The fundamental concepts. Onciples and detnelorn erten% el heat and week aro essential for your work, as such you are required to complete the Wowing tasks: 1. Doane and erplath the following concepts. Use equations and diagram; to supplement yow answer:

- Phases of Matter b. Phase Change c. Latent twat of Fusion & of Vaporisation
- In order to produce liquid water at r10%, we are mixing

- tenant amount of water at 8S% with 150g Of item -10C., assumkkg there Is no heat exchange with the surrourdIngs what should be the mass c4 the water at /SY Consider:C... e0.18610/Qat Cli Cw
- 10001(1/(kg Cl, I,. 335 Klikg; 126010/141

Two taps

- the water tank with efifferem flow rates. lap A fills the tank In 1 hour and tap B fills the tank in 1 hour If we open two taps together. find the final temperetwe Of the water in the tank.

Gas lawshelp us understand the propertOs of a working livid when it toes through certain changes within a thermodynamic system. al Your Mat task is to ciellon and explain the ideal gas processes Ia. kocharlc, isobaric, Isothermal and adiabatic use equations and MUNE; to supplement your import.

IN Three containers A. If & c are having the same volume of 1 L, Temperature of 25 'C and Pressure of 101.31tPa. Container A contan fClid;Container S fC00 end COMainer (Mid. Faid the container having the hthetit mamba of moanwithout calcadation.

A conUinec with the l'OnOwie dirneMiOnS (LS m • LS m x 0.5 ml is filled with t L of liquid nitrogen. The container is completely full, the temperature is 30Y, and the atmospheric pressure n 1 atm. Nall the Squid nitrogen evaporates it wit displace a certain volume of air. (ukulele the volume displaced in percentage. Consider the density of liquid Nitrogen as 0807 g/ml. and maw mats of Mtn: seen is 28.024 gimel

- a) A gas weighing S kg °wands within a flexible container as per it* following rule PYLarcaratant. The gaswasinitiallyat a pressure of l Maa and volume of 2 ml OW the pandas the final pfeiSure deCreaSed 5 kPa.Calculite the final volume

Your Supernser Ian her ask, sou to apply the loss of Inernxydnycionies in Owl following tasks:

a) frplain the second 13w of Mormoderunfirs, as particular how n apples to heal wide* and reverse heat engine. Use equattO1114174diAntado 10 Supplement your answer

b) The heat issuer:Awl loan ideal heat engine from boding water reservoir (lCCC) and rejected to the atmosphere how does the eff ciente of the heat engine (flange between a cold day (frCI and a hot dal (4PC/ r) A heat pump which It drhen by a heat

d) Air at 1ST and 60 kea enters thediffuser of a jet eneine steadily with a velocity of 200 mrs. the Inlet area of ilke ante, %0A m2. The ar leaves the diffuser with a velocity that is very smallcompared with the inlet velocity. Determine: a. the mass Bow lee of the air b. the temperature of the are leaving the diffuser.

The (*wept of heat transfer Is important to engineer% to understand and analyse heat transfer rates across buildings and materials

El One end of an eon poker Is placed in a ferewhere the temperature It 450 T. and the other end Is kept al a temperature ol 241 The poker is 1.2 m twig and has a radius of 5 mm. Ignoring the heat lost In convection along the lengthof the poker. fisd the amount of heal conducted from one end of the poker to the other in 5 s The thermal conductivity of the iron Is 79 Wtmli

## Phase of Matter

Thermofluid is a study which combines some of the basic principles which will be solved in the following tasks 1, task2, task 3 and task 4 , therefore it deals with heat transfer, combustion, fluid dynamics and thermodynamics. Thermofluid is very important and is basically applied in different engineering studies like mechanical engineering. In mechanical engineering it is field of development and research which is applicable at almost all generation of energy and its implementation our world today which is so hungry for energy. The thermodynamic also finds its applications in paper and pulp industry, hydrogen and fuel cell research. It is also employed in aerospace industry. Therefore the development of thermofluid permits the world to a more sustainable use of energy to meet our growing demands.

There are different states in which matter can exist, these are basically the solid, liquid, gases and in rare cases the plasma. And these different states of matter can be realized when there is variation in the temperature of the matter. For example when the temperature is increases the solid will melt and change into liquid. When the different reduces the liquid will change into solid (freezing). Again when the temperature of matter (at liquid state) is further increased the matter will change from liquid to gas. When the temperature of the gas is reduced, it can change to liquid or in some rare cases (for sublimates) it changes to solid directly. Therefore the heat required to do such changes are given names like latent heat of vaporization and latent heat of fusion which are both discussed below ;

This means physically distinctive form of mater, like solid, liquid, gas and plasma. The phase of matter is characterized by having relatively uniform chemical and physical properties.

Change one form to another by changing the properties of matter (temperature, pressure etc)

Latent heat of fusion is the heat required to convert solid into liquid

Latent heat of vaporization is the heat needed to convert liquid into vapor

Liquid-Vapor

Heat supplied by water = MCwat T

= M× 4.186kJ (85-40)

=m × 4.186kJ ×45

Heat supply= m × 188.37kJ

Heat gain= 0.150×2000× 10 + 0.150×335+ 0.150×4

= 3000+50.25+25.08

3075.33kJ

Therefore

3075.33kJ= m×188.37kJ

Mass of water =

Mass of water= 16.326kg

The graph is given below;

Assuming the mass id m then

Take the same heat transfer as shown in the graph above, Now

TA>TC>TB

From the graph

The higher T the lower C (specific heat capacity)

CB>CC>CA where C is the specific heat capacity.

Gas laws are basically the rules which influence the behavior of gases (assumed to be ideal gases) and their operations are usually determined by variations in volume of the gas, temperature of the gas, and the pressure of the gas. Therefore these laws of gas are linked to the aforementioned environmental parameters. The most common types of gas laws includes the following;

This relates the pressure and the volume and its states that “states that the volume of a given amount of gas held at constant temperature varies inversely with the applied pressure when the temperature and mass are constant. ” And the equation which relates this is shown below;

## Gas Laws and Ideal Gas Processes

=V , therefore it can be reduced to

PV=K , So P1V1=K and P2V2= K thus

P1V1=P2V2

Temperature-Volume.

This law states that ” the volume of a given amount of gas held at constant pressure is directly proportional to the Kelvin temperature.” And the equation below relates to the law,

This relates to pressure and temperature and it states that “the pressure of a given amount of gas held at constant volume is directly proportional to the Kelvin temperature” And the below equation can describe this law;

1) Isochoric process

An isochoric process is one that takes place at constant volume. In this process no work is done and the heat that flows is equal to the change in the internal energy. The heat flow is in this case

dQ = CV dT

Isobaric process

An isobaric process is carried out at constant pressure. The work is given by:

W = p (Vf - Vi)

And the heat that flows is given by:

dQ = Cp dT

Isothermal process

In this process, there is no change in the value of the gas temperature and therefore the internal energy does not change either. The work done is equal to the heat that enters or leaves the gas. We can calculate the work with:

W = nRT ln (Vf / Vi)

Adiabatic process

Adiabatic processes occur when the heat flow from or to the system to its surroundings is zero. In this case, it is fulfilled that:

pV g = cte.

where g = Cp / CV = 1.67 for an ideal gas.

2) The relationship between the critical temperature and the moles of a real gas is given by the following:

If you increase T critical = Decrease T reduced = Increase compressibility factor = Decrease the moles

The above, according to the equations:

Tr = T / Tc

Z = PV / nRT

n = PV / ZRT

That is, the compound with the lowest Critical Temperature is the compound with the most moles: CH4 is the compound with the highest amount of moles.

3) The amount of moles of added Nitrogen is determined:

n N2 = 1 L * (1000 mL / 1 L) * (0.807 g / 1 mL) * (1 mole / 28,014 g) = 28,807 moles

It is determined by law of ideal gases, the volume of nitrogen gas:

V = n R T / P = 28,807 moles * 0.082 atm * L / mol * K * 303 K / 1 atm = 715.74 L.

4) According to the relationship:

P1 * V1 = P2 * V2

P1 = 1 MPa

V1 = 2 m3

P2 = 1MPa - 0.005 MPa = 0.995 MPa

We cleared Final volume:

V2 = P1 * V1 / P2 = 1 * 2 / 0.995 = 2.01 MPa.

The laws of thermodynamics are very crucial in thermofluid applications and basically there are three main laws of thermodynamics which are knowns are the first law of thermodynamics, the second law of thermodynamics and the third laws of thermodynamics.

## Entropy and Loss of Internal Energy

This law stated that “the energy can neither be created nor destroyed but rather transformed from one energy to another”. It is basically known as the law of conservation

This law state that in all energy exchanges, if no energy enters or leaves the system, the potential energy of the state will always be less than that of the initial state." This law is always knowns as the law of entropy (Massoud, 2015).

This law states that “the entropy of a system approaches a constant value as the temperature approaches absolute zero”

(A)

A thermodynamic process is one which converts the heat energy into the mechanical energy and some of the heat into the surroundings.

During the conversion of heat energy into the mechanical energy, the total heat energy cannot be converted. Some energy will always be lost to the surrounding. During the thermodynamic process, the total energy is conserved.

The thermodynamic process that occurs in nature is an irreversible process and cannot be reversed back.

- B)

And it is also known that the efficiency of a reversible engine is

?= 1-

For a cold day ?c= 1- = 1-

?c=0.268 or 26.8%

For a hot day,

?h= 1- = 1-

?h= 0.161

Or 16.1 %

- c)

From energy balance to heat engine

Q1 =w+Q2…………………………………………………….1

For heat pump

Q3+w=Q4……………………………………………………….2

Now we know the efficiency ?= 1- = 0.25

0.75=

Q2=0.75 Q1 ……………………………………………….3

And COP =3 =

Q4= 3w …………………………………………………………4

So from equation 2 and 4

Q3+w= 3w

And from equation 1 and 3

Q1= w+0.75Q1

0.25Q1=w

Q1=4w and Q2 =3w

And from the ideal gas equation

P= ?RT

Density, ?=

?1= 0.726 kg/m^{3}

Mass flow rate, m= ?AV

- M= 0.726×4×200

M= 58.07kg/s

- From energy equation across diffuser

h1+ = h2

cp=1.005kJ/kgk

1.005×(15+273) + = 1.005×1000(T2)

T2=307.9k

Or T2= 34.9^{0}C

Heat transfer is a disciple of thermofluid which is concerned with the use, generation, and exchange and conversion heat between the physical parameters. There are three basic mechanisms of heat transfer, these include the thermal radiation, thermal convection and thermal conduction.

- Absorptivity

If dQ is the amount of radiant energy falling on the body in form of radiation in wavelength range and +d and a fraction a dQ of it is absorbed by the body and converted into heat then a is called absorptivity power or absorptivity of the body for wavelength .

- Black body

A perfect black body defined as one that absorbs all the radiations incident on it and can emit any wavelength. A black body thus neither transmits nor reflect any radiation (Nikrityuk, 2012). Such body when put to high temperature emit radiations of all wavelength and such radiations are called total radiation.

- Emissivity

The emissive power of a body for radiation of wavelength between and +d is the amount of radiation emitted per unit area of the body per second in a unit of the solid cycle in axial dilation.

- Grey body

A grey body is defined as a body with constant emissivity over all wavelength and temperature.

- e)

If Q is the amount of heat flow from hotter to the colder surface, k be the thermal conductivity, A be the crossectional area, =- be the temperature difference. l be the length of the bar and t be the time of the flow then these can be related by the following equation;

Q= kA t

Now k = 79w/mk

Radius r = 5mm = 5×10^{-3}

Then the area, =( )^{2}

?1- ?2 =0=426k

r= 1.2 m

t= 5 s

So Q= 79×t× ( )^{2 } ×6

Q13.26 Joules

Two identical rectangular rods

Let area be A and the length be L

In series, R= R1+R2

Since , L1=L2, A1=A2

Resistance in parallel KeffAeff= A1+A2=2A

A1=A2 and

For series, the rate of heat transfer is less because Keff is less and also the length is more.

For parallel (case b) rate of heat transfer is faster

The third radii is due to the insulation of 20mm.

R1= 0.07m

R2= r1+0.015 = 0.085m

R3= r1+r2+ 0.02m = 0.105m

Thf=250^{0}C

Tcf= 10^{0}C

K1= 0.5w/mk

K2= 200w/mk

- Heat loss

= Thf- Tcf

= 250-10 = 240 ^{0}C

= Rth1 + Rth 2

Rth 1= 0.4012 ^{0} C/W

Rth 2= 0.000891624 ^{0} C/W

= Rth1 +Rth 2

= 0.4012+0.000891624

= 0.40209^{0} C/W

Q= 596.8W

- Outer surface temperature

Q =

T2=T1- [Q×Rth 1]

T2= 250-[596×0.40209]

T2= 250- 239.439

T2= 10.56^{0}C

References

Massoud, M., 2015. Engineering Thermofluids: Thermodynamics, Fluid Mechanics, and Heat Transfer. 2nd ed. Baghdad: Springer Science & Business Media.

Nikrityuk, P. A., 2012. Computational Thermo-Fluid Dynamics: In Materials Science and Engineering. 2nd ed. Hull: John Wiley & Sons.

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