Page 118 - Modelling in Transport Phenomena A Conceptual Approach

98 CHAPTER 4. EVALUATION OF TRANSFER COEFFICIENTS

Substitution of this value into Eqs. (4.5-17) and (4.5-16) gives
1.1098 +(%) 0.8981
A=(*) 2.5497

= [ (4.6 x 10-5/0.2)] 1.1098 + ( 7.1490 )0.8g81 = 1.38
2.5497
191 x 103

Hence, the friction factor is
f = 4.36 x 10-~
Thus, Eq. (4.56) gives the pressure drop per unit pipe length as
PPI 32PfQ2
-- -
L n2D5
- (32) (999)(4.36 x 0.03)2 = 40 Pa/ m
-
n2(0.2)5

H Calculate &; given IAPI and D
In this case rearrangement of Eq. (4.56) gives
f = (a) 2 (4.518)

where Y is defined by
n2D5 lAPl
32 pL (4.5-19)
Substitution of Eqs. (4.510) and (4.518) into Eq. (4.59) yields
I
1 Q = -4Y log (& + S) (4.520)

Thus, the procedure to calculate the volumetric flow rate becomes:
a) Calculate Y from Eq. (4.519),
b) Substitute Y into Eq. (4.520) and determine the volumetric flow rate.

Example 4.13 What is the volumetric flow rate of water in m3/s at 2O0c
that can be delivered through a commercial steel pipe (E = 4.6 x m) 20 cm in
diameter when the pressure drop per unit length of the pipe is 40 Pa/ m?