4–2 ENERGY BALANCE FOR CLOSED SYSTEMS
Energy balance for any system undergoing any kind of process was
expressed as (see Chap. 2)
(4 –11)
Net energy transfer Change in internal, kinetic,
by heat, work, and mass potential, etc., energies
or, in the rate form,as
(4 –12)
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
(4 –13)
The energy balance can be expressed on a per unit massbasis as
(4 –14)
which is obtained by dividing all the quantities in Eq. 4 –11 by the mass m
of the system. Energy balance can also be expressed in the differential
form as
(4 –15)
For a closed system undergoing a cycle,the initial and final states are iden-
tical, and thus EsystemE 2 E 1 0. Then the energy balance for a cycle
simplifies to EinEout0 or EinEout. 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
(4 –16)
That is, the net work output during a cycle is equal to net heat input
(Fig. 4 –11).
Wnet,outQnet,in¬or¬W
#
net,outQ
#
net,in¬¬^1 for a cycle^2
dEindEoutdEsystem¬or¬deindeoutdesystem
eineout¢esystem¬¬ 1 kJ>kg 2
QQ
#
¢t,¬WW
#
¢t,¬and¬¢E 1 dE>dt 2 ¢t¬¬ 1 kJ 2
E
.
inE
.
out¬¬dEsystem>dt¬¬^1 kW^2
EinEout¬ ¬ ¢Esystem¬¬ 1 kJ 2
Chapter 4 | 173
P
V
Qnet = Wnet
FIGURE 4 –11
For a cycle E0, thus QW.
Note that the work is done by the system.
(c) The work represented by the rectangular area (region I) is done against
the piston and the atmosphere, and the work represented by the triangular
area (region II) is done against the spring. Thus,
Discussion This result could also be obtained from
Wspring^12 k 1 x^22 x^212 ^121 150 kN>m 231 0.2 m 22 02 4a
1 kJ
1 kN#m
b3 kJ
Wspring^1231320 2002 kPa 41 0.05 m^3 2a
1 kJ
1 kPa#m^3
b3 kJ
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SEE TUTORIAL CH. 4, SEC. 2 ON THE DVD.
INTERACTIVE
TUTORIAL