An Otto cycle engine is analyzed using the cold air standard method. Given the conditions at state 1, compression ratio (r), and
pressure ratio (rp). The specific heats for air are Cp = 1.0045 kJ/kg-K and Cv = 0.7175 kJ/kg-K.
T1 = 300K
P1 = 100 kPa
r = 10
rp = 1.65
a) Determine the temperature (K) at state 2.
b) Determine the pressure (kPa) at state 2.
c) Determine the pressure (kPa) at state 3.
d) Determine the temperature (K) at state 3.
e) Determine the temperature (K) at state 4.
f) Determine the pressure (kPa) at state 4.
g) Determine the net work output (kJ/kg/cycle) for the engine.
h) Determine the heat addition (kJ/kg/cycle) for the engine.
i) Determine the efficiency (%) of the engine.
Points will be given to answer showing all the work done to get right answer.
T1 = 300K
P1 = 100 kPa
r = 10
rp = 1.65
a) Determine the temperature (K) at state 2.
b) Determine the pressure (kPa) at state 2.
c) Determine the pressure (kPa) at state 3.
d) Determine the temperature (K) at state 3.
e) Determine the temperature (K) at state 4.
f) Determine the pressure (kPa) at state 4.
g) Determine the net work output (kJ/kg/cycle) for the engine.
h) Determine the heat addition (kJ/kg/cycle) for the engine.
i) Determine the efficiency (%) of the engine.
Points will be given to answer showing all the work done to get right answer.
1 answer:
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Answer:
no of unit is 17941
Explanation:
given data
fixed cost = $338,000
variable cost = $143 per unit
fixed cost = $1,244,000
variable cost = $92.50 per unit
solution
we consider here no of unit is = n
so here total cost of labor will be sum of fix and variable cost i.e
total cost of labor = $33800 + $143 n ..........1
and
total cost of capital intensive = $1,244,000 + $92.5 n ..........2
so here in both we prefer cost of capital if cost of capital intensive less than cost of labor
$1,244,000 + $92.5 n < $33800 + $143 n
solve we get
n >
n > 17941
and
cost of producing less than selling cost so here
$1,244,000 + $92.5 n < 197 n
solve it we get
n >
n > 11904
so in both we get greatest no is 17941
so no of unit is 17941
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: