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iren2701 [21]
3 years ago
8

The combustion of the paraffin oil, which is a highly refined fossil fuel product similar to kerosene composed of C14-C16 hydroc

arbons, resulted in the accumulation of black soot on the glass cylinder of you Stirling engine. When hydrocarbons are burned, some of the hydrocarbons do not fully react with the oxygen, and a complex range of compounds including PAH’s (polycyclic aromatic hydrocarbons) are created.
There is a strong correlation between the length of the hydrocarbon chain, and the amount of soot produced during burning. For instance, burning fuels with long chain hydrocarbons such as heavy fuel oil or diesel fuel produces a lot more soot than burning fuels with shorter chain hydrocarbons such as gasoline or propane.

Use molecular diagrams to explain this phenomenon.
Engineering
1 answer:
laiz [17]3 years ago
5 0

Explanation:

Paraffins or Alkanes which have a general formula of CnH2n+2 are saturated hydrocarbons (which means the carbons present have all their carbons with the maximum amount of hydrogen or there's only the presence of C-C) whole Olefins or Alkenes(with a general formula of CnH2n) and Alkynes or acetylenes (CnH2n-2) are unsaturated hydrocarbons (that os the presence of C-C and other double and triple bonds of C-C)

Sooty flames are caused by the incomplete combustion of an unsaturated hydrocarbon. When a fire is ignited to a paraffin, it burns in oxygen with a blue luminous flame which does not produce any smoke or soot but an unsaturated hydrocarbon is burned in the presence of oxygen; a yellow non-luminous flame is formed which produces a lot of smoke and soot.

Examples of paraffins are: butane(C4H10), propane(C3H8); Alkenes: ethene(C2H4), butene(C4H8) and Alkyne: hexyne(C6H10), butyne(C4H6)

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dimaraw [331]

Answer:

The drying time is calculated as shown

Explanation:

Data:

Let the moisture content be = 0.6

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total moisture of the clay  = 0.64

total drying time for the period = 8 hrs

then if the final dry and wet masses are calculated, it follows that

t = (X0+ Xc)/Rc) + (Xc/Rc)* ln (Xc/X)

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4 0
3 years ago
Consider a Carnot refrigeration cycle executed in a closed system in the saturated liquid–vapor mixture region using 1.06 kg of
Alexxandr [17]

Answer:

P_m_i_n = 442KPA

Explanation:

We are given:

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T = 22kj

Therefore we need to find coefficient performance or the cycle

COP_R = \frac {1}{(T_R/T_l) -1}

= \frac {1 }{1.2-1}

= 5

For the amount of heat absorbed:

Q_l = COP_R Wm

= 5 × 22 = 110KJ

For the amount of heat rejected:

Q_H = Q_L + W_m

= 110 + 22 = 132KJ

[tex[ q_H = \frac{Q_L}{m} [/tex];

= = \frac{132}{1.06}

= 124.5KJ

Using refrigerant table at hfg = 124.5KJ/Kg we have 69.5°c

Convert 69.5°c to K we have 342.5K

To find the minimum temperature:

T_L = \frac{T_H}{1.2};

T_L = \frac{342.5}{1.2}

= 285.4K

Convert to °C we have 12.4°C

From the refrigerant R -134a table at T_L = 12.4°c we have 442KPa

6 0
3 years ago
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Explanation:

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Answer:

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Step-by-step explanation:

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