Page 133 - Modelling in Transport Phenomena A Conceptual Approach

4.5. FLOW IN CIRCULAR PIPES 113

- (1.155 x 10A2)(5)(997) = 64,548 + Turbulent flow
-
892 x
Substitution of these values into Eqs. (3) and (2) and taking E/D x 0 gives

7.1490 0.8981

~ (;;4596)8)0.8g81 = 2.8 x 10-~

- -410g ( -- 5’0452 log A)
1
=
d7 Refi
= - 4 log [ - 5.0452 log(2.8 x f=0.0049
64,548
Hence, the power required is calculated from Eq. (1) as

(3)(2 x 10-2)(1.5)

b) The inventory rate equation fo. mass is
(Y)
-
Rate of mass in = Rate of mass out = 7iz = p(v) (4)

m = (997)(5) [ fi(2 = 0.863 kg/ s

The inventory rate equation for energy reduces to
Rate of energy in = Rate of energy out (5)
The terns in Eq. (5) are expressed by

Rate of energy in = mep(Tbi, - Tr,f) + Qw (6)
Rate of energy out = m&p(Tbout - Tr,f) (7)
where Qw is the rate of heat tmnsfer to water from the lateral surfaces of the duct.
Substitution of Eqs. (6) and (7) into Eq. (5) gives