CHEMICAL ENGINEERING

36 CHEMICAL ENGINEERING VOLUME 1 SOLUTIONS

Solution

OilflowrateD 20 , 000 ð 1000  24 ð 3600 
D 231 .5 kg/s or 231 / 960 D 0 .241 m^3 /s

Cross section area of pipeD/ 4 ð 0. 32 D 0 .0707 m^2

Oil velocity in pipeD 0. 241 / 0. 0707 D 3 .40 m/s

Reynolds number at terminalD 3. 40 ð 0. 3 ð 960 / 0. 09 D 10 , 880

Reynolds number at the refinery is twice this value or 21,760.

From equation 3.18: PD 4 R/u^2 l/du^2  (equation 3.18)

and:

P/lrefinery

P/lterminal

D

R/u^2 refinery

R/u^2 terminal

which, from Fig. 3.7: D 0. 0030 / 0. 00375 D 0. 80

Inalengthofpipedl:
dPD 4 R/u^2 dl/du^2 N/m^2
Energy dissipatedDdPQD/ 4 d^2 u 4 R/u^2 dl/du^2 Wwhereuis the velocity
in the pipe.
The heat loss to the surroundings at a distancelfrom the inlet ishTTSdlW
whereTSis the temperature of the surroundings andTis the temperature of thefluid.
Heat gained by thefluidD/ 4 d^2 uCpdTWwhereCp(J/kg K) is the specificheat
capacity of thefluid.
Thus an energy balance over the length of pipe dlgives:

R/u^2 du^3 dlDhTTsddlC/ 4 d^2 uCpdT

R/u^2 varies with temperature as illustrated in thefirst part of this problem, and hence
this equation may be written as:

AdlDBdlCCdT

or:

CdT

AB

Ddl

(whereAandBare both functions of temperature andCis a constant).
Integrating betweenl 1 andl 2 ,T 1 andT 2 gives:
∫l 2

l 1

dlD

∫T 2

T 1

[

CdT

AB

]

IfT 1 ,Ts,handTare known (20 deg K in this problem), the integral may then be
evaluated.

PROBLEM 3.22

Oil with a viscosity of 10 mNs/m^2 and density 900 kg/m^3 isflowing through a 500 mm
diameter pipe 10 km long. The pressure difference between the two ends of the pipe is