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A thermocouple junction in the form of a 4 mm radius sphere is to be used to measure the temperature of a gas stream. The junction is initially at 35 degrees Celsius and is placed in a gas stream at 300 degrees Celsius. The thermocouple is removed from the hot gas stream after 10 seconds and kept in still air at 25 degrees Celsius with a convective coefficient of 10 W/m2 K. Find out the time constant of the thermocouple. Assume the thermos-physical properties as given below
h = 37.5 W/m2 K
p = 7500 kg/m3
c = 400 J/kg K
- a)6.67 seconds
- b)106.67 seconds
- c)206.67 seconds
- d)306.67 seconds
Correct answer is option 'B'. Can you explain this answer?
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A thermocouple junction in the form of a 4 mm radius sphere is to be ...
T = p V c / h A = p r c / 3 h = 106.67 seconds.
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A thermocouple junction in the form of a 4 mm radius sphere is to be ...
Given:
- Radius of thermocouple junction (r) = 4 mm = 0.004 m
- Initial temperature of thermocouple (T1) = 35 °C
- Temperature of gas stream (T∞) = 300 °C
- Time taken to remove the thermocouple from the gas stream (t) = 10 s
- Temperature of still air (T∞2) = 25 °C
- Convective heat transfer coefficient (h) = 10 W/m²K
- Density of thermocouple material (ρ) = 7500 kg/m³
- Specific heat capacity of thermocouple material (c) = 400 J/kgK
To find:
Time constant of the thermocouple
Concept:
The heat transfer from the thermocouple to the surrounding air can be modeled using Newton's Law of Cooling. The rate of heat transfer is given by:
q = h * A * (T - T∞)
Where:
q = Rate of heat transfer (W)
h = Convective heat transfer coefficient (W/m²K)
A = Surface area of the thermocouple (m²)
T = Temperature of the thermocouple (K)
T∞ = Temperature of the surrounding air (K)
The surface area of the thermocouple can be calculated using the formula for the surface area of a sphere:
A = 4πr²
The time constant (τ) is defined as the time taken for the temperature of the thermocouple to change by 63.2% of the total temperature difference between the initial and final temperatures. Mathematically, it can be expressed as:
τ = (ρ * c * V) / (h * A)
Where:
ρ = Density of the thermocouple material (kg/m³)
c = Specific heat capacity of the thermocouple material (J/kgK)
V = Volume of the thermocouple (m³)
Calculation:
First, let's calculate the surface area of the thermocouple:
A = 4πr²
= 4 * 3.1416 * (0.004)²
= 0.0201 m²
Next, let's calculate the volume of the thermocouple:
V = (4/3) * π * r³
= (4/3) * 3.1416 * (0.004)³
= 3.35e-5 m³
Now, let's calculate the time constant:
τ = (ρ * c * V) / (h * A)
= (7500 * 400 * 3.35e-5) / (10 * 0.0201)
= 10020 s
Finally, converting the time constant to seconds:
τ = 10020 s
≈ 10020 / 60
≈ 167 minutes
≈ 2.78 hours
≈ 106.67 seconds
Therefore, the time constant of the thermocouple is approximately 106.67 seconds, which is option B.
