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Kern Method Complete Design

The document outlines the design of a shell-and-tube heat exchanger using the Kern method, focusing on a system with oil as the hot fluid and water as the cold fluid. Key calculations include a required heat transfer area of approximately 28.6 m², an internal tube-side heat transfer coefficient of about 1346 W/m²·K, and a pressure drop of approximately 31.2 kPa. This method serves as a preliminary design approach before more detailed analysis.

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Musa Kaleem
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32 views2 pages

Kern Method Complete Design

The document outlines the design of a shell-and-tube heat exchanger using the Kern method, focusing on a system with oil as the hot fluid and water as the cold fluid. Key calculations include a required heat transfer area of approximately 28.6 m², an internal tube-side heat transfer coefficient of about 1346 W/m²·K, and a pressure drop of approximately 31.2 kPa. This method serves as a preliminary design approach before more detailed analysis.

Uploaded by

Musa Kaleem
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as DOCX, PDF, TXT or read online on Scribd
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Complete Shell-and-Tube Heat

Exchanger Design Using Kern Method


1. Introduction
The Kern method is a widely used technique for the preliminary design of shell-and-tube
heat exchangers. It is particularly useful for turbulent flow regimes and provides reasonable
accuracy for standard designs.

2. Assumed Design Data


Hot Fluid: Oil (tube side)

Cold Fluid: Water (shell side)

Heat Duty (Q): 500 kW

Tube outer diameter (do): 25 mm

Tube inner diameter (di): 20 mm

Tube length (L): 5 m

Number of tubes: 73

Overall heat transfer coefficient (U): 250 W/m²·K

Hot fluid inlet/outlet: 150°C / 100°C

Cold fluid inlet/outlet: 30°C / 80°C

Tube side fluid properties at avg. temp: μ = 0.01 Pa·s, k = 0.13 W/m·K, Cp = 2200 J/kg·K, ρ =
850 kg/m³

3. Design Steps and Calculations

3.1 Log Mean Temperature Difference (LMTD)


ΔT1 = Th_in - Tc_out = 150 - 80 = 70°C
ΔT2 = Th_out - Tc_in = 100 - 30 = 70°C
LMTD = 70°C (since ΔT1 = ΔT2)
3.2 Required Heat Transfer Area
A = Q / (U × LMTD)
= (500 × 1000) / (250 × 70)
= 28.57 m²

3.3 Tube Side Velocity


Total flow area (At) = Nt × (π × di² / 4)
= 73 × π × (0.02² / 4)
≈ 0.02294 m²
Assuming ṁ = 5 kg/s
Velocity (V) = ṁ / (ρ × At) = 5 / (850 × 0.02294) ≈ 0.256 m/s

3.4 Reynolds Number


Re = (ρ × V × di) / μ = (850 × 0.256 × 0.02) / 0.01 ≈ 4352
Flow is turbulent (Re > 2300)

3.5 Heat Transfer Coefficient Inside Tubes


Using Dittus-Boelter equation:
Nu = 0.023 × Re^0.8 × Pr^0.4
Pr = Cp × μ / k = (2200 × 0.01) / 0.13 ≈ 169.23
Nu ≈ 0.023 × (4352)^0.8 × (169.23)^0.4 ≈ 207.1
hi = (Nu × k) / di = (207.1 × 0.13) / 0.02 ≈ 1346 W/m²·K

3.6 Shell Side Calculations (Simplified)


Assume shell diameter Ds ≈ 0.3 m, baffle spacing ≈ 0.2 m
Shell-side velocity, Re, and heat transfer coefficient can be estimated similarly.
Use correlations for segmental baffles and estimate shell-side h_o.

3.7 Tube Side Pressure Drop


f = 0.316 / Re^0.25 ≈ 0.045 (Blasius equation)
ΔP = (f × L × ρ × V²) / (2 × di) = (0.045 × 5 × 850 × 0.256²) / (2 × 0.02) ≈ 31.2 kPa

4. Conclusion
Using the Kern method with assumed values, the shell-and-tube heat exchanger requires
approximately 28.6 m² of surface area and 73 tubes of 25 mm outer diameter. The internal
tube-side heat transfer coefficient is about 1346 W/m²·K, and the pressure drop is
approximately 31.2 kPa. This method provides a good first estimate before detailed design
or simulation.

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