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Kalibrering av passbiter ved komparering:
Toleranser 00 1 2 L = 50 mm 0,10 0,20 0,40 0,80 L = 100 mm 0,14 0,30 0,60 1,20 β π x = πΏ πΌ x ( π‘ x β π‘ 0 ) β π s = πΏ πΌ s ( π‘ s β π‘ 0 ) β π x =0,1 π β 11,5 Β΅/K β 0,5 K = 0,58 Β΅m β π s =0,1 π β 4,25 Β΅/K β 0,5 K = 0,21 Β΅m βπ=0,37 Β΅π pumpe MΓ₯lefunksjonen: π π₯ +πΏ πΌ π₯ π‘ π₯ β π‘ 0 = π π +πΏ π π· +πΏπ+πΏ π πΆ βπΏ π π +πΏ πΌ π π‘ π β π‘ 0 π π₯ = π π +πΏ π π· +πΏπ+πΏ π πΆ βπΏ π π βπΏ πΌ πΏπ‘βπΏ πΏπΌ β π‘ Se ogsΓ₯ mΓ₯lefunksjonen i EA-4/02 S4
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Usikkerhet pga temperatur:
Toleranser 00 1 2 L = 50 mm 0,10 0,20 0,40 0,80 L = 100 mm 0,14 0,30 0,60 1,20 π π₯ = π π +πΏ π π· +πΏπ+πΏ π πΆ βπΏ π π βπΏ πΌ πΏπ‘βπΏ πΏπΌ β π‘ π’ 1 (πΏ πΌ πΏπ‘) π’ 2 (πΏ πΏπΌ β π‘ ) πΈ(πΏπ‘)=0 π’ πΏπ‘ = 50 ππΎ 3 =29 ππΎ πΈ β π‘ =0 π’ β π‘ = 0,5 πΎ 3 =0,29 πΎ π’ 1 =πΏ π’ πΌ πΏπ‘ = πΏ πΌ π’ πΏπ‘ =0,033 Β΅π πΌ π₯ = πΌ π = πΌ π π‘Γ₯π π’ 2 = πΏ π’ πΏπΌ β π‘ =0,024 Β΅π πΈ(πΏπΌ)=0 πΌ π₯ β πΌ π : π’ 1 =πΏ π’ πΌ πΏπ‘ =πΏ πΌ π’ πΏπ‘ =0,023 Β΅π πΈ πΏπΌ οΉ 0 π’ 2 =π’ πΏ πΏπΌ β π‘ =πΏ πΏπΌ π’ β π‘ =0,21 Β΅π Γ
bruke forskjellig materiale i bitene i komparatoren medfΓΈrer stor usikkerhet i lengdemΓ₯lingen pga denne usikkerhetskomponenten!
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MΓ₯lekapabilitet for passbiter med like a eller forskjellig a :
πΆ π = ππππππππ π π(95%
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Termoelement, klassisk konfigurasjon
Materiale 1 Materiale 2 T21 V+ V- Copper T21 = T22 T1 οΉ T2 T+ = T- V+ ο οΉ V- V21 ο οΉ V22 ο ο οV = V+ - V- = (V+ - V21) (V21 β V1) (V1 - V22) (V22 - V-) = SCu(T+ - T21) SM1(T21 - T1) SM2(T1 - T22) SCu(T22 - T-) Seebeck coefficient, Β΅V/K Constantan: -35 Β΅V/K Copper: +6,5 Β΅V/K = SM1(T2 - T1) β SM2(T2 - T1) = EType T(T2 - T1) βThe underlying solid-state physics is rather complicated but the phenomenon is well understood. The magnitude and sign of the Seebeck coefficient are related to an asymmetry of the electron distribution around the Fermi level. For example, the theory explains not only the different signs for metals but also why the Seebeck coefficient is much larger in semiconductors than in metals.β I beregningene forutsettes homogent materiale (samme S) langs hele lederen, og vi fΓ₯r: οV = π 1 π 2 π π ππ = π π ( π 2 β π 1 )
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MΓ₯ling av temperaturforskjell, dT = Ts β Thm, med termoelement type T
Konstantan, Γ 0,25 mm Kopper Kopper, Γ 0,25 mm Elektrisk (og termisk) isolasjon mot et elektrisk ledende bord Ts Thm Vs Vhm ο ο οV = V+ - V- = (V+ - Vhm) (Vhm - Vs) (Vs - V-) = SCu(T+ - Thm) SC(Thm - Ts) SCu(Ts - T-) Bommull = SCu(Ts - Thm) β SC(Ts - Thm) Latex V- V+ = EType T(Ts - Thm) Nanovoltmeter Papir Tilkoplingsledninger, ogsΓ₯ koppertrΓ₯d. Sensitivitet nΓ¦r 20 Β°C, type T-element: 40 Β΅V/K Det vil si: 25 mK/Β΅V Takk til Max Sievert ved Lars Gulliksen og Finn Enger!
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Typisk temperaturendring for stΓ₯l, 30 s oppvarming
Bar hud: 316 mK Latexhanske: 282 mK Papir: 151 mK Bommullshanske: 97 mK
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Typisk temperaturendring for hardmetall:
Bommullshanske: 206 mK Latexhanske: 525 mK Bar hud: 590 mK Papir: 208 mK
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Bestemmer tidskonstanten, t, for stΓ₯l:
βπ=β π ππππ ππ‘ +( π ππ‘πππ‘ β β π ππππ ππ‘ )β π β π‘ π βπ ππ‘πππ‘ =102 ππΎ β π ππππ ππ‘ =β14 ππΎ π=11,3 πππ Gjennomsnitt, t = 10,91 min Stdav til gj.snitt, s/ο3 = 0,22 min
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Bestemmer tidskonstanten, t, for hardmetall:
βπ=β π ππππ ππ‘ +( π ππ‘πππ‘ β β π ππππ ππ‘ )β π β π‘ π βπ ππ‘πππ‘ = β90 ππΎ β π ππππ ππ‘ =10 ππΎ π=7,2 πππ Gjennomsnitt, t = 6,70 min Stdav til gj.snitt, s/ο3 = 0,55 min
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Hvor lang tid tar temperaturstabiliseringen?
Lord Kelvin (1824 β 1907): Β«I often say that when you can measure what you are speaking about, and express it in numbers, you know something about it; but when you cannot measure it, when you cannot express it in numbers, your knowledge is of a meagre and unsatisfactory kind.Β»
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