We read your photo as:

Define brake power and indicated power. Explain how a dynamometer is used to measure engine performance. An engine develops 75 kW brake power with a mechanical efficiency of 82%. Calculate the indicated power and the power lost to friction.

Physics

ScanSolve AI Answer

Answer

c

91.463 kW91.463 \, \text{kW}

16.463 kW16.463 \, \text{kW}

Step-by-step solution

2 steps

Here are the answers for Q6:

Part a

  • Brake power (BP) is the actual power delivered by the engine at the crankshaft, measured at the output shaft. It represents the usable power available to do work.
  • Indicated power (IP) is the total power developed inside the engine cylinders by the combustion of fuel. It is the theoretical power generated before any losses due to friction or other mechanical inefficiencies.

Part b

A dynamometer is used to measure engine performance by applying a controlled load to the engine's output shaft. It measures the torque produced by the engine at various rotational speeds (RPM). From these measurements, the brake power (BP) can be calculated. This allows engineers to evaluate the engine's power output, fuel efficiency, and other performance characteristics under different operating conditions.

Part c

91.463 kW91.463 \, \text{kW}

  1. Calculate the indicated power (IP).

    Given brake power (BP) =75 kW= 75 \, \text{kW} and mechanical efficiency (ηmech\eta_{mech}) =82%=0.82= 82\% = 0.82.
    The mechanical efficiency is defined as:

    ηmech=Brake Power (BP)Indicated Power (IP)\eta_{mech} = \frac{\text{Brake Power (BP)}}{\text{Indicated Power (IP)}}

    Rearranging to find indicated power:

    IP=BPηmech\text{IP} = \frac{\text{BP}}{\eta_{mech}} IP=75 kW0.82\text{IP} = \frac{75 \, \text{kW}}{0.82} IP≈91.463 kW\text{IP} \approx 91.463 \, \text{kW}

    The indicated power is 91.463 kW\boxed{91.463 \, \text{kW}}.

  2. Calculate the power lost to friction (FP).

    The power lost to friction is the difference between the indicated power and the brake power:

    FP=IP−BP\text{FP} = \text{IP} - \text{BP} FP=91.463 kW−75 kW\text{FP} = 91.463 \, \text{kW} - 75 \, \text{kW} FP=16.463 kW\text{FP} = 16.463 \, \text{kW}

    The power lost to friction is 16.463 kW\boxed{16.463 \, \text{kW}}.

Was this helpful?

Try one yourself

A new problem that tests the same skill.

Still stuck on this one?

Ask a follow-up, and the answer starts from this question.

Got a different question?Ask your own question

More Physics Questions

QHere is the question, explain why 9/10 was used to multiply 500: Find out the speed at which a…QQ26. Briefly explain the following engine horse power terms: Indicated horse Power (ii) Frictional…QTeaser / foresight draft for further thoughts /self study 10 Marks to supplement the poor test…QDetermine the engine force.QA drum with a diameter of 1.5 m is rotating at 200 r/min. A brake is applied to the drum, absorbing…
Browse by topic:Mechanics
See all Physics questions →
Got a syllabus? Photograph it and I'll tell you what's due and what's worth the most.
Handwritten step-by-step solution preview

Handwritten Step-by-Step Solution

Get this as a handwritten image on WhatsApp — like a tutor wrote it for you.

Try on WhatsApp