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

74 | Thermodynamics

Control

volume Q

Mass

in

Mass

out

W

FIGURE 2–45

The energy content of a control
volume can be changed by mass flow
as well as heat and work interactions.

P

V

Qnet = Wnet

FIGURE 2–46

For a cycle E
0, thus Q
W.

Wsh, in = 100 kJ

U 1 = 800 kJ

Qout = 500 kJ

U 2 =?

Fluid

FIGURE 2–47

Schematic for Example 2–10.

⎭⎪⎬⎪⎫

⎭⎪⎪⎬⎪⎪⎫

⎭⎪⎬⎪⎫

⎭⎪⎪⎬⎪⎪⎫

Energy balance for any system undergoing any kind of process can be

expressed more compactly as

(2–35)

Net energy transfer Change in internal, kinetic,

by heat, work, and mass potential, etc., energies

or, in the rate form,as

(2–36)

Rate of net energy transfer Rate of change in internal,

by heat, work, and mass kinetic, potential, etc., energies

For constant rates, the total quantities during a time interval tare related to

the quantities per unit time as

(2–37)

The energy balance can be expressed on a per unit massbasis as

(2–38)

which is obtained by dividing all the quantities in Eq. 2–35 by the mass mof

the system. Energy balance can also be expressed in the differential form as

(2–39)

For a closed system undergoing a cycle,the initial and final states are iden-

tical, and thus Esystem
E 2 E 1 
0. Then the energy balance for a cycle

simplifies to EinEout
0 or Ein
Eout. Noting that a closed system does

not involve any mass flow across its boundaries, the energy balance for a

cycle can be expressed in terms of heat and work interactions as

(2–40)

That is, the net work output during a cycle is equal to net heat input (Fig.

2–46).

EXAMPLE 2–10 Cooling of a Hot Fluid in a Tank

A rigid tank contains a hot fluid that is cooled while being stirred by a pad-

dle wheel. Initially, the internal energy of the fluid is 800 kJ. During the

cooling process, the fluid loses 500 kJ of heat, and the paddle wheel does

100 kJ of work on the fluid. Determine the final internal energy of the fluid.

Neglect the energy stored in the paddle wheel.

Solution A fluid in a rigid tank looses heat while being stirred. The final

internal energy of the fluid is to be determined.

Assumptions 1 The tank is stationary and thus the kinetic and potential

energy changes are zero, KE 
PE 
0. Therefore, E
Uand internal

energy is the only form of the system’s energy that may change during this

process. 2 Energy stored in the paddle wheel is negligible.

Analysis Take the contents of the tank as the system(Fig. 2–47). This is a

closed systemsince no mass crosses the boundary during the process. We

observe that the volume of a rigid tank is constant, and thus there is no

moving boundary work. Also, heat is lost from the system and shaft work is

done on the system. Applying the energy balance on the system gives

Wnet,out
Qnet,in¬or¬W

#

net,out
Q

#

net,in¬¬^1 for a cycle^2

dEindEout
dEsystem¬or¬deindeout
desystem

eineout
¢esystem¬¬ 1 kJ>kg 2

Q
Q

#

¢t,¬W
W

#¬

¢t,¬and¬¢E
 1 dE>dt 2 ¢t¬¬ 1 kJ 2

E

.

inE

.

out¬
¬dEsystem>dt¬¬^1 kW^2

EinEout¬
¬ ¢Esystem¬¬ 1 kJ 2