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Ch 12: Fluid Mechanics
Young & Freedman Calc - University Physics 14th Edition
Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Not the one you use?Change textbook
Chapter 12, Problem 44

A small circular hole 6.00 mm in diameter is cut in the side of a large water tank, 14.0 m below the water level in the tank. The top of the tank is open to the air. Find (a) the speed of efflux of the water and (b) the volume discharged per second.

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To find the speed of efflux of the water, we can use Torricelli's theorem, which states that the speed \( v \) of efflux of a fluid under gravity through a hole is given by \( v = \sqrt{2gh} \), where \( g \) is the acceleration due to gravity (approximately 9.81 m/s²) and \( h \) is the height of the fluid above the hole. Here, \( h = 14.0 \) m.
Substitute the values into the equation: \( v = \sqrt{2 \times 9.81 \times 14.0} \). This will give you the speed of efflux in meters per second.
Next, to find the volume discharged per second, we need to calculate the area of the hole. The diameter of the hole is 6.00 mm, which is 0.006 m. The area \( A \) of the hole is given by the formula for the area of a circle: \( A = \pi r^2 \), where \( r \) is the radius of the hole.
Calculate the radius: \( r = \frac{0.006}{2} \) m. Then, substitute into the area formula: \( A = \pi \times (0.003)^2 \).
The volume discharged per second, also known as the flow rate \( Q \), is given by \( Q = A \times v \). Use the area calculated in the previous step and the speed of efflux from step 2 to find \( Q \).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Torricelli's Law

Torricelli's Law states that the speed of efflux of a fluid under gravity from an orifice is given by v = √(2gh), where v is the speed, g is the acceleration due to gravity, and h is the height of the fluid above the opening. This principle is crucial for determining the speed at which water exits the tank.
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Continuity Equation

The Continuity Equation in fluid dynamics expresses the conservation of mass in a flowing fluid. It states that the product of the cross-sectional area and the velocity of the fluid remains constant along a streamline. This concept helps calculate the volume of water discharged per second by relating the speed of efflux to the area of the hole.
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Hydrostatic Pressure

Hydrostatic pressure is the pressure exerted by a fluid at equilibrium due to the force of gravity. It increases with depth and is given by P = ρgh, where ρ is the fluid density, g is gravity, and h is the height of the fluid column. Understanding this concept is essential for analyzing the forces driving the water out of the tank.
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Related Practice
Textbook Question

Home Repair. You need to extend a 2.50-inch-diameter pipe, but you have only a 1.00-inch-diameter pipe on hand. You make a fitting to connect these pipes end to end. If the water is flowing at 6.00 cm/s in the wide pipe, how fast will it be flowing through the narrow one?

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Textbook Question

A soft drink (mostly water) flows in a pipe at a beverage plant with a mass flow rate that would fill 220 0.355-L cans per minute. At point 2 in the pipe, the gauge pressure is 152 kPa and the cross-sectional area is 8.00 cm2. At point 1, 1.35 m above point 2, the cross-sectional area is 2.00 cm2. Find the (b) volume flow rate. (c) flow speeds at points 1 and 2.

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Textbook Question

Water is flowing in a pipe with a varying cross-sectional area, and at all points the water completely fills the pipe. At point 1 the cross-sectional area of the pipe is 0.070 m2, and the magnitude of the fluid velocity is 3.50 m/s. (c) Calculate the volume of water discharged from the open end of the pipe in 1.00 hour.

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Textbook Question

At one point in a pipeline the water's speed is 3.00 m/s and the gauge pressure is 5.00×104 Pa. Find the gauge pressure at a second point in the line, 11.0 m lower than the first, if the pipe diameter at the second point is twice that at the first.

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Textbook Question

A pressure difference of 6.00 × 104 Pa is required to maintain a volume flow rate of 0.800m3/s for a viscous fluid flowing through a section of cylindrical pipe that has radius 0.210 m. What pressure difference is required to maintain the same volume flow rate if the radius of the pipe is decreased to 0.0700 m?

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Textbook Question

BIO. Artery Blockage. A medical technician is trying to determine what percentage of a patient's artery is blocked by plaque. To do this, she measures the blood pressure just before the region of blockage and finds that it is 1.20×104 Pa, while in the region of blockage it is 1.15×104 Pa. Furthermore, she knows that blood flowing through the normal artery just before the point of blockage is traveling at 30.0 cm/s, and the specific gravity of this patient's blood is 1.06. What percentage of the cross-sectional area of the patient's artery is blocked by the plaque?