Viscous flow — Poiseuille flow, velocity profile in pipe: Exercises

Our two master tools for this whole page:

The figure above is the mental picture behind every problem: a red parabola of speed, zero at the walls, fastest in the middle. Keep it in view.

Level 1 — Recognition

Can you name the pieces and pick the right formula?

L1.1

State which quantity in the Hagen–Poiseuille law, when doubled, changes $Q$ the most, and by what factor.

Recall Solution

$Q = \frac{π R ^{4} Δ P}{8 η L}$ . The exponents are: $R^{4}$ , $Δ P^{1}$ , $η^{- 1}$ , $L^{- 1}$ . The largest exponent is on $R$ . Doubling $R$ multiplies $Q$ by $2^{4} = 16$ . Doubling any of the others changes $Q$ by only a factor of $2$ (or $\frac{1}{2}$ ).

L1.2

A pipe has $v_{m a x} = 0.60 m/s$ . What is the mean velocity $\overset{v}{ˉ}$ , and what is the velocity right at the wall?

Recall Solution

For a parabolic profile the average is exactly half the peak: $\overset{v}{ˉ} = \frac{1}{2} v_{m a x} = \frac{1}{2} (0.60) = 0.30 m/s .$ At the wall the no-slip condition forces $v (R) = 0 m/s$ — the fluid touching the wall does not move at all.

Level 2 — Application

Plug numbers into one formula, carefully.

L2.1

Water ( $η = 1.0 \times 1 0^{- 3} Pa\cdots$ ) flows through a pipe of radius $R = 2.0 mm$ , length $L = 1.0 m$ , driven by $Δ P = 500 Pa$ . Find the flow rate $Q$ .

Recall Solution

Step 1 — pick the formula. We want $Q$ from $R, L, η, Δ P$ → Hagen–Poiseuille. Step 2 — compute $R^{4}$ . $R = 2.0 \times 1 0^{- 3} m$ , so $R^{4} = (2.0 \times 1 0^{- 3})^{4} = 1.6 \times 1 0^{- 11} m^{4}$ . Step 3 — substitute.

= \frac{\pi (8.0\times10^{-9})}{8.0\times10^{-3}}$$ $$Q = \frac{2.513\times10^{-8}}{8.0\times10^{-3}} \approx \mathbf{3.14\times10^{-6}\ \text{m}^3/\text{s}}.$$ That is about $3.1\ \text{mL/s}$.

L2.2

For the pipe in L2.1, find the maximum (centre-line) velocity.

Recall Solution

Step 1 — formula. $v_{m a x} = \frac{Δ P R ^{2}}{4 η L}$ . Step 2 — numbers. $R^{2} = (2.0 \times 1 0^{- 3})^{2} = 4.0 \times 1 0^{- 6} m^{2}$ .

= \frac{2.0\times10^{-3}}{4.0\times10^{-3}} = \mathbf{0.50\ \text{m/s}}.$$ **Cross-check:** $\bar v = \tfrac12 v_{\max}=0.25\ \text{m/s}$; and $Q=\bar v\cdot\pi R^2 = 0.25\times\pi(4.0\times10^{-6}) = 3.14\times10^{-6}\ \text{m}^3/\text{s}$ ✓ matches L2.1.

L2.3

Water flows at $Q = 2.0 \times 1 0^{- 5} m^{3} / s$ through a pipe with $R = 5.0 mm$ , $L = 2.0 m$ , $η = 1.0 \times 1 0^{- 3} Pa\cdots$ . Find $Δ P$ .

Recall Solution

Step 1 — rearrange. $Δ P = \frac{8 η L Q}{π R ^{4}}$ . Step 2 — $R^{4}$ . $(5.0 \times 1 0^{- 3})^{4} = 6.25 \times 1 0^{- 10} m^{4}$ . Step 3 — substitute.

= \frac{3.2\times10^{-7}}{1.963\times10^{-9}} \approx \mathbf{163\ \text{Pa}}.$$

Level 3 — Analysis

Compare, ratio, and reason about "what changes."

L3.1

A pipe's radius is reduced to $80%$ of its original value (a partly narrowed artery), everything else fixed. What fraction of the original flow remains?

