Page 121 - Chemical Process Equipment - Selection and Design

6.2. ENERGY BALANCE OF A FLOWING FLUID 93

EXAMPLE 6.2 = HI dnl - H2 dn, + dQ - dw,
Unsteady Flow of an Ideal Gas through a Vessel = Cp(6T, - 4T) dB + h(300 - T) de.
An ideal gas at 350 K is pumped into a 1000 L vessel at the rate of
6gmol/min anid leaves it at the rate of 4gmol/min. Initially the This rearranges into
vessel is at 310 K and 1 atm. Changes in velocity and elevation are
negligible. The contents of the vessel are uniform. There is no work dT
transfer. $~=I~(l/C,)[6CpT~+300h -(4Cp +2C, +h)T
Thermodynamic data:
U = C, T = 5T,
H=CpT=7T. r, dT h=0.
Heat transfer: The integrals are rearranged to find T,

dQ = h(300 - T) dl0
= 15(300 - T) do.

The temperature will be found as a function of time 6 with both
h = 15 and h = 0.
Some numerical values are
dn, = 6 do,
dn, = 4 de,
dn = dn, - dn, = 2 do, T2 P
n, = PflV/RTo = 1000/(0.08205)(310) = 39.32 gmol, B h=15 h=O h=45 h=O
n = no + 28, 0 31 0 310 4 1
0.2 312.7 312.9 7.02 1.02
V=?OOOP 0.5 316.5 317.0
1 322.1 323.2
T, = 350 5 346.5 354.4
n, = 6 10 356.4 370.8 1.73 1.80
co 362.26 386.84 00 00
d,=O
The pressures are calculated from
Energy balance
nRT (39.32 + 26)(0.08205)T
p=-=
d(nCJ) = n! dU + U dn = nC, dT + C, T(2 do) V 1000

Friction is introduced into the energy balance by noting that it For an incompressible fluid, integration may be performed
is a mechanical process, dWf, whose effect is the same as that of an term by term with the result
equivalent amount of heat transfer dQp Moreover, the total
effective heat transfer results in a change in entropy of the flowing AP/p + Au2/2gc + (g/g,)Az = -(K + (6.14)
liquid given by

TdS=dQ+dWf (6.11) The apparent number of variables in Eq. (6.13) is reduced by the
substitution u = V/A for unit flow rate of mass, where A is the
When the thermodynamic equivalent cross-sectional area. so that

dH = V dP + TdS (6.12) VdP + (l/gA2)VdV + (g/g,) dz = -(dW, + dq). (6.15)
and Eq. (6.11) are substituted into Eq. (6.10), the net result is Integration of these energy balances for compressible fluids under
several conditions is covered in Section 6.7.
VdP + (l/g,)U du + (g/g,) dz = -(dW, + dWf), (6.13) The frictional work loss Wf depends on the geometry of the
system and the flow conditions and is an empirical function that will
which is known as the mechanical energy balance. With the be explained later. When it is known, Eq. (6.13) may be used to
expression for friction of Eq. (6.18) cited in the next section, the find a net work effect W, for otherwise specified conditions.
mechanical energy balance becomes The first three terms on the left of Eq. (6.14) may be grouped
into a single stored energy terms as
fu"
VdP+(I./g~)udu+($/g,)dz+-dL= -dW,. (6.13')
2gJ AE = AP/p + Au2/2g, + (g/g,)Az, (6.16)