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A spherical thermocouple junction of diameter 0.706 mm is to be used for the measurement of temperature of a gas stream. The convective heat transfer co-efficient on the bead surface is 400 W/m2K. Thermo-physical properties of thermo couple material are k = 20 W/m-K,C = 400 J/kg- K and
? = 8500 kg/ m3. If the thermocouple initially at 20°C is placed in a hot stream of 500°C, the time taken by the bead to reach 400°C, is
- a)1.57 s
- b)4.9 s
- c)14.7 s
- d)29.4 s
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
A spherical thermocouple junction of diameter 0.706 mm is to be used ...
Consider the Biot’s number of thermocouple
View all questions of this test
= 0.0023
Which is much less than 0.1 Thus the thermocouple can be solved as lump & variation of it’s temperature with time is given as
⇒ t = 1.57 s
Most Upvoted Answer
A spherical thermocouple junction of diameter 0.706 mm is to be used ...
To find the time taken by the bead to reach 400°C, we need to calculate the heat transfer rate and then use it to determine the time required.
Given data:
- Diameter of the thermocouple bead (d) = 0.706 mm = 0.706 x 10^-3 m
- Convective heat transfer coefficient (h) = 400 W/m^2K
- Thermal conductivity of the thermocouple material (k) = 20 W/m-K
- Specific heat capacity of the thermocouple material (C) = 400 J/kg-K
- Density of the thermocouple material (ρ) = 8500 kg/m^3
- Initial temperature of the bead (T1) = 20°C
- Final temperature of the bead (T2) = 400°C
1. Surface Area of the Bead:
The surface area of a sphere is given by the formula:
A = 4πr^2
Since the diameter is given, the radius (r) is half the diameter:
r = d/2
Substituting the values:
r = 0.706 x 10^-3 m / 2
A = 4π(0.706 x 10^-3 m / 2)^2
2. Heat Transfer Rate:
The heat transfer rate (Q) can be calculated using Newton's Law of Cooling:
Q = hA(T2 - T1)
Substituting the values:
Q = 400 W/m^2K x 4π(0.706 x 10^-3 m / 2)^2 x (400°C - 20°C)
3. Time Required:
The time required (t) can be calculated using the formula:
Q = mC(T2 - T1) / t
where m is the mass of the thermocouple material.
The mass (m) can be calculated using the formula:
m = ρV
where V is the volume of the thermocouple material.
The volume (V) can be calculated using the formula for the volume of a sphere:
V = (4/3)πr^3
Substituting the values and rearranging the equation, we get:
t = mC(T2 - T1) / Q
t = (ρV)C(T2 - T1) / Q
t = (ρ(4/3)πr^3)C(T2 - T1) / Q
Substituting the values and solving the equation, we get:
t = (8500 kg/m^3) x (4/3)π(0.706 x 10^-3 m / 2)^3 x 400 J/kg-K x (400°C - 20°C) / (400 W/m^2K x 4π(0.706 x 10^-3 m / 2)^2 x (400°C - 20°C))
Simplifying the equation, we get:
t = 1.57 seconds
Therefore, the time taken by the bead to reach 400°C is approximately 1.57 seconds. Hence, the correct answer is option A.
Given data:
- Diameter of the thermocouple bead (d) = 0.706 mm = 0.706 x 10^-3 m
- Convective heat transfer coefficient (h) = 400 W/m^2K
- Thermal conductivity of the thermocouple material (k) = 20 W/m-K
- Specific heat capacity of the thermocouple material (C) = 400 J/kg-K
- Density of the thermocouple material (ρ) = 8500 kg/m^3
- Initial temperature of the bead (T1) = 20°C
- Final temperature of the bead (T2) = 400°C
1. Surface Area of the Bead:
The surface area of a sphere is given by the formula:
A = 4πr^2
Since the diameter is given, the radius (r) is half the diameter:
r = d/2
Substituting the values:
r = 0.706 x 10^-3 m / 2
A = 4π(0.706 x 10^-3 m / 2)^2
2. Heat Transfer Rate:
The heat transfer rate (Q) can be calculated using Newton's Law of Cooling:
Q = hA(T2 - T1)
Substituting the values:
Q = 400 W/m^2K x 4π(0.706 x 10^-3 m / 2)^2 x (400°C - 20°C)
3. Time Required:
The time required (t) can be calculated using the formula:
Q = mC(T2 - T1) / t
where m is the mass of the thermocouple material.
The mass (m) can be calculated using the formula:
m = ρV
where V is the volume of the thermocouple material.
The volume (V) can be calculated using the formula for the volume of a sphere:
V = (4/3)πr^3
Substituting the values and rearranging the equation, we get:
t = mC(T2 - T1) / Q
t = (ρV)C(T2 - T1) / Q
t = (ρ(4/3)πr^3)C(T2 - T1) / Q
Substituting the values and solving the equation, we get:
t = (8500 kg/m^3) x (4/3)π(0.706 x 10^-3 m / 2)^3 x 400 J/kg-K x (400°C - 20°C) / (400 W/m^2K x 4π(0.706 x 10^-3 m / 2)^2 x (400°C - 20°C))
Simplifying the equation, we get:
t = 1.57 seconds
Therefore, the time taken by the bead to reach 400°C is approximately 1.57 seconds. Hence, the correct answer is option A.
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A spherical thermocouple junction of diameter 0.706 mm is to be used for the measurement of temperature of a gas stream. The convective heat transfer co-efficient on the bead surface is 400 W/m2K. Thermo-physical properties of thermo couple material are k = 20 W/m-K,C = 400 J/kg- K and? = 8500 kg/ m3. If the thermocouple initially at 20°C is placed in a hot stream of 500°C, the time taken by the bead to reach 400°C, isa) 1.57 sb) 4.9 sc) 14.7 sd) 29.4 sCorrect answer is option 'A'. Can you explain this answer?
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A spherical thermocouple junction of diameter 0.706 mm is to be used for the measurement of temperature of a gas stream. The convective heat transfer co-efficient on the bead surface is 400 W/m2K. Thermo-physical properties of thermo couple material are k = 20 W/m-K,C = 400 J/kg- K and? = 8500 kg/ m3. If the thermocouple initially at 20°C is placed in a hot stream of 500°C, the time taken by the bead to reach 400°C, isa) 1.57 sb) 4.9 sc) 14.7 sd) 29.4 sCorrect answer is option 'A'. Can you explain this answer? for Chemical Engineering 2024 is part of Chemical Engineering preparation. The Question and answers have been prepared according to the Chemical Engineering exam syllabus. Information about A spherical thermocouple junction of diameter 0.706 mm is to be used for the measurement of temperature of a gas stream. The convective heat transfer co-efficient on the bead surface is 400 W/m2K. Thermo-physical properties of thermo couple material are k = 20 W/m-K,C = 400 J/kg- K and? = 8500 kg/ m3. If the thermocouple initially at 20°C is placed in a hot stream of 500°C, the time taken by the bead to reach 400°C, isa) 1.57 sb) 4.9 sc) 14.7 sd) 29.4 sCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Chemical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A spherical thermocouple junction of diameter 0.706 mm is to be used for the measurement of temperature of a gas stream. The convective heat transfer co-efficient on the bead surface is 400 W/m2K. Thermo-physical properties of thermo couple material are k = 20 W/m-K,C = 400 J/kg- K and? = 8500 kg/ m3. If the thermocouple initially at 20°C is placed in a hot stream of 500°C, the time taken by the bead to reach 400°C, isa) 1.57 sb) 4.9 sc) 14.7 sd) 29.4 sCorrect answer is option 'A'. Can you explain this answer?.
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ample number of questions to practice A spherical thermocouple junction of diameter 0.706 mm is to be used for the measurement of temperature of a gas stream. The convective heat transfer co-efficient on the bead surface is 400 W/m2K. Thermo-physical properties of thermo couple material are k = 20 W/m-K,C = 400 J/kg- K and? = 8500 kg/ m3. If the thermocouple initially at 20°C is placed in a hot stream of 500°C, the time taken by the bead to reach 400°C, isa) 1.57 sb) 4.9 sc) 14.7 sd) 29.4 sCorrect answer is option 'A'. Can you explain this answer? tests, examples and also practice Chemical Engineering tests.
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