FE Mechanical Practice Question Solution included after investigation
CASE E-054 Β· FE

The Fast-Looking Oil Flow That Was Still Laminar

The oil was moving at 2 m/s, but its Reynolds number was only 1700.

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Case E-054 engineering failure visual
HOW THIS WORKS

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Follow the investigation process used in the field β€” in five guided steps.

1

Inspect

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2

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Choose your hypothesis.

3

Prove

Run calculations and test your idea.

4

Fix

Select and validate a safe correction.

5

Sign Off

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Case Brief

SAME PRINCIPLES. HIGHER STANDARDS.
PROBLEM STATEMENT

Oil flows through a 50 mm diameter pipe at 2.0 m/s. The oil density is 850 kg/mΒ³ and dynamic viscosity is 0.05 PaΒ·s. A technician labels the flow turbulent because 2 m/s seems fast. Calculate the Reynolds number, classify the flow, and determine what velocity would correspond to Re=4000 with the same oil and pipe.

Fluid Mechanics Mechanical Design Verification Root Cause Analysis Safe Correction
E-054
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CASE CLOSED Β· ENGINEERING VERDICT

What actually failed β€” and what should change.

What failed

The flow was labeled turbulent from speed alone.

What was missed

Viscosity, density, and pipe diameter all affect the regime.

Corrected model

Use Re=ρVD/μ before classifying internal flow.

QUICK ANSWER

For ρ=850 kg/m³, V=2 m/s, D=0.05 m, and μ=0.05 Pa·s, Re=ρVD/μ=1700. The flow is therefore laminar under the usual internal-pipe-flow classification.

ENGINEERING TAKEAWAY

Pipe-flow regime depends on the dimensionless Reynolds number, not velocity alone.

CASE QUESTIONS

Questions this case should settle.

Can 2 m/s pipe flow be laminar?

Yes. Flow regime depends on Reynolds number, not velocity alone.

What velocity gives Re=4000 for this oil and pipe?

About 4.71 m/s.

NEXT FE PRACTICE QUESTION

The Pump That Needed 14 kW to Deliver 9.81 kW to the Water

Fluid Mechanics Β· 5–10 min

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WHAT YOU’LL LEARN

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