Consider the thermocouple and convection conditions of Example 1, but now allow for radiation exchange with the walls of a duct
that encloses the gas stream. If the duct walls are at 400℃ and the emissivity of the thermocouple bead is 0.9, calculate the steady-state temperature of the junction
1 answer:
Answer:
hello your question has some missing part attached below is the complete question
answer : steady state temperature = 419.713k ≈ 218.7⁰c
Time required to reach a junction ≈ 5 secs
Explanation:
The detailed solution of the given problem is attached below but the solution to the subsequent problem from which the question you asked is referenced to( problem 1 ), is not attached because it was not part of the question you asked
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Answer:
The break force that must be applied to hold the plane stationary is 12597.4 N
Explanation:
p₁ = p₂, T₁ = T₂
The heat supplied = × Heating value of jet fuel
The heat supplied = 0.5 kg/s × 42,700 kJ/kg = 21,350 kJ/s
The heat supplied = ·
= 20 kg/s
The heat supplied = 20* = 21,350 kJ/s
= 1.15 kJ/kg
T₃ = 21,350/(1.15*20) + 485.03 = 1413.3 K
p₂ = p₁ × p₂/p₁ = 95×9 = 855 kPa
p₃ = p₂ = 855 kPa
T₃ - T₄ = T₂ - T₁ = 485.03 - 280.15 = 204.88 K
T₄ = 1413.3 - 204.88 = 1208.42 K
T₅ = 1208.42*(2/2.333) = 1035.94 K
= √(1.333*287.3*1035.94) = 629.87 m/s
The total thrust = ×
= 20*629.87 = 12597.4 N
Therefore;
The break force that must be applied to hold the plane stationary = 12597.4 N.