Fins

Introduction

The module fins contains functions that can be used to find the

1. fin-efficiency, and

2. surface area

of different fin geometries.

The geometries available are:

1. cylindrical (fin with constant circular cross section)

2. rectangular (fin with constant rectangular cross section)

3. rectangularannular (circular fin around a cylindrical shape)

4. pintriangular (concial fin)

5. straightparabolic (rectangular base with parabolic cross-section profile)

Note

The formulas for fin efficiency assume an adiabatic fin tip case. If the fin tip is not adiabatic, the appropriate corrected fin length (Lc) must be used for the “length” keyword.

The corrected lengths (Lc) and corrected outer radius (roc) for the different geometries are:

Corrected fin lengths

Geometry	Corrected length
rectangular	Lc = L + t/2 (L = physical length, t = thickness, Lc = corrected length)
cylindrical	Lc = L + D/4 (L = physical length, D = diameter, , Lc = corrected length)
rectangularannular	roc = ro + t/2 (ro = outer radius, t = thickness, , roc = corrected outer radius)
pintriangular	Lc = L (no change, L = physical length, Lc = corrected length)
straightparabolic	Lc = L (no change, L = physical length, Lc = corrected length)

How to use

It is recommended that the module be imported as from pychemengg import fins as fins

The following examples demonstrate how the module fins can be used to solve heat transfer problems.

Examples

Example 1: Fins cylindrical, single fin

Example 1. A rod of diameter D = 2 cm, length= 25 cm, and thermal conductivity k = 50 W/mC is exposed to ambient temperature air at T_infinity = 20 C with a heat transfer coefficient h = 64 W/m2C. If one end of the rod is maintained at 120 C, find the heat loss from the rod. Ans: heat loss = 25.1 kW

# EXAMPLE 1
from pychemengg.heattransfer import fins as fins
# The rod is similar to a single fin.
# Model it as a cylindrical fin.
# The rod tip is assumed to be adiabatic, so length = actual length.
rod = fins.Fin(length=25e-2, diameter=2e-2, heattransfercoefficient=64, thermalconductivity=50)
# There are other parameters for the 'Fin' class, however, only the relevant ones
# need to be provided. The rest can be left as 'None' by default.
# Next call on the cylindrical method to get efficiency and surface area.
rod.cylindrical()
# The above method will run and create attributes 'efficiency' and 'surfacearea'.
max_heatrate = rod.heattransfercoefficient * rod.surfacearea * (120-20)
actual_heatrate = max_heatrate * rod.efficiency
print(f"Heat loss = {actual_heatrate: 0.1f} W")

# PRINTED OUTPUT
Heat loss =  25.1 W

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Example 2: Fins rectangular annular, many fins

Example 2. Annular fins with inner radius of 12.5 mm, outer radius of 22.5 mm and thickness of 1 mm are attached to a cylinder at 200 fins per meter. The cylinder surface is maintained at 250°C. The surrounding air temperature is 25°C. Consider the convective heat transfer coefficient (h) as 25 W/m2K and the thermal conductivity (k) of the fin as 240 W/mK.

1. Find heat loss per fin. Ans: heat loss = 13 W

2. Heat loss per unit length of cylinder. Ans: heat loss = 2961 W/m

   # EXAMPLE 2
   from pychemengg.heattransfer import fins as fins
   # Model fin as a rectangularannular fin.
   # The fin tip is not given to be adiabatic.
   # So outer radius should be corrected as correctedradius = outerfinradius + thickness/2
   correctedouter_radius = 22.5e-3 + 1e-3/2
   annularfin = fins.Fin(inner_radius=12.5e-3, outer_radius=correctedouter_radius, thickness=1e-3, heattransfercoefficient=25, thermalconductivity=240)
   # There are other parameters for the 'Fin' class, however, only the relevant ones
   # need to be provided. The rest can be left as 'None' by default.
   # Even if irrelevant ones are provided, the method only uses the relavant parameters.
   # Next call on the rectangularannular method to get efficiency and surface area.
   annularfin.rectangularannular()
   # The above method will run and create attributes 'efficiency' and 'surfacearea'.
   max_heatrateperfin = annularfin.heattransfercoefficient * annularfin.surfacearea * (250-25)
   actualheatrateperfin = max_heatrateperfin * annularfin.efficiency
   print(f"Heat loss from single fin = {actualheatrateperfin: 0.1f} W")
   # To compute heat loss from 1 m long finned cylinder
   # Total heat loss from 1m cylinder = heat loss from fins on 1m + heat loss from surface not covered by fins
   fincount=200
   total_finheatloss = fincount * actualheatrateperfin
   import math
   barecylindersurfacearea = math.pi * (12.5e-3 * 2) * 1 # area = pi * D * L
   baseareaof_singlefin = math.pi * (12.5e-3 *2) * 1e-3 # here L = 1e-3
   areacoveredby_allfins = 200 * baseareaof_singlefin
   heatlossfrom_area_without_fins = 25 * (barecylindersurfacearea-areacoveredby_allfins) * (250-25)
   totalheatloss = total_finheatloss + heatlossfrom_area_without_fins
   print(f"Heat loss for 1 m cylinder = {totalheatloss: 0.0f} W/m")
   
   # PRINTED OUTPUT
   Heat loss from single fin =  13.0 W
   Heat loss for 1 m cylinder =  2961 W/m