**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
- Amir Alimohammad
- Orders of magnitude, University of Balamand