FE Mechanical Practice Question Solution included after investigation
CASE E-056 ยท FE

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

The water received 9.81 kW. A 70% efficient pump required about 14.0 kW of shaft input.

โ–ถ INTERACTIVE CASE FILE โ€ข 5โ€“10 MIN โ€ข CLICK TO INVESTIGATE

Inspect, commit, prove, fix, and sign off.

Case E-056 engineering failure visual
HOW THIS WORKS

Think like
a real engineer.

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

1

Inspect

Review the problem, diagram, and evidence.

2

Commit

Choose your hypothesis.

3

Prove

Run calculations and test your idea.

4

Fix

Select and validate a safe correction.

5

Sign Off

See the full debrief and key takeaways.

Case Brief

SAME PRINCIPLES. HIGHER STANDARDS.
PROBLEM STATEMENT

A pump moves water at 0.050 mยณ/s against a 20 m head. Water density is 1000 kg/mยณ and pump efficiency is 70%. A sizing note multiplies hydraulic power by 0.70 and specifies a motor smaller than the useful fluid power. Determine hydraulic power and required shaft input.

Fluid Mechanics Mechanical Design Verification Root Cause Analysis Safe Correction
E-056
INVESTIGATION WORKSPACE WORK THROUGH THE CASE โ€” YOUR CHOICES MATTER.
๐Ÿ”’
DEBRIEF (LOCKED)

Complete the investigation
to unlock the full debrief.

Find the root cause, confirm the fix, and see how this connects to the exam.

๐Ÿ”’Root causeComplete the case
to reveal
๐Ÿ”’Correct fixComplete the case
to reveal
๐Ÿ”’Exam takeawayComplete the case
to reveal
CASE CLOSED ยท ENGINEERING VERDICT

What actually failed โ€” and what should change.

What failed

Efficiency was applied in the wrong direction.

What was assumed

Input power was taken as hydraulic power times efficiency.

Corrected model

Use P_hyd=ฯgQH and P_in=P_hyd/ฮท.

QUICK ANSWER

Hydraulic pump power is ฯgQH. For water at Q=0.05 mยณ/s and H=20 m, P_hyd=9.81 kW. At 70% efficiency, required shaft input is 9.81/0.70โ‰ˆ14.0 kW.

ENGINEERING TAKEAWAY

Pump efficiency is useful hydraulic output divided by shaft input, so required input is P_hyd/ฮท.

CASE QUESTIONS

Questions this case should settle.

How do you calculate hydraulic pump power?

Use P_hyd=ฯgQH.

Why divide by efficiency for pump input power?

Because pump efficiency equals useful hydraulic output divided by shaft input.

NEXT FE PRACTICE QUESTION

The 50 m Pipe That Lost 20 kPa

Fluid Mechanics ยท 5โ€“10 min

Continue Practice โ†’
WHAT YOUโ€™LL LEARN

Real failures. Lasting skills.

Each case is designed to build the judgment, analysis, and confidence you need for engineering exams โ€” and beyond.

  • Apply core engineering concepts to real-world problems
  • Practice structured troubleshooting and analysis
  • Strengthen exam-ready thinking through realistic scenarios
COMMON QUESTIONS
Do I need prior knowledge to use these cases?

Basic subject familiarity helps, but every case is designed to teach through the investigation itself.

How long does a case take?

Most cases are designed for a focused 5โ€“10 minute investigation.

Are the cases aligned with engineering exams?

Each case is mapped to a verified exam, subject, topic, and misconception before publication.

What happens after I complete a case?

The sealed debrief unlocks with the root cause, corrected reasoning, fix, and takeaway.