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Course Title : Measurements & Installations Course Code : EE Lecturer : Staff
Sheet N
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ERROR – ACCURACY - RESOLUTION
- An instrument has an accuracy of ± 0.5% FS and measures resistance from 0 to 1500 Ω. What is the resistance range in an indicated measurement of 397 Ω. Find the accuracy in the measured value of resistance R.
- A sensor has a transfer function of 5 mV / o^ C and an accuracy of ± 1%. If the temperature is known to be 60 0 C, what can be said about the output voltage.
- A temperature sensor has a transfer function of 44.5 mV / o^ C. The output voltage is measured to be 8.86 volts. What is the expected value of the temperature?
- For the same sensor in problem (2), if the output voltage was first amplified by an amplifier of gain 15 ± 0.25 and then measured by a meter of range 0 to 2 volt and accuracy ± 1.5 % FS, find the worst case of accuracy in the measured value.
- A temperature sensor has the following specifications: resolution = 2 o^ C and sensitivity = 5 mV/ o^ C. If the sensor is in contact with a gas at 100 o^ F, calculate the error in reading if its temperature is changed to 102 o^ F.
- A Bourdon tube with resolution 2 psi. If it reads 100 psi and the pressure supply varies between ± 1 psi. Find the range of reading in Pascal ( Pa).
- An input of 100 volt is applied to a voltmeter with resolution 0.1 volt. Find the reading if the input is increased or decreased in the range of ± 0.1 volt.
- If the power is given by P =I 2 R (watt), calculate the relative error in power measurement for error in resistance R and current I.
- Calculate the relative error in power P for error in voltage V and resistance R where P =V 2 /R (watt).
Course Title : Measurements & Installations Course Code : EE Lecturer : Staff
Sheet N
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PRESSURE MEASUREMENT
- Water column of height 3.3 m produces a pressure of … KPa but if it is mercury, the pressure will be …….. psi.
- A mercury manometer reading at the top of a tower is 74.15 cm and at the bottom is 76 cm. Calculate the height of the tower if the density of air between the top and the bottom is 1.258 * 10 -3^ g / cm^3.
- A bellow has an effective area of 25 cm^2 is subjected to pressure of 5 psi. Calculate the force on the bellow surface.
- A mercury manometer reading at the top of a tower is 94.5 KPa and at the bottom is 76 cm Hg. Calculate the height of the tower if the density of air between top and bottom is 1.258 * 10-3^ g / cm^3_._
- The barometric pressure is 91 KPa. Calculate the vapor pressure of the liquid and the gauge reading, figure (1). The specific gravity of oil = 0.9.
- U -tube filled with mercury shows a pressure difference proportional to 2 inches. What will be the equivalent difference when the mercury is replaced by water?
- Find the gauge reading in figure (2). Water 200 mm 3 m Mercury Air Vacuum Gauge Figure (2) 203 mm Vapor only 1.2 m Mercury Oil Figure (1)
B A
Course Title : Measurements & Installations Course Code : EE Lecturer : Staff of 5 mm diameter inclined at 30 degree to the horizontal. Find the increase in liquid level rise in the measuring tube.
- Bourdon tube registers a vacuum pressure of 310 mm Hg at 1 atm. Calculate the absolute pressure in psi.
- A Wheatstone bridge has all 3 resistors = 120 Ω and the 4th^ one is the strain gauge with its initial resistance (at no pressure) = 120 Ω and has gauge factor = 2. The bridge supply voltage is 10V and the applied pressure causes a strain = 2500 μm/m. Calculate the offset voltage of the bridge.
Course Title : Measurements & Installations Course Code : EE Lecturer : Staff
Sheet N o 3
LEVEL MEASUREMENT
- In the design of a capacitance level gauge the separation distance between the two plates is d cm and the dimensions of the plates are w meters width and h meters height. The capacitance when the tank is empty is C pF. When certain amount of insulating powder was poured in the tank the capacitance is changed to be 2C pF. T R h+ d
- In an outer type ultra sonic level gauge, the transmitter and receiver sensors are established over a liquid tank such that their height from the bottom of the tank is 2 meters. If the time taken by the signal from the transmitting instant to the receiving instant is 2 x 10 -3^ sec. Find the liquid level height in the tank, taking into consideration that the speed of signal is 350 m / sec.
