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Solution Manual For Introduction To Engineering Experimentation, 3rd Edition

Solution Manual For Introduction To Engineering Experimentation, 3rd Edition offers step-by-step solutions to help you understand tough concepts with ease.

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    1. 1 CHAPTER 1 1 . 1 (a) The weight of a person is 175 pounds which is equivalent to 778 Newtons. lbf 175 * lbf N 1 4482 . 4 = 778 Newtons (b) The magnetic force of 1,200 dynes is equivalent to 0.0027 pounds or 0.012 Newtons. dyn 1200 * dyn N 1 10 5 − = 0.012 Newtons dyn 1200 * dyn N 1 10 5 − * N lbf 4482 . 4 1 = 0.0027 pounds (c) The force exerted by the spring is 10 Newtons which is equivalent to 1,000,000 dynes or 2.248 pounds N 10 * N dyn 5 10 1 − = 1,000,000 dyne N 10 * N lbf 4482 . 4 1 = 2.248 pounds 1.2 (a) The tire pressure is 30 psig which is equivalent to 2.069 bars or 2.041 atm. Note: 1 psig is considered to be = 1 psi psig 30 * psi atm 696 . 14 1 = 2.041 atm psig 30 * psi kPa 1 895 . 6 * kPa bar 100 1 = 2.069 bars (b) The compressed air at 10 bars is equivalent to 145 psig or 9.87 atm or 295.3 inches of mercury or 102 meters of water bars 10 * bar kPa 1 100 * kPa psi 895 . 6 1 = 145 psi bars 10 * bar kPa 1 100 * kPa psi 895 . 6 1 * psi atm 696 . 14 1 = 9.87 atm bars 10 * bar kPa 1 100 * kPa psi 895 . 6 1 * psi atm 696 . 14 1 * atm cmHG 1 76 * cm in 54 . 2 1 = 295.3 in of Hg bars 10 * bar kPa 1 100 * kPa psi 895 . 6 1 * psi atm 696 . 14 1 * atm cmWater 1 2 . 1033 * cm m 100 1 = 102 meters water (c) 50 cm of water is equivalent to 0.0490 bars or 0.711 psi or 1.45 inches of mercury cmWater 50 * cmWater atm 2 . 1033 1 * atm kPa 1 325 . 101 * kPa bar 100 1 = 0.0490 bars cmWater 50 * cmWater atm 2 . 1033 1 * atm kPa 1 325 . 101 * kPa psi 895 . 6 1 * = 0.711 psi cmWater 50 * cmWater atm 2 . 1033 1 * atm cmHG 1 76 * cm in 54 . 2 1 = 1.45 in of Hg

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    1. 2 1.3 (a) The household energy use of 750 kWh is equivalent to 2,559,241.71 Btu or 645,007 kCal or 2,700,000,000 joules kWh 750 * kWh kJ 1 3600 * kJ J 1 1000 * J Btu 1055 1 = 2,559,241.71 Btu kWh 750 * kWh kJ 1 3600 * kJ J 1 1000 * J cal 186 . 4 1 * cal kcal 1000 1 = 645,007 kcal kWh 750 * kWh kJ 1 3600 * kJ J 1 1000 = 2,700,000,000 J (b) The gas water heater uses 50,000 Btu or 52,750,000 joules or 12,601.5 kcal or 38,908,386 ft.lbf Btu 50000 * Btu J 1 1055 = 52,750,000 joules Btu 50000 * Btu J 1 1055 * J cal 186 . 4 1 * cal kcal 1000 1 = 12,601.5 kcal Btu 50000 * Btu lbf ft 00128507 . 0 . 1 = 38,908,386 ft.lbf (c) The amount of heat required is 250 kCal which is equivalent to 992 Btu or 250,000 Cal or 1,046,500 joules kcal 250 * kcal cal 1 1000 * cal J 1 186 . 4 * J Btu 1055 1 = 992 Btu kcal 250 * kcal cal 1 1000 = 250,000 cal kcal 250 * kcal cal 1 1000 * cal J 1 186 . 4 = 1,046,500 J

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    1. 3 1.4 (a) The automobile rating of 150 hp is equivalent to 112 kW or 82,500 ft.lbf/sec or 106.02 Btu/sec hp 150 * hp W 1 7 . 745 * W kW 1000 1 = 112 kW hp 150 * hp s lbf ft 1 . 550 = 82,500 ft.lbf/sec hp 150 * hp W 1 7 . 745 * W s Btu 04 . 1055 1 = 106.02 Btu/sec (b) The truck rating of 400 kW is equivalent to 536 hp or 295,025 ft.lbf/sec or 379 Btu/sec kW 400 * kW W 1 1000 * W hp 7 . 745 1 = 536 hp kW 400 * kW W 1 1000 * W hp 7 . 745 1 * hp s lbf ft 1 . 550 = 295,025 ft.lbf/sec kW 400 * kW W 1 1000 * W s Btu 04 . 1055 1 = 379 Btu/sec (c) The water heater rating of 40,000 Btu/hr is equivalent to 11.72 kW hr Btu 40000 * hr Btu W 1 293 . 0 * W kW 1000 1 = 11.72 kW

