Assignment 1 - Engineering Measurement-Anandababu N

Anandababu N BITS ID: 201518BT017 Assignment 1 – Engineering Measurement Q1 . The resistance R ( θ ) of a thermostat at temperature θ K is given by R ( θ ) = α exp ( β / θ). Given that the resistance at the ice point (θ = 273.15 K) is 9.00 kΩ and the resistance at the steam point is 0.50 kΩ, find the resistance at 25 °C. Ans : From the equation R(θ ) = α exp(β/θ ). Given the resistance at the ice point (θ = 273.15 K) is 9.00 kΩ 9 kΩ = α exp ( β / 273.15 ). Ice point = 0 deg C the resistance at the steam point is 0.50 kΩ = 0.50 kΩ = α exp ( β / 373.15 ). Steam point is at 100 deg C To find the resistance at 25 °C we should be calculate the value of β and α =

= α exp ( β / 273.15 )

9 kΩ

0.50 k Ω = α exp ( β / 373.15 )

18 kΩ = exp( β / 273.15 ) / exp ( β / 373.15 ) equating both sides we get e ^ ( β / 273.15 )

= 18 kΩ

e ^ ( β / 373.15 )

( β / 273.15 )

= ln (18)

( β / 373.15 ) β 273.15

-

β 373.15

=

ln (18) (Logarithmic)

Anandababu N BITS ID: 201518BT017 Assignment 1 – Engineering Measurement 373.15 β – 273.15 β 101925.9225

= ln (18)

373.15 β – 273.15 β = ln (18) x 101925.9225 100 β = 294603.807 β = 2946.03 By using the value of β in the one of the above equation to find the value of α α = 9 kΩ / exp(2946.03 / 273.15 ) α = 1.86 × 10^−4 the resistance at 25 °C. R(298.15k) = (1.86 × 10^−4)* exp( 2944.2 / 298.15 ) R(298.15k) = 3.62 kΩ = R ( 25 °C) - Ans

Anandababu N BITS ID: 201518BT017 Assignment 1 – Engineering Measurement

Q2. A thermometer is initially at a temperature of 70°F and is suddenly placed in a liquid which is maintained at 300 °F. The thermometer indicates 200 and 270°F after intervals of 3 and 5 s, respectively. Estimate the time constant for the Thermometer. T0 = 70 deg F Tf = 300 deg F At Time interval of 3 sec, the rise in temperature T1 = 200 deg F At Time interval of 5 sec, the rise in temperature T2 = 270 deg F Find Time constant = ??? T – Tf

= e - ( t / RC)

T0 – Tf RC = time constant 200 – 300

T1 – Tf T0 – Tf

=

1 - 0.632 = 0.368 RC = 3.4 sec

= 0.4347 at 3 sec @ temp = 200 deg F

270 – 300

T2 – Tf T0 – Tf

70 - 300

=

70 - 300

= 0.1304 at 5 sec = @ temp = 270 deg F

Anandababu N BITS ID: 201518BT017 Assignment 1 – Engineering Measurement

Q 3. A thermometer has a time constant of 10 s and behaves as a first order system. It is initially at a temperature of 30°C and then suddenly subjected to a surrounding temperature of 120 °C. Calculate the 90% rise time and the time to attain 99% or the steady state temperature. T0 = 30 deg C at t = 0 ; Tf = 120 deg C at steady state Time constant - RC = 10 sec For the 90% rise time e - ( t / RC) = 0.1 and ln (0.1) = - t / RC Therefore t = ln (0.1) x RC = e (- t / RC) = 0.1 - t = ln ( 0.1 ) x RC - t = - 2.302 RC ; t = 2.302 RC, t = 23.02 sec for 99% rise steady state temperature = e (- t / RC) = 0.01 = - t / RC = ln (0.01) = - t / RC = - 4.605 , t / RC = 4.605 t = 4.605 RC & t = 4.605 x 10 t (99%) = 46.05 sec