- In a capacitance level gauge if the separation distance between the two square electrodes is 5 cm. The capacitance C when the tank is empty is 20 x 10^5 pF. Certain quantity of liquid is poured in the tank and the capacitance is dropped by 10 x 10^5 pF. If the fluid dielectric constant is 0. and the air dielectric constant is 1. Find the height of the fluid in the tank if the width of the electrodes is w. Tank shell
Course Title : Measurements & Installations Course Code : EE Lecturer : Staff R 1 =R 2 =R 3 =100 Ω and supply voltage 10 volts. Calculate the resolution of the detector to measure 1oC change in temperature.
- A resistance temperature detector ( RTD ) arranged in a bridge with each arm 100 ohm and volt supply. If the temperature of the sensor changed such that the meter indicates 0.569 volt , the material has temperature coefficient of 0.0039 o^ C-1. The resistance at 0 o^ C is 100 Ω. Find the sensor temperature and the power consumed in it.
Sheet N
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FLOW MEASUREMENT
- State the Bernoulli equation regarding the flow of an incompressible fluid.
- Water flows in a 150 mm diameter pipeline in the form of Venturi tube. The mean velocity at the entrance is 4.5 m/sec. The U tube shows a head drop of 5 cm. Calculate the diameter of the throat.
- A liquid of density 800 kg /m^3 flows in a horizontal pipe of radius 5 cm. In a section of the tube of radius 3 cm , the liquid pressure is 15 KN/m^2 less the main pipe. Calculate the velocity of the liquid in the main pipe. (take Cd = 0.9 ).
- An orifice meter ( Cd = 0.61) measuring the flow of air in a pipe is substituting by a Venturi meter ( Cd = 0.98) having the throat diameter same as that of the Orifice. for the same flow rate find the ratio of pressure drops for Venturi meter to Orifice meter.
- Water flow is controlled from 20 to 50 gal./min. The flow is measured using Venturi tube of 10 cm inlet diameter, and 5 cm throat diameter. Bellows connected to LVDT are used to measure flow such that its output is 1.8 volt / psi. Find the range of voltages that result from the flow rate.
- A nozzle is fitted in a horizontal pipe of diameter 15 cm , carrying a gas of density 1.15 kg / m^3. The diameter ratio is 0.33, the differential pressure head
Course Title : Measurements & Installations Course Code : EE Lecturer : Staff indicated by a U tube manometer containing oil of specific gravity 0.8 is 10 cm. Take the coefficient of discharge 0.8, determine the gas flow through the tube.
- A nozzle flow meter indicates a pressure drop of 20 cm of water when the flow rate of air is 100 lit /min. What would be the flow rate when the pressure drop is 40 cm of water?
Solution of Sheet N
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ERROR-ACCURACY- RESOLUTION
- An instrument has an accuracy of ± 0.5% FS and measures resistance from 0 to 1500 Ω. What is the resistance range in an indicated measurement of 397 Ω. Find the accuracy in the measured value of resistance R. The error = ± 0.5 % x 10 –2^ x 1500 = ± 7.5 Ω. The measured resistance = 397 ± 7.5 = 404.5 or 389.5 Ω. The accuracy in the measured value = [(true - measured)/ true]x 100 = [(397 –389.5)/397 ]x 100 = ± 1.9 % of the reading
- A sensor has a transfer function of 5 mV / o^ C and an accuracy of ± 1%. If the temperature is known to be 60 0 C, what can be said about the output voltage. The error = ± 1% x 5 = ± 0.05 mV / o^ C. The T.F range 5.05 to 4.95 mV / o^ C. The corresponding voltage for 60 o^ C is 60x [ 5.05 to 4.95 ] = 303 to 297 mV.
- A temperature sensor has a transfer function of 44.5 mV / o^ C. The output voltage is measured to be 8.86 volts. What is the expected value of the temperature? The expected value is T = (8.86 / 44.5)x 10^3 = 199.1 o^ C
Course Title : Measurements & Installations Course Code : EE Lecturer : Staff P= I 2 R , if an error in resistance R is ∆R and an error in current is ∆I , the corresponding error in power is ∆P. (P+∆ P) = (I+∆ I)^2 ( R+∆R ) = ( I^2 +2I∆I+∆I^2 ) ( R+∆R ) = ( I^2 R +2I∆IR+∆I^2 R + I^2 ∆R +2I∆I∆R +∆I^2 ∆R ) ∆ P ≈ 2I∆IR + I^2 ∆R
Solution of Sheet N
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PRESSURE MEASUREMENT
Law: P = ρ.g.h ρ Density Kg / m^3 , ρHg = 13600 kg/m^3. g grarity const. 9.8 m/sec^2 h depth of liquid meter. Pressure at a point in ą liquid = weight of liquid column Units: Pascal ( Pa) = 1N/m^2 = 1kg/ms^2 1 bar = 100 KPa. 1atm = 76 cm Hg = 101.325 KPa. 1 psi = 6.895 KPa. 1- P = ρ.g.h water P w = 1000* 9.8 * 3.3 = 32.34 KPa mercury P Hg =13600* 9.8 * 3.3 = 439.824 KPa / 6.895 = 63.78 psi 2- The pressure difference = weight of air column ρ of air =(1.258x10 -3^ ) x 10 -3^ / 10 6 = 1.258 kg / m^3. ∆p = (0.76 – 0.7415) * 13600 * 9.8 = 24651.
