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 m 2 , 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.

Hey everyone. So this problem is dealing with volumetric flow rates. Let's see what it gives us a section of the pipeline has a cross sectional area of 1.15 square meters. And the velocity magnitude at that point is 1.8 m per second, determine the volume delivered to a tank every hour at the pipeline's terminal. So the volume delivered every hour is a is a rate. So it's asking us to find the volumetric flow rate. So the givens from the problem, we know area is 1.15 square meters and our velocity is 1.8 m per second. And we need to recall our continuity equation Q. That's our volumetric flow rate equals A B. So it, it really is that simple. It's a plug-in chug from there. So we have 0.15 m squared, that's 1.8 m per second. And that gives us 2.07 m cubed per second. They asked us for a rate in hours. So all we have to do is a little bit of unit conversion here, 2.07 m cube per second and then there's 3600 seconds in one hour. And that gives us 7452 m cubed per hour if you look at our choices, uh and the closest one is d they just rounded it to three significant figures instead of four. And so that is our answer, 7450 m cubed per hour. That's all for this one. We'll see you guys in the next video.

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19. Fluid Mechanics

Fluid Flow & Continuity Equation

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19. Fluid Mechanics

Fluid Flow & Continuity Equation

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2:13 minutes

Problem 38c

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.

Verified step by step guidance

First, understand that the problem involves fluid flow through a pipe with varying cross-sectional areas. We need to calculate the volume of water discharged from the pipe over a given time period.

Use the principle of continuity for incompressible fluids, which states that the product of cross-sectional area (A) and fluid velocity (v) is constant along the pipe: A1 * v1 = A2 * v2. However, since we are calculating the discharge volume, we focus on the flow rate at point 1.

Calculate the flow rate at point 1 using the formula: Flow rate = A1 * v1. Substitute the given values: A1 = 0.070 m^2 and v1 = 3.50 m/s.

Determine the total volume of water discharged over 1 hour. Use the formula: Volume = Flow rate * time. Convert 1 hour into seconds (1 hour = 3600 seconds) to ensure consistent units.

Substitute the calculated flow rate and the time in seconds into the volume formula to find the total volume of water discharged from the pipe.

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Key Concepts

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

Continuity Equation

The continuity equation in fluid dynamics states that the mass flow rate of a fluid must remain constant from one cross-section of a pipe to another, assuming incompressible flow. It is expressed as A1V1 = A2V2, where A is the cross-sectional area and V is the fluid velocity. This principle helps determine how changes in pipe diameter affect fluid velocity.

Volume Flow Rate

Volume flow rate is the volume of fluid that passes through a given surface per unit time, typically measured in cubic meters per second (m³/s). It is calculated as the product of the cross-sectional area of the pipe and the fluid velocity (Q = A × V). This concept is crucial for determining the total volume of fluid discharged over a specific period.

Unit Conversion

Unit conversion is the process of converting a quantity expressed in one set of units to another. In this context, converting the time from hours to seconds is necessary to ensure consistency in units when calculating the volume of water discharged. This involves multiplying the time in hours by 3600 to convert it to seconds.

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