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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·s).

Step by step solution

01

Write down Poiseuille's law equation

The Poiseuille's law equation can be written as: \(Q = \frac{πr^4 ΔP}{8 η L}\), where: - \(Q\) is the volumetric flow rate (m³/s), - \(r\) is the radius of the pipe (m), - \(ΔP\) is the pressure difference between the two ends of the pipe (Pa or N/m²), - \(η\) is the dynamic viscosity of the fluid (Pa·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 (m²). Therefore, the pressure difference, \(ΔP\), can be expressed in the same units, as \(ΔP = \frac{ΔF}{A}\) or N/m².
03

Rearrange the equation for viscosity

Now we will solve the Poiseuille's law equation for viscosity. Rearranging the equation, we get: \(η = \frac{πr^4 ΔP}{8 Q L}\).
04

Substitute the SI units for each quantity

Substitute the SI units for each quantity in the equation for viscosity: \(η(\text{Pa·s}) = \frac{π(\text{m})^4 (\text{N/m}²)}{8 (\text{m³/s}) (\text{m})}\).
05

Simplify the expression

Simplify the expression to determine the SI units of viscosity: \(\text{Pa·s} = \frac{\cancel{\text{m}}^4 (\text{N}/\cancel{\text{m}}²)}{8 (\text{m³}/\cancel{\text{s}}) \cancel{\text{m}}}\). Dividing both the numerator and denominator of the equation by \(m³\), we get: \(\text{Pa·s} = \frac{\text{N}·\cancel{\text{m}³}}{\cancel{\text{m}³}} = \text{N·s}/\text{m}²\) Since \(1 \text{Pa} = 1 \text{N/m}²\), we can replace the \(\text{N}/\text{m}²\) with \(\text{Pa}\): \(\text{Pa·s} = \text{Pa·s}\).
06

Conclusion

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

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Most popular questions from this chapter

This table gives the terminal speeds of various spheres falling through the same fluid. The spheres all have the same radius. $$\begin{array}{llllllll}\hline m= & 5.0 & 11.3 & 20.0 & 31.3 & 45.0 & 80.0 & (\mathrm{g}) \\\ \hline v_{1}= & 1.0 & 1.5 & 2.0 & 2.5 & 3.0 & 4.0 & (\mathrm{cm} / \mathrm{s}) \\\\\hline\end{array}$$, Is the drag force primarily viscous or turbulent? Explain your reasoning.
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