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Chapter 9: Problem 53

Using Poiseuille's law [Eq. (9-15)], show that viscosity has SI units of pascal-seconds.

Short Answer

Expert verified
Answer: The SI units of viscosity are pascal-seconds (Pa路s).

Step by step solution

01

Write down Poiseuille's law equation

The Poiseuille's law equation can be written as: \(Q = \frac{蟺r^4 螖P}{8 畏 L}\), where: - \(Q\) is the volumetric flow rate (m鲁/s), - \(r\) is the radius of the pipe (m), - \(螖P\) is the pressure difference between the two ends of the pipe (Pa or N/m虏), - \(畏\) is the dynamic viscosity of the fluid (Pa路s), - \(L\) is the length of the pipe (m).
02

Express pressure difference in terms of force and area

Recall that pressure is defined as the force exerted per unit area: \(P = \frac{F}{A}\), where \(F\) is the force (N) and \(A\) is the area (m虏). Therefore, the pressure difference, \(螖P\), can be expressed in the same units, as \(螖P = \frac{螖F}{A}\) or N/m虏.
03

Rearrange the equation for viscosity

Now we will solve the Poiseuille's law equation for viscosity. Rearranging the equation, we get: \(畏 = \frac{蟺r^4 螖P}{8 Q L}\).
04

Substitute the SI units for each quantity

Substitute the SI units for each quantity in the equation for viscosity: \(畏(\text{Pa路s}) = \frac{蟺(\text{m})^4 (\text{N/m}虏)}{8 (\text{m鲁/s}) (\text{m})}\).
05

Simplify the expression

Simplify the expression to determine the SI units of viscosity: \(\text{Pa路s} = \frac{\cancel{\text{m}}^4 (\text{N}/\cancel{\text{m}}虏)}{8 (\text{m鲁}/\cancel{\text{s}}) \cancel{\text{m}}}\). Dividing both the numerator and denominator of the equation by \(m鲁\), we get: \(\text{Pa路s} = \frac{\text{N}路\cancel{\text{m}鲁}}{\cancel{\text{m}鲁}} = \text{N路s}/\text{m}虏\) Since \(1 \text{Pa} = 1 \text{N/m}虏\), we can replace the \(\text{N}/\text{m}虏\) with \(\text{Pa}\): \(\text{Pa路s} = \text{Pa路s}\).
06

Conclusion

Using Poiseuille's law, we have shown that the SI units of viscosity are pascal-seconds (Pa路s).

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Most popular questions from this chapter

In an aortic aneurysm, a bulge forms where the walls of the aorta are weakened. If blood flowing through the aorta (radius \(1.0 \mathrm{cm}\) ) enters an aneurysm with a radius of \(3.0 \mathrm{cm},\) how much on average is the blood pressure higher inside the aneurysm than the pressure in the unenlarged part of the aorta? The average flow rate through the aorta is $120 \mathrm{cm}^{3} / \mathrm{s} .$ Assume the blood is non-viscous and the patient is lying down so there is no change in height.
Water entering a house flows with a speed of \(0.20 \mathrm{m} / \mathrm{s}\) through a pipe of \(1.0 \mathrm{cm}\) inside radius. What is the speed of the water at a point where the pipe tapers to a radius of \(2.5 \mathrm{mm} ?\)
(a) What is the pressure difference required to make blood flow through an artery of inner radius \(2.0 \mathrm{mm}\) and length \(0.20 \mathrm{m}\) at a speed of \(6.0 \mathrm{cm} / \mathrm{s} ?\) (b) What is the pressure difference required to make blood flow at \(0.60 \mathrm{mm} / \mathrm{s}\) through a capillary of radius \(3.0 \mu \mathrm{m}\) and length \(1.0 \mathrm{mm} ?\) (c) Compare both answers to your average blood pressure, about 100 torr.
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