Chapter 2.8, Problem 50P

Consider a 2100-kg car cruising at constant speed of 70 km/h. Now the car starts to pass another car by accelerating to 110 km/h in 5 s. Determine the additional power needed to achieve this acceleration. What would your answer be if the total mass of the car were only 700 kg?

To determine

The additional power needed to achieve the acceleration.

If the total mass of the car is 700 kg, what would be the answer in order to achieve the acceleration?

Answer to Problem 50P

The additional power needed to achieve the acceleration is $\underset{\_}{\text{117}\text{kW}}$.

If the total mass of the car is 700 kg, the answer in order to achieve the acceleration is $\underset{\_}{\text{38}\text{.9}\text{kW}}$.

Explanation of Solution

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

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

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

Since no energy is leaving the room, ${\dot{E}}_{\text{out}}=0$.

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

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, velocity at inlet and outlet is $V_1$ and $V_2$.

If the total mass of the car is 700 kg, calculate the additional power needed to achieve the acceleration.

${\dot{W}}_{\text{in}}=\frac{m\left(V_2^2-V_1^2\right)}{2\Delta t}$ (IV)

Conclusion:

Substitute 2100 kg for m, 110 km/h for $V_2$, 70 km/h for $V_1$, and 5 s for $\Delta t$ in Equation (III).\

$\begin{array}{c}{\dot{W}}_{\text{in}}=\frac{2100\text{kg}\left({\left(110\frac{\text{km}}{\text{h}}\right)}^2-{\left(70\frac{\text{km}}{\text{h}}\right)}^2\right)}{2\times 5\text{s}}\\ =\frac{2100\text{kg}\left[7200\frac{{\text{km}}^{\text{2}}}{{\text{h}}^{\text{2}}}\right]}{10\text{s}}\\ =210\text{kg}\times 7200\frac{{\text{km}}^{\text{2}}\times \frac{{\left(1000\text{m}\right)}^2}{{\text{km}}^{\text{2}}}}{{\text{h}}^{\text{2}}\times \frac{{\left(3600\text{s}\right)}^2}{{\text{h}}^{\text{2}}}\cdot \text{s}}\\ =1512000\text{kg}\frac{\left({10}^6{\text{m}}^{\text{2}}\right)}{12.96\times {10}^6{\text{s}}^{\text{3}}}\end{array}$

               $\begin{array}{c}=116,666.7\frac{\text{kg}\cdot {\text{m}}^{\text{2}}}{{\text{s}}^{\text{3}}}\left(\frac{1\text{kJ}}{1000{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}\right)\\ =117\frac{\text{kJ}}{\text{s}}\times \left(\frac{\text{kW}}{\frac{\text{kJ}}{\text{s}}}\right)\\ =117\text{kW}\end{array}$

Thus, the additional power needed to achieve the acceleration is $\underset{\_}{\text{117}\text{kW}}$.

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

$\begin{array}{c}{\dot{W}}_{\text{in}}=\frac{700\text{kg}\left({\left(110\frac{\text{km}}{\text{h}}\right)}^2-{\left(70\frac{\text{km}}{\text{h}}\right)}^2\right)}{2\times 5\text{s}}\\ =\frac{700\text{kg}\left[7200\frac{{\text{km}}^{\text{2}}}{{\text{h}}^{\text{2}}}\right]}{10\text{s}}\\ =70\text{kg}\times 7200\frac{{\text{km}}^{\text{2}}\times \frac{{\left(1000\text{m}\right)}^2}{{\text{km}}^{\text{2}}}}{{\text{h}}^{\text{2}}\times \frac{{\left(3600\text{s}\right)}^2}{{\text{h}}^{\text{2}}}\cdot \text{s}}\\ =\left(504000\text{kg}\right)\frac{\left({10}^6{\text{m}}^{\text{2}}\right)}{12.96\times {10}^6{\text{s}}^{\text{3}}}\end{array}$

               $\begin{array}{c}=38,888.89\frac{\text{kg}\cdot {\text{m}}^{\text{2}}}{{\text{s}}^{\text{3}}}\left(\frac{1\text{kJ}}{1000{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}\right)\\ =38.9\frac{\text{kJ}}{\text{s}}\times \left(\frac{\text{kW}}{\frac{\text{kJ}}{\text{s}}}\right)\\ =38.9\text{kW}\end{array}$

Thus, if the total mass of the car is 700 kg, the answer in order to achieve the acceleration is $\underset{\_}{\text{38}\text{.9}\text{kW}}$.