Microsoft Word - Cengel and Boles TOC _2-03-05_.doc

Chapter 2 | 69

Elevator

car

Motor

FIGURE 2–34

The energy transferred to a body while

being raised is equal to the change in

its potential energy.

90 km/h

30 °

m = 1200 kg

FIGURE 2–35

Schematic for Example 2–8.

m = 900 kg

0 80 km/h

FIGURE 2–36

Schematic for Example 2–9.

EXAMPLE 2–8 Power Needs of a Car to Climb a Hill

Consider a 1200-kg car cruising steadily on a level road at 90 km/h. Now

the car starts climbing a hill that is sloped 30° from the horizontal (Fig.

2–35). If the velocity of the car is to remain constant during climbing, deter-

mine the additional power that must be delivered by the engine.

Solution A car is to climb a hill while maintaining a constant velocity. The

additional power needed is to be determined.

Analysis The additional power required is simply the work that needs to be

done per unit time to raise the elevation of the car, which is equal to the

change in the potential energy of the car per unit time:

Discussion Note that the car engine will have to produce almost 200 hp of

additional power while climbing the hill if the car is to maintain its velocity.

EXAMPLE 2–9 Power Needs of a Car to Accelerate

Determine the power required to accelerate a 900-kg car shown in Fig. 2–36

from rest to a velocity of 80 km/h in 20 s on a level road.

Solution The power required to accelerate a car to a specified velocity is to

be determined.

Analysis The work needed to accelerate a body is simply the change in the

kinetic energy of the body,

The average power is determined from

Discussion This is in addition to the power required to overcome friction,

rolling resistance, and other imperfections.

Nonmechanical Forms of Work

The treatment in Section 2–5 represents a fairly comprehensive coverage of
mechanical forms of work except the moving boundary workthat is covered
in Chap. 4. But some work modes encountered in practice are not mechani-
cal in nature. However, these nonmechanical work modes can be treated in a
similar manner by identifying a generalized force Facting in the direction

W

#

a

Wa

¢t

222 kJ

20 s

11.1 kW¬¬ 1 or 14.9 hp 2

222 kJ

Wa
^12 m 1 V^22 V^212 
^121 900 kg2ca

80,000 m

3600 s

b

2

 02 da

1 kJ>kg

1000 m^2 >s^2

b

147 kJ>s
147 kW¬¬ 1 or 197 hp 2

 1 1200 kg 21 9.81 m>s^221 90 km>h 21 sin 30°2a

1 m>s

3.6 km>h

ba

1 kJ>kg

1000 m^2 >s^2

b

W

#

g
mg¬¢z>¢t
mgVvertical