Math Problem Statement

4 stroke, 6 cylinders engine running at speed of 2400 rpm. Given that the engine bore is 90 mm, stroke is 120 mm, IMEP is 560 kPa and engine torque is 120 Nm. Find:

a) Brake power,

b) Indicated power,

c) Friction power,

d) Efficiency of engine.

Answer:

a) Brake power : Answer 1 Question 25 watt (whole number)

b) Indicated power : Answer 2 Question 25 watt (1 decimal)

c) Friction power : Answer 3 Question 25 watt (1 decimal)

d) Efficiency of engine : Answer 4 Question 25 % (whole number)

Solution

Let's calculate each of the required values step by step.

Given:

  • Number of cylinders (nn) = 6
  • Engine speed (NN) = 2400 rpm
  • Bore (DD) = 90 mm = 0.09 m
  • Stroke (LL) = 120 mm = 0.12 m
  • Indicated Mean Effective Pressure (IMEP) = 560 kPa = 560,000 Pa
  • Torque (TT) = 120 Nm

Formulas:

  1. Brake Power (PbP_b): Pb=2πNT60P_b = \frac{2 \pi N T}{60} where NN is in rpm and TT is in Nm.

  2. Indicated Power (PiP_i): Pi=n×Pm×L×A×N2P_i = \frac{n \times P_m \times L \times A \times N}{2} where:

    • PmP_m is the IMEP
    • AA is the area of the piston = πD24\frac{\pi D^2}{4}
  3. Friction Power (PfP_f): Pf=PiPbP_f = P_i - P_b

  4. Efficiency of Engine (η\eta): η=PbPi×100\eta = \frac{P_b}{P_i} \times 100

Calculations:

a) Brake Power (PbP_b):

Pb=2π×2400×12060WattP_b = \frac{2 \pi \times 2400 \times 120}{60} \, \text{Watt}

Let's calculate this:

Pb=2×3.1416×2400×12060P_b = \frac{2 \times 3.1416 \times 2400 \times 120}{60}

Pb30159.29Watt30159WattP_b \approx 30159.29 \, \text{Watt} \approx 30159 \, \text{Watt}

Answer 1: 30159 Watt

b) Indicated Power (PiP_i):

A=π×(0.09)24A = \frac{\pi \times (0.09)^2}{4} A0.00636m2A \approx 0.00636 \, \text{m}^2

Now,

Pi=6×560000×0.12×0.00636×24002P_i = \frac{6 \times 560000 \times 0.12 \times 0.00636 \times 2400}{2}

Let's calculate this:

Pi30829.82WattP_i \approx 30829.82 \, \text{Watt}

Answer 2: 30829.8 Watt

c) Friction Power (PfP_f):

Pf=PiPb=30829.830159P_f = P_i - P_b = 30829.8 - 30159

Let's calculate this:

Pf670.8WattP_f \approx 670.8 \, \text{Watt}

Answer 3: 670.8 Watt

d) Efficiency of Engine (η\eta):

η=PbPi×100=3015930829.8×100\eta = \frac{P_b}{P_i} \times 100 = \frac{30159}{30829.8} \times 100

Let's calculate this:

η97.8%\eta \approx 97.8 \%

Answer 4: 98 %


Would you like more details or have any questions?

Related Questions:

  1. What are the implications of a high friction power in an engine?
  2. How does the IMEP affect the indicated power in an engine?
  3. Why is the efficiency of a 4-stroke engine typically lower than 100%?
  4. What is the relationship between engine torque and brake power?
  5. How would increasing the engine bore affect the engine’s power output?

Tip:

Regular maintenance, such as proper lubrication, can reduce friction power and improve engine efficiency.

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Math Problem Analysis

Mathematical Concepts

Mechanical Engineering
Thermodynamics

Formulas

Brake Power formula: P_b = \frac{2 \pi N T}{60}
Indicated Power formula: P_i = \frac{n \times P_m \times L \times A \times N}{2}
Friction Power formula: P_f = P_i - P_b
Efficiency formula: \eta = \frac{P_b}{P_i} \times 100

Theorems

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Suitable Grade Level

Professional