For boiler design purposes,where the fluids are generally in turbulent flow region,the following equation may be used to determine the tube side heat transfer coefficient.

**NU=0.023Re ^{0.8}
Pr^{0.4}**

where

Nusselt
number NU=hd/12k

Reynolds
number Re=15.2Wd/m

Prandtl
number Pr=mCp/k

h=tube
side heat transfer coefficient,Btu/ft^{2}hF

d=tube
inner diameter,in

W=flow
rate per tube,lb/h

k=thermal
conductivity,Btu/fthF

Cp=specific
heat,Btu/lbF

m=viscosity,lb/fth

The
fluid properties are estimated at the average fluid temperature

Substituting
the above expressions into the equation (1) and simplifying we have:

**h = 2.44W ^{0.8}
C/d^{1.8}**

where
**C=(Cp/****m) ^{0.4}k^{0.6}**

Computation
of C requires reference to fluid properties and hence can be tedious.C
has been computed by the author for various fluids and for good engineering
estimates,the equations/data shown below may be used.
**C
values for Saturated,Superheated steam**

pressure,psia | 100 | 500 | 1000 | 2000 |

sat steam | 0.244 | 0.417 | 0.490 | 0.900 |

400 F | 0.271 | |||

500 | 0.273 | 0.360 | ||

600 | 0.281 | 0.322 | 0.413 | |

700 | 0.291 | 0.316 | 0.358 | 0.520 |

800 | 0.305 | 0.320 | 0.345 | 0.420 |

900 | 0.317 | 0.327 | 0.347 | 0.394 |

1000 F | 0.325 | 0.340 | 0.353 | 0.386 |

**Example**:
Determine the tube side heat transfer coefficient when 5000 lb/h of superheated
steam at 1000 psia and 800 F flows inside a superheater tube of inner diameter
1.75 in.
**Solution:**
C =0.345

h
=2.44 x5000^{0.8}x0.345/1.75^{1.8 }= 271 Btu/ft^{2}hF

**C
values for air,flue gases at atmospheric pressure**

temp,F | 200 | 400 | 600 | 800 | 1000 | 1200 |

C | 0.162 | 0.172 | 0.180 | 0.187 | 0.194 | 0.205 |

where t= water temperature,F

Determine specific heat,viscosity and thermal conductivity of a flue gas mixture at atmospheric pressure have the following data:

**
Table: data of gas properties**

gas | % volume | Cp | m | k | MW |

N2 | 80 | 0.286 | 0.108 | 0.03 | 28 |

O2 | 12 | 0.270 | 0.125 | 0.043 | 32 |

SO2 | 8 | 0.210 | 0.105 | 0.040 | 64 |

m

k

C

m

k

C

_{Solution:}_{}

_{Cpm
=0.286x0.8x28+0.27x0.12x32+0.21x0.08x64/(0.8x28+0.12x32+0.08x64) =0.272
Btu/lbF}

m_{m}_{
=} 0.108x28^{0.5}x0.8+0.125x32^{0.5}x0.12+0.105x64^{0.5}x0.08/(0.8x28^{0.5}+0.12x32^{0.5}+0.08x64^{0.5})=0.109
lb/fth

k_{m}_{
=} 0.03x28^{0.33}x0.8+0.043x32^{0.33}x0.12+0.04x64^{0.33}x0.08/(0.8x28^{0.33}+0.12x32^{0.33}+0.08x64^{0.33})=0.032
Btu/fthF_{}

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