wo different fluids are separated by an insulating barrier. he barrier consists of two different materials as shown . Fig. 2. The temperature of the fluids are maintained 80°C and To2 30°C. The ceady at Too1 onvection coefficients between the fluids and respective k2 arrier surfaces are h1 15 W/m²-K and h2 10 W/m²-K. Too2 Toog The thickness of the slabs are L1 0.1 m and B 2 = 0.15 m. The thermal conductivity of the slabs are 1.5 W/m-K. Find x= 0 Perfect interface espectively ki = 2 W/m-K and k2 ut the temperature distribution within the slabs (i.e., emperature as a function of x). What is the difference Figure 2: Note the indicated positive x direction for reference. f temperature between the center of slab A and slab B .e., ТА(L/2) — Тв (Li + La/2))? Fluid 2 Fluid 1 1E

Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
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Two different fluids are separated by an insulating barrier.
The barrier consists of two different materials as shown
in Fig. 2. The temperature of the fluids are maintained
80°C and To02
steady at Too1
30°C.
The
convection coefficients between the fluids and respective
k2
barrier surfaces are hi
15 W/m2.K and h2 :
10 W/m².K.
Too2
The thickness of the slabs are L1
0.1 m and
L2 =
= 0.15 m. The thermal conductivity of the slabs are
A
B
X= 0
respectively k1 = 2 W/m-K and k2
out the temperature distribution within the slabs (i.e.,
= 1.5 W/m-K. Find
Perfect interface
temperature as a function of x). What is the difference
Figure 2: Note the indicated positive
x direction for reference.
of temperature between the center of slab A and slab B
(i.e., TA(L1/2) – TB(L1 + L2/2))?
Fluid 2
Fluid 1 I
Transcribed Image Text:Two different fluids are separated by an insulating barrier. The barrier consists of two different materials as shown in Fig. 2. The temperature of the fluids are maintained 80°C and To02 steady at Too1 30°C. The convection coefficients between the fluids and respective k2 barrier surfaces are hi 15 W/m2.K and h2 : 10 W/m².K. Too2 The thickness of the slabs are L1 0.1 m and L2 = = 0.15 m. The thermal conductivity of the slabs are A B X= 0 respectively k1 = 2 W/m-K and k2 out the temperature distribution within the slabs (i.e., = 1.5 W/m-K. Find Perfect interface temperature as a function of x). What is the difference Figure 2: Note the indicated positive x direction for reference. of temperature between the center of slab A and slab B (i.e., TA(L1/2) – TB(L1 + L2/2))? Fluid 2 Fluid 1 I
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