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Roman55 [17]
3 years ago
7

Propane gas C3H8 enters a combustion chamber operating at steady state condition at 1 bar, 25ºC and is burned with 150% theoreti

cal air that enters the combustion chamber at the same state. The products leave the combustion chamber at 1 bar and 1200 K.
a) Write down the balanced chemical equation for the complete combustion.
b) Write down the balanced actual combustion reaction and calculate c) the air fuel ratio on a mass basis, d) the dew point temperature of the products in ºC, e) the heat transfer in terms of kJ/kmol of fuel.

Engineering
2 answers:
tresset_1 [31]3 years ago
7 0

Answer:

a) The balanced chemical equation is:

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

b) the balanced actual combustion reaction is:

C₃H₈ + 7.5O₂ → 3CO₂ + 4H₂O + 2.5O₂

c) The air fuel ratio on a mass basis is 23.511:1

d) The dew point temperature of the equation is equal to 100°C because the water vapor condenses at a temperature of 100°C

e) the heat transfer is -2037 kJ/mol

Explanation:

the solution is in the attached Word document

Download docx
GalinKa [24]3 years ago
6 0

Answer:

see explaination

Explanation:

Balanced equation or stoichiometry equation means in a product after reaction there is no unburned carbon compound left or we can say the oxygen is sufficient to combine with all the carbon and hydrogen moleculs to form Carbon-dioxide and water respectively.

The dew point temperature of balanced equation will be 100°c because water vapour bis present in it and it will condense at 100°c at 1 bar pressure while the other products need much lower temperatures to liquify.

See attachment.

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3 years ago
Air enters a compressor steadily at the ambient conditions of 100 kPa and 22°C and leaves at 800 kPa. Heat is lost from the comp
telo118 [61]

Answer:

a) 358.8K

b) 181.1 kJ/kg.K

c) 0.0068 kJ/kg.K

Explanation:

Given:

P1 = 100kPa

P2= 800kPa

T1 = 22°C = 22+273 = 295K

q_out = 120 kJ/kg

∆S_air = 0.40 kJ/kg.k

T2 =??

a) Using the formula for change in entropy of air, we have:

∆S_air = c_p In \frac{T_2}{T_1} - Rln \frac{P_2}{P_1}

Let's take gas constant, Cp= 1.005 kJ/kg.K and R = 0.287 kJ/kg.K

Solving, we have:

[/tex] -0.40= (1.005)ln\frac{T_2}{295} ln\frac{800}{100}[/tex]

-0.40= 1.005(ln T_2 - 5.68697)- 0.5968

Solving for T2 we have:

T_2 = 5.8828

Taking the exponential on the equation (both sides), we have:

[/tex] T_2 = e^5^.^8^8^2^8 = 358.8K[/tex]

b) Work input to compressor:

w_in = c_p(T_2 - T_1)+q_out

w_in = 1.005(358.8 - 295)+120

= 184.1 kJ/kg

c) Entropy genered during this process, we use the expression;

Egen = ∆Eair + ∆Es

Where; Egen = generated entropy

∆Eair = Entropy change of air in compressor

∆Es = Entropy change in surrounding.

We need to first find ∆Es, since it is unknown.

Therefore ∆Es = \frac{q_out}{T_1}

\frac{120kJ/kg.k}{295K}

∆Es = 0.4068kJ/kg.k

Hence, entropy generated, Egen will be calculated as:

= -0.40 kJ/kg.K + 0.40608kJ/kg.K

= 0.0068kJ/kg.k

3 0
3 years ago
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