Plan to refuel when your fuel gauge reads __________.
2 answers:
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Answer:
b) false
Explanation:
We know that Otto cycle is the ideal cycle for all petrol working engine.In Otto cycle all process are consider is ideal ,means there is no any ir-reversibility in the processes.
It consist four processes
1-2:Reversible adiabatic compression
2-3:Constant volume heat addition
3-4:Reversible adiabatic expansion
3-4:Constant volume heat rejection
Along with above 4 processes intake and exhaust processes are parallel to each other.From the P-v diagram we can see that all processes.
But actually in general we are not showing intake and exhaust line then it did not mean that in Otto cycle did not have intake and exhaust processes.
<u>Solution and Explanation:</u>
Volume of gas stream = 1000 cfm (Cubic Feet per Minute)
Particulate loading = 400 gr/ft3 (Grain/cubic feet)
1 gr/ft3 = 0.00220462 lb/ft3
Total weight of particulate matter =
Cyclone is to 80 % efficient
So particulate remaining =
emissions from this stack be limited to = 10.0 lb/hr
Particles to be remaining after wet scrubber = 10.0 lb/hr
So particles to be removed = 685.7136- 10 = 675.7136
Efficiency = output multiply with 100/input = 98.542 %
Answer: Attached below is the missing diagram
answer :
A) 1) Wr > WI, 2) Qc' > Qc
B) 1) QH' > QH, 2) Qc' > Qc
Explanation:
л = w / QH = 1 - Qc / QH and QH = w + Qc
<u>A) each cycle receives same amount of energy by heat transfer</u>
<u>(</u> Given that ; Л1 = 1/3 ЛR )
<em>1) develops greater bet work </em>
WR develops greater work ( i.e. Wr > WI )
<em>2) discharges greater energy by heat transfer</em>
Qc' > Qc
solution attached below
<u>B) If Each cycle develops the same net work </u>
<em>1) Receives greater net energy by heat transfer from hot reservoir</em>
QH' > QH ( solution is attached below )
<em>2) discharges greater energy by heat transfer to the cold reservoir</em>
Qc' > Qc
solution attached below
