assignment 1 - engineering measurement-anandababu n

8/18/2019 Assignment 1 - Engineering Measurement-Anandababu N

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Anandababu NBITS 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

=

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 )

=

9 kΩ = α exp ( β / 273.15 )

0.50 k Ω = α exp ( β / 373.15 )

β -

273.15

β

373.15

ln (18)

(Logarithmic)


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Q2. A thermometer is initially at a temperature of 70°F and is suddenly placed in aliquid 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 = ???

RC = time constant

= = 0.4347 at 3 sec @ temp = 200 deg F

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

1 - 0.632 = 0.368

RC = 3.4 sec

T1 – Tf

T0 – Tf

200 – 300

70 - 300

T2 – Tf

T0 – Tf

270 – 300

70 - 300

T – Tf = e - ( t / RC)

T0 – Tf


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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


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Q4: A force sensor has a mass of 0.5 kg, stiffness of 2×102 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)


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

) t +

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) ;

0.3

( 1- 0.32 ) x sin 20 x ( 1- 0.3

2

) t ]


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Sensor damping ratio = 14 N s m−1

/ 2

( 10 Nm-1

x 0.1 kg) = 7 ;

G (s) =1

1 / 102 s 2 + 2 x 7 / 10 s + 1

1 x 10- 1

10- 2

x s 2 + 2 x 7 s + 1

G (s) =

1 x 10 - 1

10 - 2

x s 2 + 14 s + 1

G (s) =


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Q6. A force sensor has an output range of 1 to 5 V corresponding to an input rangeof 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

=

= 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

5 – 1

2 x 105 – 0


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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 = 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

K = 20 – 0

400 – 0


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= Nl = 2.5 %

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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 = 0.5

K1 for temp T = 30 deg C

K1 = 0.55

Non linearity % = - 0.5 x 100 %

20 - 0

K = 5 – 0

10 – 0

K = 5.5 – 0

10 – 0


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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


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Q9: A pressure transducer has an input range of 0 to 104 Pa and an output range of 4to 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

K = 20 - 4104 - 0

K = 0.1538

a = Omin – KImin

a = 4

K1 = Omax – Omin Imax – Imin

K1 = 20.8 - 4.2104 - 0

K1 = 0.1596

a = Omin – KImin

a1 = 4.2

a1 = a+KI.II

4.2 = 4+10KI

KI = 0.02 mA °C-1

K1 = K + KM.IM

0.1596=0.158+10KM

KM = 6 x 10 -4 mA Pa-1 °C-1

KI = 0.02 mA °C-1

KM = 6 x 10 -4 mA Pa-1 °C-1


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Q10: Following Figure shows a block diagram of a force transducer using negativefeedback. 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

= 4.995 V

(10- + 100 x 1) x 10

1 + (10- 4 + 100 x 1) 103 x 10 x 50


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Output Voltage V0 = for Input force 50 N & Vs = 1.5 V

= 4.997 V

This means that the system output depends only on the gain K Fof the feedback element

and is independent of the gains K and K A in the forward path. Changes in K and K Adue

to modifying inputs and/or non-linear effects have negligible effect on V OUT.

(10- + 100 x 1.5) x 10

1 + (10- 4 + 100 x 1.5) 103 x 10x 50

assignment 1 - engineering measurement-anandababu n

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