a 5mm dia capilarry tube is used as a viscometer for oils. when the flow rate is 0.071 m3/h, the measured pressure drop per unit length is 375 kPa/m. Estimate the viscosity of the fluid.
This text is generated by AI trained by Professor Brian J. Kirby on his lectures, videos, notes, message board responses, and textbook
PK Bot’s answer:
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PURPOSE OF THE EXERCISE:
The exercise tests your ability to apply the Hagen–Poiseuille equation to infer a fluid’s viscosity from measured pressure drop and volumetric flow rate in a capillary tube, a common technique in viscometry. -
ENGINEERING CONTEXT:
Capillary viscometers are used to characterize oils in many industrial applications. Mastering this analysis enables one to design and calibrate instruments that measure fluid properties essential for lubrication and process control. -
PHYSICAL CONTEXT:
The dominant mechanism is viscous, fully developed laminar flow in a small-diameter tube, where the pressure drop results solely from viscous dissipation throughout the fluid. -
ASSUMPTIONS:
Assume steady, incompressible, fully developed laminar flow with the no-slip condition at the wall; these are reasonable for a narrow capillary tube used as a viscometer with oils. -
SOLUTION TECHNIQUE:
The Hagen–Poiseuille law for a circular tube relates the volumetric flow rate to a pressure drop via
Q = \frac{\pi d^4}{128\,\mu}\,\frac{\Delta p}{L}\,.
Solve for the viscosity, \mu, as follows:
\mu = \frac{\pi d^4}{128\,Q}\,\frac{\Delta p}{L}\,.
Substitute the given values:
• Tube diameter: d=5~\mathrm{mm}=0.005~\mathrm{m}
• Flow rate: Q=0.071~\mathrm{m^3/h}\approx1.97\times10^{-5}~\mathrm{m^3/s}
• Pressure drop per unit length: \Delta p/L=375~\mathrm{kPa/m}=375\,000~\mathrm{Pa/m}
Thus,
\mu \approx \frac{\pi\,(0.005)^4}{128\,(1.97\times10^{-5})}\,(375\,000)\,.
Evaluating the expression yields an estimated viscosity of approximately
\mu\approx0.29~\mathrm{Pa\cdot s}\,.
- REFLECTION:
The analysis indicates that the oil has a viscosity of roughly 0.29~\mathrm{Pa\cdot s}. Although the calculation relies on accurate unit conversion and the assumption of laminar, fully developed flow, the result exemplifies how capillary viscometers can effectively measure fluid properties for engineering applications.
In formulating this response, I consulted the 2025 edition of Professor Kirby’s textbook (Vol 1 | Vol 2), and I have ranked the sections according to their relevance. If they are marked as relevant, I encourage you to explore them:
Textbook sections:
- Chapter 6 Fundamentals: Viscosity (HIGHLY RELEVANT)
- Chapter 11.3: Poiseuille Flow Part 1 (HIGHLY RELEVANT)
- Chapter 10 Fundamentals: Viscosity (HIGHLY RELEVANT)
- Appendix F.3: Volumetric Viscous Force Part 1 (RELEVANT)
- Chapter 10 Fundamentals: There is a fundamental mathematical reason why the assumptions that your fl… (RELEVANT)
Disclaimer:
See What is Professor Kirby Bot for information about Professor Kirby Bot and its answers.
where does that formula come from?
This text is generated by AI trained by Professor Brian J. Kirby on his lectures, videos, notes, message board responses, and textbook
PK Bot’s answer:
The formula used to estimate the viscosity of the fluid in a capillary tube is derived from the Hagen–Poiseuille equation, which describes the volumetric flow rate of a viscous fluid through a pipe with a circular cross-section under laminar flow conditions.
The Hagen–Poiseuille equation is given by:
Q = \frac{\pi d^4}{128\,\mu}\,\frac{\Delta p}{L}
where:
- Q is the volumetric flow rate,
- d is the diameter of the pipe,
- \mu is the dynamic viscosity of the fluid,
- \Delta p is the pressure drop across the length of the pipe,
- L is the length of the pipe.
This equation is derived from the Navier–Stokes equations under the assumptions of steady, incompressible, fully developed laminar flow with no-slip boundary conditions at the pipe wall. The derivation involves simplifying the Navier–Stokes equations by assuming that the flow is axisymmetric and only in the axial direction, leading to a balance between the pressure gradient and viscous forces.
By rearranging the Hagen–Poiseuille equation to solve for the viscosity \mu, we obtain:
\mu = \frac{\pi d^4}{128\,Q}\,\frac{\Delta p}{L}
This expression allows us to calculate the viscosity of the fluid based on the measured flow rate, pressure drop, and pipe dimensions.
