Math Problem Statement

4. Poiseuille’s Law for Rate of Fluid Flow: Poiseuille’s law for the rate of flow of a fluid through a tube is a fourth-order polynomial that is also a power function: F = cR^4, where F is the flow rate (measured as a volume per unit time), c is a constant, and R is the radius of the tube. a. Assume that R increases by 10%. Explain why F increases by 46.41%. (Hint: Consider 1.10 raised to the fourth power.) b. What is the flow rate through a 3/4-inch pipe compared with that through a 1/2-inch pipe? c. Suppose that an artery supplying blood to the heart muscle is partially blocked and is only half its normal radius. What percentage of the usual blood flow will flow through the partially blocked artery?

Solution

Let's go through each part of the problem provided:

Given:

Poiseuille's Law for the rate of fluid flow is expressed as: F=cR4F = cR^4 where:

  • FF is the flow rate (volume per unit time),
  • cc is a constant,
  • RR is the radius of the tube.

Solution to each part:

(a) Increase in Flow Rate FF when Radius RR Increases by 10%

If RR increases by 10%, then the new radius RnewR_{\text{new}} is: Rnew=1.10×RR_{\text{new}} = 1.10 \times R

Since F=cR4F = cR^4, the new flow rate FnewF_{\text{new}} is: Fnew=c(Rnew)4=c(1.10R)4F_{\text{new}} = c(R_{\text{new}})^4 = c(1.10 \cdot R)^4

Calculating (1.10)4(1.10)^4: (1.10)41.4641(1.10)^4 \approx 1.4641

Thus: Fnew=1.4641cR4=1.4641FF_{\text{new}} = 1.4641 \cdot cR^4 = 1.4641 \cdot F

This shows that FF increases by approximately 46.41% when RR increases by 10%.


(b) Comparing Flow Rates for Different Pipe Radii (3/4-inch vs. 1/2-inch)

Let:

  • Radius R1=34R_1 = \frac{3}{4} inch
  • Radius R2=12R_2 = \frac{1}{2} inch

The flow rate F1F_1 for the 3/4-inch pipe: F1=c(34)4=c81256F_1 = c \left(\frac{3}{4}\right)^4 = c \cdot \frac{81}{256}

The flow rate F2F_2 for the 1/2-inch pipe: F2=c(12)4=c116F_2 = c \left(\frac{1}{2}\right)^4 = c \cdot \frac{1}{16}

To find the ratio F1F2\frac{F_1}{F_2}: F1F2=81256116=81×16256=1296256=5.0625\frac{F_1}{F_2} = \frac{\frac{81}{256}}{\frac{1}{16}} = \frac{81 \times 16}{256} = \frac{1296}{256} = 5.0625

So, the flow rate through the 3/4-inch pipe is approximately 5.06 times greater than that through the 1/2-inch pipe.


(c) Percentage of Blood Flow through a Partially Blocked Artery (Radius Halved)

If an artery’s radius is halved, the new radius RblockedR_{\text{blocked}} is: Rblocked=12RR_{\text{blocked}} = \frac{1}{2} R

The flow rate FblockedF_{\text{blocked}} through the blocked artery is: Fblocked=c(12R)4=c116R4=116FF_{\text{blocked}} = c \left(\frac{1}{2} R\right)^4 = c \cdot \frac{1}{16} R^4 = \frac{1}{16} F

Thus, only 6.25% of the usual blood flow will flow through the partially blocked artery.


Would you like further details on any part of these calculations or have any questions?

Related Questions:

  1. How does Poiseuille’s law change if the radius of the tube triples?
  2. What is the flow rate ratio if the radius of the pipe changes from 1 inch to 2 inches?
  3. How sensitive is the flow rate FF to small changes in radius RR?
  4. Can Poiseuille’s law apply to non-circular cross-sectional tubes?
  5. What impact does fluid viscosity have in Poiseuille’s Law?

Tip:

Remember, in Poiseuille’s Law, the flow rate FF is highly sensitive to the radius RR due to the fourth power, so small changes in RR can greatly affect FF.

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Math Problem Analysis

Mathematical Concepts

Calculus
Power Functions
Poiseuille's Law

Formulas

F = cR^4

Theorems

Poiseuille’s Law

Suitable Grade Level

Undergraduate (Introductory Physics or Engineering)