Question

Review: The mass of a car is 1 500 kg. The shape of the car’s body is such that its aerodynamic drag coefficient is D = 0.330 and its frontal area is $2.50m^2$. Assuming the drag force is proportional to$v_2$and ignoring other sources of friction, calculate the power required to maintain a speed of 100 km/h as the car climbs a long hill sloping at$3{.2}^o$.

Short answer

The power required to maintain a speed of 100 km/h as the car climbs a long hill sloping at ${3.2}^o$is $P=44.8hp$.

Step-by-step solution

Concept

Nonisolated System (Energy) The most general statement describing the behavior of a non isolated system is the conservation of energy equation:

${\mathbf{\Delta E}}_{\mathrm{system}}\mathbf{=}\sum \mathbf{T}...................................(\mathbf{8}\mathbf{.}\mathbf{1})$

$\mathbf{\Delta K}\mathbf{+}\mathbf{\Delta U}\mathbf{+}{\mathbf{\Delta E}}_{\mathrm{int}}\mathbf{=}\mathbf{W}\mathbf{+}\mathbf{Q}\mathbf{+}{\mathbf{T}}_{\mathrm{MW}}\mathbf{+}{\mathbf{T}}_{\mathrm{MT}}\mathbf{+}{\mathbf{T}}_{\mathrm{ET}}\mathbf{+}{\mathbf{T}}_{\mathrm{ER}}..................(\mathbf{8}\mathbf{.}\mathbf{2})$

Here all types of conservative or non-conservative interaction are also considered.

where,

$\Delta U=$Change in Potential energy

$\Delta K=$Change in kinetic energy

$\Delta E_{\mathrm{int}}=$Change in internal energy

Given Data

Convert the speed to metric units:

The velocity of car is $v=(100km/h)\left(\frac{1000m}{1km}\right)\left(\frac{1h}{3600s}\right)=27.8m/s$,

The mass of the car is-$m=1500\text{\hspace{0.17em}kg}$

Drag coefficient $D=0.330$,

The frontal area $A=2.50{\text{\hspace{0.17em}m}}^2$,

Slope of the hill $\theta =3.2°$,

Calculation

Write Equation 8.2 for this situation, treating the car and surrounding air as an isolated system with non conservative force acting:

$\Delta K+\Delta U_{grav}+\Delta U_{fuel}+\Delta E_{\mathrm{int}}=0$

The change in potential energy in the fuel, gives the energy that has changed to other forms-

$\Delta E_{otherforms}=-\Delta U_{fuel}$

Therefore, Power is given as,

$\begin{array}{c}P=-\frac{\Delta U_{fuel}}{\Delta t}\\ =-\frac{(-\Delta K-\Delta U_{grav}-\Delta E_{\mathrm{int}})}{\Delta t}\\ =\frac{0+(mgd\sin {3.2}^o-0)+\frac{1}{2}D\rho A{v}^{2}d}{\Delta t}\\ =mgv\sin {3.2}^o+\frac{1}{2}D\rho A{v}^3\end{array}$

where we have recognizedas$\frac{d}{\Delta t}$ the speed v of the car. Substituting given values,

$\begin{array}{c}P=\left[\begin{array}{l}(1500kg)(9.8m/s^2)(27.0m/s)sim{3.2}^o+\\ \frac{1}{2}(0.330m)(1.20kg/m^3)(2.50m^2){(27.0m/s)}^3\end{array}\right]\\ =33.4kW\\ =44.8hp\end{array}$

As there is loss in energy in overcoming the friction between the parts and the air drag, and rolling friction with the road, the power achieved will be less than required. In addition, some energy from the engine is radiated away as sound. Finally, some of the energy is used up in interacting with different sources.