Anandababu N BITS ID: 201518BT017 Assignment 1 – Engineering Measurement Q4: A force sensor has a mass of 0.5 kg, stiffness of 2×10 2 Nm−1 and a damping constant of 6.0N s m−1. (a) Calculate the steady-state sensitivity, natural frequency and damping ratio for the sensor. (b) Calculate the displacement of the sensor for a steady input force of 2 N. (c) If the input force is suddenly increased from 2 to 3 N, derive an expression for the resulting displacement of the sensor Soln : mass m = 0.5 kg, stiffness k = 2 x 102 Nm-1 ; = 0.2 Nm; damping constant  = 6.0 N sm-1 = Steady state sensitivity K = 1 / k = 1 / (2 x 102 Nm-1) = 0.005 Nm-1 undamped natural frequency = wn = √(k / m) rad / s √( 2 x 102 / 0.5) = 20 rad / s damping ratio  =  / 2√( km) ; damping ratio  = 6.0 N sm-1 / 2√( 2 x 102 x 0.5) = 0.3 under damped natural frequency = wd = wn. √( 1- 2 ) = 20 x √( 1- 0.32 ) = 19.07 rad / s b) displacement of the sensor for steady input force of 2 N x = F / k ; F = 2 N; k = 2 x 102 Nm-1 = 2 / 2 x 102 Nm-1 = 0.01 m displacement x = 0.01 m c) Resulting displacement of the sensor As damping ratio  = 0.3 < 1 (underdamped condition)

Anandababu N BITS ID: 201518BT017 Assignment 1 – Engineering Measurement

f o (t) = 1- e ^ (0.3 x 20 x t) [ cos 20 √( 1- 0.32 ) t +

0.3

x

2 sin 20 x √( 1- 0.3 ) t ]

√( 1- 0.32 ) f o (t) = 1- e ^ (6 t) x [ cos 19.07 t + 0.3145 x sin 19.07 t ] meters Steady state displacement is given below

Steady state sensitivity = 0.005 step height = 1 (change in force = 2 N to 3 N) unit step response = f o (t) Hence, steady state displacement as below = 0.005 x 1 x 1- e ^ (6 t) x [ cos 19.07 t + 0.3145 x sin 19.07 t ] meters

Q5 : An elastic force sensor has an effective seismic mass of 0.1 kg, a spring stiffness of 10N m−1 and a damping constant of 14 N s m−1. Calculate the following quantities: (i) sensor natural frequency (ii) sensor damping ratio (iii) transfer function relating displacement and force. Soln : i) sensor natural frequency, mass m = 0.1 kg, spring stiffness k = 10 Nm-1 ; = damping constant  = 14 N s m−1 undamped natural frequency wn = √(k / m) rad / s undamped natural frequency wn = √(10 Nm-1 / 0.1 kg) rad / s = 10 rad / s Sensor damping ratio  =  / 2√( km) ;

Anandababu N BITS ID: 201518BT017 Assignment 1 – Engineering Measurement

Sensor damping ratio  = 14 N s m−1 / 2√( 10 Nm-1 x 0.1 kg) = 7 ;

1

G (s) =

(1 / 10)2 s 2 +( ( 2 x 7 ) / 10 )) s + 1

1 x 10 - 1

G (s) = (10 - 2 x

s 2 +( 2 x 7 )

s+1

1 x 10 - 1

G (s) = (10

-2

x

s 2 + 14

s+1

Anandababu N BITS ID: 201518BT017 Assignment 1 – Engineering Measurement

Q6. A force sensor has an output range of 1 to 5 V corresponding to an input range of 0 to 2×105 N. Find the equation of the ideal straight line. Output Omin = 1 V ; Output Omax = 5 V ; Input I min = 0 N ; Input I Max = 2 x 105 N

5–1

=

2 x 105 – 0

= 0.00002

K = 0.00002

a = Omin – K I min = 1 - 0.00002 x 0 a=1

Oideal = KI + a Oideal = 0.00002 I + 1

Anandababu N BITS ID: 201518BT017 Assignment 1 – Engineering Measurement

Q7: A non-linear temperature sensor has an input range of 0 to 400 °C and an output range of 0 to 20 mV. The output signal at 100 °C is 4.5 mV. Find the non-linearity at 100 °C in millivolts and as a percentage of span. Solution : Input Imin = 0 deg C; Input Input Imax = 400 deg C Output Omin = 0 V ; Output Omax = 20 mV ; Output O = 4.5 mV ; Input I = 100 deg C

K

20 – 0

=

400 – 0 K = 0.05

a = Omin – K I min = 0 - 0.05 x 0 a=0

O ( I ) = 0.05 ( I ) + 0 O ( I ) = 0.05 (100 ) = 5 mV (for Input temp @ 100 deg C) O (100) = 4.5 mV N ( I ) is the difference between actual & ideal straight line behaviour

