A large tank of water has a hose connected to it (Fig....

A large tank of water has a hose connected to it (Fig. P18.61). The tank is sealed at the top and has compressed air between the water surface and the top. When the water height $h$ has the value $3.50 \mathrm{~m}$, the absolute pressure $p$ of the compressed air is $4.20 \times 10^{5} \mathrm{~Pa}$. Assume that the air above the water expands at constant temperature, and take the atmospheric pressure to be $1.00 \times 10^{\circ} \mathrm{Pa}$. (a) What is the speed with which water flows out of the hose when $h=3.50 \mathrm{~m} ?$ (b) As water flows out of the tank, $h$ decreases. Calculate the speed of flow for $h=3.00 \mathrm{~m}$ and for $h=2.00 \mathrm{~m}$. (c) At what value of $h$ does the flow stop?

A large tank of water has a hose connected to it (Fig. P18.61). The tank is sealed at the top and has compressed air between the water surface and the top. When the water height $h$ has the value $3.50 \mathrm{~m}$, the absolute pressure $p$ of the compressed air is $4.20 \times 10^{5} \mathrm{~Pa}$. Assume that the air
above the water expands at constant temperature, and take the atmospheric pressure to be $1.00 \times 10^{\circ} \mathrm{Pa}$.
(a) What is the speed with which water flows out of the hose when $h=3.50 \mathrm{~m} ?$ (b) As water flows out of the tank, $h$ decreases. Calculate the speed of flow for $h=3.00 \mathrm{~m}$ and for $h=2.00 \mathrm{~m}$.
(c) At what value of $h$ does the flow stop?

Key Concepts

Isothermal (Boyle's) Process

An isothermal process in thermodynamics is one in which the temperature of a gas remains constant while its pressure and volume change. Boyle's Law, which states that the product of pressure and volume remains constant for a given mass of an ideal gas at constant temperature, is an example of this. In a sealed tank with compressed air expanding at constant temperature, this concept explains how the compressed air pressure adjusts as the water exits the tank, impacting the driving pressure for the flow.

Bernoulli's Principle

Bernoulli's Principle is a fundamental concept in fluid dynamics that relates the pressure, velocity, and elevation in a moving fluid. It explains how an increase in the speed of a fluid occurs simultaneously with a decrease in its pressure or potential energy. In problems involving fluid flow from a tank or through a hose, this principle helps determine the speed of the fluid as it exits the tank by connecting the pressure differences to the kinetic energy of the flow.

Hydrostatic Pressure

Hydrostatic pressure is the pressure exerted by a fluid at rest due to the force of gravity. It increases with depth in the fluid, and this variation is given by the product of the fluid density, gravitational acceleration, and depth. Understanding hydrostatic pressure is essential for analyzing how differences in fluid height contribute to pressure differences, which in turn affect the flow dynamics in the tank and hose system.

Conservation of Energy in Fluid Flow

The principle of conservation of energy in fluid flow states that the total energy (including kinetic, potential, and pressure energies) remains constant along a streamline in steady flow. This concept underpins the derivation of Bernoulli's equation and is essential in connecting the energy provided by the compressed air and the gravitational potential energy of the water to the kinetic energy of the fluid flowing out of the hose.

Transcript

00:01 Apply bernoulli's equation to relate the e -flux speed of water out the hose to the height of the water in the tank and the pressure of the air above the water in the tank.

00:12 And we need to use the ideal gas equation to relate the volume of the air in the tank to the pressure of the air.

00:21 So p1 is equal to 4 .20 times 10 to the 5th pascels.

00:29 P2 is equal to pair, which is equal to 1 .00 times 10 to the 5th pascels.

00:37 And the large tank implies that v1 is equal to 0.

00:43 So if we start off with a, p1 plus pgy1 plus 1 1 1⁄2 plus 1⁄2 plus 1⁄2.

01:03 Pv2 squared.

01:06 And so 1 1�, pv2 squared is equal to p1 minus p2 plus pg times y1 minus y2.

01:17 V2 is what we are solving for.

01:21 We get the square root of 2 divided by p times p1 minus p2 plus 2g times y1 minus y2 plus 2g times y1 minus y2.

01:35 And we get 26 .2 meters per second.

01:40 For part b, h is equal to 3 .00 meters.

01:47 So we have p0, or i'm sorry, p is equal to p0 times 4 .00 meters minus h0 over 4 .00 meters minus h is equal to 4 .20 times 10 to the 5th pascal's times, 4 .00 meters minus 350, 3 .50.

02:14 3 .5 meters.

02:19 Divided by 4 .00 meters minus 3 .00 meters.

02:24 And you get 2 .10 times 10 to the 5th pascels.

02:32 And we repeat the calculation of part a, but now p2 equals what we just got.

02:38 So v2 equals the square root of 2...

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