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    1. 4 1.5 (a) 50 o F is equivalent to 10 o C ( 9 5 )(50 – 32) = 10 o C (b) 150 o C is equivalent to 302 o F ( 5 9 )(150 o C) + 32 = 302 o F (c) The water temperature increase of 40 o C is equivalent to a change of 40 K or 72 o F or 72 o R (40 o C) * C K   1 1 = 40 K (40 o C) * C F   1 8 . 1 = 72 o F (40 o C) * C R   1 8 . 1 = 72 o R (d) The air temperature change of 30 o F is equivalent to a change of 16.7 K or 16.7 o C or 30 o R (30 o F) * F K   1 556 . 0 = 16.7 K (30 o F) * F C   1 556 . 0 = 16.7 o C (30 o F) * F R   1 1 = 30 o R

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    1. 5 1.6 (a) 4 gallons is equivalent to 15.1 liter or 15,142 cm 3 or 0.535 ft 3 gal 4 * gal m 1 0037854 . 0 3 * 3 3 10 1 m Liter − = 15.1 liter gal 4 * gal m 1 0037854 . 0 3 * 3 6 3 10 1 m cm − = 15,142 cm 3 gal 4 * gal 1 0037854 . 0 * 3 3 02832 . 0 1 m ft = 0.535 ft 3 (b) 10 liters is equivalent to 2.64 gallons or 10,000 cm 3 or 0.353 ft 3 Liter 10 * liter m 1 10 3 3 − * 3 0037854 . 0 1 m gal = 2.64 gallons Liter 10 * liter m 1 10 3 3 − * 3 6 3 10 1 m cm − = 10,000 cm 3 Liter 10 * liter m 1 10 3 3 − * 3 3 02832 . 0 1 m ft = 0.353 ft 3 (c) 5 ft 3 is equivalent to 37.4 gallons or 141,600 cm 3 or 142 liters 3 5 ft * 3 13368 . 0 1 ft gal = 37.4 gallons 3 5 ft * 3 3 1 02832 . 0 ft m * 3 6 3 10 1 m cm − = 141,600 cm 3 3 5 ft * 3 3 1 02832 . 0 ft m * 3 3 10 1 m liter − = 142 liters 1.7 The air gas constant of 53.34 ft.lbf/lbm. o R is equivalent to 0.0685 Btu/lbm. o R or 287 joules/kg.K or 0.0686 kcal/kg.K R lbm lbf ft  . 34 . 53 * lbf ft Btu . 1 00128507 . 0 = 0.0685 Btu/lbm. o R R lbm lbf ft  . 34 . 53 * lbf ft Btu . 1 00128507 . 0 * Btu J 1 1055 * kg lbm 4536 . 0 1 * K R   9 5 1 = 287 J/kg.K R lbm lbf ft  . 34 . 53 * lbf ft Btu . 1 00128507 . 0 * Btu J 1 1055 * J cal 186 . 4 1 * cal kcal 1000 1 * kg lbm 4536 . 0 1 * K R   9 5 1 = K kg kcal  0685 . 0

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    1. 6 1.8 The universal gas constant is 1.986 Btu/lb mole. o R which is equivalent to 1.986 kCal.kg mole.K or 1,545 ft.lbf/lb mole. o R or 8,314 joules/kg mole.K R lbmole Btu  986 . 1 * Btu J 1 1055 * J cal 186 . 4 1 * cal kcal 1000 1 * kg lbm 4536 . 0 1 * K R   9 5 1 = 1.986 kCal.kg mole.K R lbmole Btu  986 . 1 * Btu lbf ft 00128507 . 0 . 1 = 1,545 ft.lbf/lb mole. o R R lbmole Btu  986 . 1 * Btu J 1 1055 * kg lbm 4536 . 0 1 * K R   9 5 1 = 8,314 J/kg mole.K 1.9 The thermal conductivity is 200 W/m. o C or 116 Btu/hr/ft/ o F or 0.048 kcal/sec.m. o C C m W  200 * C m W F ft hr Btu   7307 . 1 . . 1 = 116 Btu/hr/ft/ o F C m W  200 * W s Btu 04 . 1055 1 * Btu J 1 1055 * J cal 186 . 4 1 * cal kcal 1000 1 = 0.048 kcal/sec.m. o C 1.10 The thermal conductivity is 50 Btu/hr.ft. o F or 86.54 W/m. o C or 20.7 Cal/sec.m. o C F ft hr Btu  . 50 * F ft hr Btu C m W   . 1 7307 . 1 = 86.54 W/m. o C F ft hr Btu  . 50 * F ft hr Btu C m W   . 1 7307 . 1 * W s Btu 04 . 1055 1 * Btu J 1 1055 * J cal 186 . 4 1 = 20.7 cal/sec.m. o C

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    2. 1 CHAPTER 2 2.1 (a) The mechanical speedometer measures vehicle speed by measuring the angular velocity of the wheels. The angular velocity of the wheels (through some gears) causes a flexible cable to rotate. This cable causes a magnet to rotate inside a metal cup creating a circumferential drag on the cup (a torque). The drag is sensed by allowing the cup to rotate less than a complete turn against the resistance of a torsional spring. The cup is connected to a pointer which can be compared to a stationary scale. The sensing element is the magnet rotating inside the metal cup creating drag. The signal modification system is the system that allows the cup to rotate against a torsional spring. The indicator is the comparison of the pointer to a stationary scale. (b) The fuel level inside a fuel tank is measured with a mechanical float which follows the surface of the fuel (see Ch. 10). The position of the float is sensed with a connected arm which rotates a rotary variable resistor (angular potentiometer, see Ch. 8). The resistance of the variable resistor is sensed by passing applying a voltage to the resistor and measuring the current with an electromechanical gage. The sensing element is the float. The signal modification system consists of the angular potentiometer and the indicator is the electromechanical gage. (c) Most of these devices use a variable resistance device called a thermistor. The resistance of the thermistor is a strong function of it’s temperature. The resistance is sensed by passing applying a voltage to the device and sensing the current, which will be a function of temperature. The current is then converted to a digital form which is then output to the display. In this case, the thermistor is the sensing device. The signal modification system is quite complicated including the creation of the current and the conversion to digital form. Finally, the indicator is the liquid crystal display. Wheel Gears Flex Cable Rot . Magnet Cup Pointer

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    2. 2 2.2 True Value = 0.5000 inches Determination of Bias Error. First, the average of readings must be calculated: Average of Readings = in 4821 . 0 48214 . 0 10 / ) 4822 . 0 4820 . 0 4822 . 0 4821 . 0 4822 . 0 4820 . 0 4821 . 0 4821 . 0 4824 . 0 4821 . 0 ( = = + + + + + + + + + Bias Error = Average of Readings - True Value = − = − = − 0 4821 0 5000 0 01786 0 0179 . . . . in Determination of Maximum Precision Error: Maximum Precision Error = Largest difference between a single reading and the Average of Readings       = − = 0 4824 0 4821 0 0003 . . . in 2.3 Readings: 20.2, 20.2, 20.6, 20.0, 20.4, 20.2, 20.0, 20.6, 20.0, 20.2 (lb) First determine the average of the reading: Average W = 20.2 lb For bias error, Bias Error = Average Value - True Value = 20.24 - 20.0 = 0.2 lb For maximum precision error, we need the reading with the greatest deviation from the average reading (20.6 lb). Therefore, Maximum Precision Error = 20.6 - 20.2 = 0.4 lb 2.4 (a) Intrusive; The thermometer causes a loading error. (b) Non - intrusive; The photography does not affect the speed of the bullet at any time. (c) Non - intrusive; Optical thermal radiation device would yield a non - intrusive measurement as long as it is insulated from the furnace. (d) Non - intrusive; The speed of the car is unaffected by waves measured by the radar gun.

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    2. 3 2.5 (a) A single conducting wire induces a small magnetic field around itself and if it is alternating current, will induce an alternating magnetic in the clamp on ammeter. The clamp on ammeter will have a negligible effect on the current in the wire and for all practical purposes is non - intrusive. (b) The orifice meter (see Ch.10) measures fluid flowrate by obstructing the flow in pipe and measuring the resultant pressure drop. The pressure drop is significant and this device is intrusive. (c) This device passes a beam of infrared radiation through the gases which absorb some of the radiation (see Ch. 10). This measurement has no effect on the composition of the gases and negligible effect on the gas temperature. It is non - intrusive. (d) This device measures rotational speed by shining a pulsing light on a mark on the shaft and adjusting the pulsing rate until the mark appears stationary (see Ch. 8). The light has negligible effect on the rotation of the shaft and is non - intrusive. 2.6 (a) Bias Error; The output will consistently deviate from the true value. (b) Precision Error; The speedometer output shows data scatter. (c) If the difference is consistent with time, then it is a bias error - either of calibration or spatial error. If the difference varies with time, it is precision. 2.7 (a) In most cases, this error will by systematic since repeated measurements at the same time will produce the same error. However, if measurements are made over a long period of time and the temperature varies randomly, the error will be random. (b) This error will be the same each time the measurement is made by the same person with the same procedure and hence is systematic. However, if the measurement is made by several people using different procedures, it may appear to be random. (c) This error will be the same each time the measurement is made and is always systematic. 2.8 (a) This error is usually considered systematic if the readings are all made at the same ambient temperature. However, if the readings are taken over a period of time and the ambient temperature varies randomly, then the error will appear random. (b) This is always random since the fields normally vary in a random manner. (c) Since this is a malfunction, it is not predictable in occurrence so it would be considered random. 2.9 (a) This error is usually considered random even if the readings are all made at the same conditions since corn - growing conditions are highly dependent on various factors. (b) The deterioration of asphalt/concrete in a highway is a combination of factors that needs further analysis since the same - grade concrete will deteriorate at a same rate which gives a systematic error, but given the various conditions at which the different portions of the highway is exposed to the elements, each section will have a random error. (c) The variation of height of the same type of tree in an orchard qualifies as a random error since the height of each tree is highly dependent various growth factors. (d) The variation of drying time of concrete columns of a highway is subjected to both systematic and random error; only if the drying conditions are constant for all columns and the concrete grade is uniform throughout the highway will the error be limited to systematic.

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    2. 4 2.10 (a) The variation in access for a popular website would usually be considered a random error since it is dependent on unknown factors. (b) The variation in average access per day of a popular website would also be considered a random error since it is due to uncontrollable factors. (c) The variation in the rider - ship of a bus or train line would usually be considered a random error since it is dependent on unknown factors. (d) The variation in the rider - ship of a bus or train would also be considered a random error since it is due to uncontrollable factors. 2.11 Resolution or readability does not necessarily give any information about accuracy so we cannot make any statement about accuracy. However, the digital device can be read to only 1 part in 999 of the full scale reading. It may be possible to interpolate between divisions on the analog device giving an effective resolution that is better. 2.12 The span is 50 - 0 = 50 m/s. 2.13 (a) The span is 50 - 5 = 45 psig (b) 70 cm vacuum is taken to be 70 cm of mercury which is equivalent to 93 kPa. Thus, the span is 200 - 93 = 107 kPa (c) The span is 150 - 10 = 140 kPa 2.14 Device (D) would be the best. Device (C) is really the closest in its range. However, measurement errors might cause device (C) to be over range for some measurements producing meaningless results. 2.15 Manufacturer Accuracy =  2.0% of full scale =  0.02(30V) =  0.6V % uncertainty of accuracy =  (0.6V/5V)(100) =  12% The resolution of the device is 0.1 Volts. With a reading of 5V, % uncertainty of resolution = (0.1V/5V)(100) = 2% (or  1%) 2.16 (a) The maximum reading for each range will be 2.999, 29.99, 299.9 and 2999. and the resolution uncertainty will be 1 in the least significant digit. So the resolution uncertainty will be 0.001V, ).01V, 0.1V and 1V for the three ranges. This could also be viewed as  0.0005V,  0.005V,  0.05V and  0.5V (b) The uncertainties will be 2% of full scale. This is .02*3 for the lowest scale or  0.06V. Similarly for the higher ranges, the uncertainties will be  0.6V,  6V and  60V. (c) The resolution uncertainty is negligible compared to the accuracy. Hence we can use the results of part (b). For the 30 V range the relative uncertainy will be 0.06/25 =  2.4%. For the higher ranges, the uncertainties are  24% and  240%.

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    2. 5 2.17 Since the device reads 0.5 psi when it should read zero, it has a zero offset of 0.5 psi which will affect all readings. Zero offset is not a component of accuracy. The accuracy specification of 0.2% of full scale gives an uncertainty of  0.002  50 =  0.1 psi. This means that we can have an expected error in any reading of 0.5  0.1 psi. For an applied pressure of 20 psi, the reading would be expected to be in the range 20.4 to 20.6 psi. We can reduce the expected error by either adjusting the zero (if possible) or by subtracting 0.5 psi from each reading. It may be possible to reduce the error due to the accuracy specification by a calibration of the gage. 2.18 With 2V reading when leads are shorted together, Error 1 = 2V Error 2 =  4%(100) =  4V  Maximum Total Error = +6/ - 2V = (6V/80V)(100) = +7.5% With 0V reading when leads are shorted together, Error =  4%(100) =  4V  Maximum Percent Error =  (4V/80V)(100) =  5% 2.19 The range of both temperature gages will allow the intended 300 C measurement. The uncertainty for each of the two gages is 2% of its span; this gives an uncertainty of 8 C and 18 C for the temperature gage with the smaller and larger span respectively. Thus, the temperature gage with range of 100 C to 500 C should be selected since there is smaller uncertainty. 2.20 The sensitivity is  output/  input. This is (125 - 5)/1000 - 100) = 0.1333 mV/kPa. 2.21 The sensitivity is  output/  input. This is (150 - 10)/100 - 10) = 1.556 mV/psi 2.22 The relationship between pressure and temperature is given by: PV mRT P mRT V = = The sensitivity is given by: dP dT mR V P T R m and V cons t i i = = ; , , tan  We see the sensitivity dP dT is proportional to the initial pressure and is changed when the initial filling pressure is changed. 2.23 (a) The sensitivity from A to C is not a constant and gets smaller from A to C. (b) If a high degree of sensitivity is required, use A - B. For most purposes, B - C would not be recommended due to the sensitivity approaching zero at C. 2.24 Usually, the maximum output increases proportionally with increasing range so the sensitivity will be unchanged. 2.25 Installing an amplifier will increase the sensitivity.

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    2. 6 2.26 (a) For this device, the output is  P and the input is Q. So the sensitivity is simply the derivative of  P with respect to Q. Solving for  P we get: 2       =  C Q P then 2 2 C Q dQ P d =  (b) The sensitivity increases with flowrate and with pressure drop. (c) This device is best for values of  P which are high relative to the design value. At 10% of the design Q, the pressure drop will be only 1% of the design pressure drop. 2.27 Ideal sensitivity = 0.1 cm/N Input span = 200 N  Ideal output span = Input span  Ideal Sensitivity = 200N  0.1cm/N = 20 cm Actual sensitivity = 0.105 cm/N  Actual output span = Input span  Actual sensitivity = 200N  0.105cm/N =21 cm  %error of output span = ( ) Actual Ideal Ideal − 100 ( ) = − = 21 20 20 100 5%

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    2. 7 2.28 0 20 40 60 80 100 120 Gage Reading - psi 0 20 40 60 80 100 True Pressure - psi Plot of Data for Problem 2.13 The best fit line is: Reading = 2.9 + 0.978 (True Pressure) psi True Deviation (Readings - Best Fit) (psi) Pressure (psi) Cycle 1 Cycle 2 Cycle 3 Cycle 4 Cycle 5 20 - 2 - 3 - 2 - 2 - 3 40 - 3 - 2 - 3 - 3 - 3 60 - 3 - 4 - 3 - 4 - 2 80 - 1 - 1 - 2 - 2 - 1 100 0 - 1 - 1 0 1 80 3 2 3 3 3 60 1 1 1 0 0 40 1 0 1 1 2 20 2 2 2 2 2 0 2 1 1 3 1 The accuracy can be evaluated from the extreme deviation values. The values are +3 and - 4 psi. For a span of 100 psi, these values result in an accuracy of +3% and - 4% of full scale. The repeatability can be evalued from Table P2.13 in the problem statement. The maximum deviation for a given measurand and direction is 2 psi. The repeatability can then be expressed as  1% of full scale. Hysteresis is the maximum difference between the up and down readings for any measurand value in the same calibration cycle. The maximum difference is 5 psi, occuring several times (Cycle 2, 20 psi is an example). This can be expressed as 5% of span. -4 -3 -2 -1 0 1 2 3 Reading minus Best Fit - psi 0 20 40 60 80 100 True Pressure - psi Deviation Data for Problem 2.13

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    2. 8 2.29 -200 0 200 400 600 800 1000 1200 Gage Reading -kPa 0 200 400 600 800 1000 True Pressure - kPa Plot of Data for Problem 2.14 Best Fit Line Best Fit Line: Reading = - 4.5 + 1.017 (True Pressure) Deviations based on difference between readings and best fit line. True Deviations (Readings - Best Fit) (N) Force (N) Cycle 1 Cycle 2 Cycle 3 Cycle 4 Cycle 5 200 - 7 - 8 - 5 - 6 - 7 400 - 11 - 13 - 11 - 10 - 9 600 - 10 - 9 - 9 - 13 - 10 800 - 2 - 3 - 4 - 6 - 4 1000 10 10 9 10 8 800 7 5 7 7 8 600 0 - 3 - 3 - 3 - 2 400 - 3 1 0 - 1 1 200 4 2 3 1 6 0 15 17 14 16 15 The accuracy can be determined from the deviation data. The maximum positive deviation is 17 Pa and the largest negative deviation is - 13 Pa. For a span of 1000 Pa, this translates to an accuracy of +1.7% and - 1.3% of full scale. Note: this accuracy only applies when the readings are corrected using the above curve fit to the data. The repeatability can be evaluated from the deviations given in Table P2.14. This occurs at the 200 Pa "down" reading and has a value of 5 Pa. This translates to  0.5% of full scale. The hysteresis error is given by the maximum deviation between the up and the down readings for any value of the measurand in one cycle. The value is 14 Pa which occurs at 400 Pa in Cycle 2. -15 -10 -5 0 5 10 15 20 Reading minus 0 200 400 600 800 1000 True Pressure -k Pa Deviation Data for Problem 2.14 Best Fit - kPa

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    2. 9 2.30 -10 0 10 20 30 40 50 Output - mV 0 20 40 60 80 100 Force - N Plot of Data for Problem 2.15 Best Fit Line Equations of the best fit line: mV = - 0.27 + 0.408 Force or Force = 0.662 + 2.45 mV True Deviations (Readings - Best Fit ) (mV) Force (N) Cycle 1 Cycle 2 Cycle 3 Cycle 4 Cycle 5 20 - 0.05 0.04 0.07 - 0.06 0.00 40 - 0.09 - 0.05 - 0.14 - 0.20 - 0.12 60 - 0.14 - 0.17 - 0.05 - 0.04 - 0.17 80 - 0.04 0.01 0.05 - 0.07 0.07 100 0.26 0.29 0.33 0.19 0.30 80 - 0.06 - 0.08 - 0.06 0.00 0.03 60 - 0.05 - 0.11 - 0.18 - 0.08 - 0.15 40 - 0.08 - 0.18 - 0.10 - 0.16 - 0.11 20 - 0.07 - 0.02 0.09 - 0.02 - 0.05 0 0.28 0.25 0.35 0.32 0.20 The accuracy is determined from the maximum and minimum deviation values. From the deviation table, these values are +0.35 mv and - 0.20 mV. For an output span of 40 mV, this corresponds to +0.9% and - 0.5% of full scale. The repeatability can be determined directly from Table P2.15 in the problem statement. The maximum deviation for any reading (taken in the same direction of varying force) is 0.16 mV at 20 N (down reading). This can be stated as  0.2% of full scale. -0.2 -0.1 0 0.1 0.2 0.3 0.4 Reading minus Best Fit - mV 0 20 40 60 80 100 Force - N Deviation Data for Problem 2.15

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    2. 10 2.31 Static calibration refers to calibration processes where the measurand is changed slowly while allowing the device to come to equilibrium. On the other hand, dynamic calibration refers to processes of a more complicated nature often used for devices where the measured variable is changing rapidly. Types of recommended calibration: (a) Oral thermometer - Static: always allowed to come into equilibrium. (b) Pressure gage used in water line: Static for most purposes. Dynamic effects need more sophisticated instruments. (c) Car speedometer: Static and Dynamic. Dynamic effects are important during acceleration. 2.32 In this case, temperature is the measurand and since it is varying over time, technically a dynamic measurement. However, since the room temperature varies so relatively slowly at 6 C/h, we can consider its measurement to be static. The reading will be an accurate representation of room temperature because the time compared to the time constant. 2.33 In this case, temperature is the measurand and since it is varying over time, technically a dynamic measurement. However, since the room temperature varies so relatively slowly at 1 F/h, we can consider its measurement to be static. The reading will be an accurate representation of room temperature because the time compared to the time constant. 2.34 Both thermometers are dipped in ice water; the smaller one will reach equilibrium faster. The time constant is mc/hA. Since m is proportional to volume, the volume to surface area will be smaller for the smaller thermometer and hence the smaller thermometer will have the smaller time constant.

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    2. 11 2.3 5 (a) The concepts of time constant, response time, rise time, and settling time do not apply to zero order systems due to their instantaneous responses. All of the concepts would have values approaching zero. (b) First order systems as shown in Response A of Figure 2.8 (b) can most effectively be represented by the concept of time constant in Eq. 2.3: y y e at t e t = − = = − 1 0 632   . The preceding equation is a numerical specification of the transient response of the first order system. Although the time constant is the most appropriate concept, the response time and rise time concepts can also be used. (c) For overdamped second order systems also shown in Response A of Figure 2.8(b), the time constant concept of Eq. 2.3 is not really applicable. Therefore, use of the response time and rise time terms is more appropriate occurring when y y e values are 0.95 and 0.1 to 0.9, respectively. (d) Underdamped second order systems as in Response B of Figure 2.8(b) are oscillatory responses which can best be represented by the settling time concept. This concept is the time until the amplitude of the oscillations are less than fraction of the equilibrium response. 2.3 6 (a) The light intensity is changing slowly relative to the response of the meter. Each time a measurement is made, it is essentially a static measurement. This is not a dynamic measurement. (b) The cycling occurs at 1500/60 = 25 times per second. A pressure transducer with a response time of 2 seconds will just measure some kind of average value and will not respond to the pressure variations. The response time must be much shorter than the time for each cycle (1/25 = 0.04 sec) to show the variation in pressure. The response time should probably be less than 1% of 0.04 seconds or 0.0004 seconds. (c) No, it would not measure the power accurately, It is possible to construct a scenario in which the power if off every time the power is measured. It would be necessary to measure the power every minute or more frequently to get a reasonable measurement. There are better ways to measure the power consumption. For example, the on - off time of the heater can be measured over time and separately the power consumption when it is on.

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    2. 12 2.3 7 The response is shown below: 0 3000 Time rpm reading Since the responses for the tachometer resembles that of Response B of Figure 2.8(b), the measuring system may be characterized as second order. To eliminate oscillations, the damping should be increased. 2.3 8 Assuming this to be a first order system, y y e e t = − − 1  where t = 5 sec; and  = 2 sec We want y = 0 at t = 0. Hence, define y = T − T i = T − 20. Then y e = T  − T i = 80 − 20 = 60 This results in T e − − = − − 20 80 20 1 5 2 Solving, we get T = 75.1C 2. 39 Assuming this to be a first order system, y y e e t = − − 1  where t = 4 sec; and  = 2 sec We want y = 0 at t = 0. Hence, define y = T − T i = T − 75. Then y e = T  − T i = 180 − 75 = 105 This results in 2 4 1 75 180 75 − − = − − e T Solving, we get T = 165.8 F 2.4 0 Since this is a first order system, y y e e t = − − 1  where y e = 80 − 20 = 60  C; and  = 4 sec. Since the rise time is the time it takes y/y e to increase from 0.1 to 0.9, 0 1 1 0 42 0 9 1 9 21 9 21 0 42 8 79 4 1 4 2 2 1 . . sec . . sec . . . sec = − → = = − → = = − = − = − − e t e t t t t t t  Therefore the rise time is 8.79 seconds and the 90% response time is 9.21 seconds.

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    2. 13 2.4 1 Since this is a first order system, y y e e t = − − 1  where y e = 180 − 75 = 105  F; and  = 4 sec. Since the rise time is the time it takes y/y e to increase from 0.1 to 0.9, sec 79 . 8 42 . 0 21 . 9 sec 21 . 9 1 9 . 0 sec 42 . 0 1 1 . 0 1 2 2 4 1 4 = − = − =  = → − = = → − = − − t t t t e t e t t

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    2. 14 2.4 2 (a) For an input of 20 lb, the nominal output voltage is calculated below: ( ) Output mV = − = 20 100 0 30 6 (b) Linearity Uncert.=  0.1%(full scale) ( ) ( ) =  =  =  =  0 001 30 0 03 0 03 6 100 0 5% . . . . mV mV mV mV of reading Hysteresis Uncert. =  0.08%(full scale) ( ) ( ) reading of mV mV mV mV % 40 . 0 100 6 024 . 0 024 . 0 30 0008 . 0  =  =  =  = Repeatability Uncert. =  0.03%(full scale) ( ) ( ) =  =  =  =  0 0003 30 0 009 0 009 6 100 0 15% . . . . mV mV mV mV of reading Zero Balance Uncert. =  2%(full scale) ( ) ( ) reading of mV mV mV mV % 10 100 6 6 . 0 6 . 0 30 02 . 0  =  =  =  = Temp Effect Uncert. on Zero: Max. temp. variation = +25, - 20  F Max. + error : ( ) ( ) reading of mV mV mV F mV % 25 . 0 100 6 015 . 0 015 . 0 ) 25 ( 30 00002 . 0 = = = = Max. - error  0.012 mV or 0.20% of reading Temp effect uncert. on span: same values as uncertainty on zero Summary Of Results: Type Uncertainty, mV Uncertainty, % of reading Linearity  0.03  0.5 Hysteresis  0.024  0.4 Repeatability  0.009  0.15 Zero Balance  0.6  10 Temp Eff. (Zero) +0.015/ − 0.012 +0.25/ − 0.20 Temp Eff. (Span) +0.015/ − 0.012 +0.25/ − 0.20

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    2. 15 2.4 3 (a) For an input of 20 lb, the nominal output voltage is calculated below: ( ) mV Output 8 . 4 24 0 1000 200 = − = (b) Linearity Uncert.=  0.1%(full scale) ( ) ( ) reading of mV mV mV mV % 5 . 0 100 8 . 4 024 . 0 024 . 0 24 001 . 0  =  =  =  = Hysteresis Uncert. =  0.08%(full scale) ( ) ( ) reading of mV mV mV mV % 40 . 0 100 8 . 4 0192 . 0 0192 . 0 24 0008 . 0  =  =  =  = Repeatability Uncert. =  0.05%(full scale) ( ) ( ) reading of mV mV mV mV % 25 . 0 100 8 . 4 012 . 0 012 . 0 24 0005 . 0  =  =  =  = Zero Balance Uncert. =  1%(full scale) ( ) ( ) reading of mV mV mV mV % 5 100 8 . 4 24 . 0 24 . 0 15 ( 24 01 . 0  =  =  =  = Temp Effect Uncert. on Zero: Max. temp. variation = +15, - 10  C Max. + error : ( ) ( ) reading of mV mV mV C mV % 15 . 0 100 8 . 4 0072 . 0 0072 . 0 ) 15 ( 24 00002 . 0 = = = = Max. - error  0.0048 mV or 0.1% of reading Temp effect uncert. on span: same values as uncertainty on zero Summary Of Results: Type Uncertainty, mV Uncertainty, % of reading Linearity  0.024  0.5 Hysteresis  0.0192  0.4 Repeatability  0.012  0.25 Zero Balance  0.24  5 Temp Eff. (Zero) +0.0072/ − 0.0048 +0.15/ − 0.1 Temp Eff. (Span) +0.0072/ − 0.0048 +0.15/ − 0.1

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    3. 1 CHAPTER 3 3.1 Output = 5 Volts = V o Input = 5  V = 5  10 - 6 volts = V i Gain G V V G G dB o i dB = = =  = = = = − 5 5 10 10 20 20 10 120 6 6 10 10 6 log log 3.2 G dB =60dB V i =3mV=3  10 - 3 volts G dB =60dB=20log 10 G 3dB= log 10 G  G = 10 3  G = V o /V i V o = G  V i = 10 3 (3  10 - 3 ) = 3 volts 3.3 Eq. 3.2 applies. For G =10, GdB = log10(10) = 20. Similarly, for G = 100 and 500, the decibel gains are 40 and 54. 3.4 The circuit resembles Fig. 3.9 (a). For this problem we want the voltage drop across the resistor Rs to be 0.01xV s . The current in the loop is ) /( i s s R R V I + = and the voltage drop across the resistor is V drop = I s xR s . Combining these: 120 ) 120 /( ) /( 01 . 0 i s s i s s s R V R R R V V + =  + =  . Solving for R i , we get R i = 11,880  . 3.5 The circuit resembles Fig. 3.9(a). The input voltage, V i , is IxR i . The current is V s /(R s +R i ). Combining, V i =R i xV s /(R s +R i ). In the first case: 0.005=5x10 6 xV s /(R s +5x10 6 ) For the second case: 0.0048=10,000xV s /(R s +10,000) These can be solved simultaneously to give R s = 416  .

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    3. 2 3.6 a) From Eq. 3.14, G R R R R R R = + = + = 1 100 1 99 2 1 2 1 2 1 Since R 1 and R 2 typically range from 1k  to 1M  , we arbitrarily choose: R 2 =99k   R 1 = 1k  b) f = 10 kHz = 10 4 Hz GPB = 10 6 Hz for 741 G = 100 From Eq. 3.15, f GPB G Hz Hz c = = = 10 100 10 6 4 This is the corner frequency so signal is - 3dB from dc gain. dc gain = 100 = 40dB. Gain at 10 4 Hz is then 37 dB. From Eq. 3.16,   = −       = −       = − = − − − tan tan 1 1 4 4 10 10 4 45 f f c 

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    3. 3 3.7 G R R R R R R = + = + = 1 100 1 99 2 1 2 1 2 1 Selecting R 1 = 1k  , R 2 can be evaluated as 99k  .. Since GBP = 1 MHz = 100(Bandwidth)  bandwidth = 10 kHz = f c Gain will decrease 6dB from DC value for each octave above 10 kHz. The phase angle can be determined from Eq. 3.16,  = −       − tan 1 f f c f(Hz) 0 5k 10k 100k  0 - 26.6 - 45 - 84.3

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    3. 4 3.8 G R R R R = = + = 1000 1 999 2 1 2 1 Selecting R 2 = 999 k  , R 1 can be evaluated as 1 k  . Since GBP = 1MHz for the  A741C op - amp and G = 1000 at low frequencies, GBP = 1MHz = 1000(Bandwidth)  Bandwidth = 1 kHz = f c If f = 10 kHz and f c = 1 kHz, we must calculate the number of times f c doubles before reaching f. f f x c x x  =  =  = 2 1000 2 10000 3 32 . Now the gain can be calculated knowing that for each doubling the gain decreases by 6dB (i.e. per octave) Gain dB dB dB ( ) log . ( ) = − = 20 1000 3 32 6 40 10 From Eq. 3.1 6 ,  = −       − tan 1 f f c = −       = − − tan . 1 10000 1000 84 3 

