Thermodynamics: An Engineering Approach
Thermodynamics: An Engineering Approach
9th Edition
ISBN: 9781259822674
Author: Yunus A. Cengel Dr., Michael A. Boles
Publisher: McGraw-Hill Education
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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?

Expert Solution & Answer
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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 117kW_.

If the total mass of the car is 700 kg, the answer in order to achieve the acceleration is 38.9kW_.

Explanation of Solution

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

E˙inE˙out=dEsystem/dt (I)

Here, the rate of energy transfer entering and leaving the system is E˙in and E˙out, rate of change in internal, kinetic, and potential energy per unit time is dEsystem/dt.

Since no energy is leaving the room, E˙out=0.

Substitute 0 for E˙out in Equation (I).

E˙in0=dEsystem/dtE˙in=dEsystem/dtE˙in=ΔEsysΔt (II)

Here, the change in the internal energy of the system is ΔEsys.

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

W˙in=ΔKEΔt=m(V22V12)2Δt (III)

Here, velocity at inlet and outlet is V1 and V2.

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

W˙in=m(V22V12)2Δt (IV)

Conclusion:

Substitute 2100 kg for m, 110 km/h for V2, 70 km/h for V1, and 5 s for Δt in Equation (III).\

W˙in=2100kg((110kmh)2(70kmh)2)2×5s=2100kg[7200km2h2]10s=210kg×7200km2×(1000m)2km2h2×(3600s)2h2s=1512000kg(106m2)12.96×106s3

               =116,666.7kgm2s3(1kJ1000m2/s2)=117kJs×(kWkJs)=117kW

Thus, the additional power needed to achieve the acceleration is 117kW_.

Substitute 700 kg for m, 110 km/h for V2, 70 km/h for V1, and 5 s for Δt in Equation (III).\

W˙in=700kg((110kmh)2(70kmh)2)2×5s=700kg[7200km2h2]10s=70kg×7200km2×(1000m)2km2h2×(3600s)2h2s=(504000kg)(106m2)12.96×106s3

               =38,888.89kgm2s3(1kJ1000m2/s2)=38.9kJs×(kWkJs)=38.9kW

Thus, if the total mass of the car is 700 kg, the answer in order to achieve the acceleration is 38.9kW_.

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Chapter 2 Solutions

Thermodynamics: An Engineering Approach

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