Chapter 2.8, Problem 52P

To determine

The additional power required to achieve the desired acceleration.

The additional power required to achieve the desired acceleration when the total mass of the car is $\text{700}\text{kg}$.

Answer to Problem 52P

The additional power required to achieve the desired acceleration is $\text{77}\text{.8}\text{kW}$.

The additional power required to achieve the desired acceleration when the total mass of the car is $\text{700}\text{kg}$ is $\text{38}\text{.9}\text{kW}$.

Explanation of Solution

Write the energy balance equation for a control volume when the car cruising is taken as the system.

${\dot{E}}_{\text{in}}-{\dot{E}}_{\text{out}}=d{E}_{\text{system}}/dt$ (I)

Here, rate of energy transfer entering the system is ${\dot{E}}_{\text{in}}$, rate of energy transfer leaving the system is ${\dot{E}}_{\text{out}}$, rate of change in internal, kinetic, and potential energies per unit time is $d{E}_{\text{system}}/dt$.

Since no energy is leaving the system, the rate of energy transfer leaving the system is zero.

${\dot{E}}_{\text{out}}=0$

Substitute 0 for ${\dot{E}}_{\text{out}}$ in Equation (I).

$\begin{array}{l}{\dot{E}}_{\text{in}}-0=d{E}_{\text{system}}/dt\\ {\dot{E}}_{\text{in}}=d{E}_{\text{system}}/dt\\ {\dot{E}}_{\text{in}}=\frac{\Delta E_{\text{sys}}}{\Delta t}\end{array}$ (II)

Here, change in the internal energy of the system is $\Delta E_{\text{sys}}$ and change in time is $\Delta t$.

Rewrite Equation (II) when rate of work done on the system.

$\begin{array}{c}{\dot{W}}_{\text{in}}=\frac{\Delta KE}{\Delta t}\\ =\frac{m\left(V_2^2-V_1^2\right)}{2\Delta t}\end{array}$ (III)

Here, change in kinetic energy is $\Delta KE$, additional power required to achieve the acceleration is ${\dot{W}}_{\text{in}}$, mass of the car is m and initial and final velocities of the car are $V_1$ and $V_2$ respectively.

Conclusion:

Substitute $1400\text{kg}$ for m, $110\text{km}/\text{h}$ for $V_2$, $70\text{km}/\text{h}$ for $V_1$, and $5\text{s}$ for $\Delta t$ in Equation (III).

$\begin{array}{c}{\dot{W}}_{\text{in}}=\frac{\left(1400\text{kg}\right)\left({\left(110\text{km}/\text{h}\right)}^2-{\left(700\text{km}/\text{h}\right)}^2\right)}{2\left(5\text{s}\right)}\\ =\frac{\left(1400\text{kg}\right)\left({\left(110\text{km}/\text{h}\times \frac{1\text{m}/\text{s}}{3.6\text{km}/\text{h}}\right)}^2-{\left(70\text{km}/\text{h}\times \frac{1\text{m}/\text{s}}{3.6\text{km}/\text{h}}\right)}^2\right)}{\left(10\text{s}\right)}\\ =\frac{\left(1400\text{kg}\right)\left(555.56{\text{m}}^2/{\text{s}}^2\right)}{\left(10\text{s}\right)}\left(\frac{1\text{kJ}/\text{kg}}{1000{\text{m}}^2/{\text{s}}^2}\right)\\ =77.8\text{kJ}/\text{s}\end{array}$

$\begin{array}{c}=77.8\text{kJ}/\text{s}\left(\frac{1\text{kW}}{1\text{kJ}/\text{s}}\right)\\ =77.8\text{kW}\end{array}$

Thus, the additional power required to achieve the desired acceleration is $\text{77}\text{.8}\text{kW}$.

Substitute $700\text{kg}$ for m, $110\text{km}/\text{h}$ for $V_2$, $70\text{km}/\text{h}$ for $V_1$, and $5\text{s}$ for $\Delta t$ in Equation (III).

$\begin{array}{c}{\dot{W}}_{\text{in}}=\frac{\left(700\text{kg}\right)\left({\left(110\text{km}/\text{h}\right)}^2-{\left(70\text{km}/\text{h}\right)}^2\right)}{2\left(5\text{s}\right)}\\ =\frac{\left(700\text{kg}\right)\left({\left(110\text{km}/\text{h}\times \frac{1\text{m}/\text{s}}{3.6\text{km}/\text{h}}\right)}^2-{\left(70\text{km}/\text{h}\times \frac{1\text{m}/\text{s}}{3.6\text{km}/\text{h}}\right)}^2\right)}{\left(10\text{s}\right)}\\ =\frac{\left(700\text{kg}\right)\left(555.56{\text{m}}^2/{\text{s}}^2\right)}{\left(10\text{s}\right)}\left(\frac{1\text{kJ}/\text{kg}}{1000{\text{m}}^2/{\text{s}}^2}\right)\\ =38.9\text{kJ}/\text{s}\end{array}$

$\begin{array}{c}=38.9\text{kJ}/\text{s}\left(\frac{1\text{kW}}{1\text{kJ}/\text{s}}\right)\\ =38.9\text{kW}\end{array}$

Thus, additional power required to achieve the desired acceleration when the total mass of the car is $\text{700}\text{kg}$ is $\text{38}\text{.9}\text{kW}$.