Page 485 - Bird R.B. Transport phenomena

§15.5 Use of the Macroscopic Balances to Solve Unsteady-State Problems 465

or

f
-In ^ + 513 ^j - 1 + 132 = 0 (15.4-20)
By solving this equation with p } = 100 psia, we obtain p 2 = 86 psia.
(b) Flow through the compressor. We are now ready to apply the mechanical energy bal-
ance to the compressor. We start by putting Eq. 15.2-2 into the form

2
= \{v\ -v 2) + j-dp (15.4-21)
W m + E v
P
J P2
To evaluate the integral in this equation, we assume that the compression is adiabatic and
further that E v between planes 2 and 3 can be neglected. We may use Eq. 15.2-5 to rewrite Eq.
15.2-21 as

in which W M is the energy required of the compressor. By substituting numerical values into
Eq. 15.4-22, we get

-
= 9 + 7834 = 7825 ft lb //lb m (15.4-23)
The power required to compress the fluid is

= 277,000 ft %/s = 504 hp (15.4-24)
The power required would be virtually the same if the flow in the pipeline were adiabatic (see
Problem 15A.2).
The assumptions used here—assuming the compression to be adiabatic and neglecting
the viscous dissipation—are conventional in the design of compressor-cooler combinations.
Note that the energy required to run the compressor is greater than the calculated work, W m,
by (i) E v between planes 2 and 3, (ii) mechanical losses in the compressor itself, and (iii) errors
in the assumed p-p path. Normally the energy required at the pump shaft is at least 15 to 20%
greater than W nr

USE OF THE MACROSCOPIC BALANCES TO SOLVE
UNSTEADY-STATE PROBLEMS AND PROBLEMS
WITH NONFLAT VELOCITY PROFILES
In Table 15.5-1 we summarize all five macroscopic balances for unsteady state and non-
flat velocity profiles, and for systems with multiple entry and exit ports. One practically
never needs to use these balances in this degree of completeness, but it is convenient to
have the entire set of equations collected in one place. We illustrate their use in the ex-
amples that follow.