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Chapter 14: Problem 129

Fluid originally flows through a tube at a rate of \(100 \mathrm{cm}^{3} / \mathrm{s} .\) To illustrate the sensitivity of flow rate to various factors, calculate the new flow rate for the following changes with all other factors remaining the same as in the original conditions. (a) Pressure difference increases by a factor of \(1.50 .\) (b) A new fluid with 3.00 times greater viscosity is substituted. (c) The tube is replaced by one having 4.00 times the length. (d) Another tube is used with a radius 0.100 times the original. (e) Yet another tube is substituted with a radius 0.100 times the original and half the length, and the pressure difference is increased by a factor of 1.50 .

Short Answer

Expert verified
The new flow rates for the given changes are: (a) When the pressure difference increases by a factor of 1.50, the new flow rate is \(150 \mathrm{cm}^3/\mathrm{s}\). (b) When the viscosity increases by a factor of 3.00, the new flow rate is \(33.33 \mathrm{cm}^3/\mathrm{s}\). (c) When the tube length increases by a factor of 4.00, the new flow rate is \(25 \mathrm{cm}^3/\mathrm{s}\). (d) When the tube radius decreases by a factor of 0.100, the new flow rate is \(100,000 \mathrm{cm}^3/\mathrm{s}\). (e) When the tube has a radius 0.100 times the original, half the length, and a pressure difference 1.50 times the original, the new flow rate is \(30,000 \mathrm{cm}^3/\mathrm{s}\).

Step by step solution

01

(a) Pressure difference increases by a factor of 1.50

Calculate the new flow rate when the pressure difference, 螖P, increases by a factor of 1.50: \(Q_{new} = \frac{\pi \cdot 1.50 \Delta P r^4}{8\eta L}\) Since all the other factors remain constant, we can rewrite the equation with respect to the initial flow rate: \(Q_{new} = 1.50Q\) Therefore, the new flow rate will be: \(Q_{new} = 1.50 \cdot 100 \mathrm{cm}^3/\mathrm{s} = 150 \mathrm{cm}^3/\mathrm{s}\)
02

(b) A new fluid with 3.00 times greater viscosity is substituted

Calculate the new flow rate when the fluid viscosity, 畏, increases by a factor of 3.00: \(Q_{new} = \frac{\pi \Delta P r^4}{8\cdot 3.00 \eta L}\) Since all the other factors remain constant, we can rewrite the equation with respect to the initial flow rate: \(Q_{new} = \frac{Q}{3.00}\) Therefore, the new flow rate will be: \(Q_{new} = \frac{100 \mathrm{cm}^3/\mathrm{s}}{3.00} = 33.33 \mathrm{cm}^3/\mathrm{s}\)
03

(c) The tube is replaced by one having 4.00 times the length

Calculate the new flow rate when the tube length, L, increases by a factor of 4.00: \(Q_{new} = \frac{\pi \Delta P r^4}{8\eta \cdot 4.00 L}\) Since all the other factors remain constant, we can rewrite the equation with respect to the initial flow rate: \(Q_{new} = \frac{Q}{4.00}\) Therefore, the new flow rate will be: \(Q_{new} = \frac{100 \mathrm{cm}^3/\mathrm{s}}{4.00} = 25 \mathrm{cm}^3/\mathrm{s}\)
04

(d) Another tube is used with a radius 0.100 times the original

Calculate the new flow rate when the tube radius, r, decreases by a factor of 0.100: \(Q_{new} = \frac{\pi \Delta P (0.100r)^4}{8\eta L}\) Since all the other factors remain constant, we can rewrite the equation with respect to the initial flow rate: \(Q_{new} = \frac{Q}{(0.100)^4}= \frac{Q}{(0.0001)^2}\) Therefore, the new flow rate will be: \(Q_{new} = \frac{100 \mathrm{cm}^3/\mathrm{s}}{(0.0001)^2} = 100,000 \mathrm{cm}^3/\mathrm{s}\)
05

(e) A tube with a r of 0.100 times the original, L of half the original, and 螖P of 1.50 times the original

Calculate the new flow rate when: -radius (r) decreases by a factor of 0.100 -length (L) decreases by a factor of 0.50 -pressure difference (螖P) increases by a factor of 1.50 \(Q_{new} = \frac{\pi (1.50\Delta P) (0.100r)^4}{8\eta (0.50 L)}\) Considering the changes in the equation, we can rewrite the equation with respect to the initial flow rate: \(Q_{new} = \frac{1.50(0.100)^4}{(0.50)}Q\) Therefore, the new flow rate will be: \(Q_{new} = \frac{1.50(0.0001)^2}{0.50} \cdot 100 \mathrm{cm}^3/\mathrm{s} = 30,000 \mathrm{cm}^3/\mathrm{s}\)

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Key Concepts

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

Understanding Pressure Difference in Fluid Flow
In the realm of fluid dynamics, the pressure difference acts as a driving force that propels fluid through a tube or conduit. When considering the flow rate of a fluid, an increase in pressure difference logically results in a greater flow rate, assuming other factors such as fluid viscosity, tube length, and tube radius remain constant.

