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Chapter 2: Problem 41

The pistons of a V-6 automobile engine develop 226 hp. If the engine driveshaft rotational speed is \(4700 \mathrm{RPM}\) and the torque is \(248 \mathrm{ft} \cdot \mathrm{lbf}\), what percentage of the developed power is transferred to the driveshaft? What accounts for the difference in power? Does an engine this size meet your transportation needs? Comment.

Short Answer

Expert verified
Percentage transferred: 98.8%. The difference is due to internal losses. The engine is adequate for standard transportation needs.

Step by step solution

01

Convert Rotational Speed to Radians per Second

First, convert the rotational speed from RPM (revolutions per minute) to radians per second using the formula: \[ \text{Rotational Speed} (\text{rad/sec}) = \frac{2\pi \times \text{RPM}}{60} \]Plug in the given values: \[ \text{Rotational Speed} = \frac{2\pi \times 4700}{60} \]Simplify to get approximately \[ \text{Rotational Speed} \approx 492.6 \text{ rad/sec} \]
02

Calculate the Power Transferred to the Driveshaft

Use the torque and rotational speed to calculate the power transferred to the driveshaft using the formula: \[ P_{\text{driveshaft}} = \text{Torque} \times \text{Rotational Speed} \]Substitute the known values: \[ P_{\text{driveshaft}} = 248 \text{ ft} \times \text{lbf} \times 492.6 \text{ rad/sec} \]Simplify to get: \[ P_{\text{driveshaft}} \approx 122,804.8 \text{ ft} \times \text{lbf/sec} \]
03

Convert Power to Horsepower

Convert the power from ft lbf/sec to horsepower using the conversion factor 1 hp = 550 ft lbf/sec: \[ P_{\text{driveshaft (hp)}} = \frac{122,804.8}{550} \]Simplify to get: \[ P_{\text{driveshaft (hp)}} \approx 223.3 \text{ hp} \]
04

Calculate the Percentage of Power Transferred

Calculate the percentage of power transferred to the driveshaft using the formula: \[ \text{Percentage} = \frac{P_{\text{driveshaft}} \times 100}{P_{\text{eng}}} \]Substitute the known values: \[ \text{Percentage} = \frac{223.3 \times 100}{226} \]Simplify to get approximately: \[ \text{Percentage} \approx 98.8\text{%} \]
05

Discuss the Difference in Power

The small difference in power (1.2%) is likely due to internal losses such as friction and heat within the engine components and transmission losses.
06

Comment on Transportation Needs

For a typical V-6 engine developing 226 hp and transferring about 98.8% of its power to the driveshaft, this engine is generally powerful enough for most standard transportation needs such as daily commuting and highway driving.

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Key Concepts

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

Torque and Rotational Speed
Torque and rotational speed are fundamental concepts in understanding engine power calculations. Torque is the measure of the force that can cause an object to rotate about an axis. It is calculated in units like pound-feet (ft路lb).
Rotational speed, on the other hand, is how fast the driveshaft or any rotating component is spinning. It is typically measured in revolutions per minute (RPM).
To get a clearer picture, imagine turning a wrench to tighten a bolt. The force you apply on the wrench is torque, and how fast you can make the bolt rotate is the rotational speed.
When we combine these two鈥攖orque and rotational speed鈥攚e can calculate the power that an engine delivers. The power is basically a product of how much torque an engine can apply and how fast it can spin the driveshaft.
Mathematically, we use the formula:
\[ P_{\text{driveshaft}} = \text{Torque} \times \text{Rotational Speed} \] In this given exercise, we converted the rotational speed from RPM to radians per second to match the standard units in the formula. The conversion uses:
\[ \text{Rotational Speed} (\text{rad/sec}) = \frac{2\text{蟺} \times \text{RPM}}{60} \] This ensures consistency in the units, making the calculations accurate.
Horsepower Conversion
Understanding horsepower conversion is crucial for engine power calculations. Horsepower (hp) is a unit of measure for power, which gauges how much work an engine can perform over a period of time. Historically, 1 horsepower was defined as the power needed to lift 550 pounds by one foot in one second.
In the context of this exercise, the power derived from torque and rotational speed is initially calculated in foot-pounds per second (ft路lb/sec). To convert this to horsepower, a conversion factor is used: 1 hp = 550 ft路lb/sec.
This conversion is necessary because engine power is typically expressed in horsepower rather than ft路lb/sec, making it easier to understand and compare engine output.
Here's the conversion formula we used:
\[ P_{\text{driveshaft (hp)}} = \frac{P_{\text{driveshaft}}}{550} \] By dividing the calculated power in ft路lb/sec by 550, we get the equivalent power in horsepower. In this case, we found that the driveshaft transfers approximately 223.3 hp from the 226 hp generated by the engine.
Power Transfer Efficiency
Power transfer efficiency is about how effectively power generated by the engine gets transferred to the driveshaft. Efficiency is critical because some power can be lost due to factors such as friction, heat, and mechanical inefficiencies within the engine components.
To determine efficiency, the power transferred to the driveshaft is compared to the total power generated by the engine. The formula is:
\[ \text{Percentage} = \frac{P_{\text{driveshaft}} \times 100}{P_{\text{eng}}} \] For instance, in this exercise, we calculated the efficiency to be approximately 98.8%. This means 98.8% of the engine's power is effectively reaching the driveshaft, while 1.2% is lost.
These losses could be due to:
  • Friction within the engine and transmission components
  • Heat produced during combustion and transmission
Understanding and improving power transfer efficiency is essential for designing more efficient and powerful engines.
Notably, for most general transportation needs, an efficiency of 98.8% is remarkably high, indicating a well-engineered engine configuration.

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Most popular questions from this chapter

A heat pump cycle delivers energy by heat transfer to a dwelling at a rate of \(40,000 \mathrm{Btu} / \mathrm{h}\). The coefficient of performance of the cycle is \(2.8\). (a) Determine the power input to the cycle, in hp. (b) Evaluating electricity at \(\$ 0.085\) per \(\mathrm{kW} \cdot \mathrm{h}\), determine the cost of electricity during the heating season when the heat pump operates for 2000 hours.

Carbon dioxide \(\left(\mathrm{CO}_{2}\right)\) gas within a piston-cylinder assembly undergoes a process from a state where \(p_{1}=\) \(5 \mathrm{lbf} / \mathrm{in}^{2}, V_{1}=2.5 \mathrm{ft}^{3}\) to a state where \(p_{2}=20 \mathrm{lbf} / \mathrm{in}^{2}, V_{2}=\) \(0.5 \mathrm{ft}^{3}\). The relationship between pressure and volume during the process is given by \(p=23.75-7.5 \mathrm{~V}\), where \(V\) is in \(\mathrm{ft}^{3}\) and \(p\) is in \(\mathrm{lbf} / \mathrm{in}^{2}\) Determine the work for the process, in Btu.

A gas is contained in a vertical piston-cylinder assembly by a piston with a face area of 40 in. \(^{2}\) and weight of \(100 \mathrm{lbf}\). The atmosphere exerts a pressure of \(14.7 \mathrm{lbf} / \mathrm{in}^{2}\) on top of the piston. A paddle wheel transfers 3 Btu of energy to the gas during a process in which the elevation of the piston increases by \(1 \mathrm{ft}\). The piston and cylinder are poor thermal conductors, and friction between them can be neglected. Determine the change in internal energy of the gas, in Btu.

Oxygen \(\left(\mathrm{O}_{2}\right)\) gas within a piston-cylinder assembly undergoes an expansion from a volume \(V_{1}=0.01 \mathrm{~m}^{3}\) to a volume \(V_{2}=0.03 \mathrm{~m}^{3}\). The relationship between pressure and volume during the process is \(p=A V^{-1}+B\), where \(\mathrm{A}=\) \(0.06\) bar \(\cdot \mathrm{m}^{3}\) and \(B=3.0\) bar. For the \(\mathrm{O}_{2}\), determine (a) the initial and final pressures, each in bar, and (b) the work, in \(\mathrm{kJ}\).

A heat pump cycle operating at steady state receives energy by heat transfer from well water at \(10^{\circ} \mathrm{C}\) and discharges energy by heat transfer to a building at the rate of \(1.2 \times 10^{5} \mathrm{~kJ} / \mathrm{h}\). Over a period of 14 days, an electric meter records that \(1490 \mathrm{~kW}\) - \(\mathrm{h}\) of electricity is provided to the heat pump. These are the only energy transfers involved. Determine (a) the amount of energy that the heat pump receives over the 14-day period from the well water by heat transfer, in \(\mathrm{kJ}\), and (b) the heat pump's coefficient of performance.
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