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Chapter 12: Problem 67

A capillary tube of radius \(R\) and length \(L\) is connected in series with another tube of radius \(R / 2\) and length \(L / 4\). If the pressure difference across the two tubes taken together is \(p\), then the ratio of pressure difference across the first tube to that across the second tube is (a) \(1: 4\) (b) \(1: 1\) (c) \(4: 1\) (d) \(2: 1\)

Short Answer

Expert verified
The pressure difference ratio is \(4:1\).

Step by step solution

01

Understand the Problem

We have two capillary tubes connected in series. The first tube has radius \(R\) and length \(L\), and the second tube has radius \(\frac{R}{2}\) and length \(\frac{L}{4}\). We need to find the ratio of the pressure difference across the first tube to the pressure difference across the second tube given that the total pressure difference across both is \(p\).
02

Apply Poiseuille's Law

According to Poiseuille's Law, the flow rate \(Q\) through a tube is given by the formula: \[ Q = \frac{\pi R^4 \Delta P}{8 \eta L} \]Where \(\Delta P\) is the pressure difference, \(\eta\) is the viscosity of the fluid, and \(R\) and \(L\) are the radius and length of the tube respectively. For tubes in series, the flow rate \(Q\) is the same through both tubes.
03

Calculate Pressure Difference for Tube 1

Let \(\Delta P_1\) be the pressure difference across the first tube. Applying Poiseuille's Law for the first tube:\[ Q = \frac{\pi R^4 \Delta P_1}{8 \eta L} \]
04

Calculate Pressure Difference for Tube 2

Let \(\Delta P_2\) be the pressure difference across the second tube, with radius \(\frac{R}{2}\) and length \(\frac{L}{4}\). The flow rate \(Q\) for the second tube is:\[ Q = \frac{\pi \left(\frac{R}{2}\right)^4 \Delta P_2}{8 \eta \left(\frac{L}{4}\right)} = \frac{\pi \frac{R^4}{16} \Delta P_2}{2 \eta L} \] Simplifying, it becomes:\[ Q = \frac{\pi R^4 \Delta P_2}{32 \eta L} \]
05

Equate Flow Rates of Both Tubes

Since the flow rate is the same for both tubes in series:\[ \frac{\pi R^4 \Delta P_1}{8 \eta L} = \frac{\pi R^4 \Delta P_2}{32 \eta L} \]
06

Simplify and Solve for Pressure Ratio

Cancelling out the common factors and simplifying gives:\[ \frac{\Delta P_1}{8} = \frac{\Delta P_2}{32} \]This solves to:\[ \Delta P_1 = 4 \Delta P_2 \]
07

Determine Required Ratio

From Step 6, we have \( \Delta P_1 = 4 \Delta P_2 \), so the ratio of the pressure difference across the first tube to that across the second tube is \(4:1\).

