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In an experiment to determine the thermal conductivity of a long solid 2.5 cm diameter rod, its base is place
in a furnace with a large portion of it projecting in to the room air at 22°C. Aft er stea dy state condition
prevail, t he temp eratrue at two point s, 10 cm apar t are found to be 110°C and 85°C respectively. T he
convective heat t ransfer coefficient between the rod surface and the surrounding air is 28.4 W/m2K.
Determine the t hermal conductivit y of the rod material _________________ W/mK
in a furnace with a large portion of it projecting in to the room air at 22°C. Aft er stea dy state condition
prevail, t he temp eratrue at two point s, 10 cm apar t are found to be 110°C and 85°C respectively. T he
convective heat t ransfer coefficient between the rod surface and the surrounding air is 28.4 W/m2K.
Determine the t hermal conductivit y of the rod material _________________ W/mK
Correct answer is between '490,520'. Can you explain this answer?
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°C. The temperature of the rod is measured at different points along its length, and the heat flow rate through the rod is also measured. The data is then used to calculate the thermal conductivity of the rod.
The thermal conductivity of a material is defined as the rate at which heat flows through a unit area of the material when a temperature gradient exists. In this experiment, the temperature gradient is created by heating one end of the rod in a furnace and allowing the other end to cool in the room air. The heat flow rate through the rod is measured using a heat flux sensor, which measures the rate at which heat flows per unit area.
To calculate the thermal conductivity of the rod, the following equation is used:
k = Q/(AΔT)
Where k is the thermal conductivity, Q is the heat flow rate, A is the cross-sectional area of the rod, and ΔT is the temperature difference across the length of the rod.
The cross-sectional area of the rod can be calculated using the formula for the area of a circle:
A = πr^2
Where r is the radius of the rod, which is half the diameter. In this case, the diameter is 2.5 cm, so the radius is 1.25 cm.
A = π(1.25 cm)^2 = 4.91 cm^2
The temperature difference across the length of the rod can be calculated by subtracting the temperature at the cool end from the temperature at the hot end. For example, if the temperature at the hot end is 200°C and the temperature at the cool end is 50°C, then the temperature difference is:
ΔT = 200°C - 50°C = 150°C
Finally, the thermal conductivity can be calculated by dividing the heat flow rate by the product of the cross-sectional area and the temperature difference:
k = Q/(AΔT)
For example, if the heat flow rate is 100 watts, then the thermal conductivity is:
k = 100 W/(4.91 cm^2 × 150°C) = 0.013 W/(cm·°C)
Therefore, the thermal conductivity of the rod is 0.013 W/(cm·°C).
The thermal conductivity of a material is defined as the rate at which heat flows through a unit area of the material when a temperature gradient exists. In this experiment, the temperature gradient is created by heating one end of the rod in a furnace and allowing the other end to cool in the room air. The heat flow rate through the rod is measured using a heat flux sensor, which measures the rate at which heat flows per unit area.
To calculate the thermal conductivity of the rod, the following equation is used:
k = Q/(AΔT)
Where k is the thermal conductivity, Q is the heat flow rate, A is the cross-sectional area of the rod, and ΔT is the temperature difference across the length of the rod.
The cross-sectional area of the rod can be calculated using the formula for the area of a circle:
A = πr^2
Where r is the radius of the rod, which is half the diameter. In this case, the diameter is 2.5 cm, so the radius is 1.25 cm.
A = π(1.25 cm)^2 = 4.91 cm^2
The temperature difference across the length of the rod can be calculated by subtracting the temperature at the cool end from the temperature at the hot end. For example, if the temperature at the hot end is 200°C and the temperature at the cool end is 50°C, then the temperature difference is:
ΔT = 200°C - 50°C = 150°C
Finally, the thermal conductivity can be calculated by dividing the heat flow rate by the product of the cross-sectional area and the temperature difference:
k = Q/(AΔT)
For example, if the heat flow rate is 100 watts, then the thermal conductivity is:
k = 100 W/(4.91 cm^2 × 150°C) = 0.013 W/(cm·°C)
Therefore, the thermal conductivity of the rod is 0.013 W/(cm·°C).
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In an experiment to determine the thermal conductivity of a long solid 2.5 cm diameter rod, its base is placein a furnace with a large portion of it projecting in to the room air at 22°C. Aft er stea dy state conditionprevail, t he temp eratrue at two point s, 10 cm apar t are found to be 110°C and 85°C respectively. T heconvective heat t ransfer coefficient between the rod surface and the surrounding air is 28.4 W/m2K.Determine the t hermal conductivit y of the rod material _________________ W/mKCorrect answer is between '490,520'. Can you explain this answer?
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In an experiment to determine the thermal conductivity of a long solid 2.5 cm diameter rod, its base is placein a furnace with a large portion of it projecting in to the room air at 22°C. Aft er stea dy state conditionprevail, t he temp eratrue at two point s, 10 cm apar t are found to be 110°C and 85°C respectively. T heconvective heat t ransfer coefficient between the rod surface and the surrounding air is 28.4 W/m2K.Determine the t hermal conductivit y of the rod material _________________ W/mKCorrect answer is between '490,520'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about In an experiment to determine the thermal conductivity of a long solid 2.5 cm diameter rod, its base is placein a furnace with a large portion of it projecting in to the room air at 22°C. Aft er stea dy state conditionprevail, t he temp eratrue at two point s, 10 cm apar t are found to be 110°C and 85°C respectively. T heconvective heat t ransfer coefficient between the rod surface and the surrounding air is 28.4 W/m2K.Determine the t hermal conductivit y of the rod material _________________ W/mKCorrect answer is between '490,520'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In an experiment to determine the thermal conductivity of a long solid 2.5 cm diameter rod, its base is placein a furnace with a large portion of it projecting in to the room air at 22°C. Aft er stea dy state conditionprevail, t he temp eratrue at two point s, 10 cm apar t are found to be 110°C and 85°C respectively. T heconvective heat t ransfer coefficient between the rod surface and the surrounding air is 28.4 W/m2K.Determine the t hermal conductivit y of the rod material _________________ W/mKCorrect answer is between '490,520'. Can you explain this answer?.
In an experiment to determine the thermal conductivity of a long solid 2.5 cm diameter rod, its base is placein a furnace with a large portion of it projecting in to the room air at 22°C. Aft er stea dy state conditionprevail, t he temp eratrue at two point s, 10 cm apar t are found to be 110°C and 85°C respectively. T heconvective heat t ransfer coefficient between the rod surface and the surrounding air is 28.4 W/m2K.Determine the t hermal conductivit y of the rod material _________________ W/mKCorrect answer is between '490,520'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about In an experiment to determine the thermal conductivity of a long solid 2.5 cm diameter rod, its base is placein a furnace with a large portion of it projecting in to the room air at 22°C. Aft er stea dy state conditionprevail, t he temp eratrue at two point s, 10 cm apar t are found to be 110°C and 85°C respectively. T heconvective heat t ransfer coefficient between the rod surface and the surrounding air is 28.4 W/m2K.Determine the t hermal conductivit y of the rod material _________________ W/mKCorrect answer is between '490,520'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In an experiment to determine the thermal conductivity of a long solid 2.5 cm diameter rod, its base is placein a furnace with a large portion of it projecting in to the room air at 22°C. Aft er stea dy state conditionprevail, t he temp eratrue at two point s, 10 cm apar t are found to be 110°C and 85°C respectively. T heconvective heat t ransfer coefficient between the rod surface and the surrounding air is 28.4 W/m2K.Determine the t hermal conductivit y of the rod material _________________ W/mKCorrect answer is between '490,520'. Can you explain this answer?.
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