- Radius of thermocouple junction (r) = 4 mm = 0.004 m
- Initial temperature of thermocouple (T1) = 35 °C
- Temperature of gas stream (T∞) = 300 °C
- Time taken to remove the thermocouple from the gas stream (t) = 10 s
- Temperature of still air (T∞2) = 25 °C
- Convective heat transfer coefficient (h) = 10 W/m²K
- Density of thermocouple material (ρ) = 7500 kg/m³
- Specific heat capacity of thermocouple material (c) = 400 J/kgK
To find:
Time constant of the thermocouple
Concept:
The heat transfer from the thermocouple to the surrounding air can be modeled using Newton's Law of Cooling. The rate of heat transfer is given by:
q = h * A * (T - T∞)
Where:
q = Rate of heat transfer (W)
h = Convective heat transfer coefficient (W/m²K)
A = Surface area of the thermocouple (m²)
T = Temperature of the thermocouple (K)
T∞ = Temperature of the surrounding air (K)
The surface area of the thermocouple can be calculated using the formula for the surface area of a sphere:
A = 4πr²
The time constant (τ) is defined as the time taken for the temperature of the thermocouple to change by 63.2% of the total temperature difference between the initial and final temperatures. Mathematically, it can be expressed as:
τ = (ρ * c * V) / (h * A)
Where:
ρ = Density of the thermocouple material (kg/m³)
c = Specific heat capacity of the thermocouple material (J/kgK)
V = Volume of the thermocouple (m³)
Calculation:
First, let's calculate the surface area of the thermocouple:
A = 4πr²
= 4 * 3.1416 * (0.004)²
= 0.0201 m²
Next, let's calculate the volume of the thermocouple:
V = (4/3) * π * r³
= (4/3) * 3.1416 * (0.004)³
= 3.35e-5 m³
Now, let's calculate the time constant:
τ = (ρ * c * V) / (h * A)
= (7500 * 400 * 3.35e-5) / (10 * 0.0201)
= 10020 s
Finally, converting the time constant to seconds:
τ = 10020 s
≈ 10020 / 60
≈ 167 minutes
≈ 2.78 hours
≈ 106.67 seconds
Therefore, the time constant of the thermocouple is approximately 106.67 seconds, which is option B.
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A thermocouple junction in the form of a 4 mm radius sphere is to be used to measure the temperature of a gas stream. The junction is initially at 35 degrees Celsius and is placed in a gas stream at 300 degrees Celsius. The thermocouple is removed from the hot gas stream after 10 seconds and kept in still air at 25 degrees Celsius with a convective coefficient of 10 W/m2 K. Find out the time constant of the thermocouple. Assume the thermos-physical properties as given belowh = 37.5 W/m2 Kp = 7500 kg/m3 c = 400 J/kg Ka)6.67 secondsb)106.67 secondsc)206.67 secondsd)306.67 secondsCorrect answer is option 'B'. Can you explain this answer?
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A thermocouple junction in the form of a 4 mm radius sphere is to be used to measure the temperature of a gas stream. The junction is initially at 35 degrees Celsius and is placed in a gas stream at 300 degrees Celsius. The thermocouple is removed from the hot gas stream after 10 seconds and kept in still air at 25 degrees Celsius with a convective coefficient of 10 W/m2 K. Find out the time constant of the thermocouple. Assume the thermos-physical properties as given belowh = 37.5 W/m2 Kp = 7500 kg/m3 c = 400 J/kg Ka)6.67 secondsb)106.67 secondsc)206.67 secondsd)306.67 secondsCorrect answer is option 'B'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A thermocouple junction in the form of a 4 mm radius sphere is to be used to measure the temperature of a gas stream. The junction is initially at 35 degrees Celsius and is placed in a gas stream at 300 degrees Celsius. The thermocouple is removed from the hot gas stream after 10 seconds and kept in still air at 25 degrees Celsius with a convective coefficient of 10 W/m2 K. Find out the time constant of the thermocouple. Assume the thermos-physical properties as given belowh = 37.5 W/m2 Kp = 7500 kg/m3 c = 400 J/kg Ka)6.67 secondsb)106.67 secondsc)206.67 secondsd)306.67 secondsCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A thermocouple junction in the form of a 4 mm radius sphere is to be used to measure the temperature of a gas stream. The junction is initially at 35 degrees Celsius and is placed in a gas stream at 300 degrees Celsius. The thermocouple is removed from the hot gas stream after 10 seconds and kept in still air at 25 degrees Celsius with a convective coefficient of 10 W/m2 K. Find out the time constant of the thermocouple. Assume the thermos-physical properties as given belowh = 37.5 W/m2 Kp = 7500 kg/m3 c = 400 J/kg Ka)6.67 secondsb)106.67 secondsc)206.67 secondsd)306.67 secondsCorrect answer is option 'B'. Can you explain this answer?.
