Air at p=1 atm enters a long tube of length 2.5 m and diameter of 12 mm at an inlet temperature of Tm,i=100oC and mass flowrate
of 300 x 10-6 kg/s. A constant heat flux is applied to the air from the tube surface. If the tube surface temperature at the exit is Ts,o=180oC, determine the heat rate entering the tube. Evaluate properties at T=400 K.
1 answer:
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Answer: Pi= 4 - 4/3 + 4/5 - 4/7 + 4/9 ...
Explanation:
Is the same as the example,
If Π/4 = 1 - 1/3 + 1/5 - 1/7 + 1/9 ...
Then
(Π/4 )*4= 4*(1 - 1/3 + 1/5 - 1/7 + 1/9 ...)
Π =4 - 4/3 + 4/5 - 4/7 + 4/9 ...
The way to write this is
Sum(from n=0 to n=inf) of (-1)^n 4/(2n+1)
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The composition of gas in the feed, the percentage conversion and the
theoretical yield are combined to give the product stream composition.
Response:
The composition of gas in the product stream are;
- HCl: 0.4 kmol/h, Cl₂: 1.6 kmol/h, H₂O: 1.6 kmol/h, O₂: 0.5 kmol/h
The amount of oxygen used = 30% exceeding the theoretical amount
Number of moles of hydrochloric acid = 4 kmol/h
Percentage conversion = 80%
Required:
The composition of the gas in the product feed.
Solution;
The given reaction is; 4HCl + O₂ 2Cl₂ + 2H₂O
Which gives;
Moles of limiting reactant reacted = 4 kmol/h × 0.80 = 3.6 kmol/h
Which gives;
Number of moles of HCl in the stream = 4 kmol/h - 3.6 kmol/h = 0.4 kmol/h
Number of moles of Cl₂ produced = 2 kmol/h × 0.8 = 1.6 kmol/h
Similarly;
Number of moles of H₂O produced = 2 kmol/h × 0.8 = 1.6 kmol/h
Number of moles of O₂ in the product stream = 30% × 1 kmol/h + 20% × 1 kmol/h = 0.5 kmol/h
The composition of the production stream is therefore;
- <u>HCl: 0.4 kmol/h</u>
- <u>Cl₂: 1.6 kmol/h</u>
- <u>H₂O: 1.6 kmol/h</u>
- <u>O₂: 0.5 kmol/h</u>
Learn more about theoretical and actual yield here: