/*! 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 100 A bug from South America known a... [FREE SOLUTION] | 91影视

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Chapter 9: Problem 100

A bug from South America known as Rhodnius prolixus extracts the blood of animals. Suppose Rhodnius prolixus extracts \(0.30 \mathrm{cm}^{3}\) of blood in 25 min from a human arm through its feeding tube of length \(0.20 \mathrm{mm}\) and radius \(5.0 \mu \mathrm{m} .\) What is the absolute pressure at the bug's end of the feeding tube if the absolute pressure at the other end (in the human arm) is 105 kPa? Assume the viscosity of blood is 0.0013 Pa-s. [Note: Negative absolute pressures are possible in liquids in very slender tubes.]

Short Answer

Expert verified
Answer: The absolute pressure at the bug's end of the feeding tube is approximately 67,700 Pa.

Step by step solution

01

Calculate the flow rate of the blood

To find the flow rate, we need to divide the volume of blood extracted by the time taken. The formula for the flow rate (Q) is: $$Q = \frac{V}{t}$$ Where V is the volume extracted and t is the time taken. Plugging in the values, we get: $$Q = \frac{0.30 \, \mathrm{cm}^3}{25 \, \mathrm{min}}$$ Let's first convert the units to SI so that we have: $$Q = \frac{0.30 \times 10^{-6} \, \mathrm{m}^3}{25 \times 60 \, \mathrm{s}}$$ Now, we can calculate the flow rate: $$Q = 2.0 \times 10^{-10} \, \mathrm{m}^3/\mathrm{s}$$
02

Apply Poiseuille's Law to find the pressure drop

Poiseuille's Law relates the pressure drop across a tube to the flow rate of the liquid inside it. The formula is: $$螖P = \frac{8畏LQ}{蟺R^4}$$ Where 螖P is the pressure drop, 畏 is the viscosity of the liquid, L is the length of the tube, Q is the flow rate, and R is the radius of the tube. Plugging in the values, we get: $$螖P = \frac{8(0.0013 \, \mathrm{Pa \cdot s})(0.20 \times 10^{-3} \, \mathrm{m})(2.0 \times 10^{-10} \, \mathrm{m}^3/\mathrm{s})}{蟺 (5.0脳10^{-6} \, \mathrm{m})^4}$$ Solving, we find the pressure drop: $$螖P \approx -37,300 \, \mathrm{Pa}$$ Note that the pressure drop is negative. This is expected because the bug is sucking blood out of the arm.
03

Calculate the absolute pressure at the bug's end of the feeding tube

To determine the absolute pressure at the bug's end of the feeding tube, we will subtract the pressure drop from the known pressure at the other end of the tube (105 kPa). The formula is: $$P_\mathrm{bug} = P_\mathrm{arm} + 螖P$$ Plugging in the values, we get: $$P_\mathrm{bug} = 105,000 \, \mathrm{Pa} - 37,300 \, \mathrm{Pa}$$ Calculating the absolute pressure, we find: $$P_\mathrm{bug} \approx 67,700 \, \mathrm{Pa}$$ The absolute pressure at the bug's end of the feeding tube is approximately 67,700 Pa.

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