Bentley-chapter 5 problems.pdf

# Bentley-chapter 5 problems.pdf - ELEN417-Measurement...

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Unformatted text preview: ELEN417-Measurement Systems Homework 4 – Solution Chapter 5 Problem 1: { { 59 9 = 10 Ω 0 15 0 100 = 100Ω { , = 1 0 = 100Ω a) Rin of buffer? Recorder scale for accurate reading? Set pH = 15 → VR = 100 mV = (59 ) × × × + 0 + 100 = (59 ) × 100 1 ×1× = (59 × 15)() × 9 × + 100 + 100 10 + 2 109 2 × 100 = → = 292 Ω 109 + 59 × 15 = 15 = 0.15 / 100 b) pH = 7 and RT = 2×109 Ω 292 × 106 1 = (0.15 ⁄ ) × (59 ⁄ × 7) ( ) × = 3.95 9 6 2 × 10 + 292 × 10 2 ELEN417-Measurement Systems University of Balamand Problem Solutions = 3.95 − 7 = −3.05 → %. . . = −3.05 × 100% = −20.4 %. . . 15 Problem 2: Requirements: ̂ . . . ≤ 2% Length of potentiometer = 25 cm = 5 W 250Ω → 2500Ω Available Potentiometer { 250Ω From derivation in the book (“Example of Thévenin equivalent circuit calculation: potentiometric displacement sensor”, section 5.1.2, Equation 5.12): ̂ . . . = 400 27 27 × 5000 ≤ 2 → ≤ (2 =2 = 665 Ω) 27 400 400 Choose available potentiometer = 500 Ω 2 Power consideration: ≤ 5 → ≤ √5 × 500 = 50 Maximum sensitivity = 50 = 2 / 25 ELEN417-Measurement Systems University of Balamand Problem Solutions Problem 3: : 0 − 104 { : = 105 Ω 4 20 { 1 5 : 0 − 104 Differential transmitter: output current-input pressure relation () = 4 + 1.6 () Indicator: display pressure-input voltage relation () = 2.5 × 103 [ () − 1] For an input pressure of 5×103 Pa: () = 4 + 1.6 (5 ) = 12 Corresponding current passing into the indicator: = × 105 = 12 × 5 = 11.91 + 750Ω 10 + 750 Corresponding indicator voltage: = × = 250Ω × 11.91 = 2.9775 Corresponding pressure displayed: () = 2.5 × 103 [2.9775 − 1] = 4.944 Corresponding pressure error: = 4944 − 5000 = −56 ELEN417-Measurement Systems University of Balamand Problem Solutions Problem 4: = 0.1 = 103 ⁄ = 10 { = 5 { = 102 ⁄ = 20 / (1) ∆() − ∆ () = ( s + + (2) ∆ () = ( s + + ) ∆̇ () 1 ) ∆̇ () → ∆̇ () = ∆ () s + + Replace (2) in (1), we get: s + + (s) (3) ∆() − ∆ () = ∆ () = ∆ () (s) s + + (4) ∆() = [1 + (s) ∆ () (s) The system transfer function becomes: (5) ∆() (s) 0.1s + 10 + 1000⁄ 0.1s2 + 10 + 1000 = = = ∆ () (s) + (s) 5.1s + 30 + 1100⁄ 5.1s2 + 30 + 1100 ELEN417-Measurement Systems University of Balamand Problem Solutions Problem 7: SOLUTION SOLUTION ETh = VS x V L E Th RL RTh R L RTh = Rp x (1 ‒ x) x = d/dT VS x RL R P x 1 x R L 10 x 1 x 1 x 1 Displacement (cm) x = d/lp RTh (kΩ) ETh (V) VL (V) 2 5 8 0.2 0.5 0.8 1.6 2.5 1.6 2 5 8 1.72 4.00 6.90 Over the full-range, at 1 cm increment: d (cm) x VL (V) 0 0.0 0.00 1 0.1 0.92 2 0.2 1.72 ELEN417-Measurement Systems 3 0.3 2.48 4 0.4 3.23 5 0.5 4.00 6 0.6 4.84 University of Balamand 7 0.7 5.79 8 0.8 6.90 9 0.9 8.26 10 1.0 10.00 Problem Solutions 10.00 8.00 6.00 4.00 2.00 0.00 0 ELEN417-Measurement Systems 1 2 3 4 5 6 University of Balamand 7 8 9 10 Problem Solutions Problem 9: SOLUTION Input: 0 – 104 Pa Output: 0 – 1cm Length = 1 cm RP = 10 kΩ VS = 10 V Pin = 5 kpa a) Pressure-Displacement relation: d (cm) 10 4 Pin (Pa ) For Pin = 5000 Pa: D 10 4 5000 0.5 cm b) c) Transducer open-circuit output voltage: ETh Vs d l p 10 0.5 5 V Output voltage on voltmeter: RTh R p x1 x 10 0.5 0.5 2.5 k 10 RL 5 4 V VL ETh 2.5 10 RTh RL ELEN417-Measurement Systems University of Balamand Problem Solutions ...
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• Winter '15
• Orders of magnitude, University of Balamand

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