Alberta Einstein teaches a business class at Podunk University. To evaluate the students in this class, she has given three test
s. It is now the end of the semester and Alberta asks you to create a program that inputs each student’s test scores and outputs the average score for each student and the overall class average. (Hint: The outer loop should allow for Ms. Einstein to input all the students, one by one, and the inner loop should accept the three exam scores and compute the average for each student.)
1 answer:
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Answer:
33.56 ft^3/sec.in
Explanation:
Duration = 6 hours
drainage area = 185 mi^2
constant baseflow = 550 cfs
<u>Derive the unit hydrograph using the inverse procedure </u>
first step : calculate for the volume of direct runoff hydrograph using the details in table 2 attached below
Vdrh = sum of drh * duration
= 29700 * 6 hours ( 216000 secs )
= 641,520,000 ft^3.
next step : Calculate the volume of runoff in equivalent depth
Vdrh / Area = 641,520,000 / 185 mi^2
= 1.49 in
Finally derive the unit hydrograph
Unit of hydrograph = drh / volume of runoff in equivalent depth
= 50 ft^3 / 1.49 in = 33.56 ft^3/sec.in
Answer:
Mechanical Efficiency = 83.51%
Explanation:
Given Data:
Pressure difference = ΔP=1.2 Psi
Flow rate =
Power of Pump = 3 hp
Required:
Mechanical Efficiency
Solution:
We will first bring the change the units of given data into SI units.
Now we will find the change in energy.
Since it is mentioned in the statement that change in elevation (potential energy) and change in velocity (Kinetic Energy) are negligible.
Thus change in energy is
As we know that Mass = Volume x density
substituting the value
Energy = Volume * density x ΔP / density
Change in energy = Volumetric flow x ΔP
Change in energy = 0.226 x 8.274 = 1.869 KW
Now mechanical efficiency = change in energy / work done by shaft
Efficiency = 1.869 / 2.238
Efficiency = 0.8351 = 83.51%