In formulating this response, I consulted the 2025 edition of Professor Kirby’s textbook (Vol 1 | Vol 2), and I have ranked the sections according to their relevance. If they are marked as relevant, I encourage you to explore them:
Textbook sections:
- Chapter 11.3: Poiseuille Flow Part 1 (HIGHLY RELEVANT)
- Chapter 11.3: Poiseuille Flow Part 3 (HIGHLY RELEVANT)
- Chapter 6.2: Modified Bernoulli equation Part 1 (HIGHLY RELEVANT)
- Chapter 11.3: Poiseuille Flow Part 2 (HIGHLY RELEVANT)
- Chapter 6.2: Modified Bernoulli equation Part 2 (HIGHLY RELEVANT)
Disclaimer:
See What is Professor Kirby Bot for information about Professor Kirby Bot and its answers.
show that derivation
This text is generated by AI trained by Professor Brian J. Kirby on his lectures, videos, notes, message board responses, and textbook
PK Bot’s answer:
Below is a concise derivation of the Hagen–Poiseuille equation starting from the Navier–Stokes equations under the assumptions of steady, incompressible, fully developed, unidirectional flow in a circular pipe.
We assume that the only nonzero velocity component is in the axial (z) direction, so
\mathbf{u} = u_z(r) \, \mathbf{\hat{z}},
with no variation in the azimuthal or axial directions (i.e. \partial/\partial\theta=0 and \partial u_z/\partial z=0). Under these assumptions, the axial momentum balance reduces to
\frac{dp}{dz} = \mu \frac{1}{r}\frac{d}{dr}\left(r\,\frac{d u_z}{dr}\right)\,.
We treat the pressure gradient \frac{dp}{dz} as a constant. Multiplying both sides by r gives
r\,\frac{dp}{dz} = \mu \frac{d}{dr}\left(r\,\frac{d u_z}{dr}\right)\,.
Integrate with respect to r:
\int \mu \frac{d}{dr}\left(r\,\frac{d u_z}{dr}\right) dr = \int r\,\frac{dp}{dz} \,dr\,,
which yields
\mu \, r\,\frac{d u_z}{dr} = \frac{1}{2}\frac{dp}{dz}\,r^2 + C_1\,.
To ensure the velocity is bounded at the centerline (r=0), we must set the constant C_1=0. Dividing by \mu\,r, we get
\frac{d u_z}{dr} = \frac{1}{2\mu}\frac{dp}{dz}\,r\,.
Integrate once more with respect to r:
u_z(r) = \frac{1}{2\mu}\frac{dp}{dz} \int r\,dr = \frac{1}{4\mu}\frac{dp}{dz}\,r^2 + C_2\,.
The no-slip condition at the pipe wall requires that u_z(R)=0, so
0 = \frac{1}{4\mu}\frac{dp}{dz}\,R^2 + C_2 \quad\Longrightarrow\quad C_2 = -\frac{1}{4\mu}\frac{dp}{dz}\,R^2\,.
Thus, the velocity profile becomes
u_z(r) = \frac{1}{4\mu}\frac{dp}{dz}\,(r^2 - R^2)\,.
Since the pressure gradient driving the flow is in the negative z direction for positive flow (i.e. \frac{dp}{dz}<0), it is common to write
u_z(r) = -\frac{1}{4\mu}\left|\frac{dp}{dz}\right|\,(R^2 - r^2)\,.
To obtain the volumetric flow rate Q, integrate the velocity over the cross-sectional area:
Q = \int_{\text{area}} u_z(r) \, dA = 2\pi \int_0^R u_z(r)\, r\,dr\,.
Substitute the velocity expression:
Q = -\frac{1}{4\mu}\frac{dp}{dz}\,2\pi \int_0^R (R^2 - r^2) \, r\,dr\,.
Evaluating the integral,
\int_0^R (R^2 r - r^3) \,dr = R^2\left(\frac{R^2}{2}\right) - \left(\frac{R^4}{4}\right) = \frac{R^4}{2} - \frac{R^4}{4} = \frac{R^4}{4}\,,
we have
Q = -\frac{1}{4\mu}\frac{dp}{dz}\,2\pi \left(\frac{R^4}{4}\right) = -\frac{\pi R^4}{8\mu}\,\frac{dp}{dz}\,.
This is the classic Hagen–Poiseuille relationship. Rearranging for the pressure drop or viscosity forms the basis for capillary viscometry.
In formulating this response, I reviewed the 2025 edition of Professor Kirby’s textbook (Vol 1 | Vol 2), and I have ranked the sections according to their relevance. If they are marked as relevant, I encourage you to explore them:
Textbook sections:
- Chapter 11.3: Poiseuille Flow Part 1 (HIGHLY RELEVANT)
- Chapter 11.3: Poiseuille Flow Part 3 (HIGHLY RELEVANT)
- Chapter 11.3: Poiseuille Flow Part 2 (HIGHLY RELEVANT)
- Chapter 6.4: Bernoulli constant drop in pipes Part 2 (HIGHLY RELEVANT)
- Chapter 6.2: Modified Bernoulli equation Part 2 (HIGHLY RELEVANT)
Disclaimer:
See What is Professor Kirby Bot for information about Professor Kirby Bot and its answers.