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

or, on a unit mass basis,

(2–3)

where Vdenotes the velocity of the system relative to some fixed reference

frame. The kinetic energy of a rotating solid body is given by Iv^2 where I

is the moment of inertia of the body and vis the angular velocity.

The energy that a system possesses as a result of its elevation in a gravita-

tional field is called potential energy(PE) and is expressed as

(2–4)

or, on a unit mass basis,

(2–5)

where gis the gravitational acceleration and zis the elevation of the center

of gravity of a system relative to some arbitrarily selected reference level.

The magnetic, electric, and surface tension effects are significant in some

specialized cases only and are usually ignored. In the absence of such

effects, the total energy of a system consists of the kinetic, potential, and

internal energies and is expressed as

(2–6)

or, on a unit mass basis,

(2–7)

Most closed systems remain stationary during a process and thus experi-

ence no change in their kinetic and potential energies. Closed systems

whose velocity and elevation of the center of gravity remain constant during

a process are frequently referred to as stationary systems.The change in

the total energy Eof a stationary system is identical to the change in its

internal energy U. In this text, a closed system is assumed to be stationary

unless stated otherwise.

Control volumes typically involve fluid flow for long periods of time, and

it is convenient to express the energy flow associated with a fluid stream in

the rate form. This is done by incorporating the mass flow ratem

.

, which is

the amount of mass flowing through a cross section per unit time. It is

related to the volume flow rateV

.

, which is the volume of a fluid flowing

through a cross section per unit time, by

Mass flow rate: (2–8)

which is analogous to m
rV. Here ris the fluid density,Acis the cross-

sectional area of flow, and Vavgis the average flow velocity normal to Ac.

The dot over a symbol is used to indicate time ratethroughout the book.

Then the energy flow rate associated with a fluid flowing at a rate of m

.

is

(Fig. 2–4)

Energy flow rate: (2–9)

which is analogous to E
me.

E

#


m

#

e¬¬ 1 kJ>s or kW 2

m#
rV

#


rAcVavg¬¬ 1 kg>s 2

e
ukepe
u

V^2

2

gz¬¬ 1 kJ>kg 2

E
UKEPE
Um

V^2

2

mgz¬¬ 1 kJ 2

pe
gz¬¬ 1 kJ>kg 2

PE
mgz¬¬ 1 kJ 2

1

2

ke

V^2

2

¬¬ 1 kJ>kg 2

54 | Thermodynamics

D

Steam

Vavg

Ac = pD^2 /4

m = rAcVavg

E = me

• 
  


• •

FIGURE 2–4

Mass and energy flow rates associated
with the flow of steam in a pipe of
inner diameter Dwith an average
velocity of Vavg.