Velocity and temperature profiles for laminar flow in a tube of radius ro = 10 mm have the form: u(r) = 0.15[1 − (r/ro ) 2 ] T(r
) = 314.8 + 75.0(r/ro ) 2 − 18.8(r/ro ) 4 with units of m/s and K, respectively. Determine the corresponding value of the mean (or bulk) temperature, Tm, at this axial position.
1 answer:
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Answer:
Explanation:
There are a total of 6 states and 3 bits in this problem. Whenever the Reset button is pressed, RESET state is called otherwise the state according to the diagram is called. For the combination to be "01011", the input sequence has to be in the same order. If 0 is pressed instead of 1 in state "010", the last state of output ending with 0 will be called and likewise in all the states that follow.
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: