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The E.M.F (E) in a thermocouple at any temperature T is given as E=0.2T−5×10−1T2 where T is temperature in ( C). Show that at steam point (100∘C) the thermocouple will generate an emf ( E) of 15mV.?
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The E.M.F (E) in a thermocouple at any temperature T is given as E=0.2...
Given Information:
The EMF (E) in a thermocouple at any temperature T is given as E = 0.2T - 5×10^(-1)T^2, where T is the temperature in degrees Celsius.
Required:
To show that at the steam point (100°C), the thermocouple will generate an EMF (E) of 15mV.
Calculation:
To find the EMF (E) at the steam point (100°C), we substitute T = 100°C into the given equation.
E = 0.2T - 5×10^(-1)T^2
E = 0.2(100) - 5×10^(-1)(100)^2
E = 20 - 5×10^(-1)×10000
E = 20 - 5×1000
E = 20 - 5000
E = -4980mV
Discussion:
The calculated EMF at the steam point is -4980mV. However, the required EMF is 15mV, which is a positive value. Therefore, it seems that there might have been an error in the calculation or the given equation.
Correction:
Upon reviewing the given equation, it appears that there was a mistake in the equation. The correct equation for the EMF (E) in a thermocouple at any temperature T should be:
E = 0.2T - 5×10^(-3)T^2
Let's recalculate the EMF (E) at the steam point (100°C) using the corrected equation:
E = 0.2T - 5×10^(-3)T^2
E = 0.2(100) - 5×10^(-3)(100)^2
E = 20 - 5×10^(-3)×10000
E = 20 - 5×10^(-3)×10000
E = 20 - 50
E = -30mV
Conclusion:
After recalculating using the correct equation, we find that the EMF (E) at the steam point (100°C) is -30mV, which is not equal to the required value of 15mV. This indicates that there might be a mistake in the given equation or some other factor affecting the EMF generation. Further analysis or clarification is needed to determine the correct EMF at the steam point.
The EMF (E) in a thermocouple at any temperature T is given as E = 0.2T - 5×10^(-1)T^2, where T is the temperature in degrees Celsius.
Required:
To show that at the steam point (100°C), the thermocouple will generate an EMF (E) of 15mV.
Calculation:
To find the EMF (E) at the steam point (100°C), we substitute T = 100°C into the given equation.
E = 0.2T - 5×10^(-1)T^2
E = 0.2(100) - 5×10^(-1)(100)^2
E = 20 - 5×10^(-1)×10000
E = 20 - 5×1000
E = 20 - 5000
E = -4980mV
Discussion:
The calculated EMF at the steam point is -4980mV. However, the required EMF is 15mV, which is a positive value. Therefore, it seems that there might have been an error in the calculation or the given equation.
Correction:
Upon reviewing the given equation, it appears that there was a mistake in the equation. The correct equation for the EMF (E) in a thermocouple at any temperature T should be:
E = 0.2T - 5×10^(-3)T^2
Let's recalculate the EMF (E) at the steam point (100°C) using the corrected equation:
E = 0.2T - 5×10^(-3)T^2
E = 0.2(100) - 5×10^(-3)(100)^2
E = 20 - 5×10^(-3)×10000
E = 20 - 5×10^(-3)×10000
E = 20 - 50
E = -30mV
Conclusion:
After recalculating using the correct equation, we find that the EMF (E) at the steam point (100°C) is -30mV, which is not equal to the required value of 15mV. This indicates that there might be a mistake in the given equation or some other factor affecting the EMF generation. Further analysis or clarification is needed to determine the correct EMF at the steam point.
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The E.M.F (E) in a thermocouple at any temperature T is given as E=0.2T−5×10−1T2 where T is temperature in ( C). Show that at steam point (100∘C) the thermocouple will generate an emf ( E) of 15mV.?
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The E.M.F (E) in a thermocouple at any temperature T is given as E=0.2T−5×10−1T2 where T is temperature in ( C). Show that at steam point (100∘C) the thermocouple will generate an emf ( E) of 15mV.? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about The E.M.F (E) in a thermocouple at any temperature T is given as E=0.2T−5×10−1T2 where T is temperature in ( C). Show that at steam point (100∘C) the thermocouple will generate an emf ( E) of 15mV.? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The E.M.F (E) in a thermocouple at any temperature T is given as E=0.2T−5×10−1T2 where T is temperature in ( C). Show that at steam point (100∘C) the thermocouple will generate an emf ( E) of 15mV.?.
The E.M.F (E) in a thermocouple at any temperature T is given as E=0.2T−5×10−1T2 where T is temperature in ( C). Show that at steam point (100∘C) the thermocouple will generate an emf ( E) of 15mV.? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about The E.M.F (E) in a thermocouple at any temperature T is given as E=0.2T−5×10−1T2 where T is temperature in ( C). Show that at steam point (100∘C) the thermocouple will generate an emf ( E) of 15mV.? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The E.M.F (E) in a thermocouple at any temperature T is given as E=0.2T−5×10−1T2 where T is temperature in ( C). Show that at steam point (100∘C) the thermocouple will generate an emf ( E) of 15mV.?.
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