Q : 1 Define the following and give SI units

STERADIAN;

A steradian can be defined as the solid angle subtended at the center of a unit sphere by a unit area on its surface. For a general sphere of radius r, any portion of its surface with area A = r2 subtends one steradian.

ß = swept area L²/ L²

(radius)²

In SI units its unti is steradian and symbol is "sr"

RADIAN :

The term radian arises from the fact that the length of a circular arc, corresponding to an angle of one radian, is equal to the radius of the arc. This is shown in the illustration. Point p represents the center of the circle. The angle q, representing one radian, is such that the length of the subtended circular arc is equal to the radius, r, of the circle.

One Radian is 180/π degrees, or about 57.296°

DENSITY:

Density is a measure of how much mass is contained in a given unit volume. It is usually expressed in kg/m³ .Put simply, if mass is a measure of how much stuff there is an object density is a measure of how tightly that stuff is packed together.

LUMINANCE:

Luminance is a photometric measure of the luminous intensity per unit area of light travelling in a given direction. It describes the amount of light that passes through or is emitted from a particular area, and falls within a given solid angle. The SI unit for luminance is candela per square meter (cd/m2).

Q:2 Elaborate the definition of

(a) Pound force kilogram force and Gram force (b) Convert 27º C to Kelvin Fahrenheit and Rankine values

POUND FORCE

The pound-force is equal to the gravitational force exerted on a mass of one avoirdupois

pound on the surface of Earth. Since the 18th century, the unit has been used in low-

precision measurements, for which small changes in Earth's gravity (which varies from

place to place by up to half a percent) can safely be neglected.

The 20th century, however, brought the need for a more precise definition. A

standardized value for acceleration due to gravity was therefore needed. Today, in

accordance with the General Conference on Weights and Measures, standard gravity is

usually taken to be 9.80665 m/s2 (32.174 049 ft/s2).

The acceleration of the standard gravitational field (gn) and the international avoirdupois

pound (lbm) define the pound-force as

This definition can be rephrased in terms of the slug. A slug has a mass unit

equivalent of 32.174049 lbm. A pound-force is the amount of force required to

accelerate a slug at a rate of 1 ft/s2, so:

KILOGRAM FORCE

The kilogram-force (kgf or kgF), or kilopond (kp, from Latin pounds meaning weight),

is a gravitational metric unit of force. It is equal to the magnitude of the force exerted by

one kilogram of mass in a 9.80665 m/s2 gravitational field (standard gravity, a

conventional value approximating the average magnitude of gravity on Earth). Therefore

one kilogram-force is by definition equal to 9.80665 N. Similarly, a gram-force

is 9.80665 mN, and a milligram-force is 9.80665 µN.

One kilogram-force is approximately 2.204622 pounds.

GRAM FORCE

The gram-force is a metric unit of force (gf). The gram-force is equal to a mass of one

gram multiplied by the standard acceleration due to gravity on Earth, which is defined as

exactly 9.80665 meter per second². Then one (1) gram-force is equal to 0.001 kg ×

9.80665 meter per second² = 0.00980665 kilogram × meter per second² = 0.00980665

Newton (1N).

(b) Convert 27º C to Kelvin Fahrenheit and Rankine values

27ºC = K?

27ºC = 27+273k

27ºC = 300 Kelvin

27ºC = ºF ?

ºF=1.8(ºC)+32

ºF= 1.8(27)+32

27ºC =80.6 ºF

Q:3 Give three rules of extensive and intensive quantities?

INTENSIVE QUANTITY

In the physical sciences, an intensive property (also called a bulk property, intensive quantity, or intensive variable), is a physical property of a system that does not depend on the system size or the amount of material in the system: it is scale invariant.

By contrast, an extensive property (also extensive quantity, extensive variable, or extensive parameter) is one that is additive for independent, no interacting subsystems. It is directly proportional to the amount of material in the system.

For example, density is an intensive property of a substance because it does not depend on the amount of that substance; mass and volume, which are measures of the amount of the substance, are extensive properties. In general the ratio of two extensive properties (such as mass and volume) that scale in the same way is scale-invariant, and hence an intensive property (such as density)

EXTENSIVE QUANTITY

An extensive property is defined by the IUPAC Green Book as a physical quantity which

is the sum of the properties of separate no interacting subsystems that compose the

entire system. The value of such an additive property is proportional to the size of

the system it describes, or to the quantity of matter in the system.

Extensive properties are the counterparts of intensive properties, which are intrinsic to a

particular subsystem. Dividing one type of extensive property by a different type of

extensive property will in general give an intensive value. For

example, mass (extensive) divided by volume (extensive) gives density (intensive).

Three rules applicable for extensive and intensive quantities.

1. The ratio of two extensive quantities is intensive. For example mass m is extensive volume v is intensive but density ρ is intensive.

Intensive= ρ=m/v=EXTENSIVE /EXTENSIVE

2. An equation with an extensive quantity on the left-hand side must also have extensive quantity on the right- hand side an as illustration consider the perfect gas equation

PV=nRT (Extensive form)

The left hand side contains the extensive quantity V and the right-hand side contains the extensive quantity moles n.

3. The extensive form of an equation can be transformed in to an intensive form by dividing both side by an extensive variable for e.g.

P V/N=RT

PV=RT (Intensive form)

Where V=V/n is the specific volume.

Q:4 Methane goes through combustion by consuming Oxygen through the formula

CH4+2O2→CO2+ 2H2O

If inputs are CH4 = 1 KG, O2 = 4 KG, N2= 13.2 KG PROVE LAW OF

CONSERVATION CO2 =2.75 KG, H2O = 2,25 KG

CH4 0=input -0+0 – consumption

Consumption = input =1.0 kgCH4 /s

O2 : 0=input -0+0 – consumption

Input = consumtion

= 1.0 kgCH4 x kmol CH4 x 2 kmol O2 x 32 kg O2 = 4.0 kgO2

S 16kg CH4 1 kmol CH4 k mol O2 s

Air input =4.0 kgO2 x kmol O2 x 1.00 kmol air x 29 kg air = 17.3 kg air

S 32 kg O2 0.21 kmol O2 kmol air s

CO2 ;0=0 – output +generation -0

Output = Gen

=1.0 kgCH4 x kmol CH4 x 1 kmol CO2 x 44 kg CO2 = 2.75 kgO2

S 16kg CH4 1 kmol CH4 k mol CO2 s

H2 O : 0=0 – output +generation – 0

Output = Gen

=1.0 kgCH4 x kmol CH4 x 2 kmol H2 O x 18 kg H2 O = 2.25 kgH2 O

S 16kg CH4 1 kmol CH4 k mol H2O s

N2 : 0 = Input – Output + 0 – 0

Output = Input

= 17.3 kg air x kmol air x 0.79 kmol N2 x 28 kg N2 = 13.2 kg N2

S 29 kg air 1 kmol air k mol N2 s

Total : 0= Input – Output + 0-0

Input = Output

(CH4) + (O2) + (N2) = (CO2) + (H2O) + (N2)

1.0kg/s + 4.0kg/s + 13.2 kg/s = 2.75 kg/s + 2.25 kg/s + 13.2 kg/s

18.2 kg/s = 18.2 kg/s

Q : 5 The time period of simple pendulum is give by

T= 2π √LG

Find T if L=20 cm G=9.8m/s

Soloution

T= 2 22/7 √0.209.8

T = 2 × 3.142 ×0.1428

T = 0.892 sec

Q : 6 write universal accounting equation for

(1) Positive and nagitive charges

(2) Derive the equation for

(q+ - q-) final - (q+ - q-)initial = (q+ - q-) in - (q+ - q-)out

From above equation

Positive and negative charges

Electric charge is a physical property of matter that causes it to experience a force when

near other electrically charged matter. Electric charge comes in two types,

called positive and negative. Two positively charged substances, or objects, experience

a mutual repulsive force, as do two negatively charged objects. Positively charged

objects and negatively charged objects experience an attractive force. The SI unit of

electric charge is the coulomb (C), although in electrical engineering it is also common

to use the ampere-hour (Ah). The study of how charged substances interact is classical,

which is accurate insofar as quantum effects can be ignored.

The electric charge is a fundamental conserved property of some subatomic particles,

which determines their electromagnetic interaction. Electrically charged matter is

influenced by, and produces, electromagnetic fields. The interaction between a moving

charge and an electromagnetic field is the source of the electromagnetic force, which is

one of the four fundamental forces (See also: magnetic field).

Twentieth-century experiments demonstrated that electric charge is quantized; that is, it

comes in multiples of individual small units called the elementary charge, e,

approximately equal to 1.602×10−19 coulombs (except for particles called quarks, which

have charges that are multiples of ⅓e). The proton has a charge of e, and

the electron has a charge of −e. The study of charged particles, and how their

interactions are mediated by photons, is quantum electrodynamics.

Universal account ion equation for each charge is given below

q+ final - q+initial = q+ in - q+

out + q+gen - q+

cons (General)

q- final – q-initial = q- in – q-

out + q-gen - q-

cons (General)

q+ final - q+initial = q+ in - q+

out (Ordinary)

q- final – q-initial = q- in – q- (Ordinary)

Derive the equation for

(q+ - q-) final - (q+ - q-)initial = (q+ - q-) in - (q+ - q-)out

From above equation

we have two type of charges to count , one is positive charge and other is negative

charge the symbol q+ is used for positive charge and the symbol q- is used for negative

charge . The universal accounting for each type of charge is given below

q+ final - q+initial = q+ in - q+

out + q+gen - q+

cons (General) →1-1

q- final – q-initial = q- in – q-

out + q-gen - q-

cons (General) → 1-2

q+ final - q+initial = q+ in - q+

out (Ordinary) →1-3

q- final – q-initial = q- in – q- (Ordinary) →1-4

Equation 1-3 and 1-4 is not valid in nuclear reaction where mass is converted to energy

and vice versa. In this case ,charge also can be created or consume

For example consider the reaction of an electron e- with the positive e+ which forms

pure energy with no mass left over

e-+ e+ → Energy

in this reaction positive and negative charge is generated that is amount of positive and

negative charge in the universe is now greater

if we subtract the equation 1-2 from 1-1 we obtain

(q+ - q-) final - (q+ - q-)initial =

(q+ - q-) in - (q+ - q-)out +(q+ - q-)gen- (q+ - q-)cons →1-5

In all the years physics have studded matter they have always observer that when a

positive charge is generated a negative charge also generated with it so if positive

charge consume so negative also consume so in mathematical form we write it as

q+gen = q-

gen →1-6

q+cons = q-

cons →1-7

when equations 1-6,1-7 substituted in 1-5 so we obtain

(q+ - q-) final - (q+ - q-)initial = (q+ - q-) in - (q+ - q-)out

Q : 7 An inventor develops a 20-lbm cart which catches ball thrown in to when

mass of 1.0 lbm travelling of 132 ft/sec if initial velocity is zero what is the velocity

after 10 balls impact the cart.

Solution

Final amount = initial amount +input – output

(m cart + 10 m ball) Vf = m cart Vi + 10( m ball V ball ) – 0

Vf = m cart Vi + 10( m ball V ball ) – 0 / (m cart + 10 m ball)

Vf= (20 lbm) (0 ft/sec) + 10(1.0 lbm x 132 ft/sec)

20 lbm +10 (1.0 lbm)

Vf = 44 ft/sec

Q : 8 A rock is fixed to a 3.00 meter long string. Rock is revolving two circle per

sec what is the

1. Frequency

2. Period

3. Angular Velocity

4. Centripetal acceleration

Solution

Frequency = f= 2 rev/sec

Period =p= 1/ f = sec/2 rev = 0.5 rev/sec

Angular velocity = ⱳ= 2πf = 2π rad × 2 rev =12.6 rad/sec

Rev s

Speed = v= r ⱳ = (3.00 m) (12.6 rad/sec) = 37.3 m/s

Centripetal acceleration = a= v2 /r = ( 37.3 m/s )2 = 463 m/s

Q : 9 How much is the energy stored in 20µf capictor that has a 20v differnce

acroos the plates?

