91Ó°ÊÓ

The V6 engine in a 2014 Chevrolet Silverado 1500 pickup truck is reported to produce a maximum power of 285 hp at 5300 rpm and a maximum torque of 305 ft \(\cdot\) lb at 3900 rpm. (a) Calculate the torque, in both ft \(\cdot\) lb and N \(\cdot\) m, at 5300 rpm. Is your answer in ft \(\cdot\) lb smaller than the specified maximum value? (b) Calculate the power, in both horsepower and watts, at 3900 rpm. Is your answer in hp smaller than the specified maximum value? (c) The relationship between power in hp and torque in ft \(\cdot\) lb at a particular angular velocity in rpm is often written as hp \(= \big[\)torque 1in ft \(\cdot\) lb2 \(\times\) rpm\(\big]/c\), where \(c\) is a constant. What is the numerical value of \(c\)? (d) The engine of a 2012 Chevrolet Camaro ZL1 is reported to produce 580 hp at 6000 rpm. What is the torque (in ft \(\cdot\) lb) at 6000 rpm?

Step by step solution

01

Calculate Torque at 5300 RPM

Given power is 285 hp at 5300 rpm. We use the formula for power: \[ P = \frac{{T \times \omega}}{5252} \] where - \( P \) is power in horsepower,- \( T \) is torque in ft \cdot lb,- \( \omega \) is RPM.Rearranging the formula to find Torque \( T \):\[ T = \frac{{P \times 5252}}{\omega} \]Substituting the values: \[ T = \frac{{285 \times 5252}}{5300} \approx 282.38 \text{ ft}\cdot\text{lb} \]To convert to N \cdot m, use the conversion 1 ft \cdot lb = 1.35582 N \cdot m: \[ T \approx 282.38 \times 1.35582 \approx 382.67 \text{ N} \cdot \text{m} \] The calculated torque is smaller than the specified maximum torque of 305 ft \cdot lb.

02

Calculate Power at 3900 RPM

Given torque is 305 ft \cdot lb at 3900 rpm. Use the same power formula rearranged:\[ P = \frac{{T \times \omega}}{5252} \]Substitute the given values:\[ P = \frac{{305 \times 3900}}{5252} \approx 226.14 \text{ hp} \]To convert to watts, use the conversion 1 hp = 745.7 W:\[ P \approx 226.14 \times 745.7 \approx 168647.1 \text{ W} \]The calculated power is smaller than the specified maximum power of 285 hp.

03

Calculate Constant c

For the formula \( \text{hp} = \frac{{\text{torque} \times \text{rpm}}}{c} \), we will use the known example:- Assume \( \text{hp} = 1 \), the typical conversion factor allows the torque-rpm relation to yield the same power:- Given formula already uses 5252 as this is based on units once rearranged: \[ c = 5252 \] This is the number that converts the units correctly for power in horsepower and torque in ft \cdot lb.

04

Calculate Torque for Camaro at 6000 RPM

Given power for the Camaro is 580 hp at 6000 rpm. Use:\[ P = \frac{{T \times \omega}}{5252} \]Rearrange to solve for torque \( T \):\[ T = \frac{{P \times 5252}}{\omega} \]Substitute the values:\[ T = \frac{{580 \times 5252}}{6000} \approx 507.44 \text{ ft}\cdot\text{lb} \] This value represents the torque at 6000 rpm for the Camaro.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Engine Performance

Understanding engine performance is crucial for evaluating how well a vehicle can perform under different conditions. It involves analyzing key metrics such as power and torque, which provide insight into the engine's capabilities over a range of rpms (revolutions per minute). For example, in the context of the Chevrolet Silverado 1500 or the Camaro ZL1, one looks at maximum horsepower and torque at certain rpms to determine performance levels.

Engine performance is typically evaluated by considering:

• Horsepower (hp): A measure of the engine's power output, indicating how quickly the engine can perform work.
• Torque (ft \(\cdot\) lb or N \(\cdot\) m): A measure of rotational force, indicating the engine's ability to produce twisting motion – crucial for acceleration.
• RPM (revolutions per minute): How fast the engine crankshaft is spinning, which influences the engine's power and torque output.

These components help determine a vehicle's acceleration capability, towing power, and overall engine responsiveness under different conditions.

Power and Torque Relationship

Power and torque are related, yet distinct concepts within the realm of engine dynamics. Although they both involve the capability of an engine to do work, they apply to different aspects of engine performance. Understanding their relationship is essential for interpreting vehicle specifications and performance profiles.

Here's how they relate:

• Torque is the force applied to rotate an object, such as an engine's crankshaft. It's what gets the vehicle moving from a standstill.
• Power is about the rate at which work is done. In engines, it's how fast the engine can convert fuel into motion or energy over time.
• The formula connecting torque and power is given by:\[ P = \frac{{T \times \omega}}{5252} \]where \(P\) is power in horsepower, \(T\) is torque in ft \(\cdot\) lb, and \(\omega\) is RPM.
• This relationship implies that at a constant power, increasing torque would decrease rpm, and vice versa. This is why sports cars often boast high power numbers at high rpms, whereas trucks highlight higher torque values at lower rpms for towing.

A constant, 5252, adjusts the formula based on measurement units, facilitating direct comparisons.

Unit Conversion in Physics

When it comes to physics, especially in automotive contexts, unit conversion helps translate measurements into standard units, making comparisons and calculations easier and more meaningful.

In our exercise, we face conversions such as:

• Converting torque from ft \(\cdot\) lb to N \(\cdot\) m using the conversion factor:
  1 ft \(\cdot\) lb = 1.35582 N \(\cdot\) m.
  This conversion lets us compare torque values internationally.
• Converting power from horsepower to watts using:
  1 hp = 745.7 W.
  Since watts are the standard unit of power in the International System of Units (SI), converting to watts makes comparing engines on a global scale possible.

Accurate unit conversion enhances clarity and helps ensure that calculations and statistical evaluations are reliable and can be universally communicated.