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Hello, I need some help with a few instrumenttation questions. Here is the first,,, One of the advantages of the millivoltmeter ovr the potentiometer in measuring thermocouple emf is that,,, B- millivoltmeter require no calibration in terms of internal or external resistance,,,C- readings on millivoltmeter scales are indicated by pointer position without adjustment,,,,, D- millivoltmeter scales are directly proportional to the length of the slide-wires. The second question is,,,A null-balance potentiometer indicates the temperature of a thermocouple by,, A- measuring the current with a galvanometer,,, B- measuring the voltage across the thermocouple refernce junction,,,, C- making the slide wire voltage equal to the voltage from the thermocouple. I think the answer to the second question is C,, the first question i am lost on. Can anyone help me out on these?😔0
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Member • Apr 29, 2010
you are correct,the 2nd one correct one because both are measuring meters only but in potentiometer we need extra meters to note down the values from that meter values we calculate the voltage value,so it is indirect way.so its better for to use mili voltmeter it simply shows the reading.
is there any wrong thing in my answer please do reply with reasons
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Member • Apr 30, 2010
for second one it is 'C' , for first one it is 'C' since emf generated at the junction of thermocouple is in Millivolts,which can be directly read by millivoltmeter without need of balancing as in case of potentiometer, please contradict if am wrong!!!Are you sure? This action cannot be undone. -
Member • Apr 30, 2010
Thank You for your input, I beleive you both are right on the first two, I have another question to get some input on,,,, Eletrostatic interference can be nullified by,,, A- connecting the thermocouple ground to the positive side of the junction,,,, B- twisting and grounding the thermocouple wires,,, or C- sheilding and grounding the signal leads..Are you sure? This action cannot be undone. -
Member • Apr 30, 2010
To minimize noise pickup, a metal over braid or twisted shielded wire is commonly used.I think 'C' is correctAre you sure? This action cannot be undone. -
Member • May 15, 2010
I think you might be right but i would like to get a few more thoughts on this, need some more input on this fellow's! Also how would i calculate the amount of thermal energy radiated by an oxidized cast- iron surface at 100 degre c. The surface area of the body is 15 square meters.sauravgoswamiTo minimize noise pickup, a metal over braid or twisted shielded wire is commonly used.I think 'C' is correctAre you sure? This action cannot be undone. -
Member • May 21, 2010
Anyone out there?? Really need some help with this!!😔Are you sure? This action cannot be undone. -
Member • May 1, 2011
The millivoltmeter used must be a high input impedance one for accurate readings. Otherwise the meter itself will draw some current which will lower the indicated millivolts and therefore the temperature.trexHello, I need some help with a few instrumenttation questions. Here is the first,,, One of the advantages of the millivoltmeter ovr the potentiometer in measuring thermocouple emf is that,,, B- millivoltmeter require no calibration in terms of internal or external resistance,,,C- readings on millivoltmeter scales are indicated by pointer position without adjustment,,,,, D- millivoltmeter scales are directly proportional to the length of the slide-wires. The second question is,,,A null-balance potentiometer indicates the temperature of a thermocouple by,, A- measuring the current with a galvanometer,,, B- measuring the voltage across the thermocouple refernce junction,,,, C- making the slide wire voltage equal to the voltage from the thermocouple. I think the answer to the second question is C,, the first question i am lost on. Can anyone help me out on these?😔
All modern digital meters have an impedance of >11 megohms and so are suitable. If you ask me, other than the nuisance of adjusting slide wires and the large unwieldy size the slide wire potentiometer is quite accurate.
BioramaniAre you sure? This action cannot be undone. -
Member • May 1, 2011
The ambient temperature has a bearing on this. As a first degree approximation assuming 25 C as the ambient the loss will be about 10 KCal/hour/sq.mtr./degC temp diff between surface and ambient.trexI think you might be right but i would like to get a few more thoughts on this, need some more input on this fellow's! Also how would i calculate the amount of thermal energy radiated by an oxidized cast- iron surface at 100 degre c. The surface area of the body is 15 square meters.
BioramaniAre you sure? This action cannot be undone. -
Member • Jul 18, 2012
right ............Are you sure? This action cannot be undone. -
Member • Aug 12, 2012
Hi everyone..
i will like to put my experience with thermocouples and the electrostatic dangers.
i was in rajsthan in a cement plant when there was erection and commissioning going on both instrumentation side and DCS side. there was this particular section of thermocouples installed in the cooler section i guess which were showing erroneous values after 2-3 days of its commissioning. we had a lot of pressure saying that it was DCS side mistake and customer was shouting on us like anything. then the next day we went to the thermocouples junction box to remove and check the wiring. the junction box was having the electrostatic charge and my senior had put his tester to open the bolts (maybe the tester had some charge) when suddenly there was a minor explosive sound and me and my senior was thrown to the ground. that was dead fast and we couldnt understand as to what the hell happened. then we shut down the area and directly connected a ground cable tot he JB and then opened it only to see the terminal blocks were blackened by the electrostatic discharge which shorted through the JB.
always use teflon/teflon/metal braided cable for the thermocouples till the JB or till its first termination point. (even if you use isolators).Are you sure? This action cannot be undone.