USSC 5660 - Minimum Clearances and Floor Protection

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INSERTPELLETSTOVEINSTALLATION‐continued
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MinimumClearancetoCombustiblesandFloorProtection
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A:Sidewall:16”tosideofappliance
B:TopTrim:5”abovetopofappliance
C:Sidetrim:10.5”tosideofappliance
D:Mantle:10”abovetopofappliance
FloorProtection
WheninMASONRYfireplaces,use½inchthicknon‐combustiblematerial.
WheninFactory‐builtfireplaces,usea1inch,non‐combustiblesurfacewithakfactorof0.84.Formultiplelayers,
addR‐valuesofeachlayertodeterminetheoverallR‐value.TheRvaluefortherequired
boardis1.2.
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ConvertspecificationtoR‐value:
k‐factorisgivenwitharequiredthickness(T)ininches:R=1/kxT.TheC‐factorisgiven:R=1/C
Example:
Ifthefloorprotectoris4”brickwithaC‐factorof1.25 over1/8”mineralboardwitha“k”factorof
0.29thetotalR‐
valueofthesystemis:
4”brickC=1.25,R=1/1.25=0.8
1/8”mineralboardK=0.29,R=1/0.29x0.125=0.431
TotalR=Rbrick+Rmineral=0.8+0.431=1.231
TotalRisgreaterthan1.2, thesystemisacceptable.
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Figure20
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