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Date	May 2018	Marks available	1	Reference code	18M.3.HL.TZ2.10
Level	Higher level	Paper	Paper 3	Time zone	2
Command term	Estimate	Question number	10	Adapted from	N/A

Question

Two tubes, A and B, are inserted into a fluid flowing through a horizontal pipe of diameter 0.50 m. The openings X and Y of the tubes are at the exact centre of the pipe. The liquid rises to a height of 0.10 m in tube A and 0.32 m in tube B. The density of the fluid = 1.0 × 10³ kg m^–3.

M18/4/PHYSI/HP3/ENG/TZ2/10

The viscosity of water is 8.9 × 10^–4 Pa s.

Show that the velocity of the fluid at X is about 2 ms^–1, assuming that the flow is laminar.

[3]

a.

Estimate the Reynolds number for the fluid in your answer to (a).

[1]

b.i.

Outline whether your answer to (a) is valid.

[1]

b.ii.

Markscheme

$\frac{1}{2} ρ v_{X}^{2} = p_{Y} - p_{X} = ρ g Δ h$ $\frac{1}{2}\rho v_{\text{X}}^2 = {p_{\text{Y}}} - {p_{\text{X}}} = \rho g\Delta h$

v_X = $\sqrt{2 \times 9.8 \times (0.32 - 0.10)}$ $\sqrt {2 \times 9.8 \times (0.32 - 0.10)}$

v_x = 2.08 «ms^–1»

[3 marks]

a.

R = « $\frac{v r ρ}{η} = \frac{2.1 \times 0.25 \times 10^{3}}{8.9 \times 10^{- 4}}$ $\frac{{vr\rho }}{\eta } = \frac{{2.1 \times 0.25 \times {{10}^3}}}{{8.9 \times {{10}^{ - 4}}}}$ » 5.9 × 10⁵

[1 mark]

b.i.

(R > 1000) flow is not laminar, so assumption is invalid

OWTTE

[1 mark]

b.ii.

Examiners report

[N/A]

a.

[N/A]

b.i.

[N/A]

b.ii.

Syllabus sections

Option B: Engineering physics » Option B: Engineering physics (Additional higher level option topics) » B.3 – Fluids and fluid dynamics (HL only)

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