Page 119 - Modelling in Transport Phenomena A Conceptual Approach

4.5. FLOW IN CIRCULAR PIPES 99

Solution

Physical properties
p = 999 kg/ m3
For water ut 20 "C (293 K) :
I-1 = lool 10-6 kg/ m.
Analysis
Substitution of the given values into Eq. (4.5-19) yields

n2D5 lAPl
32 pL

7r2 (0.2)5 (40)
= \i (32)(999) = 1.99 x 10-~
Hence, Eq. (4.5-20) gives the volumetric flow mte as

Q=-lYlog(*+@) 3.7065 pY

(1001 x io-
= - (4)(1.99 x log [ (4.6 x 10-5/0.2) + (999)(1.99 x 10-3)
3.7065
= 0.03 m3/ s
H Calculate D; given Q and lAPl
Swamee and Jain (1976) and Cheng and Turton (1990) presented explicit equa-
tions to solve problems of this type. These equations, however, are unnecessarily
complex. A simpler equation can be obtained by using the procedure suggested by
Tosun and Akgahin (1993) as follows. Equation (4.56) can be rearranged in the
form
f = (DN)5 (4.521)
where N is defined by

(4.522)
For turbulent flow, the value of f changes between 0.00025 and 0.01925. Using an
average value of 0.01 for f gives a relationship between D and N as
0.4
D=- (4.5-23)
N
Substitution of Eq. (4.521) to the left-hand side of Eq. (4.59), and substitution
of Eqs. (4.5-10) and (4.5-23) to the right-hand side of Eq. (4.5-9) gives
D = - [log(sN) + 5.806 ( - 0.l'i'lY) -115

I
'
)
({
0.575
N PQN (4.524)