Solution

E= 1/2 q (V2 - V1) = 1/2 C(V2 - V1) (V2 - V1) = C(V2 - V1)2

=1/2 (20 × 10-6F) (C2/J) x (20)2 x (J/C)2

F V

= 4.0 X 10-4 J

Q : 10 Describe reversible and irreversible compression and expansion of gas

with example ?

Expansion of a gas

For an adiabatic free expansion process, the gas is contained in an insulated container

and a vacuum. The gas is then allowed to expand in the vacuum. The work done by or

on the system is zero, because the volume of the container does not change. The First

Law of Thermodynamics then implies that the net internal energy change of the system

is zero. For an ideal gas, the temperature remains constant because the internal energy

only depends on temperature in that case. Since at constant temperature, the entropy is

proportional to the volume, the entropy increases in this case, therefore this process is

irreversible

Example of compression

Let's now look at a common example of adiabatic compression- the compression stroke

in a gasoline engine. We will make a few simplifying assumptions: that the

uncompressed volume of the cylinder is 1000cc's (one liter), that the gas within is nearly

pure nitrogen (thus a diatomic gas with five degrees of freedom and so   = 7/5), and

that the compression ratio of the engine is 10:1 (that is, the 1000 cc volume of

uncompressed gas will compress down to 100 cc when the piston goes from bottom to

top). The uncompressed gas is at approximately room temperature and pressure (a

warm room temperature of ~27 degC or 300 K, and a pressure of 1 bar ~ 100,000 Pa,

or about 14.7 PSI, or typical sea-level atmospheric pressure).

so our adiabatic constant for this experiment is about 1.58 billion.

The gas is now compressed to a 100cc volume (we will assume this happens quickly

enough that no heat can enter or leave the gas). The new volume is 100 ccs, but the

constant for this experiment is still 1.58 billion:

so solving for P:

or about 362 PSI or 24.5 atm. Note that this pressure increase is more than a simple

10:1 compression ratio would indicate; this is because the gas is not only compressed,

but the work done to compress the gas has also heated the gas and the hotter gas will

have a greater pressure even if the volume had not changed.

We can solve for the temperature of the compressed gas in the engine cylinder as well,

using the ideal gas law. Our initial conditions are 100,000 pa of pressure, 1000 cc

volume, and 300 K of temperature, so our experimental constant is:

We know the compressed gas has V = 100 cc and P = 2.50E6 pascals, so we can solve

for temperature by simple algebra:

That's 751 Kelvin's, or 477 °C, or 892 °F. This is why a high compression engine

requires fuels specially formulated to not self-ignite (which would cause engine

knocking when operated under these conditions of temperature and pressure), or that

a supercharger and intercooler to provide a lower temperature at the same pressure

would be advantageous. A diesel engine operates under even more extreme conditions,

with compression ratios of 20:1 or more being typical, in order to provide a very high

gas temperature which ensures immediate ignition of injected fuel

mathematical equation for an ideal gas undergoing a reversible (i.e., no entropy

generation) adiabatic process is

where P is pressure, V is specific or molar volume, and

 being the specific heat for constant pressure,   being the specific heat for

constant volume,   is the adiabatic index, and  is the number of degrees of

freedom (3 for monatomic gas, 5 for diatomic gas).

For a monatomic ideal gas,  , and for a diatomic gas (such

as nitrogen and oxygen, the main components of air)  .[4] Note that the

above formula is only applicable to classical ideal gases and not Bose–

Einstein or Fermi gases.

For reversible adiabatic processes, it is also true that

where T is an absolute temperature.

This can also be written as