Answer:
If the heat engine operates for one hour:
a) the fuel cost at Carnot efficiency for fuel 1 is $409.09 while fuel 2 is $421.88.
b) the fuel cost at 40% of Carnot efficiency for fuel 1 is $1022.73 while fuel 2 is $1054.68.
In both cases the total cost of using fuel 1 is minor, therefore it is recommended to use this fuel over fuel 2. The final observation is that fuel 1 is cheaper.
Explanation:
The Carnot efficiency is obtained as:

Where
is the atmospheric temperature and
is the maximum burn temperature.
For the case (B), the efficiency we will use is:

The work done by the engine can be calculated as:
where Hv is the heat value.
If the average net power of the engine is work over time, considering a net power of 2.5MW for 1 hour (3600s), we can calculate the mass of fuel used in each case.

If we want to calculate the total fuel cost, we only have to multiply the fuel mass with the cost per kilogram.

Answer:
This doesn't represent an equilibrium state of stress
Explanation:
∝ = 1 , β = 1 , y = 1
x = 0 , y = 0 , z = 0 ( body forces given as 0 )
Attached is the detailed solution is and also the conditions for equilibrium
for a stress state to be equilibrium all three conditions has to meet the equilibrum condition as explained in the attached solution
Answer:
2.0%
Explanation:
Percentage of aggregate = 94%
Specific gravity = 2.65
Specific gravity of asphalt = 1.9
Density of mix = 147pcf = 147lb/ft³
Total weight of mix: (volume = 1ft³)
= (147lb/ft³)(1ft³)
= 147lb
Percentage weight of asphalt in<u> mix:</u>
100% - 94%
= 6%
Weight of asphalt binders
= 6% x 147lb
= 8.82lb
Weight of aggregate in mix:
= 94% x 147
= 138.18lb
Specific weight of asphalt binder:
(Gab)(Yw)
Yw = specific Weight of water
= 62.4lb
Gab = specific gravity of asphalt binder
= 1.0
(62.4lb)(1.0)
= 62.4 lb/ft³
Volume of asphalt in binder:
8.82/62.4
= 0.14ft³
Specific weight of binder in mix:
2.65 x 62.4lb/ft³
= 165.36 lb/ft³
Volume of aggregate:
= 138.18/165.36
= 0.84ft³
Volume of void in the mix:
1ft³ - 0.84ft³ - 0.14ft³
= 0.02ft³
<u>The percentage of void in total mix:</u>
VTM = (0.02ft³/1ft³)100
= 2.0%
An automobile travels along a straight road at 15.65 m/s through a 11.18 m/s speed zone. A police car observed the automobile. At the instant that the two vehicles are abreast of each other, the police car starts to pursue the automobile at a constant acceleration of 1.96 m/s2 . The motorist noticed the police car in his rear view mirror 12 s after the police car started the pursuit and applied his brakes and decelerates at 3.05 m/s2
Find the total time required for the police car to over take the automobile.
Answer:
15.02 sec
Explanation:
The total time required for the police car to overtake the automobile is related to the distance covered by both cars which is equal from instant point of abreast.
So; we can say :

By using the second equation of motion to find the distance S;





where ;
u = 0





Recall that:



= 46.68 - 7.85 t -2.505 t² = 0
Solving by using quadratic equation;
t = -6.16 OR t = 3.02
Since we can only take consideration of the value with a positive integer only; then t = 3.02 secs
From the question; The motorist noticed the police car in his rear view mirror 12 s after the police car started the pursuit;
Therefore ; the total time required for the police car to over take the automobile = 12 s + 3.02 s
Total time required for the police car to over take the automobile = 15.02 sec