Answer:
80.7lbft/hr
Explanation:
Flow rate of water in the system = 3.6x10^-6
The height h = 100
1s = 1/3600h
This implies that
Q = 3.6x10^-6/[1/3600]
Q = 0.0000036/0.000278
Q = 0.01295
Then the power is given as
P = rQh
The specific weight of water = 62.3 lb/ft³
P = 62.3 x 0.01295 x 100
P = 80.675lbft/h
When approximated
P = 80.7 lbft/h
This is the average power that could be generated in a year.
This answers the question and also corresponds with the answer in the question.
Answer:
b. The pirating streams are eroding headwardly to intersect more of the other streams’ drainage basins, causing water to be diverted down their steeper gradients.
Explanation:
From the Kaaterskill NY 15 minute map (1906), this shows two classic examples of stream capture.
The Kaaterskill Creek flow down the east relatively steep slopes into the Hudson River Valley. While, the Gooseberry Creek is a low gradient stream flowing down the west direction which in turn drains the higher parts of the Catskills in this area.
However, there is Headward erosion of Kaaterskill Creek which resulted to the capture of part of the headwaters of Gooseberry Creek.
The evidence for this is the presence of "barbed" (enters at obtuse rather than acute angle) tributary which enters Kaaterskill Creek from South Lake which was once a part of the Gooseberry Creek drainage system.
It should be noted again, that there is drainage divide between the Gooseberry and Kaaterskill drainage systems (just to the left of the word Twilight) which is located in the center of the valley.
As it progresses, this divide will then move westward as Kaaterskill captures more and more of the Gooseberry system.
Answer: Hello the question is incomplete below is the missing part
Question: determine the temperature, in °R, at the exit
answer:
T2= 569.62°R
Explanation:
T1 = 540°R
V2 = 600 ft/s
V1 = 60 ft/s
h1 = 129.0613 ( value gotten from Ideal gas property-air table )
<em>first step : calculate the value of h2 using the equation below </em>
assuming no work is done ( potential energy is ignored )
h2 = [ h1 + ( V2^2 - V1^2 ) / 2 ] * 1 / 32.2 * 1 / 778
∴ h2 = 136.17 Btu/Ibm
From Table A-17
we will apply interpolation
attached below is the remaining part of the solution
Answer:
a) 246.56 Hz
b) 203.313 Hz
c) Add more springs
Explanation:
Spring constant = 12000 N/m
mass = 5g = 5 * 10^-3 kg
damping ratio = 0.4
<u>a) Calculate Natural frequency </u>
Wn = √k/m = 
= 1549.19 rad/s ≈ 246.56 Hz
<u>b) Bandwidth of instrument </u>
W / Wn = 
W / Wn = 0.8246
therefore Bandwidth ( W ) = Wn * 0.8246 = 246.56 * 0.8246 = 203.313 Hz
C ) To increase the bandwidth we have to add more springs
Answer:
P=11 kW
Explanation:
Given that
Number of poles= 8
I.E.C. 180L motor frame
From data book , for 8 poles motor at 50 Hz
Speed = 730 rpm
Power factor = 0.75
Efficiency at 100 % load= 89.3 %
Efficiency at 50 % load= 89.1 %
Output power = 11 kW
Therefore the rated output power of 8 poles motor will be 11 kW. Thus the answer will be 11 kW.
P=11 kW