only attempt once Two different fluids are separated by an insulating barrier.The barrier consists of two different materials as shownin Fig. 2. The temperature of the fluids are maintained80°C and To02steady at Too130°C.Theconvection coefficients between the fluids and respectivek2barrier surfaces are hi15 W/m2.K and h2 :10 W/m².K.Too2The thickness of the slabs are L10.1 m andL2 == 0.15 m. The thermal conductivity of the slabs areABX= 0respectively k1 = 2 W/m-K and k2out the temperature distribution within the slabs (i.e.,= 1.5 W/m-K. FindPerfect interfacetemperature as a function of x). What is the differenceFigure 2: Note the indicated positivex direction for reference.of temperature between the center of slab A and slab B(i.e., TA(L1/2) – TB(L1 + L2/2))?Fluid 2Fluid 1 I

Answered: Image /qna-images/answer/1cfd90e6-16db-42cf-a8ee-039732cda70c.jpg

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

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

Principles of Heat Transfer (Activate Learning with these NEW titles from Engineering!)

8th Edition

ISBN:9781305387102

Author:Kreith, Frank; Manglik, Raj M.

Publisher:Kreith, Frank; Manglik, Raj M.

Chapter2: Steady Heat Conduction

Section: Chapter Questions

Problem 2.32P

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Heat transfer