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Chapter 14: Problem 23

A long tube delivers \(10.0\) liters of alcohol in \(10.0 \mathrm{~min}\). What is the value of the flow rate? [Hint: Recall that \(1.00\) liter \(=1000 \mathrm{~cm}\) 3. ]

Short Answer

Expert verified
The flow rate is 1000 cm鲁/min.

Step by step solution

01

Understanding Flow Rate

Flow rate is the volume of fluid passing through a point in a period of time. It is generally expressed in volume per unit time, such as liters per minute (L/min) or cubic centimeters per minute (cm鲁/min).
02

Identify Given Values

We are given that the tube delivers 10.0 liters of alcohol in 10.0 minutes. So, the volume (V) is 10.0 liters and the time (t) is 10.0 minutes.
03

Convert Liters to Cubic Centimeters

Using the conversion provided, 1 liter is equivalent to 1000 cubic centimeters. Therefore, 10.0 liters is equal to \( 10.0 \times 1000 = 10000 \) cm鲁.
04

Calculate the Flow Rate

The flow rate \( Q \) can be calculated using the formula: \( Q = \frac{V}{t} \). Substitute \( V = 10000 \text{ cm}^3 \) and \( t = 10 \text{ min} \) into the equation, we get: \[ Q = \frac{10000 \text{ cm}^3}{10 \text{ min}} = 1000 \text{ cm}^3/ ext{min}. \]
05

Express the Flow Rate in Desired Units

Since we calculated in cm鲁/min, which is a common unit for flow rate in technical contexts, there's no need for further conversion unless another unit is specified.

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Key Concepts

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

Unit Conversion
Understanding how to convert between different units is a crucial skill in solving mathematics problems, particularly in fields like fluid dynamics. In our exercise, we were working with liters and needed to convert them to cubic centimeters to make further calculations easier.
  • 1 liter is equivalent to 1000 cubic centimeters (or cm鲁). This conversion factor comes from the metric system, where volume units follow the conversion: 1 cm鲁 equals the volume of a cube with a 1 cm side.
  • Converting units correctly ensures that calculations are accurate, as mathematical operations must use compatible units. If units are mismatched, it can lead to erroneous outcomes.
Remember to always double-check your unit conversions to avoid common pitfalls that can throw off your entire calculation. Particularly in scientific contexts, precision and correctness in unit conversion can be pivotal.
Fluid Dynamics
Fluid dynamics is the study of fluids (liquids and gases) in motion. Many practical applications depend on understanding how fluids flow through different environments, such as in pipes, air ducts, or even around aircraft. In the provided exercise, the concept of flow rate is central, which describes how much fluid passes through a point in a given time.
  • Flow rate is often denoted by the symbol \( Q \) and reflects the efficiency and speed of fluid movement.
  • In a steady state, the flow rate through a system helps designers and engineers predict pressure changes and optimize performance.
  • When dealing with liquids like alcohol in a tube, understanding the flow helps to ensure safe and efficient delivery or transfer.
Fluid dynamics emphasizes the importance of considering viscosity, temperature, and forces affecting the fluid to fully understand its behavior.
Mathematics Problem-Solving
Solving any mathematical problem involves a clear understanding of the problem statement and careful application of known formulas and concepts. In our exercise, this involved several steps:
  • Firstly, identify what the problem is asking: A flow rate. We need to calculate how fast a fluid is moving through the tube by determining its flow rate.
  • We are given initial values for volume and time, which are critical to solving the problem since they allow us to use the flow rate formula, \( Q = \frac{V}{t} \).
  • Then, perform unit conversion when necessary, like converting liters to cubic centimeters, to ensure compatibility with the formula used.
  • Finally, the calculation step involves substituting the known values into the equation and carrying out the division to find the solution.
Problem-solving in this context underscores the need to grasp each part of the procedure to achieve an accurate and reliable outcome.
Measurement Units
Measurement units are the building blocks for any scientific, technical, or engineering calculations. They provide a standard language that scientists and engineers use worldwide to communicate quantitative data.
  • In fluid dynamics and similar fields, measurement units like liters, minutes, and cubic centimeters are common. These units must be consistent throughout a calculation to achieve an accurate result.
  • The system of units used should align with the problem requirements. For instance, using liters to express volumes in metric systems helps maintain consistency with international conventions.
  • Flow rate, measured in units such as cm鲁/min or L/min, allows insight into the dynamics of fluid movement and is key to performance analysis in systems handling liquids or gases.
Choosing appropriate units and being adept in converting between them is crucial for clear communication and meaningful analysis of experimental data in any scientific endeavor.

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Most popular questions from this chapter

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, \(P_{i}-P_{o}=\rho g h=\left(13600 \mathrm{~kg} / \mathrm{m}^{3}\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(0.0500 \mathrm{~m})=6660 \mathrm{~N} / \mathrm{m}^{2}\right.\) 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, $$ J=\frac{\pi r^{4}\left(P_{-}-P_{p}\right)}{8 \eta L}=\frac{\pi\left(7.5 \times 10^{-4} \mathrm{~m}\right)^{4}\left(66660 \mathrm{~N} / \mathrm{m}^{2}\right)}{8\left(8.01 \times 10^{-4} \mathrm{~kg} / \mathrm{m} \cdot \mathrm{s}(0.200 \mathrm{~m})\right.}=5.2 \times 10^{-6} \mathrm{~m}^{3} / \mathrm{s}=5.2 \mathrm{~mL} / \mathrm{s} $$ In \(30.0 \mathrm{~s}\), the quantity that would flow out of the tube is \((5.2 \mathrm{~mL} / \mathrm{s})\) \((30 \mathrm{~s})=1.6 \times 10^{2} \mathrm{~mL}\)

A pump lifts water at the rate of \(9.0\) liters/s from a lake through a 5.0-cm-i.d. pipe and discharges it into the air at a point \(16 \mathrm{~m}\) above the level of the water in the lake. What are the theoretical (a) velocity of the water at the point of discharge and \((b)\) power delivered by the pump.

A wind tunnel is to be used with a 20-cm-high model car to approximately reproduce the situation in which a \(550-\mathrm{cm}\) -high car is moving at \(15 \mathrm{~m} / \mathrm{s}\). What should be the wind speed in the tunnel? Is the flow likely to be turbulent? We want the Reynolds number \(N_{R}\) to be the same in both cases, so that the situations will be similar. That is, $$ N_{R}=\left(\frac{\rho v D}{\eta}\right)_{\text {tunnel }}=\left(\frac{\rho v D}{\eta}\right)_{\text {air }} $$ Both \(\rho\) and \(\eta\) are the same in the two cases, hence, $$ v, D_{i}=v_{a} D_{a} \text { from which } v_{t}=v_{a} \frac{D_{a}}{D_{i}}=(15 \mathrm{~m} / \mathrm{s})(550 / 20)=0.41 \mathrm{~km} / \mathrm{s} $$ To investigate turbulence, evaluate \(N_{R}\) using \(\rho=1.29 \mathrm{~kg} / \mathrm{m}^{3}\) and \(\eta\) \(=1.8 \times 10^{-5} \mathrm{~Pa} \cdot \mathrm{s}\) for air. Consequently \(N_{R}=5.9 \times 10^{6}\), a value far in excess of that required for turbulent flow. The flow will certainly be turbulent.

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.

An artery in a person has been reduced to half its original inside diameter by deposits on the inner artery wall. By what factor will the blood flow through the artery be reduced if the pressure differential across the artery has remained unchanged? The relationship governing flow rate, pressure differential, and opening radius is Poiseuille's Law, wherein \(J \propto r^{4}\). Therefore,
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