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Key Concepts

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

Poiseuille's Law
Poiseuille's Law is fundamental when studying the flow of fluids through pipes and tubes. It describes how factors like the radius and length of a tube affect the fluid flow rate.
According to Poiseuille's Law, the flow rate \(Q\) is given by:
  • \(Q = \frac{\pi R^4 \Delta P}{8 \eta L}\)
Here, \(\Delta P\) is the pressure difference, \(R\) is the tube's radius, \(\eta\) is the viscosity of the fluid, and \(L\) is the length of the tube.
Important points:
  • Flow rate is directly proportional to the pressure difference.
  • Flow rate is also proportional to the fourth power of the radius. This means small changes in radius can have dramatic effects on flow rate.
  • It is inversely proportional to the length of the tube and the fluid's viscosity.
Using Poiseuille's Law allows us to predict how changing the dimensions or fluid properties of the tube affects the flow.
Capillary Action
Capillary action is the ability of a liquid to flow in narrow spaces without the assistance of external forces like gravity. This phenomenon occurs in narrow tubes, like capillaries, and is essential in natural processes.
Here鈥檚 how capillary action works:
  • Adhesion: This is the attraction between the liquid molecules and the tube's surface. This force draws the liquid up the sides of the tube.
  • Cohesion: This is the attraction between the liquid molecules themselves. It helps in supporting the column of liquid that moves upwards.
Factors affecting capillary action include:
  • The diameter of the tube: Smaller tubes exhibit more noticeable capillary action.
  • The type of liquid: Liquids with lower surface tension display stronger capillary action.
Understanding capillary action is crucial, especially in biological and chemical contexts where liquid transport in tiny structures is necessary.
Pressure Difference
Pressure difference is a critical parameter in fluid dynamics. It represents the difference in pressure between two points in a fluid system. When a fluid flows between two points, such as in a tube, it moves from high pressure to low pressure.
Key elements:
  • Driving Force: The pressure difference is the driving force that moves fluid through a system according to Poiseuille's Law.
  • Flow Rate Impact: A larger pressure difference often results in a higher flow rate, as long as other factors remain constant.
  • Measurement: It's essential to measure the pressure difference to understand and predict fluid behavior in various systems.
In our exercise, the total pressure difference \(p\) is divided between two tubes connected in series, emphasizing the importance of pressure regulation in systems design.
Series Connection of Tubes
The series connection of tubes is a common setup in fluid dynamics, where multiple tubes are connected end-to-end. This configuration implies that the same fluid flows through each tube sequentially.
Important aspects:
  • Uniform Flow Rate: In a series connection, the flow rate is consistent across all tubes because the total mass of fluid entering the system equals the mass exiting.
  • Pressure Distribution: The total pressure difference across the complete system is the sum of the pressure differences across each tube. This is crucial in our given problem, where different dimensions of the tubes affect this distribution.
  • Scaling Law: Applying Poiseuille鈥檚 Law can help determine how variations in parameters like radius and length influence the pressure drop across each tube.
This insight assists in engineering applications, where understanding flow characteristics in complex systems is necessary for effective design.

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Most popular questions from this chapter

In a test experiment on a model aeroplane in a wind tunnel, the flow speeds on the upper and lower surfaces of the wing are \(70 \mathrm{~m} / \mathrm{s}\) and \(63 \mathrm{~m} / \mathrm{s}\) respectively. What is the lift on the wing, if its area is \(2.5 \mathrm{~m}^{2} ?\) Take the density of air to be \(1.3 \mathrm{~kg} / \mathrm{m}^{3}\). (a) \(5.1 \times 10^{2} \mathrm{~N}\) (b) \(6.1 \times 10^{2} \mathrm{~N}\) \(\left.\begin{array}{ll}(2] & 6\end{array}\right] 0^{3} \mathrm{~N}\)

A rectangular plate \(2 \mathrm{~m} \times 3 \mathrm{~m}\) is immersed in water in such a way that its greatest and least depth are \(6 \mathrm{~m}\) and \(4 \mathrm{~m}\) respectively, from the water surface. The total thrust on the plate is (a) \(294 \times 10^{3} \mathrm{~N}\) (b) \(294 \mathrm{~N}\) (c) \(100 \times 10^{7} \mathrm{~N}\) (d) \(400 \times 10^{1} \mathrm{~N}\)

A water film is made between two straight parallel wires of length \(10 \mathrm{~cm}\) separated by \(5 \mathrm{~mm}\) from each other. If the distance between the wires is increased by \(2 \mathrm{~mm}\). How much work will be done? Surface tension for water is 72 dyne \(\mathrm{cm}^{-1}\). (a) \(288 \mathrm{erg}\) (b) \(72 \mathrm{erg}\) (c) \(144 \mathrm{crg}\) (d) \(216 \mathrm{crg}\)

The surface tension of a liquid at its boiling point (a) becomes zero (b) becomes infinity (c) is equal to the value at room temperature (d) is half to the value at the room temperature

A metallic sphere of mass \(M\) falls through glycerine with a terminal velcity \(v\). If we drop a ball of mass \(8 M\) of same metal into a column of glycerine, the terminal velocity of the ball will be (a) \(2 \underline{\underline{v}}\) (b) \(4 \underline{\mathrm{v}}\) (c) \(8 v\) (d) \(16 \mathrm{v}\)
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