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    3. 5 3.9 G = 100 (Actually - 100 since signal inverted) Input impedance = 1000   R 1 From Eq. 3.17, G R R R R k = − − =  = 2 1 2 2 100 1000 100  Since GPB noninv = 10 6 Hz, from Eq. 3.18, ( ) GPB R R R GPB Hz inv noninv = + = + =  2 1 2 6 5 100000 1000 100000 10 9 9 10 . From Eq. 3 . 1 5, f GPB G kHz c = =  = 9 9 10 100 9 9 5 . . 3.10 G = 10 (Actually - 10 since output inverted) Input impedance = 10 k  = 10000   R 1 From Eq. 3.17, G R R R R k = − − =  = 2 1 2 2 10 10000 100  Since GPB noninv = 10 6 Hz, from Eq. 3.18, ( ) GPB R R R GPB kHz inv noninv = + = + = 2 1 2 6 100000 10000 100000 10 909 From Eq. 3.15, f GPB G kHz c = =  = 0 909 10 10 90 9 3 . .

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    3. 6 3.11 (a) 10 = 2 N . N = ln10/ln2 = 3.33 (b) dB/decade = NxdB/octave = 3.33x6 = 20 dB/decade 3.12 The gain of the op - amp itself is V o =g(V p − V n ) [A] V p is grounded so V p = 0 [B] The current through the loop including V i , R 1 , R 2 , and V o is I V R V V R R L i o = = − +   1 2 V n can then be evaluated as ( ) V V I R V R V V R R n i L i i o = − = − − + 1 1 1 2 [C] Substituting [C] and [B] into [A] ( ) V g V R V V R R o i i o = − + − +       1 1 2 Rearranging: ( ) ( ) V R R gR V R R R g V V G R g R R gR o i o i 1 2 1 1 2 1 2 1 2 1 + + = − − + = = − + + Noting the g is very large G R R = − 2 1

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    3. 7 3.13 The complete circuit is as follows, For a loading error of 0.1%, the voltage drop across R s should be 90  0.001 = 0.09 V. The current through R s is then: I V R A R R s S s = = = 0 09 10 0 009 . . I Rs also flows through R 1 and the combination of R 2 and R i . For R 2 and R i , we have: V IR R R R k R o i = = = +             = +             = 10 0 009 1 1 1 0 009 1 1 1 100 1124 2 2 2 . .  The voltage drop across R 1 =90 − 0.09 − 10= 79.91V R V I 1 79 91 0 009 8879 0 = = = . . . 

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    3. 8 3.14 a) If we ignore the effects of R s and R 0 , we can use Eq. 3.19: V V R R R R R R o i = + = +  = 2 1 2 2 2 2 8 120 100000 7142 9 .  (If we include R s and R 0 ,the value of R 2 is 7193  , less than 1% different.) b) I V R V R R A s = = + = + =  1 2 120 100000 7142 9 0 00112 . . (neglecting load effects)  P = I 2 R = (0.0012) 2 (100000 + 7142.9) = 0.13 W c) I V R A A = = + + +           = + + =  120 0 5 100000 1 1 7142 9 1 10 120 0 5 100000 7092 2 0 00112 6 . . . . . Voltage drop across line load resistor V I R V small A = =  = 0 00112 0 5 0 00056 . . . ( ) 3.15 If f 1 = 7600 Hz and f 2 = 2100 Hz then the following equation may be used, f 2  2 x = f 1 where x = # octaves Substituting, 2100 2 7600 2 3 619 2 3 619 1856  = = =  = x x x x octaves . log log . . R s R 1 R 2 R 0 I A 120V

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    3. 9 3.16 f c = 1kHz = 1000Hz , Butterworth Rolloff = 24 dB/octave A V f kHz Hz f kHz Hz out 1 1 2 0 10 3 3000 20 20000 = = = = = . Since Rolloff = 24 dB/octave = 6n dB/octave,  n = 4 From Eq. 3.20, ( ) ( ) G f f c n 1 1 2 2 4 1 1 1 1 3000 1000 0 01234 = + = + =  .  A A G V A in 1in 1out 1 2 010 0 01234 81 = = = = . . . From Eq. 3.20, ( ) G 2 2 4 1 1 20000 1000 0 00000625 = + =  .  A G A mV out in 2 2 2 0 00000625 81 0 051 = = = . ( . ) . 3.17 Using Eq. 3.2, ) 6 . 5 / ( log 20 2 10 o V = − . Solving, V o = 4.45 3.18 We want a low - pass filter with a constant gain up to 10 Hz but a gain of 0.1 at 60 Hz. Using Eq. 3.20: ( ) ( ) G f f c n n = + = + 1 1 0 1 1 1 60 10 1 2 2 . Solving for n, we get 1.28. Since this is not an integer, we select n = 2. With this filter, the 10 Hz signal will be attenuated 3 dB. If this is a problem, then a higher corner frequency and possibly a higher filter order might be selected.

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