Page 359 - Analysis, Synthesis and Design of Chemical Processes, Third Edition

assumptions made in Step 1. This is the critical step in learning from experience.
Example 11.1

Evaluate the heat transfer coefficient for water at 93°C (200°F) flowing at 3.05 m/s (10 ft/s) inside a 38
mm (1.5 inch) diameter tube. From previous experience, you know that the heat transfer coefficient for
2
water, at 21°C (70°F) and 1.83 m/s (6 ft/s), in these tubes is 5250 W/m °C. Follow the PAR process to
establish the heat transfer coefficient at the new conditions.
Step 1—Predict: Assume that the velocity and temperature have no effect. Predicted Heat Transfer
2
Coefficient = 5250 W/m °C
Step 2—Authenticate/Analyze: Using the properties given below we find that the Reynolds
number for the water in the tubes is
–4
3
Re = uρD pipe /μ = (1.83)(997.4)(1.5)(0.0254)/(9.8 × 10 ) = 71 × 10 → Turbulent Flow

Use the Sieder-Tate equation [2] to check the prediction:

(11.1)

Take the ratio of Equation 11.1 for the two conditions given above, and rearrange and substitute
numerical values. Using ′ to identify the new condition at 93°C, we get

(11.2)

(11.3)

2
2
h′ = (2.725)(5250) W/m °C = 14,300 W/m °C
The initial assumption that the velocity and temperature do not have a significant effect is incorrect.
Equation (11.3) reveals a velocity effect of a factor of 1.5 and a viscosity effect of a factor of 1.73. All
other factors are close to 1.0.