A rod is made of two different layers:chromium and granite; as shown in the diagram below. Both layers have the same length (L = 0.28 m) and the same area ( A = 0.13 cm?). In the considered system, chromium is joined to the left hand which is at Tjeg = 60.1 C, whereas granite is in physical contact with the right end at Tight = 13.5 C. The thermal conductivity constants of the materials are given as k, = 93.7 W/ (m K) and k, = 2 W/ (m K). Assuming steady state is eventually achieved, answer the following questions. a) Evaluate the junction temperature between the two layers. Tundtion = 59,12 "c b) Calculate the rate of heat flow (power) through conduction in Sl units. P= 42.3614 X J/s Tieft Layer 1 Layer 2 Tright

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A rod is made of two different layers:chromium and granite; as shown in the diagram below. Both layers have the same length (L = 0.28 m) and the same area (
A = 0.13 cm?). In the considered system, chromium is joined to the left hand which is at Tjeg = 60.1 C, whereas granite is in physical contact with the right end at
Tight = 13.5 C. The thermal conductivity constants of the materials are given as k, = 93.7 W/ (m K) and k, = 2 W/ (m K). Assuming steady state is eventually
achieved, answer the following questions.
a) Evaluate the junction temperature between the two layers. Tundtion = 59,12
b) Calculate the rate of heat flow (power) through conduction in Sl units. P= 42.3614
X J/s
Tieft
Layer 1
Layer 2
Tright
Transcribed Image Text:A rod is made of two different layers:chromium and granite; as shown in the diagram below. Both layers have the same length (L = 0.28 m) and the same area ( A = 0.13 cm?). In the considered system, chromium is joined to the left hand which is at Tjeg = 60.1 C, whereas granite is in physical contact with the right end at Tight = 13.5 C. The thermal conductivity constants of the materials are given as k, = 93.7 W/ (m K) and k, = 2 W/ (m K). Assuming steady state is eventually achieved, answer the following questions. a) Evaluate the junction temperature between the two layers. Tundtion = 59,12 b) Calculate the rate of heat flow (power) through conduction in Sl units. P= 42.3614 X J/s Tieft Layer 1 Layer 2 Tright
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