Answer:
Bending stress at point 3.96 is \sigma_b = 1.37 psi
Explanation:
Given data:
Bending Moment M is 4.176 ft-lb = 50.12 in- lb
moment of inertia I = 144 inc^4
y = 3.96 in

putting all value to get bending stress

Bending stress at point 3.96 is
= 1.37 psi
Answer:
The velocity of flow is 10.0 m/s.
Explanation:
We shall use Manning's equation to calculate the velocity of flow
Velocity of flow by manning's equation is given by

where
n = manning's roughness coefficient
R = hydraulic radius
S = bed slope of the channel
We know that for an asphalt channel value of manning's roughness coefficient = 0.016
Applying values in the above equation we obtain velocity of flow as

Answer:
The problem is that the pumps would consume more energy than the generators would produce.
Explanation:
Water has a potential energy associated with the height it is at. The higher it is, the higher the potential energy. When water flows down into the turbines that energy is converted to kinetic energy and then into electricity.
A pump uses electricity to add energy to the water to send it to a higher potential energy state.
Ideally no net energy woul be hgenerate or lost, because the generators would release the potential energy and pumps would store it again in the water. However the systems are not ideal, everything has an efficiency and losses. The losses would accumulate and the generator would be generating less energy than the pumps consume, so that system wastes energy.
What should be done is closing the floodgates to keep the water up in the dam at night producing only the power that is needed and releasing more water during the day.
Answer:
See step by step explanations for answer.
Explanation:
600 megawatts =
568 690.272 btu / second
thermal eficiency=work done/Heat supllied
0.38=568690.272/Heat supplied
Heat supplied=1496553.35btu /s
heat emmitted to the atmosphere=heat supplied -work done=(1496553.35-568690.272)=927863.1 btu/s
feed rate=(1496553.35)/12000=124.71 lb/s =10775184.1056 lb/day=5 387.472 ton / day
sulphur content released=(0.03*124.71)/(1.496553)=2.5 lb SO2/million Btu of heat input
so
the degree (%) of sulfur dioxide control needed to meet an emission standard=(2.5/0.15)*100=1666.67 %
the CO2 emission rate=220*(1.496553) =329.241 lb/s =12 903.0802 metric ton / day
As there are 10 V, for Vp1, that is the peak-voltage of the source:

Then, transformer's theory says that the relation of transformations is:
V1/V2=a
Where V1 is the voltage in the primary and V2 in the secondary.
V1=14.14 V
V2=8.55 V
a=1.65
Then, with the 8.5 V, we find the real peak-voltage, taking in account that in the diodes we have a drop of 0.7 V each, so:
8.5 -1.4=7.1 V
And this will be called VpL
Now we proceed to calculate the mean voltage:

Where Vr is the ripple voltage, we asume that is 1 V
So, Vmean = 6.6 V
Then we have
Vmean/R= I mean
We have that R=1000 Ohm
Imedia=6.6 V/1000 Ohm
Imedia=6.6 mAmps
Finally, we can calculate the capacitor:
C=Q/Vr
C=Imean/(Vr*2f)
Where f is 60Hz
C=6.6mA/(1V*120)
C=5.5 uFarads
Therefore:
C=5.5 uFarads that works at 12 V