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A steel plate of thermal conductivity 50 W/m-K and thickness 10 cm passes a heat flux by conduction of 25 kW/m2. If the temperature of the hot surface of the plate is 100°C then what is the temperature of the cooler side of the plate?
- a)30°C
- b)40°C
- c)50°C
- d)60°C
Correct answer is option 'C'. Can you explain this answer?
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A steel plate of thermal conductivity 50 W/m-K and thickness 10 cm pas...
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A steel plate of thermal conductivity 50 W/m-K and thickness 10 cm pas...
°C and the cold surface temperature is 20°C, then the thermal resistance of the plate can be calculated as follows:
First, we need to convert the heat flux from kW/m2 to W/m2:
25 kW/m2 = 25,000 W/m2
Next, we can use Fourier's law of heat conduction to calculate the heat transfer rate through the plate:
q = kA(dT/dx)
where q is the heat transfer rate, k is the thermal conductivity, A is the cross-sectional area, and (dT/dx) is the temperature gradient across the plate.
Assuming steady-state heat transfer and a uniform temperature gradient, we can simplify this equation to:
q = kA(T1 - T2)/L
where T1 and T2 are the temperatures of the hot and cold surfaces, respectively, and L is the thickness of the plate.
Rearranging this equation, we get:
R = (T1 - T2)/(q/A)
where R is the thermal resistance of the plate.
Plugging in the given values, we get:
R = (100°C - 20°C)/(25,000 W/m2 / 1 m2 / 50 W/m-K / 0.1 m)
= 0.16 K/W
Therefore, the thermal resistance of the steel plate is 0.16 K/W.
First, we need to convert the heat flux from kW/m2 to W/m2:
25 kW/m2 = 25,000 W/m2
Next, we can use Fourier's law of heat conduction to calculate the heat transfer rate through the plate:
q = kA(dT/dx)
where q is the heat transfer rate, k is the thermal conductivity, A is the cross-sectional area, and (dT/dx) is the temperature gradient across the plate.
Assuming steady-state heat transfer and a uniform temperature gradient, we can simplify this equation to:
q = kA(T1 - T2)/L
where T1 and T2 are the temperatures of the hot and cold surfaces, respectively, and L is the thickness of the plate.
Rearranging this equation, we get:
R = (T1 - T2)/(q/A)
where R is the thermal resistance of the plate.
Plugging in the given values, we get:
R = (100°C - 20°C)/(25,000 W/m2 / 1 m2 / 50 W/m-K / 0.1 m)
= 0.16 K/W
Therefore, the thermal resistance of the steel plate is 0.16 K/W.
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A steel plate of thermal conductivity 50 W/m-K and thickness 10 cm passes a heat flux by conduction of 25 kW/m2. If the temperature of the hot surface of the plate is 100°C then what is the temperature of the cooler side of the plate?a)30°Cb)40°Cc)50°Cd)60°CCorrect answer is option 'C'. Can you explain this answer? for Mechanical Engineering 2025 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A steel plate of thermal conductivity 50 W/m-K and thickness 10 cm passes a heat flux by conduction of 25 kW/m2. If the temperature of the hot surface of the plate is 100°C then what is the temperature of the cooler side of the plate?a)30°Cb)40°Cc)50°Cd)60°CCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A steel plate of thermal conductivity 50 W/m-K and thickness 10 cm passes a heat flux by conduction of 25 kW/m2. If the temperature of the hot surface of the plate is 100°C then what is the temperature of the cooler side of the plate?a)30°Cb)40°Cc)50°Cd)60°CCorrect answer is option 'C'. Can you explain this answer?.
A steel plate of thermal conductivity 50 W/m-K and thickness 10 cm passes a heat flux by conduction of 25 kW/m2. If the temperature of the hot surface of the plate is 100°C then what is the temperature of the cooler side of the plate?a)30°Cb)40°Cc)50°Cd)60°CCorrect answer is option 'C'. Can you explain this answer? for Mechanical Engineering 2025 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A steel plate of thermal conductivity 50 W/m-K and thickness 10 cm passes a heat flux by conduction of 25 kW/m2. If the temperature of the hot surface of the plate is 100°C then what is the temperature of the cooler side of the plate?a)30°Cb)40°Cc)50°Cd)60°CCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A steel plate of thermal conductivity 50 W/m-K and thickness 10 cm passes a heat flux by conduction of 25 kW/m2. If the temperature of the hot surface of the plate is 100°C then what is the temperature of the cooler side of the plate?a)30°Cb)40°Cc)50°Cd)60°CCorrect answer is option 'C'. Can you explain this answer?.
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