Recall Solution

Step 1 — proportionality. At fixed $Δ P, L, η$ : $Q \propto R^{4}$ . Step 2 — ratio. New radius $= 0.80 R$ , so $\frac{Q _{new}}{Q _{old}} = (0.80)^{4} = 0.4096 \approx 0.41 .$ Only about 41% of the flow survives a mere 20% narrowing — a dramatic loss. This is exactly why arterial narrowing is so serious (Blood flow and circulatory system).

L3.2

Two pipes carry the same fluid under the same $Δ P$ . Pipe B is twice as long and has $1.5 \times$ the radius of pipe A. Find $Q_{B} / Q_{A}$ .

Recall Solution

Step 1 — collect dependences. $Q \propto \frac{R ^{4}}{L}$ (with $Δ P, η$ fixed). Step 2 — ratio.

= \frac{5.0625}{2} = \mathbf{2.53}.$$ The $R^4$ boost (factor $5.06$) beats the length penalty (factor $\tfrac12$), so pipe B still carries about $2.5\times$ more.

L3.3

At what radial position $r$ is the local velocity equal to the mean velocity $\overset{v}{ˉ}$ ? Give $r$ as a fraction of $R$ .

Recall Solution

Step 1 — write both. $v (r) = v_{m a x} (1 - \frac{r ^{2}}{R ^{2}})$ and $\overset{v}{ˉ} = \frac{1}{2} v_{m a x}$ . Step 2 — set equal.

\;\Rightarrow\; 1-\frac{r^2}{R^2} = \tfrac12 \;\Rightarrow\; \frac{r^2}{R^2}=\tfrac12.$$ **Step 3 — solve.** $$r = \frac{R}{\sqrt2} \approx \mathbf{0.707\,R}.$$ So the flow moves at its *average* speed about 71\% of the way out to the wall — see the green line in the figure below.

Level 4 — Synthesis

Combine Poiseuille with another idea (continuity, viscosity's law, power).

L4.1

A fluid enters a wide feeder pipe (radius $R_{1} = 6 mm$ ) at mean speed $\overset{v}{ˉ}_{1} = 0.20 m/s$ and passes into a narrow pipe (radius $R_{2} = 2 mm$ ). Using the Equation of continuity, find the mean speed $\overset{v}{ˉ}_{2}$ in the narrow pipe.

Recall Solution

Step 1 — continuity. Volume is conserved: $Q = \overset{v}{ˉ}_{1} π R_{1}^{2} = \overset{v}{ˉ}_{2} π R_{2}^{2}$ . Step 2 — solve.

= 0.20\times 9 = \mathbf{1.8\ \text{m/s}}.$$ **Note:** continuity fixes the *mean* speed from area; Poiseuille then tells you the pressure drop needed to *drive* that mean speed. They answer different questions — pair them, don't confuse them.

L4.2

Find the shear stress $τ$ that the fluid exerts on the wall ( $r = R$ ) of the pipe in L2.1, using Newton's law of viscosity, $τ = η ∣ d v / d r ∣$ (see Viscosity and Newton's law of viscosity). Recall $R = 2.0 mm$ , $Δ P = 500 Pa$ , $L = 1.0 m$ .

Recall Solution

Step 1 — differentiate the profile. $v (r) = \frac{Δ P}{4 η L} (R^{2} - r^{2}) \Rightarrow \frac{d v}{d r} = - \frac{Δ P}{2 η L} r$ . Step 2 — evaluate at the wall $r = R$ and take magnitude.

= \eta\cdot\frac{\Delta P}{2\eta L}R = \frac{\Delta P\,R}{2L}.$$ Notice $\eta$ cancels! **Step 3 — numbers.** $$\tau_{\text{wall}} = \frac{(500)(2.0\times10^{-3})}{2(1.0)} = \mathbf{0.50\ \text{Pa}}.$$ **Why $\eta$ cancels:** a stickier fluid has a *steeper* gradient near the wall by exactly the amount that offsets its extra stickiness, for a *given* pressure push.

L4.3

Find the hydraulic power the pump must deliver to sustain the flow in L2.1, given by $P_{pump} = Δ P \cdot Q$ . Use $Q = 3.14 \times 1 0^{- 6} m^{3} / s$ and $Δ P = 500 Pa$ .

Recall Solution

Step 1 — why this product? Power = force × velocity = (pressure × area) × velocity = pressure × (area × velocity) = $Δ P \cdot Q$ . The units confirm it: $Pa \cdot m^{3} / s = \frac{N}{m ^{2}} \cdot \frac{m ^{3}}{s} = \frac{N\cdotm}{s} = W$ . Step 2 — multiply.

