Automotive Engine’s Power At full throttle, a four-cylinder automobile engine produces 149 ft ? lb of torque at a speed of 3600 rpm. In the units of hp, how much power does the engine produce? Approach We can use Equation (8.5) to fi nd the engine’s instantaneous power output. As an intermediate

Question

Automotive Engine’s Power

At full throttle, a four-cylinder automobile engine produces 149 ft.lb of torque at a speed of 3600 rpm. In the units of hp, how much power does the engine produce?

Approach
We can use Equation (8.5)

[latex] P = T\omega (torque) [/latex] (8.5)

to find the engine’s instantaneous power output. As an intermediate step, the rotational speed first must be converted into the consistent units of rad/s by applying the conversion factor 1 rpm = 0.1047 rad/s from Table 8.1.

Table 8.1
Conversion Factors between Various Units for Angular Velocity

rpm	rps	deg/s	rad/s
1	[latex]1.667×10^{-2}[/latex]	6	0.1047
60	1	360	6.283
0.1667	[latex]2.777×10^{-3}[/latex]	1	[latex]1.745×10^{-2}[/latex]
9.549	0.1592	57.30	1

Accepted Answer

In dimensionally consistent units, the engine’s speed is

[latex] \omega = (3600 rpm) \left(0.1047 \frac{rad/s}{rpm} ight) [/latex]

[latex]= 376.9(\cancel{rpm})\left(\frac{rad/s}{\cancel{rpm}} ight) [/latex]

[latex]= 376.9 \frac{rad}{s} [/latex]

The power that the engine develops is

[latex] P =(149 ft . lb)\left(376.9 \frac{rad}{s} ight) \longleftarrow \left[P = T\omega ight] [/latex]

[latex] =5.616 imes 10^4 (ft. lb)\left(\frac{\cancel{rad}}{s} ight) [/latex]

[latex]=5.616 imes 10^4\frac{ft.lb}{s} [/latex]

Here we eliminated the radian unit in the angular velocity term because it is dimensionless. Although (ft.lb)/s is an acceptable dimension for power in the USCS, the derived unit of horsepower is more conventional when expressing the power in engines and automotive applications. In the final step, we apply the conversion factor from Table 7.2:

Table 7.2 Conversion Factors Between Various Units for Power in the USCS and SI

(ft . lb)/s	W	hp
1	1.356	[latex]1.818×10^{-3}[/latex]
0.7376	1	[latex]1.341×10^{-3}[/latex]
550	745.7	1

[latex] P=\left(5.616 imes 10^4 \frac{ft. lb}{s} ight)\left(1.818 imes 10^{-3} \frac{hp}{(ft.lb)/s} ight) [/latex]

[latex]=102.1 \left(\cancel{\frac{ft.lb}{s}} ight)\left(\frac{hp}{\cancel{(ft.lb)/s}} ight) [/latex]

[latex]=102.1 hp [/latex]

Discussion
This value is typical for the maximum power that an automotive engine can develop. In normal driving situations, however, the engine’s power output is considerably less and will be determined by the throttle setting, the engine’s speed, and the torque transferred to the transmission.

P = 102.1 hp

Question 8.3: Automotive Engine’s Power At full throttle, a four-cylinder ...

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