Math Problem Statement

4 stroke, 4 cylinder engine running at 2200 rpm develops 70 kW of brake power. Given that brake thermal efficiency is 40%, CV is 44 MJ/kg, engine bore is 130 mm, stroke is 110 mm and efficiency,η is 80%. Find:

a) Indicated power,

b) IMEP,

c) Mass flow rate of fuel, mfuel,

d) Indicated thermal efficiency.

Answer:

a) Indicated power : Answer 1 Question 26 kW (1 decimal)

b) IMEP : Answer 2 Question 26 kPa (1 decimal)

c) Mass flow rate of fuel, mfuel : Answer 3 Question 26 kg/s (3 decimal)

d) Indicated thermal efficiency : Answer 4 Question 26 % (1 decimal)

Solution

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

Given:

  • Number of cylinders (nn) = 4
  • Engine speed (NN) = 2200 rpm
  • Brake power (PbP_b) = 70 kW = 70,000 W
  • Brake thermal efficiency (ηb\eta_b) = 40% = 0.40
  • Calorific value (CV) = 44 MJ/kg = 44,000 kJ/kg
  • Engine bore (DD) = 130 mm = 0.13 m
  • Stroke (LL) = 110 mm = 0.11 m
  • Mechanical efficiency (ηm\eta_m) = 80% = 0.80

Formulas:

  1. Indicated Power (PiP_i): Pi=PbηmP_i = \frac{P_b}{\eta_m}

  2. Indicated Mean Effective Pressure (IMEP): IMEP=2×Pin×L×A×NIMEP = \frac{2 \times P_i}{n \times L \times A \times N} where:

    • AA is the area of the piston = πD24\frac{\pi D^2}{4}
  3. Mass Flow Rate of Fuel (m˙fuel\dot{m}_{\text{fuel}}): m˙fuel=Pbηb×CV\dot{m}_{\text{fuel}} = \frac{P_b}{\eta_b \times \text{CV}}

  4. Indicated Thermal Efficiency (ηi\eta_i): ηi=Pim˙fuel×CV×100\eta_i = \frac{P_i}{\dot{m}_{\text{fuel}} \times \text{CV}} \times 100

Calculations:

a) Indicated Power (PiP_i):

Pi=Pbηm=700.80kWP_i = \frac{P_b}{\eta_m} = \frac{70}{0.80} \, \text{kW}

Let's calculate this:

Pi=87.5kWP_i = 87.5 \, \text{kW}

Answer 1: 87.5 kW

b) Indicated Mean Effective Pressure (IMEP):

First, calculate the area AA of the piston: A=π×(0.13)240.01327m2A = \frac{\pi \times (0.13)^2}{4} \approx 0.01327 \, \text{m}^2

Now calculate the IMEP: IMEP=2×87.5×10004×0.11×0.01327×2200IMEP = \frac{2 \times 87.5 \times 1000}{4 \times 0.11 \times 0.01327 \times 2200}

Let's calculate this:

IMEP278.48kPaIMEP \approx 278.48 \, \text{kPa}

Answer 2: 278.5 kPa

c) Mass Flow Rate of Fuel (m˙fuel\dot{m}_{\text{fuel}}):

m˙fuel=70×10000.40×44000kg/s\dot{m}_{\text{fuel}} = \frac{70 \times 1000}{0.40 \times 44000} \, \text{kg/s}

Let's calculate this:

m˙fuel0.00398kg/s\dot{m}_{\text{fuel}} \approx 0.00398 \, \text{kg/s}

Answer 3: 0.004 kg/s

d) Indicated Thermal Efficiency (ηi\eta_i):

ηi=87.5×10000.004×44000×100\eta_i = \frac{87.5 \times 1000}{0.004 \times 44000} \times 100

Let's calculate this:

ηi49.7%\eta_i \approx 49.7\%

Answer 4: 49.7 %


Would you like more details or have any questions?

Related Questions:

  1. How does mechanical efficiency affect the indicated power of an engine?
  2. What factors influence the indicated mean effective pressure (IMEP)?
  3. Why is brake thermal efficiency typically lower than indicated thermal efficiency?
  4. How does fuel mass flow rate correlate with brake power and efficiency?
  5. What impact does increasing the engine bore have on IMEP and power output?

Tip:

Regular monitoring of engine performance parameters like IMEP and fuel consumption can help in maintaining the engine's efficiency and detecting any potential issues early.

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Math Problem Analysis

Mathematical Concepts

Thermodynamics
Internal combustion engines
Power calculations

Formulas

Indicated Power formula
IMEP formula
Mass flow rate formula
Indicated thermal efficiency formula

Theorems

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Suitable Grade Level

Advanced Engineering