Page 436 - Handbook of Energy Engineering Calculations

(127.0 mm). Only the shellside product must be adjusted since only it is
affected by the baffle spacing. Therefore, the shell-side factor of the previous
trial is multiplied by the ratio of the baffle spacing to the 0.6 power:
0.6
(3.5/5.0) × 0.367 = 0.296
The sum of the products (SOP) for this trial is 1.056. The shellside pressure
3
2
dropΔP = 3.9 × (5.0/3.5) = 11.4 lb/in (78.5 kPa).
a
Because we have now reached the point where the assumed design nearly
satisfies our conditions, tube-layout tables can be used to find a standard
shell-size containing the next increment above 109 tubes. A 10-in-dia (254
mm) shell in a fixed-tubesheet design contains 110 tubes.

Again, correcting the products of the heat-transfer factors from the
previous trial:

Tubeside pressure drop:
1.8
2
ΔP = 10 × (109/110) = 9.8Ib/in (67.5 KPa)
i
The shellside pressure drop is now corrected for the actual shell diameter
of 10 in (25.4 cm) instead of 9 in (22.86 cm).
ΔP × (9/10) = 10.1 Psi (69.6 KPa)
o
Any value of SOP between 0.95 and 1.05 is satisfactory as this gives a
result within the accuracy range of the basic equations; unknowns in selecting

the fouling factor do not justify further refinement. Therefore, the above is a
satisfactory design for heat transfer and is within the pressure-drop
restrictions specified. The surface area of the heat exchanger is A = 110 × 12
2
2
× 0.1636 = 216 ft (20.1 m ). The design overall coefficient is U =
2
2
2,391,000/(30.4 × 1.8 × 216) = Btu/(h) (ft ) (°F) (35.6 W/m .°C).
The foregoing example shows that the essence of the design procedure is
selecting tube configurations and baffle spacings that will satisfy heat-

transfer requirements within the pressure-drop limitations of the system.

Related Calculations. The preceding procedure was for rating a heat