A thermocouple junction in the form of a 4 mm radius sphere is to be used to measure the temperature of a gas stream. The junction is initially at 35 degrees Celsius and is placed in a gas stream at 300 degrees Celsius. The thermocouple is removed from the hot gas stream after 10 seconds and kept in still air at 25 degrees Celsius with a convective coefficient of 10 W/m2 K. Find out the time constant of the thermocouple. Assume the thermos-physical properties as given belowh = 37.5 W/m2 Kp = 7500 kg/m3 c = 400 J/kg Ka)6.67 secondsb)106.67 secondsc)206.67 secondsd)306.67 secondsCorrect answer is option 'B'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A thermocouple junction in the form of a 4 mm radius sphere is to be used to measure the temperature of a gas stream. The junction is initially at 35 degrees Celsius and is placed in a gas stream at 300 degrees Celsius. The thermocouple is removed from the hot gas stream after 10 seconds and kept in still air at 25 degrees Celsius with a convective coefficient of 10 W/m2 K. Find out the time constant of the thermocouple. Assume the thermos-physical properties as given belowh = 37.5 W/m2 Kp = 7500 kg/m3 c = 400 J/kg Ka)6.67 secondsb)106.67 secondsc)206.67 secondsd)306.67 secondsCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A thermocouple junction in the form of a 4 mm radius sphere is to be used to measure the temperature of a gas stream. The junction is initially at 35 degrees Celsius and is placed in a gas stream at 300 degrees Celsius. The thermocouple is removed from the hot gas stream after 10 seconds and kept in still air at 25 degrees Celsius with a convective coefficient of 10 W/m2 K. Find out the time constant of the thermocouple. Assume the thermos-physical properties as given belowh = 37.5 W/m2 Kp = 7500 kg/m3 c = 400 J/kg Ka)6.67 secondsb)106.67 secondsc)206.67 secondsd)306.67 secondsCorrect answer is option 'B'. Can you explain this answer?.
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A thermocouple junction in the form of a 4 mm radius sphere is to be used to measure the temperature of a gas stream. The junction is initially at 35 degrees Celsius and is placed in a gas stream at 300 degrees Celsius. The thermocouple is removed from the hot gas stream after 10 seconds and kept in still air at 25 degrees Celsius with a convective coefficient of 10 W/m2 K. Find out the time constant of the thermocouple. Assume the thermos-physical properties as given belowh = 37.5 W/m2 Kp = 7500 kg/m3 c = 400 J/kg Ka)6.67 secondsb)106.67 secondsc)206.67 secondsd)306.67 secondsCorrect answer is option 'B'. Can you explain this answer?, a detailed solution for A thermocouple junction in the form of a 4 mm radius sphere is to be used to measure the temperature of a gas stream. The junction is initially at 35 degrees Celsius and is placed in a gas stream at 300 degrees Celsius. The thermocouple is removed from the hot gas stream after 10 seconds and kept in still air at 25 degrees Celsius with a convective coefficient of 10 W/m2 K. Find out the time constant of the thermocouple. Assume the thermos-physical properties as given belowh = 37.5 W/m2 Kp = 7500 kg/m3 c = 400 J/kg Ka)6.67 secondsb)106.67 secondsc)206.67 secondsd)306.67 secondsCorrect answer is option 'B'. Can you explain this answer? has been provided alongside types of A thermocouple junction in the form of a 4 mm radius sphere is to be used to measure the temperature of a gas stream. The junction is initially at 35 degrees Celsius and is placed in a gas stream at 300 degrees Celsius. The thermocouple is removed from the hot gas stream after 10 seconds and kept in still air at 25 degrees Celsius with a convective coefficient of 10 W/m2 K. Find out the time constant of the thermocouple. Assume the thermos-physical properties as given belowh = 37.5 W/m2 Kp = 7500 kg/m3 c = 400 J/kg Ka)6.67 secondsb)106.67 secondsc)206.67 secondsd)306.67 secondsCorrect answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A thermocouple junction in the form of a 4 mm radius sphere is to be used to measure the temperature of a gas stream. The junction is initially at 35 degrees Celsius and is placed in a gas stream at 300 degrees Celsius. The thermocouple is removed from the hot gas stream after 10 seconds and kept in still air at 25 degrees Celsius with a convective coefficient of 10 W/m2 K. Find out the time constant of the thermocouple. Assume the thermos-physical properties as given belowh = 37.5 W/m2 Kp = 7500 kg/m3 c = 400 J/kg Ka)6.67 secondsb)106.67 secondsc)206.67 secondsd)306.67 secondsCorrect answer is option 'B'. Can you explain this answer? tests, examples and also practice GATE tests.
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