= 1.57\ \text{mW}.$$ All of this power is dissipated as heat by viscous friction — it is the "cost" of pushing a sticky fluid.

Level 5 — Mastery

Multi-step, symbolic, or edge-case reasoning with no formula handed to you.

L5.1

Show algebraically that the mean velocity of Poiseuille flow is exactly half the maximum, i.e. $\overset{v}{ˉ} = \frac{1}{2} v_{m a x}$ , by integrating the profile over the cross-section.

Recall Solution

Step 1 — set up. $\overset{v}{ˉ} = \frac{Q}{π R ^{2}}$ where $Q = \int_{0}^{R} v (r) 2 π r d r$ and $v (r) = v_{m a x} (1 - \frac{r ^{2}}{R ^{2}})$ . Step 2 — integrate.

= 2\pi v_{\max}\left[\frac{r^2}{2}-\frac{r^4}{4R^2}\right]_0^R = 2\pi v_{\max}\left(\frac{R^2}{2}-\frac{R^2}{4}\right)=2\pi v_{\max}\frac{R^2}{4}.$$ So $Q = \dfrac{\pi R^2 v_{\max}}{2}$. **Step 3 — divide.** $$\bar v = \frac{Q}{\pi R^2} = \frac{v_{\max}}{2}. \qquad\blacksquare$$

L5.2

A "power-law" pipe designer wants to triple the flow rate $Q$ but must keep $Δ P$ , $L$ and $η$ fixed. By what percentage must the radius be increased?

Recall Solution

Step 1 — proportionality. $Q \propto R^{4} \Rightarrow \frac{R _{new}}{R _{old}} = (\frac{Q _{new}}{Q _{old}})^{1/4} = 3^{1/4}$ . Step 2 — evaluate. $3^{1/4} = 1.3161$ . Step 3 — percentage increase. $(1.3161 - 1) \times 100% \approx 31.6% .$ A tiny 32% widening triples the flow — the $R^{4}$ law is astonishingly generous to wide pipes.

L5.3

The centre-line fluid moves fastest, yet Newton's law says shear stress is proportional to the gradient $d v / d r$ . Explain what the shear stress is at the exact centre $r = 0$ , and where it is largest. Support your claim with the algebra.

Recall Solution

Step 1 — gradient. From $v (r) = \frac{Δ P}{4 η L} (R^{2} - r^{2})$ , $\frac{d v}{d r} = - \frac{Δ P}{2 η L} r .$ Step 2 — read off the centre. At $r = 0$ : $\frac{d v}{d r} = 0$ , so $τ (0) = η \frac{d v}{d r} = 0 .$ Step 3 — where is it largest? $∣ d v / d r ∣ \propto r$ , which is biggest at the wall $r = R$ . So shear stress grows linearly from zero at the centre to a maximum at the wall — this is the case checked numerically in L4.2 ( $τ_{wall} = 0.50 Pa$ ). The subtlety: "fastest" (high $v$ ) is not "most sheared" (high $d v / d r$ ). At the centre the parabola is momentarily flat (its slope is zero), so neighbouring layers there move at nearly the same speed — no rubbing, no shear. All the internal friction lives near the wall.

Active Recall

Recall One-line takeaways (cover the answers!)

Doubling $R$ multiplies $Q$ by? ::: $2^{4} = 16$ .
Radius cut to 80%: flow becomes? ::: $(0.8)^{4} \approx 0.41$ , i.e. 41%.
Local speed equals the mean at $r =$ ? ::: $R / 2 \approx 0.71 R$ .
Wall shear stress formula? ::: $τ_{wall} = \frac{Δ P R}{2 L}$ ( $η$ cancels).
Shear stress at the pipe centre? ::: Zero — the profile is flat there.
Pump power to sustain flow? ::: $P = Δ P \cdot Q$ .

Connections

Parent: Poiseuille flow (Hinglish) — the derivations these exercises drill.
Viscosity and Newton's law of viscosity — used in L4.2 and L5.3 for wall shear.
Equation of continuity — used in L4.1 for the speed-up in narrowing pipes.
Reynolds number and turbulence — every answer here assumes laminar flow.
Blood flow and circulatory system — the $R^{4}$ artery result of L3.1.
Bernoulli's principle — the frictionless contrast to these viscous losses.