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Fluid Mechanics

#fluid-mechanics
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Worked examples

Worked example: apparent viscosity change with shear rate

In notes →

Question. For a power law fluid with flow behaviour index n=0.6n = 0.6, if the shear rate is increased by a factor of 3232, by what factor does the apparent viscosity change?

Solution.

  1. Start from μa=Kγ˙ n−1\mu_a = K \dot{\gamma}^{\,n-1}.
  2. KK is fixed for a given fluid, so it cancels in a ratio: μa2μa1=(γ˙2γ˙1)n−1\dfrac{\mu_{a2}}{\mu_{a1}} = \left(\dfrac{\dot\gamma_2}{\dot\gamma_1}\right)^{n-1}.
  3. Substitute γ˙2/γ˙1=32\dot\gamma_2/\dot\gamma_1 = 32 and n−1=0.6−1=−0.4n - 1 = 0.6 - 1 = -0.4.
  4. 32−0.4=(25)−0.4=2−2=1432^{-0.4} = (2^5)^{-0.4} = 2^{-2} = \dfrac{1}{4}.
  5. Sanity check: n<1n < 1 is pseudoplastic (shear-thinning), so apparent viscosity must fall as shear rate rises — a ratio below 11 is consistent.

Answer: Apparent viscosity decreases by a factor of 44.

Worked example: reading nn off a log-log plot

In notes →

Question. In a log-log plot of shear stress (τ\tau) versus shear rate (γ˙\dot{\gamma}), the slope of the line equals nn, the flow behaviour index. For a shear-thinning (pseudoplastic) fluid, what is the value of nn?

Solution.

  1. Start from the power law equation: τ=Kγ˙n\tau = K \dot{\gamma}^n.
  2. Take the log of both sides: log⁡τ=log⁡K+nlog⁡γ˙\log \tau = \log K + n \log \dot{\gamma}.
  3. This is a straight line in log⁡τ\log \tau vs. log⁡γ˙\log \dot{\gamma} space, with slope nn and yy-intercept log⁡K\log K — so the slope read off a log-log plot of τ\tau against γ˙\dot{\gamma} is exactly nn.
  4. A shear-thinning (pseudoplastic) fluid is defined, per the table above, by n<1n < 1: apparent viscosity falls as shear stress rises.

Answer: n<1n < 1 — the log-log line is shallower than the n=1n = 1 Newtonian line (still positive slope, just less steep).

Worked example: time-dependent vs. time-independent classification (GATE-style)

In notes →

Question. Select the CORRECT statements regarding time-dependent and time-independent non-Newtonian fluids:

(A) Pseudoplastic fluids exhibit shear-thinning behaviour independent of the duration of shear. (B) Thixotropic fluids show a time-dependent decrease in apparent viscosity at a constant shear rate. (C) Rheopectic fluids show a time-dependent decrease in viscosity under constant shear. (D) Bingham plastics do not flow until the applied shear stress exceeds the threshold yield stress.

Solution.

  1. Pseudoplastic fluids are time-independent: shear-thinning (μa↓\mu_a \downarrow as γ˙↑\dot\gamma \uparrow) depends only on the instantaneous shear rate, not on how long shear has been applied — (A) is correct.
  2. Thixotropic fluids are defined by exactly this: apparent viscosity falls over time at a fixed shear rate — (B) is correct.
  3. Rheopectic fluids are the opposite of thixotropic: apparent viscosity increases over time at a fixed shear rate, not decreases — (C) is incorrect (it describes thixotropic behaviour, mislabeled as rheopectic).
  4. Bingham plastics stay solid until τ\tau exceeds the yield stress τy\tau_y, then flow like a Newtonian fluid — (D) is correct, matching τ=τy+μpγ˙\tau = \tau_y + \mu_p \dot\gamma.

Answer: A, B, and D are correct; C is incorrect.

Worked example: rheology models across fluid mechanics topics (GATE-style)

In notes →

Question. Which of the following statements regarding fluid rheology models is/are TRUE?

(a) The Casson equation is given by τ=τy+Kcγ˙\sqrt{\tau} = \sqrt{\tau_y} + \sqrt{K_c \dot\gamma}. (b) For human blood, rouleaux formation at low shear rates contributes to yield stress behaviour. (c) Dilatant fluids show an increase in apparent viscosity with decreasing shear rate. (d) In pipe flow, turbulent flow has a flatter velocity profile across the tube diameter compared to laminar flow.

Solution.

  1. (a) matches the Casson equation exactly (τ=τ0+Kcγ˙\sqrt\tau = \sqrt{\tau_0} + \sqrt{K_c\dot\gamma}, with τy\tau_y here used interchangeably with τ0\tau_0 for the yield stress) — correct.
  2. (b) is the physical mechanism behind blood's yield-stress behaviour: red blood cells stack into rouleaux at low shear rates, and disrupting that structure needs a threshold stress — correct.
  3. (c) inverts the definition of dilatant behaviour: n>1n > 1 means τ↑→μa↑\tau \uparrow \rightarrow \mu_a \uparrow, i.e. viscosity rises with increasing shear rate, not decreasing — incorrect.
  4. (d) matches the velocity-profile comparison above: turbulent eddies redistribute momentum across the cross-section, flattening the profile into a near-plug shape versus laminar flow's parabolic profile — correct.

Answer: (a), (b), and (d) are correct; (c) is incorrect.

04 examples