Heat Transfer Project

Introduction/Background

The purpose of this project was to perform a radiation analysis of a storage tank of liquefied methane and its distribution. The net radiative transfer to the liquid without a radiation shield was found and compared to the radiative transfer with a shield to understand the effects of the shield. The calculated flow rate for each case was then compared to the boil-off gas compressor requirement of .2% full tank capacity per day in order to ensure that the set-up was feasible and desirable from a safety and engineering standpoint. For the first case, three radiated surfaces were considered: a black liquid, a gray cylindrical sidewall (emissivity 0.2), and a black roof. The second case included a gray radiation shield (emissivity 0.25) between the sidewall and the roof.

Theoretical Development (see attached need circuits, diagrams, etc)

The following equations were used in the numerical evaluation:

The constants are shown in the table below and are the same for each case:

CONSTANTS

Property | Value | Units |

diam1 | 50 | m |

diam4 | 50 | m |

height | 40 | m |

Area1 | 1963.50 | m^2 |

Area2 | 1963.50 | m^2 |

ε3 | 0.2 | |

ε4 | 0.25 | |

T1 | 113.15 | K |

T2 | 353.15 | K |

T3 | 233.15 | K |

σ | 5.67E-08 | W/(m^2*K^4) |

Eb1 | 9.29 | W/m^2 |

Eb2 | 881.90 | W/m^2 |

Eb3 | 167.54 | W/m^2 |

In each case, the values of Ri and Rj and S were the same, and were found using the equations for coaxial parallel disks:

Ri= riL

Rj= rjL

S=1+ 1+Rj2Ri2

From this, the first value of F for use in the equivalent circuits could be found:

Fij= 12 { S-[S2-4(rjri)2]12 }

Then, for Case 1, using this first value of F, the remaining were found and allowed for the complete equivalent resistances to be calculated:

F13 =1- F12

F31= A1*F13/A3

F32=F31

R12=1A1*F12

R13=1A1*F13

R23=1A3*F32

R3=1- ∈3∈3*A3

In case 2, the process was similar, but because of the extra surface the process was altered somewhat:

F14= 12 {...