Page 852 - Fundamentals of Water Treatment Unit Processes : Physical, Chemical, and Biological

Appendix D: Fluid Mechanics—Reviews of Selected Topics 807

After substitution of Equation D.55 in D.57, with STP defined,
p(stp) is the standard pressure, 101,325 Pa, or 1.00 atm
" #
(k 1)=k T(stp) is the standard temperature of gas, 08C or 273.15 K
p 1 k p 2
P ¼ Qr 1 (D:58) V(stp) is the volume of gas at STP, which is 0.0224 m for
3
r k 1
1 p 1
mass of 1.00 mol
in which
Substituting Equation D.61 in Equation D.59 gives
Q is the flow of gas at given temperature and pressure
3
(m =s) " #
(k 1)=k
r is the density of gas at given temperature and pressure P(stp) ¼ Q(stp) r(stp) k p 2 1 (D:62)
3
(kg=m ) r 1 p 1 k 1 p 1
Now, if Q and r are for state ‘‘1,’’ defined here as the
Equation D.62 is one method to obtain P(stp). An equivalent
compressor intake condition, the r’s cancel, giving, after
form is obtained by the sequence of steps that follow. First,
applying the subscript ‘‘1’’ for Q,
multiply r(stp)=r 1 by MW=MW to give r(molar, stp)=
r(molar) 1 . Then, substitute from the gas law,
" #
(k 1)=k
k p 2
P ¼ Q 1 p 1 1 (D:59)
k 1 p 1
p 1 ¼ r(molar) RT 1 (D:63)
1
This form is workable and is the actual power of the com-
to give
pressions under the conditions stated, i.e., whatever the con-
ditions are that define states ‘‘1’’ and ‘‘2.’’
r(molar, stp) k
P(stp) ¼ Q(stp) r(molar) 1 RT 1
D.3.5.2 Power for Compressible Fluid Flow—In r(molar) 1 k 1
General " (k 1)=k #
p 2
The power expended for any kind of fluid compression is, in 1 (D:64)
p 1
general,
The r(molar) 1 terms cancel to give
P ¼ Q(gas) r(gas) w (D:60)
k
in which w is the work of compression or expansion per unit P(stp) ¼ Q(stp)r(molar, stp)ŠRT 1 k 1
½
mass, whether adiabatic, isothermal, or polytropic (J=kg) " (k 1)=k #
Equation D.60 applies to any kind of compression, p 2 1 (D:65)

whether adiabatic, isothermal, or polytropic in which w is p 1
known. It applies also to irreversible processes, assuming w
is known. and since
D.3.5.3 Power for Standard Temperature
n p(stp)
and Pressure r(molar, stp) ¼ ¼ (D:66)
V RT(stp)
In some cases, conversion to standard temperature and pres-
sure (stp) is needed. [As a note, ‘‘standard’’ conditions may be
substituting Equation D.66 in D.65 gives
defined in accordance with the conventions of a particular
group. There is not a universally adopted STP.] " #
(k 1)=k
A key principle for compressible fluid flow is that the mass n(stp) k p 2
P(stp) ¼ Q(stp) RT 1 1 (D:67)
flow is constant from section to section (for steady state con- V(stp) k 1 p 1
ditions). Therefore, the mass flow at any section, ‘‘1,’’ equals
the mass flow at any other section, ‘‘2.’’ If ‘‘2’’ is for ‘‘standard’’ 3
Now, since at STP, n ¼ 1.00 mol and V ¼ 0.0224 m ,
conditions, then, the continuity relation is
" #
(k 1)=k
Q 1 r ¼ Q(stp)r(stp) (D:61) 1:00 mol 8:321 J k p 2
1 P(stp) ¼ Q(stp) T 1 1
3
0:0224 m mol K k 1 p 1
in which
(D:68)
Q(stp) is the flow of gas at standard temperature and
3
pressure (m =s) " (k 1)=k #
k p 2
r(stp) is the density of gas at standard temperature and ¼ 321Q(stp)T 1 k 1 1 (D:69)
3
pressure (kg=m ) p 1