of Bernoulli’s principle stating that the speed of efflux of a fluid under gravity from a sharp-edged hole is equal to the speed that a body would acquire in falling freely from the same height, given by v = ?(2gh). Torricelli’s law is vital for calculating the discharge velocity of water exiting an orifice when the water surface is open to the atmosphere.

Gauge Pressure

Gauge pressure is the pressure of a fluid relative to the ambient atmospheric pressure rather than absolute pressure. In many practical applications, including flow dynamics in hoses and nozzles, it is the gauge pressure that is measured and used to evaluate the force acting on surfaces. It is defined as the difference between the absolute pressure and the atmospheric pressure.

Hydrostatic Pressure

This concept refers to the pressure exerted by a fluid at equilibrium due to the force of gravity. In a tank with a free surface, the pressure at any given depth increases linearly with that depth according to the relation P = ?gh, where ? is the fluid density, g is the gravitational acceleration, and h is the depth below the free surface. This principle is fundamental in determining the force on objects submerged or in contact with an opening at a given depth.

Bernoulli's Equation

Bernoulli's equation is a statement of energy conservation for flowing fluids. It relates the pressure, kinetic energy per unit volume, and potential energy per unit volume along a streamline. In problems involving water flow from a tank, it is used to connect the hydrostatic pressure in the tank with the kinetic energy of the jet at the outlet, allowing the determination of the velocity of the exiting water.

Transcript

00:01 So here in this problem we are given this large tank and water can drain from this large trunk through this section and the diameter of this section we are given as d1 is 6 .60 centimeter.

00:16 Now the nozzle is fitted at the bottom of the tank and the nozzle diameter d2 we are given as 2 .20 centimeter.

00:28 Now there are three parts in the equation in the a part.

00:31 A rubber stopper is inserted into the nozzle as you can see here and we have to calculate the friction force exerted on the stopper by the nozzle.

00:41 So for that we'll be first calculating the pressure at this point.

00:47 Now we know that pressure at this point will be given by the expression p is equals to row g h, where row is the density of water which is 1000 kg per meter cube times acceleration due to gravity.

01:06 Is 9 .81 meter per second square times this distance h which is 7 .5 so from here we'll get the pressure as 73 ,575 pascal or newton per meter square.

01:28 Now this is the pressure which is exerted at this point.

01:32 Now the force exerted due to this pressure would be equals to force is equal to pressure times the area.

01:39 So pressure we this much 73 ,575 newton per meter square and cross -sectional area of this section is pi -by -4 d t2 square so pi -by -four d2 square now we can plug the values this is as it is 735 newton per meter square times pi by four d two so diameter the d2 we are given as 2 .20 centimeter but we have to take the values in meters so we'll be converting into meter square now we're going to simplify this and we'll get the force as 27 .9 newton so this is the force acting on the nozzle by the water towards this side so this force should be equal to the frictional force so that it can counterbalance this force that means we can conclude that the friction force acting on the rubber stopper by the nozzle is 27 .9 newton.

02:54 So this is the answer for the a part.

02:56 Now next we'll be solving the b part...

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