ME 322: Instrumentation Lecture 17 February 27, 2015 Professor Miles Greiner Temperature measurements, thermocouple circuits, thermocouple demo Announcements/Reminders • HW 6 due now • HW 7 due Friday • Lab 6 next week – Only 4 wind tunnels (we are constructing a 5th) – Watch your WebCampus to find when your group is scheduled to attend lab. • Bring Excel from HW 6 and use it to process the data you acquire. • This will help check the data as you take it and allow you to complete the data acquisition phase of the lab in one hour Midterm I Scores • Average 75, St. Dev 18 – In 2014 it was 74 and 18 (very similar) • Solutions posted outside PE 213 – I will only consider revising scores before Wednesday, March 4, 2015 Phenomena used to Measure Temperatures • Liquid density change (in glass thermometer) • Metal Deformation (Coil, bimetallic strips) • Gas Pressure • Wire resistance • Problem – All devices act line fins and affect the temperature of the locations that they are measuring Thermocouples π΄ π΅ • Employ the Seebeck Effect – When two dissimilar metals (A & B) are in contact, a small electrical potential (voltage) is produced that depends on the junction temperature. • Probes can consist of two wires and be inexpensive • Rugged shielded probes can be expensive Demonstration (three junctions) Iron 2 + VOUT 1 3 Put into Ice VOUT 1 Fe/Con down 2 Cu/Fe little change 3 Con/Cu up Constantan Ni/Cu • For demo use type-J thermocouple pair (Iron/Constant) connected to a copper (Cu) digital voltmeter • Output is in the 10’s of microvolts – 10mV = 0.01 mV = 0.00001 V • VOUT depends on all three junction temperatures – The sensitivity of VOUT to temperature is not the same for all the junctions. Thermocouple Circuit Metal C + TS TT = Terminal Block Temp ≈ uniform VOUT - TS HE WOUT TR TR • Four junctions, including reference • Let VCA(T) be voltage decrease going from C to A at junction temperature T – VCA(T) = VC(T) - VA(T) • πππ’π‘ = π+ − π− = VCA(ππ ) + VAB(ππ )+ VBA(ππ )+ VAC(ππ ) – How are these voltage related? • ππΆπ΄ ππ = ? ππ΄πΆ ππ • ππ΅π΄ ππ = ? ππ΄π΅ ππ • πππ’π‘ = ππ΄π΅ ππ − ππ΄π΅ ππ (transfer function, ππ desired, ππ undesired) – If terminal block is isothermal, then πππ’π‘ not dependent on Temperature TT or metal C – How to find ππ΄π΅ π ? • 2nd Law of Thermodynamics (heat engine) – If TS = TR, then VOUT = ? Standardization • Industry uses standard wire material pairs (page 276) • The composition of the two wires must be well-controlled and sufficiently-different to give predictable (small uncertainty) and useful (sensitive) voltages • Different wire pairs have different operating ranges and sensitivities, S = dVTC/dT = d(Reading)/d(Measurand) How to find VAB(T) VOUT T TR= 0°C • Material Science Calculations, or • Calibration: • Put reference junction is pure water/Ice Slurry, TR = 0°C • Measure VOUT for a range of T • See Page 277 for results • Not really linear • Different sensitivities (slopes) • Standard wire uncertainty: – Larger of 2.2°C or 0.7% of measurement Circuits without a Reference Junction TT TS TT ? ? • πππ’π‘ = ππΆπ΄ ππ + ππ΄π΅ ππ + ππ΅πΆ ππ = ? – Problem, we have data for wire pair AB, but not CA or CB • “Thought” experiment: If TS = TT , then by 2nd law ? ππΆπ΄ ? – πππ’π‘ = ππ + ππ΄π΅ ππ + ππ΅πΆ ππ = ? – So ππΆπ΄ ππ + ππ΅πΆ ππ = −ππ΄π΅ ππ (effect of C cancels out) • πππ’π‘ = ππ΄π΅ ππ − ππ΄π΅ ππ – Don’t need VCA(T) or VBC(T) data to find this transfer function! Problem 9.22 A type E thermocouple is placed in an oven and connected to a computer data-acquisition system. The junction box temperature is independently measured to be 30°C. The thermocouple voltage is found to be 37.0 mV. What is the temperature of the oven? Thermocouple Signal Conditioner TS (°C) 0 400 VSC (V) 0 10 ? Out of range Transfer Function 10 Reading VSC [V] πππΆ = 0 0 • Measurand, T [°C] In lab use Omega DRE–TC-J; for Type J (Iron/Constantan) thermocouples ππππΆ ππ 400 – http://wolfweb.unr.edu/homepage/greiner/teaching/MECH322Instrumentation/Labs/Lab%2007 %20Boiling%20Water%20Temperature/Lab7%20Index.htm – Wiring: Iron (white insulation) goes to +Tc; Constantan (red stripe) goes to (-Tc) – Transfer Function: πππΆ = 10π • • πππΆ = ππππΆ ππ = 10π 400°πΆ ππ 400β π π β = 0.0025 ππ = = 0.025 °πΆ ; πππΆ = ππππΆ ππ – Inverted transfer function: TS = (40°C/V)*VSC Conditioner Provides – – – – – π π = 0.00005 °πΆ ; Gπππ = πππΆ = 500 ππΆ Reference Junction Compensation Amplification Low Pass Filter (RF noise rejection) Linearization Galvanic Isolation (avoid ground loops even in water)