Instrumentation Introduction

1. Units
There are only 6 basic units used in engineering. These are

mass	kilogramme (kg)
length	metre (m)
time	seconds (s)
current	Ampere (A)
temperature	Kelvin (K)
luminous intensity	Candela (Cd)

All other quantities used are multiples of these units.
2. Fundamental mechanical laws
VELOCITY


Linear	Velocity = rate of change of distance (m/s) v = distance moved/time taken for constant velocity or x/t. v = $\frac{dx}{dt}$ for instantaneous velocity.
Angular	Angular velocity = rate of change of angle (rad/s) $\omega$ = angle turned/time taken for constant velocity or q/t. $\omega$ = $\frac{dq}{dt}$ for varying conditions.

ACCELERATION


Linear	Acceleration = rate of change of velocity $\frac{m}{s^2}$. a = $\frac{dv}{dt}$ = $\frac{d^2v}{dt^2}$
Angular	Angular acceleration = rate of change of velocity (rad/s2) a = $\frac{dw}{dt}$ = $\frac{d^2q}{dt^2}$

NEWTON'S 2nd LAW OF MOTION This concerns the force required to overcome the inertia of a body. Inertia (or mass) is that property of a body which opposes changes to its motion.


Linear Force = Mass x acceleration F = $M(\frac{dv}{dt})$ F = $M(\frac{d^2x}{dt^2})$

Angular Torque = Moment of inertia x angular acceleration T = $I(\frac{d\omega}{dt})$ T = $I(\frac{d^2\theta}{dt^2})$

SPRING

Linear	Force = Constant * Change in length (N) F = kx
Angular	Torque = Constant * angle (N m) T = k*$\theta$

Pressure
Pressure = Force per unit area ($\frac{N}{m^2}$)
p = $\frac{F}{A}$
1 pascal = 1 $\frac{N}{m^2}$
1 bar = $10^5 Pa$
FLOW RATE IN PIPES
Flow rate = Cross sectional area * mean velocity
Q = $A \frac{dx}{dt}$

3. Fundamental Electrical Laws

Current: $i = \frac{dq}{dt}$ (coulomb/s or Amperes)
RESISTANCE: $V = IR$ (Ohm's Law)
INDUCTANCE: This law is equivalent to the 2nd. law of motion. Inductance (L Henries) is that property of a component that resists changes to the flow of current. The emf (E) produced is $E = L \frac{di}{dt}$
$E = L\frac{d^2q}{dt^2}$
CAPACITANCE : This law is equivalent to the spring law. Capacitance (C Farads) is the property of a component which enables it to store electric charge.
$Q=CV$(Coulombs)


4. Stress and Strain

DIRECT STRESS ($\sigma$): When a force F acts directly on an area A as shown in figure , the resulting direct stress is the force per unit area and is given as $\sigma$ = $\frac{F}{A}$
where
F is the force normal to the area in Newtons
A is the area in m2
and $\sigma$ (sigma) is the direct stress in $\frac{N}{m^2}$ or Pascals.
Since 1 Pa is a small unit kPa , MPa and GPa are commonly used also.
If the force pulls on the area so that the material is stretched then it is a tensile force and stress and this is positive.
If the force pushes on the surface so that the material is compressed, then the force and stress is compressive and negative.
DIRECT STRAIN ($\varepsilon$):Consider a piece of material of length L as shown in figure. The direct stress produces a change in length DL. The direct strain produced is $\varepsilon$ (epsilon) defined as $\varepsilon = \frac{DL}{L}$
The units of change in length and original length must be the same and the strain has no units.
Strains are normally very small so often to indicate a strain of $10^{-6}$ we use the name micro strain and write it as $\mu \varepsilon$.
For example we would write a strain of 7 x 10^-6 as 7$\mu \varepsilon$.
Tensile strain is positive and compressive strain is negative.
MODULUS OF ELASTICITY (E) : Many materials are elastic up to a point. This means that if they are deformed in any way, they will spring back to their original shape and size when the force is released. It has been established that so long as the material remains elastic, the stress and strain are related by the simple formula $E = \frac{\sigma}{\varepsilon }$
and E is called the MODULUS OF ELASTICITY. The units are the same as those of stress.

Also Read

• Weight or Force measurement module
• Calibration of pressure gauge using dead weight tester
• Study & Calibration of LVDT Transducer for Displacement measurement

ELASTIC LIMIT: A material is only elastic up to a certain point. If the elastic limit is exceeded, the material becomes permanently stretched. The stress-strain graph for some metals are shown below. The modulus of elasticity does not apply above the elastic limit. Strain gauges should not be stretched beyond the elastic limit of the strain gauge material which is approximately 3000$\mu \varepsilon$.


A typical value for the elastic limit of strain gauges is 3000 $\mu \varepsilon$.

POISSONS' RATIO:
Consider a piece of material in 2 dimensions as shown in figure. The stress in the y direction is ${{\sigma }_{y}}$ and there is no stress in the x direction. When it is stretched in the y direction, it causes the material to get thinner in all the other directions at right angles to it. This means that a negative strain is produced in the x direction. For elastic materials it is found that the applied strain (${{\varepsilon }_{y}}$) is always directly proportional to the induced strain (${{\varepsilon }_{x}}$) such that
$\frac{\varepsilon_{x}}{\varepsilon_{y}}= -\nu$
where $\nu$ (Nu)is an elastic constant called Poissons' ratio. The strain produced in the x direction is $\varepsilon_x = -\nu*\varepsilon_y$

Resistivity

RESISTIVITY : The resistance of a conductor increases with length L and decreases with cross sectional area A so we may say R is directly proportional to L and inversely proportional to A.
R = Constant *$\frac{L}{A}$
The constant is the resistivity of the material $\rho$ hence R = $\frac{\rho*L}{A} Ohms

Temperature coefficient of Resistance

:
The resistance of conductors changes with temperature. This is a problem when strain gauge devices are used. Usually the resistance increases with temperature. The amount by which the resistance changes per degree per ohm of the original resistance is called the temperature coefficient of resistance and is denoted $\alpha$. The units are Ohms per Ohm per degree.
Let the resistance of a conductor be $R_0$ at 0⁰C.
Let the resistance be $R_1$ at $\theta_1$ ⁰C. The change in resistance = $\alpha \theta_1 R_0$.
The new resistance is $R_1 = R_0+\alpha \theta_1 R_0$.
Let the resistance be $R_2$ at $\theta_2$ ⁰C. The change in resistance = $\alpha \theta_2 R_0$.
The new resistance is $R_2 = R_0+ \alpha \theta_2 R_0$.
If the temperature changes from $\theta_1$ to $\theta_2$ the resistance changes by
$\delta R = R_2 - R_1 = (R_0+\alpha \theta_2 R_0)-(R_0 + \alpha \theta_1 R_0)$
$\delta R = R_0 \alpha \delta \theta$