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P6.17 A capillary viscometer measures the time required for a specified volume ...
Jun 3, 2024
Solution
a
To derive the formula for the time required for a specified volume vv of liquid to flow through the capillary tube, we start by using the Hagen-Poiseuille equation for laminar flow in a cylindrical tube: Q=πD4ΔP128μL Q = \frac{\pi D^4 \Delta P}{128 \mu L} where QQ is the volumetric flow rate, DD is the diameter of the tube, ΔP\Delta P is the pressure difference, μ\mu is the dynamic viscosity, and LL is the length of the tube
b
The volumetric flow rate QQ is also given by: Q=vt Q = \frac{v}{t} where vv is the volume of the liquid and tt is the time. Equating the two expressions for QQ, we get: vt=πD4ΔP128μL \frac{v}{t} = \frac{\pi D^4 \Delta P}{128 \mu L} Solving for tt, we obtain: t=128μLvπD4ΔP t = \frac{128 \mu L v}{\pi D^4 \Delta P}
c
For part (b), we need to find the capillary diameter DD that results in a transit time tt of 6 seconds. Given: L=12 cmL = 12 \, \text{cm}, l=2 cml = 2 \, \text{cm}, v=8 cm3v = 8 \, \text{cm}^3, and the fluid is water at 20∘C20^\circ \text{C} with μ=0.001 Pa⋅s\mu = 0.001 \, \text{Pa} \cdot \text{s}. The pressure difference ΔP\Delta P can be approximated by the hydrostatic pressure difference: ΔP=ρgl \Delta P = \rho g l where ρ\rho is the density of water (≈1000 kg/m3\approx 1000 \, \text{kg/m}^3) and gg is the acceleration due to gravity (9.81 m/s29.81 \, \text{m/s}^2). Substituting the values, we get: ΔP=1000×9.81×0.02=196.2 Pa \Delta P = 1000 \times 9.81 \times 0.02 = 196.2 \, \text{Pa} Now, substituting all known values into the formula for tt: 6=128×0.001×0.12×8πD4×196.2 6 = \frac{128 \times 0.001 \times 0.12 \times 8}{\pi D^4 \times 196.2} Solving for DD, we get: D4=128×0.001×0.12×8π×196.2×6 D^4 = \frac{128 \times 0.001 \times 0.12 \times 8}{\pi \times 196.2 \times 6} D4=0.122883703.2 D^4 = \frac{0.12288}{3703.2} D4=3.32×10−5 D^4 = 3.32 \times 10^{-5} D=(3.32×10−5)1/4 D = (3.32 \times 10^{-5})^{1/4} D≈0.023 cm D \approx 0.023 \, \text{cm}
Answer
D≈0.023 cmD \approx 0.023 \, \text{cm}
Key Concept
Hagen-Poiseuille equation for laminar flow in a cylindrical tube: Q=πD4ΔP128μL Q = \frac{\pi D^4 \Delta P}{128 \mu L}
Explanation
The time required for a specified volume of liquid to flow through a capillary tube can be derived using the Hagen-Poiseuille equation, which relates the volumetric flow rate to the pressure difference, viscosity, and dimensions of the tube. By equating the volumetric flow rate to the volume divided by time, we can solve for the transit time.
可以讲的详细些吗
Solution
a
使用Hagen-Poiseuille方程来推导液体通过毛细管所需的时间公式。Hagen-Poiseuille方程描述了圆柱形管道中层流的体积流量：Q=πD4ΔP128μL Q = \frac{\pi D^4 \Delta P}{128 \mu L} 其中，QQ是体积流量，DD是管道直径，ΔP\Delta P是压力差，μ\mu是流体粘度，LL是管道长度。
b
体积流量QQ也可以表示为：Q=vt Q = \frac{v}{t} 其中，vv是流体体积，tt是流体通过管道所需的时间。将这个表达式代入Hagen-Poiseuille方程：vt=πD4ΔP128μL \frac{v}{t} = \frac{\pi D^4 \Delta P}{128 \mu L}
c
解出时间tt：t=128μLvπD4ΔP t = \frac{128 \mu L v}{\pi D^4 \Delta P} 这是液体通过毛细管所需时间的近似公式。
d
现在，使用给定的数据计算毛细管直径DD。已知L=12 cmL=12 \mathrm{~cm}，l=2 cml=2 \mathrm{~cm}，v=8 cm3v=8 \mathrm{~cm}^{3}，水在20∘C20^{\circ} \mathrm{C}的粘度μ=0.001 Pa⋅s\mu = 0.001 \mathrm{~Pa \cdot s}，时间t=6t=6秒。假设ΔP=ρgh\Delta P = \rho g h，其中ρ\rho是水的密度，gg是重力加速度，hh是液柱高度。
e
计算ΔP\Delta P：ΔP=ρgh=1000 kg/m3⋅9.8 m/s2⋅0.12 m=1176 Pa \Delta P = \rho g h = 1000 \mathrm{~kg/m^3} \cdot 9.8 \mathrm{~m/s^2} \cdot 0.12 \mathrm{~m} = 1176 \mathrm{~Pa}
f
将所有已知值代入时间公式，解出DD：6=128⋅0.001⋅0.12⋅8πD4⋅1176 6 = \frac{128 \cdot 0.001 \cdot 0.12 \cdot 8}{\pi D^4 \cdot 1176} D4=128⋅0.001⋅0.12⋅8π⋅1176⋅6 D^4 = \frac{128 \cdot 0.001 \cdot 0.12 \cdot 8}{\pi \cdot 1176 \cdot 6} D4≈1.32×10−5 D^4 \approx 1.32 \times 10^{-5} D≈(1.32×10−5)1/4 D \approx (1.32 \times 10^{-5})^{1/4} D≈0.023 cm D \approx 0.023 \, \text{cm}
Answer
D≈0.023 cmD \approx 0.023 \, \text{cm}
Key Concept
Hagen-Poiseuille方程描述了圆柱形管道中层流的体积流量。公式为：Q=πD4ΔP128μL Q = \frac{\pi D^4 \Delta P}{128 \mu L}
Explanation
通过使用Hagen-Poiseuille方程，我们可以推导出液体通过毛细管所需的时间公式，并利用给定的数据计算出毛细管的直径。
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