 # Thermodynamics Report Anonymous

### Question Description

i want a perfect paraphrase of the report and do some changes of the organisation if needed to. you know it is important to make your report different from the one i gave you but with the same content.

the report i gave you to paraphrase is without the graphs, tables and equations to shorten the number of pages. however, i will upload the original one if you need to understand the content more.

i prefer to paraphrase from the one that has the graphs, tables and equations, so it become more easier to understand the content.

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ProfHenryM
School: UCLA  Hello buddy, your assignment is complete. Goodbye and welcome again in the future.

Thermodynamics Report
Student’s Name
Institution Affiliation

1

THERMODYNAMICS REPORT

2
Thermodynamics Report

Q1
The model demonstrating the behaviour of oxygen gas can be illustrated by several equations.
These include the Peng-Robinson equation, Van der Waals, Ideal gas law and the Redlichkwong. Out of the four equations, the Peng-Robinson was found to be the most useful because it
is appropriate in representing real gases such as oxygen (Sandler, 2017). As opposed to the other
methods, the equation is also well established within manufacturing industries (Ashour et,al
2011. According to Tsai & Chen, the method also covers several vital physical characteristics,
hich include vapour-liquid equilibriums (1998). As such, it illustrates reduced variables, which
are a result of deriving dimensionless quantities. For instance, (𝑇𝑟 , 𝑃𝑟 , 𝑉𝑟) has been derived from
𝑇

𝑃

𝑉

𝑐

𝑐

𝑐

(𝑇 , 𝑃 , 𝑉 ). Therefore, Dam & Donald assert that the method guarantees precise results similar to
the actual experimental behaviour (2014).
According to Goodwin, et al, The Peng-Robinson equation can be illustrated as follows (2010)

𝑃=

𝑅𝑇
𝑎𝜑

(𝐸𝑞 1)
𝑉𝑚 − 𝑏 𝑉𝑚 (𝑉𝑚 + 𝑏) + 𝑏(𝑉𝑚 − 𝑏)

In this case, P represents Pressure,
R is the constant representing the ideal gas
T stands for temperature
Vm stands for molar volume
a represents 0.45724
b stands for 0.07780

𝑅2 𝑇𝑐 2

𝑅𝑇𝑐
𝑃𝑐

𝑃𝑐

(𝐸𝑞 )

𝜑 = [1 + (1 − 𝑇𝑟 0.5 )(0.37464 + 1.54226𝜔 − 0.26992𝜔2 )]2

THERMODYNAMICS REPORT

3

Here Tc represents the critical temperature, while P c stands for Pressure. Tr and 𝜔 represesent the
reduced temperature and the acentric factor respectively.
The above equations were then used in plotting the P-V-T Diagram after the identiiction of the
thermodynamic properties of oxygen. Additionally, a MATLAB Code was established to help in
the plotting of isotherms sets for various molar volumes and temperature ranges. The values for
the temperature range was at the critical temperature, above it or below it (𝑇 ≤ Tc ≤ 𝑇).

The following is the data representation in the form of a table
Thermodynamic properties of

Values

Oxygen

The

The gas constant (R)

83.14 cm3 bar/mol K

Critical temperature (Tc )

154.60 K

Critical Pressure ( Pc )

50.80 bar

Acentric factor (𝜔)

0.02220

graphical representation of
Oxygen Isotherms on a P-V
diagram ranging on the

temperatures of 110, 140, 154, 180, and 210 are demonstrated as follows.

THERMODYNAMICS REPORT

4

Oxygen isotherm at 154 K is shown as follows with the plotting of equilibrium lines against the
line of saturation and isotherms of oxygen at the critical point.

The complete P-V-T diagram for the five oxygen isotherms at varied temperatures can be shown
as follows.

THERMODYNAMICS REPORT

5

Material

C1

C2

C3

C4 (1E-02)

C5

Oxygen

51.2450

-1200.20

-6.43610

2.84050

1.0

Q2
According to Strahle (1993), in scenarios where fuel burns without any loss of heat, the resulting
mixture of gases gets to the optimum temperature known as adiabatic flame temperature.
However, the flam temperature varies, depending on several factors or conditions. The
following are some of the points for consideration.

Adiabatic combustion when the pressure is constant

Strahle further argues that when the pressure is constant, a certain amount of energy is lost to
expansion processes, but there is no heat loss to the surroundings. Therefore, the first law of
thermodynamics applies in the following scenario
∆𝑈 = 𝑄𝐶𝑜𝑛𝑠𝑡𝑃 − 𝑊𝐶𝑜𝑛𝑠𝑡𝑃 = 𝑄𝐶𝑜𝑛𝑠𝑡𝑃 − 𝑃∆𝑉 (𝐸𝑞 2)
Here, 𝑈 represents internal energy
𝑄𝐶𝑜𝑛𝑠𝑡𝑃 Represents the heat at constant pressure

THERMODYNAMICS REPORT

6

𝑊𝐶𝑜𝑛𝑠𝑡𝑃 Stands for work at constant pressure
Enthalpy (H) is equal to 𝑈 + 𝑃𝑉

Where, 𝐻 refers to the enthalpy. Therefore, any change in the enthalpy when the pressure is
constant results to the following

∆𝐻 = ∆𝑈 + 𝑃∆𝑉 + 𝑉∆𝑃 = ∆𝑈 + 𝑃∆𝑉

The representation of the adiabatic process is

𝑄𝐶𝑜𝑛𝑠𝑡𝑃 = 0 𝑎𝑛𝑑 ∆𝐻 = 0

However, at equilibrium, the balance of energy is

∑ 𝐻𝑅𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠 (𝑇𝑖 , 𝑃) = ∑ 𝐻𝑃𝑟𝑜𝑑𝑢𝑐𝑡𝑠,𝐼𝑛𝑒𝑟𝑡𝑠 (𝑇, 𝑃) (𝐸𝑞2.2)

𝑇𝑓

∑ 𝐻𝑅𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠 (𝑇𝑖 , 𝑃) = ∑ 𝑛𝑖 ∫ 𝐶𝑃𝑖 𝑑𝑇 + 𝑄𝑉𝐿 (𝐸𝑞2.2.1)
𝑇𝑖

In this case, 𝑛𝑖 stands for moles number of 𝑖 𝑡ℎ species,
𝑇𝑖 Stands for initial temperature 𝑤ℎ𝑖𝑐ℎ 𝑖𝑠 (𝐾)
𝑇𝑓 Illustrates final temperature

THERMODYNAMICS REPORT

7...

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