Using the relationship provided in the exercise, where the flow rate is directly proportional to the pressure difference (Q 鈭� 螖P), it becomes clear why an increase in pressure difference by a factor of 1.50 leads to a flow rate that is also increased by 1.50 times. This demonstrates the sensitivity of fluid flow to changes in pressure, which is a core principle in applications ranging from engineering to medicine.
The Impact of Fluid Viscosity on Flow Rate
The viscosity of a fluid is a measure of its resistance to deformation and flow. In layman's terms, it's akin to the 'thickness' of the fluid; honey has a higher viscosity than water, for example. The relationship between fluid viscosity and flow rate is inversely proportional. This means that as the viscosity increases, it becomes harder for the fluid to flow, resulting in a lower flow rate.

So when the viscosity in our exercise is increased by a factor of 3.00, the flow rate decreases to one-third of the original rate. This concept is vital for understanding fluid behavior in both industrial processes and biological systems, where viscosity can play a major role in the effectiveness of fluid transport.
How Tube Length Influences Fluid Flow Rate
The length of the tube through which a fluid flows is a factor of great importance for flow rate. In our example scenario, when the tube's length is quadrupled, the flow rate is quartered. This is a direct consequence of the fact that a longer path increases the resistance against the fluid flow, demanding more pressure to maintain the same flow rate.

For students trying to understand why this occurs, imagine pushing a fluid through a long straw versus a short one. The longer the straw, the more effort is required to push the same amount of fluid through.
The Significance of Tube Radius in Fluid Flow
Finally, let's delve into the influence of the tube radius on fluid flow. Tube radius has an exponential relationship with flow rate, specifically to the fourth power as dictated by the Poiseuille's Law. A decrease in radius has a dramatic effect on flow as seen in the example, leading to a much lower flow rate 鈥� much greater than linearly proportional changes seen with other factors.

In practical terms, this means a slight decrease in the tube's radius can cause a significant decrease in the flow rate. If a condition requires the radius to be smaller, compensatory measures such as increasing pressure are often necessary to achieve a desired flow rate.

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Most popular questions from this chapter

Explain why the fluid reaches equal levels on either side of a manometer if both sides are open to the atmosphere, even if the tubes are of different diameters.

Bird bones have air pockets in them to reduce their weight-this also gives them an average density significantly less than that of the bones of other animals. Suppose an ornithologist weighs a bird bone in air and in water and finds its mass is \(45.0 \mathrm{g}\) and its apparent mass when submerged is \(3.60 \mathrm{g}\) (assume the bone is watertight). (a) What mass of water is displaced? (b) What is the volume of the bone? (c) What is its average density?

Water supplied to a house by a water main has a pressure of \(3.00 \times 10^{5} \mathrm{N} / \mathrm{m}^{2}\) early on a summer day when neighborhood use is low. This pressure produces a flow of \(20.0 \mathrm{L} / \mathrm{min}\) through a garden hose. Later in the day, pressure at the exit of the water main and entrance to the house drops. and a flow of only \(8.00 \mathrm{L} / \mathrm{min}\) is obtained through the same hose. (a) What pressure is now being supplied to the house, assuming resistance is constant? (b) By what factor did the flow rate in the water main increase in order to cause this decrease in delivered pressure? The pressure at the entrance of the water main is \(5.00 \times 10^{5} \mathrm{N} / \mathrm{m}^{2},\) and the original flow rate was \(200 \mathrm{L} /\) min. (c) How many more users are there, assuming each would consume \(20.0 \mathrm{L} / \mathrm{min}\) in the morning?

The inside volume of a house is equivalent to that of a rectangular solid \(13.0 \mathrm{m}\) wide by \(20.0 \mathrm{m}\) long by \(2.75 \mathrm{m}\) high. The house is heated by a forced air gas heater. The main uptake air duct of the heater is \(0.300 \mathrm{m}\) in diameter. What is the average speed of air in the duct if it carries a volume equal to that of the house's interior every 15 minutes?

Verify that pressure has units of energy per unit volume.
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