In car engines, write out what SI stands for.
A four-stroke 2 -cylinder internal combustion engine running at the speed
of , has a bore and stroke length of and respectively.
The engine develops a brake torque of , while the volumetric
efficiency is 0,85 . The air-fuel ratio by volume is . The fuel used in this engine
has a calorific value of .
Calculate:
5.2.1 The engine’s brake power in kW
5.2.2 The mean speed of the engine’s pistons in
5.2 The engine’s swept volume in
The engine’s induced volume in
The brake thermal efficiency

Ask by John Parry. in South Africa

Dec 06,2024

Question

In car engines, write out what SI stands for.
A four-stroke 2 -cylinder internal combustion engine running at the speed
of , has a bore and stroke length of and respectively.
The engine develops a brake torque of , while the volumetric
efficiency is 0,85 . The air-fuel ratio by volume is . The fuel used in this engine
has a calorific value of .
Calculate:
5.2.1 The engine’s brake power in kW
5.2.2 The mean speed of the engine’s pistons in
5.2 The engine’s swept volume in
The engine’s induced volume in
The brake thermal efficiency

Ask by John Parry. in South Africa

Dec 06,2024

UpStudy AI · Accepted Answer

### 5.2.1 The Engine's Brake Power in kW

The brake power (BP) of an engine can be calculated using the formula:

$$
BP = \frac{2 \pi N T}{60}
$$

where:
- $$ N $$ = engine speed in revolutions per minute (rpm)
- $$ T $$ = brake torque in Nm

Given:
- $$ N = 240 \, \text{r/min} $$
- $$ T = 11,860 \, \text{kNm} = 11,860,000 \, \text{Nm} $$

Substituting the values:

$$
BP = \frac{2 \pi (240) (11,860,000)}{60}
$$

Calculating:

$$
BP = \frac{2 \pi (240) (11,860,000)}{60} \approx \frac{2 \times 3.14159 \times 240 \times 11,860,000}{60}
$$

$$
BP \approx \frac{2 \times 3.14159 \times 240 \times 11,860,000}{60} \approx 1,507,000 \, \text{W} \approx 1507 \, \text{kW}
$$

### 5.2.2 The Mean Speed of the Engine's Pistons in $$ \mathrm{m/s} $$

The mean speed of the pistons can be calculated using the formula:

$$
\text{Mean Speed} = \frac{2 \times \text{Stroke Length} \times N}{60}
$$

Given:
- Stroke Length = $$ 0.585 \, \text{m} $$
- $$ N = 240 \, \text{r/min} $$

Substituting the values:

$$
\text{Mean Speed} = \frac{2 \times 0.585 \times 240}{60}
$$

Calculating:

$$
\text{Mean Speed} = \frac{2 \times 0.585 \times 240}{60} = \frac{280.8}{60} \approx 4.68 \, \text{m/s}
$$

### 5.2.3 The Engine's Swept Volume in $$ \mathrm{m}^3 $$

The swept volume (V_s) of a cylinder can be calculated using the formula:

$$
V_s = \frac{\pi}{4} \times \text{Bore}^2 \times \text{Stroke}
$$

Given:
- Bore = $$ 0.380 \, \text{m} $$
- Stroke = $$ 0.585 \, \text{m} $$

Calculating the swept volume for one cylinder:

$$
V_s = \frac{\pi}{4} \times (0.380)^2 \times (0.585)
$$

Calculating:

$$
V_s = \frac{3.14159}{4} \times 0.1444 \times 0.585 \approx 0.067 \, \text{m}^3
$$

Since there are 2 cylinders:

$$
\text{Total Swept Volume} = 2 \times V_s \approx 2 \times 0.067 \approx 0.134 \, \text{m}^3
$$

### 5.2.4 The Engine's Induced Volume in $$ \mathrm{m}^3/s $$

The induced volume (V_i) can be calculated using the formula:

$$
V_i = V_s \times N \times \text{Volumetric Efficiency}
$$

Given:
- Volumetric Efficiency = $$ 0.85 $$
- $$ N = 240 \, \text{r/min} $$

Calculating the induced volume per second:

$$
V_i = 0.134 \, \text{m}^3 \times \frac{240}{60} \times 0.85
$$

Calculating:

$$
V_i = 0.134 \times 4 \times 0.85 \approx 0.456 \, \text{m}^3/s
$$

### 5.2.5 The Brake Thermal Efficiency

The brake thermal efficiency ($$ \eta_b $$) can be calculated using the formula:

$$
\eta_b = \frac{BP}{\dot{Q}}
$$

Where $$ \dot{Q} $$ is the heat input, calculated as:

$$
\dot{Q} = V_i \times \text{Air-Fuel Ratio} \times \text{Calorific Value}
$$

Given:
- Air-Fuel Ratio = $$ 7:1 $$
- Calorific Value = $$ 38600 \, \text{kJ/m}^3 $$

Calculating $$ \dot{Q} $$:

$$
\dot{Q} = V_i \times \left( \frac{1}{7} \right) \times 38600
$$

Substituting $$ V_i $$:

$$
\dot{Q} = 0.456 \, \text{m}^3/s \times \left( \frac{1}{7} \right) \times 38600
$$

Calculating:

$$
\dot{Q} \approx 0.456 \times 0.142857 \times 38600 \approx 0.456 \times 5514.29 \approx 2515.71 \, \text{kW}
$$

Now, substituting $$ BP $$ and $$ \dot{Q} $$ into the efficiency formula:

$$
\eta_b = \frac{1507}{2515.71} \approx 0.597 \text{ or } 59.7\%
$$

### Summary of Results
- **Brake Power**: $$ 1507 \, \text{kW} $$
- **Mean Speed of Pistons**: $$ 4.68 \, \text{m/s} $$
- **Swept Volume**: $$ 0.134 \, \text{m}^3 $$
- **Induced Volume**: $$ 0.456 \, \text{m}^3/s $$
- **Brake Thermal Efficiency**: $$ 59.7\% $$
### 5.2.1 The Engine's Brake Power in kW The brake power (BP) of an engine is approximately 1507 kW. ### 5.2.2 The Mean Speed of the Engine's Pistons in $$ \mathrm{m/s} $$ The mean speed of the pistons is approximately 4.68 m/s. ### 5.2.3 The Engine's Swept Volume in $$ \mathrm{m}^3 $$ The engine's swept volume is approximately 0.134 m³. ### 5.2.4 The Engine's Induced Volume in $$ \mathrm{m}^3/s $$ The engine's induced volume is approximately 0.456 m³/s. ### 5.2.5 The Brake Thermal Efficiency The brake thermal efficiency is approximately 59.7%.

Upstudy AI Solution

Answer

5.2.1 The Engine’s Brake Power in kW

5.2.2 The Mean Speed of the Engine’s Pistons in

5.2.3 The Engine’s Swept Volume in

5.2.4 The Engine’s Induced Volume in

5.2.5 The Brake Thermal Efficiency

Solution

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