Week 9 Proses Konduksi Pada Dinding Berlapis
Week 9 Proses Konduksi Pada Dinding Berlapis
Week 9 Proses Konduksi Pada Dinding Berlapis
BERLAPING
PERTEMUAN 9
By: Didik Nurhadi, M.Pd., Ph.D., CPHCM., CIQaR.
Fourier’s Law of Conduction
∆T
q=kA
∆x
Where (British Unit):
q= rate of heat transfer (Btu/hr)
A = cross-sectional area of heat transfer (ft2)
x = thickness of slab (ft)
r = thickness of cylindrical wall (ft)
T = temperature difference (°F)
k = thermal conductivity of slab (Btu/ft-hr-°F)
Latihan 1: Satu Lapisan Dinding
∆T Where:
1 Lapis dinding: q = k A q
∆x Q = A = Heat Flux (Btu/hr-ft2)
T = Temperature Difference (oF)
∆T
Dinding berlapis: Q = ∆x
Rth = k = Thermal Resistance (hr-ft2-oF/Btu)
Rth
Metode Analogi Tahanan Pada Diding Berlapis
Contoh Dinding Berlapis
Diketahui:
Dinding yang terbuat dengan 4 lapisan
∆x gipsum = 12 mm = 0,12 m
∆x fiberglass = 75 mm = 0,75 m
∆x plywood = 20 mm = 0,2 m
∆x hardboard = 20 mm = 0,2 m
k1 gipsum = 0,176 W/m 0K
k2 fiberglass = 0,115 W/m 0K
k3 plywood = 0,036 W/m 0K
k4 hardboard = 0,215 W/m 0K
t0 = 20 0C = 293 0K
t1 = -10 0C = 263 0K
q
Ditanya : Q = = …. ?
A
Contoh Dinding Berlapis
Jawab:
Latihan 3: Dinding Berlapis