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chapter1soultions

chapter1soultions - PROBLEM 1.12 KN OWN Sequenca...

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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% l-lfl ”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 semi-log format. A linear curve results. This suggests y = ae . Plotting y vs it in semi-log 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. l-lfi 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 indicate-d 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 root-sum-square {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. 1-20 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 calibraLiG-n. 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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