Propane burns at an equivalence ratio (ER) of 0.6, determine actual air-fuel ratio. If excess air is 5%, what will be the actual
1 answer:
Answer:
when 5% excess air is supplied, moles of air supplied/moles of fuel =
Explanation:
Equivalence ratio = 0.6
Equivalence ratio = Actual air to fuel ratio (AAFR)/ stoichiometric air to fuel ratio SAFR
combustion reaction of propane is
From above reaction, 1 mole of propane, from the reaction, 5 moles of oxygen required,
we know that air contains 21% O_2 and 79% N_2,
Therefore, moles of air based on stoichiometry
Theoretical air to fuel ratio
Given
Actual Air Fuel Ratio
when 5% excess air is supplied, moles of air supplied/moles of fuel =
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Answer:
The pressure difference across hatch of the submarine is 3217.68 kpa.
Explanation:
Gauge pressure is the pressure above the atmospheric pressure. If we consider gauge pressure for finding pressure differential then no need to consider atmospheric pressure as they will cancel out. According to hydrostatic law, pressure varies in the z direction only.
Given:
Height of the hatch is 320 m
Surface gravity of the sea water is 1.025.
Density of water 1000 kg/m³.
Calculation:
Step1
Density of sea water is calculated as follows:
Here, density of sea water is, surface gravity is S.G and density of water is
.
Substitute all the values in the above equation as follows:
kg/m³.
Step2
Difference in pressure is calculated as follows:
pa.
Or
kpa.
Thus, the pressure difference across hatch of the submarine is 3217.68 kpa.
Answer: ε₁+ε₂+ε₃ = 0
Explanation: Considering the initial and final volume to be constant which gives rise to the relation:-
l₀l₀l₀=l₁l₂l₃
taking natural log on both sides
Considering the logarithmic Laws of division and multiplication :
ln(AB) = ln(A)+ln(B)
ln(A/B) = ln(A)-ln(B)
Use the image attached to see the definition of true strain defined as
ln(l1/1o)= ε₁
which then proves that ε₁+ε₂+ε₃ = 0