Course Title : Measurements & Installations Course Code : EE Lecturer : Staff = 1.258 * 9.8 * H H = 200 m 3- The force F = P*A 5 psi = 5 * 6.895 KPa = 34475 Pa. F = 34475 * = 86.1875 Newton 4- 76 cm Hg = 100 KPa ∆p = (101.325 – 94.5) * 10^3 = 6.825x10 3 Pa = ρ* g (^) * H = 1.258 * 9.8 .H H = 553.6 m 5- Pressure at A = pressure at B (^91) * 103 = (0.203 (^) * 9.8 (^) * 13600) + (1.2 (^) * 9.8 (^) *0.9 (^) * 1000) + Pvap
6- Mercury P = ρHg * h * g
= 13.600 * 2 * 2.54* 10 -2^ * 9.
Water = 1000 * h * 9. h = 13.6 * 2 = 27.2 inch 7- Gauge pressure = water pressure + Pv = Pw + Pv Pw = 1000 * 9.8 * 3 = 32.34 KPa Pv = Patm – P (^) Hg = 101.325 * 10^3 – 0.2 * 13600 * 9. = 101.325 * 10^3 – 26.656 = 74.469 KPa Gauge pressure = 106.809 KPa 8- P = ρ (^) * h (^) * g = ρ (^) * ( y 1 + y 2 ) (^) * g. = 1.256 * 10-3^ m^2 y 1 =5 (^) * 10 -3^ m 50 x 10^3 = 13600 (y 1 + y 2 )* 9.8 = 133.28 10^3 (y 1 + y 2 ). (y 1 + y 2 ) = 0.375 m = 375.150 mm. Continuity equation , the same volume A 1 y 1 =A 2 y 2 Π (d 1 / 2)^2 y 1 = Π (d 2 / 2)^2 y 2
Course Title : Measurements & Installations Course Code : EE Lecturer : Staff 13- Absolute pressure = Gauge pressure + Atmosphere pressure. Pabs = Pg +Patm absolute press = 310 mmHg + 760 mmHg =1.070 m Hg = 1.070 * 13600 * 9.8 Pa = psi ≈ 20683psi 14- Gauge Factor = (ΔR/Ro)/strain = 2 = (ΔR/120)/(250010-6) So ΔR = 0.6 Ω ΔV (voltage offset) = VS( ) = - 0.0124 V.
Solution of Sheet N
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LEVEL MEASUREMENT
- In an outer type ultra sonic level gauge, the transmitter and receiver sensors are established over a liquid tank such that their height from the bottom of the tank is 2 meters. If the time taken by the signal from the transmitting instant to the receiving instant is 2 x 10 - 3^ sec. Find the liquid level height in the tank, taking into consideration that the speed of signal is 350 m / sec.
- h + d = 2m
- The time t = 2 x 10 - 3^ sec.
- The speed of signal is 350 m / sec. Solution: d = 0.35m h = 2-0.35 = 1.65m
- In a capacitance level gauge if the separation distance between the two square electrodes is 5 cm. The capacitance C when the tank is empty is 20 x 10^5 pF. Certain quantity of liquid is poured in the tank and the capacitance is dropped by 10 x 10^5 pF. T R h+ d
Course Title : Measurements & Installations Course Code : EE Lecturer : Staff If the fluid dielectric constant is 0.1 and the air dielectric constant is 1. Find the height of the fluid in the tank if the width of the electrodes is w meter. Solution. The total capacitance is: C = Kεo A / d Where: K = Dielectric constant εo = Permittivity = 8.85 pF / m (معامل النفاذ الكهربى) A = Plate common area. d = Plate separation =5 cm For fluid K = 0.1 & for air K = 1 Plate width = w meter Area A = w^2 For air C = 1 x 8.85 pF / m x w^2 m^2 / 0. 05 m = 20 x 10 5 pF So W^2 = 11299.435 m^2. W = 106.298 m. For fluid C = (0.1 x 8.85 pF / m x 106.298L 1 m^2 / 0. 05 m ) + (8.85 pF / m x 106.298L 2 m^2 / 0. 05 m ) = 10 x 10 5 pF 0.0188h 2 + (1.8814 x 10-3)h 1 = 1 h 1 + h 2 = w = 106. h 1 = 59.0121 m and h 2 = 47.2858 m.
