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.
Ch 12: Fluid Mechanics
Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Not the one you use?Change textbook
Chapter 12, Problem 50
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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Identify the relationship between pressure difference, volume flow rate, and radius for a viscous fluid in a cylindrical pipe using Poiseuille's Law: \( Q = \frac{\pi \cdot \Delta P \cdot r^4}{8 \cdot \eta \cdot L} \), where \( Q \) is the volume flow rate, \( \Delta P \) is the pressure difference, \( r \) is the radius, \( \eta \) is the dynamic viscosity, and \( L \) is the length of the pipe.
Recognize that the volume flow rate \( Q \) and the dynamic viscosity \( \eta \) remain constant, and the length \( L \) of the pipe is unchanged. Therefore, the relationship between pressure difference and radius can be expressed as \( \Delta P_1 \cdot r_1^4 = \Delta P_2 \cdot r_2^4 \).
Substitute the known values into the equation: \( 6.00 \times 10^4 \text{ Pa} \cdot (0.210 \text{ m})^4 = \Delta P_2 \cdot (0.0700 \text{ m})^4 \).
Solve for the new pressure difference \( \Delta P_2 \) by rearranging the equation: \( \Delta P_2 = \frac{6.00 \times 10^4 \text{ Pa} \cdot (0.210 \text{ m})^4}{(0.0700 \text{ m})^4} \).
Calculate the value of \( \Delta P_2 \) using the given values to find the required pressure difference for the reduced radius.

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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Poiseuille's Law
Poiseuille's Law describes the volumetric flow rate of a viscous fluid through a cylindrical pipe. It states that the flow rate is directly proportional to the pressure difference and the fourth power of the pipe's radius, and inversely proportional to the fluid's viscosity and the pipe's length. This relationship highlights the significant impact of the pipe's radius on flow rate.
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Gauss' Law
Viscosity
Viscosity is a measure of a fluid's resistance to deformation or flow. In the context of fluid dynamics, it quantifies the internal friction within the fluid, affecting how easily it can move through a pipe. Higher viscosity means greater resistance to flow, requiring a larger pressure difference to maintain the same flow rate.
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Continuity Equation
The Continuity Equation in fluid dynamics states that the mass flow rate must remain constant from one cross-section of a pipe to another, assuming incompressible flow. This principle implies that if the pipe's radius decreases, the fluid velocity must increase to maintain the same volume flow rate, affecting the required pressure difference.
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Related Practice
Textbook Question
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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
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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