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Unformatted text preview: PROBLEM 1.12 KN OWN: Sequenca calibration data set ufTahla LE
n=5mV
%=5mV
FITJI}: $ﬁ[eh}wux
SULUTION
By impaction aftha data, the maximum hysteresis occurs at x = 31}. For this case, 6it = (Ehlhux = Yup ' Yduqm
= {12 HIV or %(eh]m = 1130 I ('02 Itl'WS HIV)
= 4% llﬂ ”1““ PROBLEM 1.16 SOLUTION The data are plotted below. The slope of a line passing thmugh the data is 1.365 and the 3f intercept is 2.12. The data can be ﬁt to the line 3; = 1.355}: +2.12. Therefore, the static
sensitivity' is K = 1.365 for all x. 1—111 PROBLEM 1.18 KNOWN: Calibration data
Fl ND: Plot data. Estimate K. SOLUTION The data are plotted below in semilog format. A linear curve results. This suggests y = ae . Plotting y vs it in semilog format is equivalent to
plotting
logy=loga + bx From the plot, a = 5 and b = wl. Hence, the data describe y = 5e". Now, K =
dyidx Ix, so that X [psi] K
[1.135 4.?6
0.1 4.52
I15 3.l]3
Lt} 4.34 The magnitude ofthe static sensitivity decreases with x. The negative sign indicates that y
will decrease as :It increases COMMENT An aspect of this problem is to draw attention to the fact that many measurement systems
may have a static sensitivity that is dependent on input value. The operatin g principle of
many systems will determine how K behaves. llﬁ PROBLEM 1.19 KNOWN: A bulb thermometer is used to measure outside temperature. Fl ND: Extraneous variabies that might inﬂuence thermometer output. SOLUTION A thermometer's indicated temperature will be inﬂuenced by the temperature of solid objects
to which it is in contact, and radiation exchange with bodies at different temperatures
[including the sky or sun, buildings, people and ground) within its line of sight. Hence,
location should be carefully selected and even randomized. We know that a bulb
thermometer does not respond quickly,r to temperature changes, so that a sufﬁcient period of
time needs to be allowed for the instrument to adjust to new temperatures. By replication of
the measurement, effects due to instnuneot hysteresis and instrument and procedural repeatability can be randomised. Because of limited resolution in such an instrument, different competent temperature observers might record different indicated temperatures even if the instrument output were
ﬁxed. Either observers should be randomized or, if not, the test replicated. It is interesting to
note that such a randomization will bring about a predictable scatter in recorded data of about '31» the resolution of the instrument scale. 1—1? PROBLEM 1.22 KNUWN: FSD = lﬂﬂﬂ N
FIND: e;
SOLUTION From the given speciﬁcations, the elemental errors are estimated by: eL= o_oo1 x IUUUN E1N
eH = ﬂﬂlﬂl 3: WHEN =1N
on = llﬂﬂlﬁ x lilIUGN =1.5N
ez = {lﬂﬂﬁ 1r IDDDN = EN The overall instrument error is estimated as:
e= [12 +12 +1.52 +22)”2 =2.9N CEMENT This rootsumsquare {R33} method provides a "probable“ estimate (i.e. the most liloe11.F
estimate) of the instrument error possible in an}Ir given measurement. "Possible" is a big word
here as error values will most likely change between measurements. 120 PROBLEM 1 .33 KNOWN: Water at 20°C Q =flCﬁ1de}
C = are; D = 1 1n ZﬂQﬁlﬂemm FIND: Expected ealihratinn curve SOLUTION Part of a test matrix is to specify the range of the independent variable and to anticipate the
range resulting in the dependent 1.rariahle. In this ease, the pressure drop will be measured se
that it is the dependent variable during a static calibraLiGn. TD anticipate the autism range of
the calibration then: Rearranging the known relation,
ap = [qrcafptz
Fer p= 993 1(ng (Appendix C}, and A = 71:sz4, we ﬁnd:
Q {emm} dp (Wm2 2 1.6
5 lﬂ
11] 4t] This is plotted below. It is clear that K will not he a constant as K = {{Q}. 1—29 ...
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 Spring '08
 Singh,V

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