- In the design of a capacitance level gauge the separation distance between the two plates is d cm and the dimensions of the plates are w meters width and h meters height. The capacitance when the tank is empty is C pF. When certain amount of insulating powder was poured in the tank the capacitance is changed to be 2C pF. If the dielectric constant of the powder is 3 times that of the air. Compute the ratio of the height of the powder to the total height of the tank h. Solution. The total capacitance is: C = Kεo A / d Tank shell Tank shell
Course Title : Measurements & Installations Course Code : EE Lecturer : Staff 3- Thermo couple linear operation up to 1100 oC
0 ÷ 1100 o^ C → 45.14 mv
E = K ( Th – Tc ) → E (45.14 mv ) = K (1100 - 0 )
K = 45.14 / 1100 mv / oC 4- RTD The RTD law: R = Ro (1+ αT) related to 0 oC.
ρN i = 8.7 * 10-8^ Ω.cm = 8.7 * 10 –10^ Ω.m at 20 oC
d = 0.002 mm = 2 * 10-3^ mm A = π = 3.14 * 10 - 12^ m^2. Given: Ro = 100 Ω and α = 0.0068 oC- R 20 = Ro ( 1+ α x 20 ) = 100 ( 1+ 0.0068 x 20 ) = 113.6 Ω 113.6 = ρ 20 x( l 20 / A) = 8.7 x 10 -7^ x ( l 20 / 3.14 x 10-6^ ) l 20 = 0.41 m When the temperature changes from T 1 to T 2 : R 2 = R 1 + Ro α (T 2 – T 1 ) R 100 = R 20 + Ro x α ( 100 – 20 ) = 113.6 + 100 x 0.0068 x 80 = 168 Ω 5- RTD In a Wheatstone bridge. The condition of bridge balance is: Given: R 2 = R 3 = 25 Ω This means that the resistance of RTD = 37.36 Ω at T oC temperature. 37.36 = Ro ( 1 + αT ) = 25 ( 1 + 0.00392 x T ). R 2 R 3 R 1 R
Course Title : Measurements & Installations Course Code : EE Lecturer : Staff 6- Thermistor Given: The temperature 212 oF is equal to ( 212 – 32 ) x = 100 oC + 273 = 373 oK R 373 = 20 KΩ , β = 3650 and RT = 500 KΩ T =? T = 280.67 oK – 273 = 7.674 o^ C x ( 9 / 5 ) + 32 = 45.81 oF 7 - RTD Given: α = 0.004 oC –1^ R 20 = 106 Ω R 1 = R 2 = R 3 = 100 Ω and Us = 10 v Required the RTD resolution to measure 1 oC change. 1 oC change means that: T 2 = 21 oC , T 1 = 20 oC and To = 0 o^ C R 21 = R 20 + α ( 21 – 20 ) x Ro R 20 = Ro ( 1 + α x 20 ) → Ro = 98.148 Ω R 21 = 106 + 0.004 x 98.148 = 106.392 Ω Resolution means the difference between the measuring device readings at T 1 = 20 oC and T 2 = 21 oC The reading of the measuring device Vab b R 2 R 3 R 1 RT a 10 V R 500 20 0
- 2 0 0 25 50 75 100 T (oC)
Course Title : Measurements & Installations Course Code : EE Lecturer : Staff
Solution of Sheet N O^5
FLOW MEASUREMENT
Notes K = A 1 A 2 Gallon = 3.7854 liters 1- constant kinetic energy + potential energy + pressure energy = constant. 2- V 1 A 1 = V 2 A 2 3- K = A 1 A 2 4- orifice venturi 5- Flow range from 20 to 150 gal/min For 20 gal/min = 1.2618 10 -3^ m^3 /sec = 9.05 10 -
Course Title : Measurements & Installations Course Code : EE Lecturer : Staff For 50 gal/min = 3.1545 10 -3^ m^3 /sec = 9.05 10 - = 194.3944 Pa = 0.02819 PSI = 1214.965 Pa = 0.17620 PSI 1 psi = 6.895 KPa. Lower range = 0.02819 1.8 = 0.0507 volt Upper range = 0.17620 1.8 = 0.3171 volt 6- , diameter ratio = 0.33 Cd = 0.8 d 1 = 15 cm 7- When