N ( I ) = 4.5 – ( 0.05 x 100 + 0); where N ( I ) is non linearity N ( 4 ) = - 0.5

Anandababu N BITS ID: 201518BT017 Assignment 1 – Engineering Measurement

Non linearity % =

=

- 0.5

x 100 %

Nl = 2.5 %

20 - 0

----------------------------------------------------------------------------------------------------------------------

Q8: A force sensor has an input range of 0 to 10 kN and an output range of 0 to 5V at a standard temperature of 20 °C. At 30 °C the output range is 0 to 5.5 V. Quantify this environmental effect. Solution : Input Imin = 0 ; Input Imax = 10 KN ; Output Omin

= 0 ; Output Omax = 5 v @

standard temperature T = 20 deg C, @ temperature = 30 deg C ; Output Omin = 0 v ; Output Omax = 5.5 V

K for temp T = 20 deg C K

=

5–0

K = 0.5

10 – 0

K1 for temp T = 30 deg C K

=

5.5 – 0 10 – 0

K1 = 0.55

Anandababu N BITS ID: 201518BT017 Assignment 1 – Engineering Measurement K1 = K+KM x IM K = sensitivity KM = Modifying sensitivity w.r.t environment , temp variation IM = modifying input i e., change in temp = 20 – 30 = 10 0.55 = 0.5 + 10 KM KM = 0.005 V KN-1 C-1

a = 0 – 0.5 x 0 = 0

O (ideal) = 0.5 x 30 + 0 = 15

N ( I ) = 15 – (0.5 x 30 + 0) N(I)=0

As no shift of the curve to the axis I or O, then: KI = 0

Anandababu N BITS ID: 201518BT017 Assignment 1 – Engineering Measurement

Q9: A pressure transducer has an input range of 0 to 104 Pa and an output range of 4 to 20 mA at a standard ambient temperature of 20 °C. If the ambient temperature is increased to 30 °C, the range changes to 4.2 to 20.8 mA. Find the values of the environmental sensitivities KI and KM. Solution Omax = 20 mA ; Omin = 4 mA ; Imax = 104 ; Imin = 0 ; @ temp = 20 deg C Omax = 20.8 mA ; Omin = 4.2 mA @ temp = 30 °C IM = 30-20 = 10 °C

T ambient = 20°C

T ambient = 30°C

II = 30-20 = 10°C

IM = 30-20 = 10°C

K = Omax – Omin Imax – Imin

K1 = Omax – Omin Imax – Imin

a1 = a+KI.II

K1 = K + KM.IM

4.2 = 4+10KI

0.1596=0.158+10KM

K = 20 - 4 104 - 0

K1 = 20.8 - 4.2 104 - 0

KI = 0.02 mA °C-1

KM = 6 x 10 -4 mA Pa-1 °C-1

K = 0.1538

K1 = 0.1596

a = Omin – KImin

a = Omin – KImin

a=4

a1 = 4.2

KI = 0.02 mA °C-1 KM = 6 x 10 -4 mA Pa-1 °C-1

Anandababu N BITS ID: 201518BT017 Assignment 1 – Engineering Measurement

Q10: Following Figure shows a block diagram of a force transducer using negative feedback. The elastic sensor gives a displacement output for a force input; the displacement sensor gives a voltage output for a displacement input. VS is the supply voltage for the displacement sensor. Calculate the output voltage V0 when (i) VS=1.0 V, F =50 N (ii) VS =1.5 V, F =50 N.

K = sensing element

=

Force transducer

= 10-4

KA = Amplifier gain

=

Amplifier

= 103

Kf = Feedback element =

Coil and magnet

= 10

Km x Im = Modifying Input =

Displacement sensor

= 100 x Vs

Fin = Input Force

=

= 50 N

= Output Voltage V0 = = Output Voltage V0 = for Input force 50 N & Vs = 1.0 V (10-4 + 100 x 1) x 103 1 + (10- 4 + 100 x 1) 103 x 10

x 50

= 4.995 V

Anandababu N BITS ID: 201518BT017 Assignment 1 – Engineering Measurement

Output Voltage V0 = for Input force 50 N & Vs = 1.5 V (10-4 + 100 x 1.5) x 103 1 + (10- 4 + 100 x 1.5) 103 x 10

x 50 = 4.997 V

This means that the system output depends only on the gain KF of the feedback element and is independent of the gains K and KA in the forward path. Changes in K and KA due to modifying inputs and/or non-linear effects have negligible effect on VOUT.