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.lb at 3900 rpm. (a) Calculate the torque, in both ft.lb and N.m, at 5300 rpm. Is your answer in ft.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.lb at a particular angular velocity in rpm is often written as hp = [torque (in ft.lb) x rpm]/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.lb) at 6000 rpm?

It is given that the maximum power produced by the engine as $P_{max}=285\mathrm{hp}$at 5300rpm and the maximum torque produced by the engine as ${\tau }_{max}=305\mathrm{ft}.\mathrm{lb}$ at 3900rpm.

The relationship between power in hp and torque in ft.lb at a particular angular velocity in rpm can be written as $hp=\frac{torque\times rpm}{c}$, where c is a constant.

The torque produced by the engine is given by the formula $\tau =\frac{P_{max}}{\omega }$. Calculate $\tau$ as follows:

$$\begin{gathered} \tau =\frac{P_{max}}{\omega } \\ =\frac{\left(285\mathrm{hp}\right)\left(\frac{745.7\mathrm{W}}{1\mathrm{hp}}\right)}{\left(5300\mathrm{rpm}\right)\left(\frac{2\pi \mathrm{rad}}{1\mathrm{rev}}\right)\left(\frac{1\min }{60\mathrm{s}}\right)} \\ =\frac{212524.5\mathrm{W}}{\left(530\mathrm{rpm}\right)\left(\frac{\pi \mathrm{rad}}{1\mathrm{rev}}\right)\left(\frac{1\min }{3\mathrm{s}}\right)} \\ =382.9\mathrm{N}\cdot \mathrm{m} \end{gathered}$$

Convert the torque into the units of $ft·lb$as follows:

$$\begin{gathered} \tau =382.9\mathrm{N}\cdot \mathrm{m}\left(\frac{1\mathrm{b}}{4.44822\mathrm{N}}\right)\left(\frac{1\mathrm{ft}}{0.3048\mathrm{m}}\right) \\ =382.9\mathrm{N}\cdot \mathrm{m}\left(\frac{1\mathrm{ft}\cdot \mathrm{b}}{1.35581746\mathrm{N}\cdot \mathrm{m}}\right) \\ =\frac{382.9\mathrm{ft}\cdot \mathrm{lb}}{1.35581746} \\ =282\mathrm{ft}\cdot \mathrm{b} \end{gathered}$$

Therefore, the value of torque is $\tau =382.9N·m,\tau =282ft·lb$and the torque in $ft·lb$ is less than the maximum specified value.

The power produced by the engine is given by $P={\tau }_{max}\omega$. Calculate $P$ as follows:

$$\begin{gathered} P=(305\mathrm{ft}.\mathrm{lb})\left(\frac{4.44822\mathrm{N}}{1\mathrm{b}}\right)\left(\frac{0.3048\mathrm{m}}{1\mathrm{ft}}\right)\left(3900\mathrm{rpm}\right)\left(\frac{2\pi \mathrm{rad}}{1\mathrm{rev}}\right)\left(\frac{1\min }{60\mathrm{s}}\right) \\ =(305\mathrm{ft}.\mathrm{lb})\left(\frac{1.35581746\mathrm{N}\cdot \mathrm{m}}{1\mathrm{ft}\cdot \mathrm{lb}}\right)\left(130\mathrm{rpm}\right)\left(\frac{\pi \mathrm{rad}}{1\mathrm{rev}}\right)\left(\frac{1\min }{1\mathrm{s}}\right) \\ =1.69\times {10}^5\mathrm{W} \end{gathered}$$

Convert the power into the units of hp as follows:

$$\begin{gathered} P=1.69\times {10}^{5}W\left(\frac{1hp}{745.7W}\right) \\ =226hp \end{gathered}$$

Therefore, the required value of the power is $P=1.69\times {10}^{5}W$, $P=226\mathrm{hp}$ and the power in hp is smaller than the specified maximum value.

The power produced by a rotating engine is given by .$P=\tau \omega$ Find the power in hp, torque in ft.lb and angular velocity in rpm as follows:

$$\begin{gathered} 1\mathrm{hp}=\tau (\mathrm{ft}\cdot \mathrm{lb})\left(\frac{4.44822\mathrm{N}}{1\mathrm{b}}\right)\left(\frac{0.3048\mathrm{m}}{1\mathrm{ft}}\right)\omega \left(\mathrm{rmm}\right)\left(\frac{2\pi \mathrm{rad}}{1\mathrm{rev}}\right)\left(\frac{1\min }{60\mathrm{s}}\right)\left(\frac{1\mathrm{hp}}{745.7\mathrm{W}}\right) \\ =\tau (\mathrm{ft}\cdot \mathrm{lb})\left(\frac{4.44822\mathrm{N}}{1\mathrm{b}}\right)\left(\frac{0.3048\mathrm{m}}{1\mathrm{ft}}\right)\omega \left(\mathrm{rpm}\right)\left(\frac{2\pi \mathrm{rad}}{1\mathrm{rev}}\right)\left(\frac{1\min }{60\mathrm{s}}\right)\left(\frac{1\mathrm{hp}}{745.7\mathrm{N}\cdot \mathrm{m}/\mathrm{s}}\right) \\ =\frac{\tau (\mathrm{ft}\cdot \mathrm{lb})\omega \left(\mathrm{rmm}\right)}{5254} \end{gathered}$$

Therefore, the value of constant is $c=5254$.

The torque produced by the engine is given by $\tau =\frac{5254P}{\omega }$. Here, $P=580\mathrm{hp}$ and $\omega =6000\mathrm{rpm}$ then,

$$\begin{gathered} \tau =\frac{5254\left(580\mathrm{hp}\right)}{6000\mathrm{rpm}} \\ =\frac{3047320\mathrm{hp}}{6000\mathrm{rpm}} \\ =508\mathrm{ft}.\mathrm{lb} \end{gathered}$$

Therefore, the torque produced by the engine is $\tau =508\mathrm{ft}.\mathrm{lb}$.