Answer:
a) 42.08 ft/sec
b) 3366.33 ft³/sec
c) 0.235
d) 18.225 ft
e) 3.80 ft
Explanation:
Given:
b = 80ft
y1 = 1 ft
y2 = 10ft
a) Let's take the formula:
1 + 8f² = (20+1)²
= 8f² = 440
f² = 55
f = 7.416
For velocity of the faster moving flow, we have :
V1 = 42.08 ft/sec
b) the flow rate will be calculated as
Q = VA
VA = V1 * b *y1
= 42.08 * 80 * 1
= 3366.66 ft³/sec
c) The Froude number of the sub-critical flow.
V2.A2 = 3366.66
Where A2 = 80ft * 10ft
Solving for V2, we have:
= 4.208 ft/sec
Froude number, F2 =
F2 = 0.235
d) 
= 18.225ft
e) for critical depth, we use :
![y_c = [\frac{(\frac{3366.66}{80})^2}{32.2}]^1^/^3](https://tex.z-dn.net/?f=%20y_c%20%3D%20%5B%5Cfrac%7B%28%5Cfrac%7B3366.66%7D%7B80%7D%29%5E2%7D%7B32.2%7D%5D%5E1%5E%2F%5E3%20)
= 3.80 ft
Answer:
True
Explanation:
A semicircular or circular torch movement should be used when depositing weld beads.
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.
Explanation:
≈4.8
There really isn't an elegant way to express it. Just plug and chug for irrationals raised to other irrationals.
Answer:
t = 2244.3 sec
Explanation:
calculate the thermal diffusivity
Temperature at 28 mm distance after t time = = 50 degree C
we know that
![\frac{ 50 -25}{300-25} = erf [\frac{28\times 10^{-3}}{2\sqrt{1.34\times 10^{-5}\times t}}]](https://tex.z-dn.net/?f=%5Cfrac%7B%2050%20-25%7D%7B300-25%7D%20%3D%20erf%20%5B%5Cfrac%7B28%5Ctimes%2010%5E%7B-3%7D%7D%7B2%5Csqrt%7B1.34%5Ctimes%2010%5E%7B-5%7D%5Ctimes%20t%7D%7D%5D)
from gaussian error function table , similarity variable w calculated as
erf w = 0.909
it is lie between erf w = 0.9008 and erf w = 0.11246 so by interpolation we have
w = 0.08073
![erf 0.08073 = erf[\frac{3.8245}{\sqrt{t}}]](https://tex.z-dn.net/?f=erf%200.08073%20%3D%20erf%5B%5Cfrac%7B3.8245%7D%7B%5Csqrt%7Bt%7D%7D%5D)
solving fot t we get
t = 2244.3 sec
