Answer:
a) 0.684
b) 0.90
Explanation:
Catalyst
EO + W → EG
<u>a) calculate the conversion exiting the first reactor </u>
CAo = 16.1 / 2 mol/dm^3
Given that there are two stream one contains 16.1 mol/dm^3 while the other contains 0.9 wt% catalyst
Vo = 7.24 dm^3/s
Vm = 800 gal = 3028 dm^3
hence Im = Vin/ Vo = (3028 dm^3) / (7.24dm^3/s) = 418.232 secs = 6.97 mins
next determine the value of conversion exiting the reactor ( Xai ) using the relation below
KIm = 
------ ( 1 )
make Xai subject of the relation
Xai = KIm / 1 + KIm --- ( 2 )
<em>where : K = 0.311 , Im = 6.97 ( input values into equation 2 )</em>
Xai = 0.684
<u>B) calculate the conversion exiting the second reactor</u>
CA1 = CA0 ( 1 - Xai )
therefore CA1 = 2.5438 mol/dm^3
Vo = 7.24 dm^3/s
To determine the value of the conversion exiting the second reactor ( Xa2 ) we will use the relation below
XA2 = ( Xai + Im K ) / ( Im K + 1 ) ----- ( 3 )
<em> where : Xai = 0.684 , Im = 6.97, and K = 0.311 ( input values into equation 3 )</em>
XA2 = 0.90
<u />
<u />
<u />
Explanation:
Air fuel ratio:
Air fuel ratio is the ratio of mass of air to the mass of fuel.So we can say that
As we know that fuel burn in the presence of air that is why we have to maintain a proper amount of air fuel ratio.
When we need more power then we have supply more fuel and to burn this fuel ,require a specified amount of air.So for different loading condition of engine different air fuel ratio is required.
When air is less and fuel is more then it is called rich air fuel ratio .when air is more and fuel is less then it is called poor air fuel ratio.
Answer:
The maximum power that can be generated is 127.788 kW
Explanation:
Using the steam table
Enthalpy at 20 bar = 2799 kJ/kg
Enthalpy at 2 bar = 2707 kJ/kg
Change in enthalpy = 2799 - 2707 = 92 kJ/kg
Mass flow rate of steam = 5000 kg/hr = 5000 kJ/hr × 1 hr/3600 s = 1.389 kg/s
Maximum power generated = change in enthalpy × mass flow rate = 92 kJ/kg × 1.389 kg/s = 127.788 kJ/s = 127.788 kW
The maintenance is in charge of controlling that all the machines of a company are constantly running in order to avoid damages that cause loss of money when the machines fail.
The maintenance based on vibration monitoring allows to predict failures in some rotating machines such as:
1. worn bearings
2.alignment
3.balance
4. affected gears
5. bent shafts
6. rocks
7.gags
8. eccentricity
9. failures of electrical origin
Answer:
there's no photo? but I'm willing to help
