/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 1 Oil flows through a pipe \(8.0 \... [FREE SOLUTION] | 91Ó°ÊÓ

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Oil flows through a pipe \(8.0 \mathrm{~cm}\) in diameter, at an average speed of \(4.0 \mathrm{~m} / \mathrm{s}\). What is the flow rate, \(J\), in \(\mathrm{m}^{3} / \mathrm{s}\) and \(\mathrm{m}^{3} / \mathrm{h}\) ? $$ \begin{aligned} J &=A v=\pi(0.040 \mathrm{~m})^{2}(4.0 \mathrm{~m} / \mathrm{s})=0.020 \mathrm{~m}^{3} / \mathrm{s} \\ &=\left(0.020 \mathrm{~m}^{3} / \mathrm{s}\right)(3600 \mathrm{~s} / \mathrm{h})=72 \mathrm{~m}^{3} / \mathrm{h} \end{aligned} $$

Short Answer

Expert verified
The flow rate is 0.020 m³/s and 72 m³/h.

Step by step solution

01

Calculate the Cross-sectional Area of the Pipe

The flow rate can be calculated using the formula \( J = A v \), where \( A \) is the cross-sectional area of the pipe and \( v \) is the velocity of the oil. First, calculate the area \( A \) using the radius of the pipe, obtained from the diameter (8.0 cm). Convert the diameter into meters: \( 8.0 \text{ cm} = 0.08 \text{ m} \), and thus the radius is \( 0.08 / 2 = 0.04 \text{ m} \). The area \( A \) is given by \( A = \pi r^2 \). Therefore, \( A = \pi (0.04)^2 \approx 0.005024 \text{ m}^2 \).
02

Calculate Flow Rate in Cubic Meters per Second

Now, use the formula \( J = A v \) to find the flow rate in cubic meters per second. Substitute \( A = 0.005024 \text{ m}^2 \) and \( v = 4.0 \text{ m/s} \): \( J = 0.005024 \times 4.0 = 0.020096 \text{ m}^3/s \). For simplicity in reporting, this is rounded to \( 0.020 \text{ m}^3/s \).
03

Convert Flow Rate to Cubic Meters per Hour

Convert the flow rate from \( \text{m}^3/s \) to \( \text{m}^3/h \). Since there are 3600 seconds in an hour, multiply the flow rate by 3600: \( J = 0.020 \times 3600 = 72 \text{ m}^3/h \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Cross-sectional Area
The cross-sectional area is an essential part of flow rate calculations in fluid dynamics. It represents the slice of space through which the fluid moves. Imagine taking a knife and cutting straight through a pipe. The shape you see, typically a circle, is the cross-section.This area helps us understand how much fluid can pass through a given section of the pipe.
To calculate this area for a circular pipe, use the formula:
  • The cross-sectional area, \( A \), is calculated as \( A = \pi r^2 \), where \( r \) is the radius of the pipe.
First, convert your diameter to radius by dividing it by two. In this exercise, the diameter is 8.0 cm, which converts to 0.08 meters (since 1 cm = 0.01 m), making the radius 0.04 meters. Then simply square the radius and multiply by \( \pi \) (approximately 3.14159) to find the area in square meters.
Fluid Dynamics
Fluid dynamics allows us to predict and calculate the movement of fluids like oils, water, and gases through various systems. This field applies principles from physics to analyze how fluids flow and interact with their surroundings.
Flow rate, represented as \( J \) or sometimes \( Q \), is a key concept. It's a measure of the volume of fluid moving through a given cross-sectional area per unit of time.In our scenario, the flow rate \( J \) is calculated using the formula:
  • \( J = A \times v \), where \( A \) is the cross-sectional area and \( v \) is the velocity of the fluid.
Understanding fluid dynamics is crucial for designing systems that involve the transfer of fluids, ensuring pipes are adequately sized and functioning efficiently.
Unit Conversion
Unit conversion is a vital aspect of many mathematical and scientific calculations. It helps to ensure that our calculations use consistent units, making comparisons and calculations meaningful.
In this exercise, we're tasked with converting the flow rate from cubic meters per second to cubic meters per hour:
  • The key is recognizing the time conversion factor of seconds to hours.
Since there are 3600 seconds in an hour (60 seconds/minute * 60 minutes/hour), you multiply the flow rate in \( \text{m}^3/\text{s} \) by 3600 to get it in \( \text{m}^3/\text{h} \).This conversion ensures the representation of flow rate in a more practical, often used, unit for measuring large fluid volumes over time.
Pipe Diameter
Pipe diameter is a simple but crucial dimension that tells us how wide the pipe is. It directly affects the flow rate and the speed at which fluid moves through the pipe.
The flow rate through a pipe is heavily dependent on its diameter, as it determines the size of the cross-sectional area.
  • For round pipes, the cross-sectional area is calculated using the radius, which is half of the diameter.
  • So, the diameter must first be converted to meters for typical industrial or scientific calculations.
In this instance, with an 8.0 cm diameter, we directly impact both the area calculation and ultimately the flow rate, reinforcing why attention to dimensional accuracy is key.

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Most popular questions from this chapter

Compute the average speed of water in a pipe having an i.d. of \(5.0 \mathrm{~cm}\) and delivering \(2.5 \mathrm{~m}^{3}\) of water per hour.

A 14 -cm inner diameter (i.d.) water main furnishes water to a \(1.00 \mathrm{~cm}\) i.d. (i.e., inner diameter) faucet pipe. If the average speed in the faucet pipe is \(3.0 \mathrm{~cm} / \mathrm{s}\), what will be the average speed it causes in the water main? The two flows are equal. From the Continuity Equation, $$J=A_{1} v_{1}=A_{2} v_{2}$$ Letting 1 be the faucet and 2 be the water main, we have $$ v_{2}=v_{1} \frac{A_{1}}{A_{2}}=v_{1} \frac{\pi r_{1}^{2}}{\pi r_{2}^{2}}=(3.0 \mathrm{~cm} / \mathrm{s})\left(\frac{1}{14}\right)^{2}=0.015 \mathrm{~cm} / \mathrm{s}$$.

Exactly \(250 \mathrm{~mL}\) of fluid flows out of a tube whose inner diameter is \(7.0 \mathrm{~mm}\) in a time of \(41 \mathrm{~s}\). What is the average speed of the fluid in the tube? From \(J=A v\), since \(1 \mathrm{~mL}=10^{-6} \mathrm{~m}^{3}\), $$ v=\frac{J}{A}=\frac{\left(250 \times 10^{-6} \mathrm{~m}^{3}\right) /(41 \mathrm{~s})}{\pi(0.0035 \mathrm{~m})^{2}}=0.16 \mathrm{~m} / \mathrm{s} $$

A hypodermic needle of length \(3.0 \mathrm{~cm}\) and i.d. \(0.45 \mathrm{~mm}\) is used to draw blood \((\eta=4.0 \mathrm{mPl})\). Assuming the pressure differential across the needle is \(80 \mathrm{cmHg}\), how long does it take to draw \(15 \mathrm{~mL}\) ?

How much water will flow in \(30.0\) s through \(200 \mathrm{~mm}\) of capillary tube of \(1.50 \mathrm{~mm}\) i.d., if the pressure differential across the tube is \(5.00 \mathrm{~cm}\) of mercury? The viscosity of water is \(0.801 \mathrm{cP}\) and \(\rho\) for mercury is \(13600 \mathrm{~kg} / \mathrm{m}^{3}\). We shall make use of Poiseuille's Law, \(J=\pi r^{4}\left(P_{i}-P_{o}\right) / 8 \eta L\), and therefore, $$P_{i}-P_{o}=\left(\rho g h=\left(13600 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(0.0500 \mathrm{~m})=6660 \mathrm{~N} / \mathrm{m}^{2}\right.$$ The viscosity expressed in \(\mathrm{kg} / \mathrm{m} \cdot \mathrm{s}\) is $$\eta=(0.801 \mathrm{cP})\left(10^{-3} \frac{\mathrm{kg} / \mathrm{m} \cdot \mathrm{s}}{\mathrm{cP}}\right)=8.01 \times 10^{-4} \mathrm{~kg} / \mathrm{m} \cdot \mathrm{s}$$ Thus, $$J=\frac{\pi r^{4}\left(P_{i}-P_{o}\right)}{8 \eta L}=\frac{\pi\left(7.5 \times 10^{-4} \mathrm{~m}\right)^{4}\left(6660 \mathrm{~N} / \mathrm{m}^{2}\right)}{8\left(8.01 \times 10^{-4} \mathrm{~kg} / \mathrm{m} \cdot \mathrm{s}\right)(0.200 \mathrm{~m})}=5.2 \times 10^{-6} \mathrm{~m}^{3} / \mathrm{s}=5.2 \mathrm{~mL} / \mathrm{s}$$ In \(30.0 \mathrm{~s}\), the quantity that would flow out of the tube is \((5.2 \mathrm{~mL} / \mathrm{s})(30 \mathrm{~s})=1.6 \times 10^{2} \mathrm{~mL}\).

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