30 CHEMICAL ENGINEERING VOLUME 1 SOLUTIONS
IfR/ u^2 D 0. 04 /Re^0.^25 , the head loss per unit lengthlis:
hf/lD 8 0. 04 /Re^0.^25 l/du^2 / 2 gD 0. 016 u^2 /d / ud^0.^25D 0. 016 u^1.^75 0.^25 / ^0.^25 d^1.^25 The velocityuDG/ ADG/ / 4 d^2 D 1. 27 G/ d^2
∴ hf/lD 0. 016 1. 27 G/ d^2 ^1.^75 0.^25 / ^0.^25 d^1.^25 D 0. 024 G^1.^75 0.^25 / ^2 d^4.^75
The power required for pumping if the pump efficiency is*is:
PDGg 0. 024 G^1.^75 0.^25 / ^2 d^4.^75 /*If*D 0 .5,PD 0. 47 G^2.^75 0.^25 / ^2 d^4.^75 W
IfcDpower cost/W, the cost of pumping is given by: 0. 47 cG^2.^75 0.^25 / ^2 d^4.^75
The total annual cost is thenDˇaCˇbdC,G^2.^75 0.^25 0.^25 / ^2 d^4.^75 where,D
0. 47 c
Differentiating the total cost with respect to the diameter gives:
dC/ddDˇb 4. 75 ,G^2.^75 0.^25 / ^2 d^5.^75For minimum cost, dC/ddD0,d^5.^75 D 4. 75 ,G^2.^75 0.^25 / ^2 ˇbanddDKG^0.^48 0.^43 / ^0.^35
where: KD 4. 75 ,ˇb^0.^174
PROBLEM 3.15
A heat exchanger is to consist of a number of tubes each 25 mm diameter and 5 m long
arranged in parallel. The exchanger is to be used as a cooler with a rating of 4 MW and
the temperature rise in the water feed to the tubes is to be 20 deg K.
If the pressure drop over the tubes is not to exceed 2 kN/m^2 , calculate the minimum
number of tubes that are required. Assume that the tube walls are smooth and that entrance
and exit effects can be neglected. Viscosity of waterD1mNs/m^2.
Solution
Heat loadDmassflowðspecificheatðtemperature rise, or: 4000Dmð 4. 18 ð 20
and: mD 47 .8 kg/s
Pressure dropD2kN/m^2 D 2000 / 1000 ð 9. 81 D 0 .204 m of water.
From equation 3.23,R/ u^2 Re^2 DPfd^3 / 4 l ^2
D 2000 ð 0. 253 ð 1000 / 4 ð 5 ð 10 ^6 D 1. 56 ð 106
If the tubes are smooth, then from Fig. 3.8:ReD 2. 1 ð 104.
∴water velocityD 2. 1 ð 104 ð 10 ^3 / 1000 ð 0. 025 D 0 .84 m/s.
Cross-sectional area of each tubeD/ 4 0. 252 D 0 .00049 m^2.