Answer:
Explained below
Explanation:
The isohyetal method is one used in estimating Rainfall whereby the mean precipitation across an area is gotten by drawing lines that have equal precipitation. This is done by the use of topographic and other data to yield reliable estimates.
Whereas, the arithmetic method is used to calculate true precipitation by the way of getting the arithmetic mean of all the points or arial measurements that will be considered in the analysis.
Answer:
C
Explanation:
I COULD be wrong, i'm not sure but im confident its c
Answer:
See explaination
Explanation:
Lets first consider the term Isentropic efficiency. The isentropic efficiency of a compressor or pump is defined as the ratio of the work input to an isentropic process, to the work input to the actual process between the same inlet and exit pressures. IN practice, compressors are intentionally cooled to minimize the work input.
Please kindly check attachment for the step by step solution of the given problem.
Answer:
a) ![\mathbf{\sigma _ 1 = 4800 psi}](https://tex.z-dn.net/?f=%5Cmathbf%7B%5Csigma%20_%201%20%3D%204800%20psi%7D)
![\mathbf{ \sigma _2 = 0}](https://tex.z-dn.net/?f=%5Cmathbf%7B%20%5Csigma%20_2%20%3D%200%7D)
b)![\mathbf{\sigma _ 1 = 6000 psi}](https://tex.z-dn.net/?f=%5Cmathbf%7B%5Csigma%20_%201%20%3D%206000%20psi%7D)
![\mathbf{ \sigma _2 = 3000 psi}](https://tex.z-dn.net/?f=%5Cmathbf%7B%20%5Csigma%20_2%20%3D%203000%20psi%7D)
Explanation:
Given that:
diameter d = 12 in
thickness t = 0.25 in
the radius = d/2 = 12 / 2 = 6 in
r/t = 6/0.25 = 24
24 > 10
Using the thin wall cylinder formula;
The valve A is opened and the flowing water has a pressure P of 200 psi.
So;
![\sigma_{hoop} = \sigma _ 1 = \frac{Pd}{2t}](https://tex.z-dn.net/?f=%5Csigma_%7Bhoop%7D%20%3D%20%5Csigma%20_%201%20%3D%20%5Cfrac%7BPd%7D%7B2t%7D)
![\sigma_{long} = \sigma _2 = 0](https://tex.z-dn.net/?f=%5Csigma_%7Blong%7D%20%3D%20%5Csigma%20_2%20%3D%200)
![\sigma _ 1 = \frac{Pd}{2t} \\ \\ \sigma _ 1 = \frac{200(12)}{2(0.25)}](https://tex.z-dn.net/?f=%5Csigma%20_%201%20%3D%20%5Cfrac%7BPd%7D%7B2t%7D%20%5C%5C%20%5C%5C%20%5Csigma%20_%201%20%3D%20%5Cfrac%7B200%2812%29%7D%7B2%280.25%29%7D)
![\mathbf{\sigma _ 1 = 4800 psi}](https://tex.z-dn.net/?f=%5Cmathbf%7B%5Csigma%20_%201%20%3D%204800%20psi%7D)
b)The valve A is closed and the water pressure P is 250 psi.
where P = 250 psi
![\sigma_{hoop} = \sigma _ 1 = \frac{Pd}{2t}](https://tex.z-dn.net/?f=%5Csigma_%7Bhoop%7D%20%3D%20%5Csigma%20_%201%20%3D%20%5Cfrac%7BPd%7D%7B2t%7D)
![\sigma_{long} = \sigma _2 = \frac{Pd}{4t}](https://tex.z-dn.net/?f=%5Csigma_%7Blong%7D%20%3D%20%5Csigma%20_2%20%3D%20%5Cfrac%7BPd%7D%7B4t%7D)
![\sigma _ 1 = \frac{Pd}{2t} \\ \\ \sigma _ 1 = \frac{250*(12)}{2(0.25)}](https://tex.z-dn.net/?f=%5Csigma%20_%201%20%3D%20%5Cfrac%7BPd%7D%7B2t%7D%20%5C%5C%20%5C%5C%20%5Csigma%20_%201%20%3D%20%5Cfrac%7B250%2A%2812%29%7D%7B2%280.25%29%7D)
![\mathbf{\sigma _ 1 = 6000 psi}](https://tex.z-dn.net/?f=%5Cmathbf%7B%5Csigma%20_%201%20%3D%206000%20psi%7D)
![\sigma _2 = \frac{Pd}{4t} \\ \\ \sigma _2 = \frac{250(12)}{4(0.25)}](https://tex.z-dn.net/?f=%5Csigma%20_2%20%3D%20%5Cfrac%7BPd%7D%7B4t%7D%20%5C%5C%20%5C%5C%20%20%5Csigma%20_2%20%3D%20%5Cfrac%7B250%2812%29%7D%7B4%280.25%29%7D)
![\mathbf{ \sigma _2 = 3000 psi}](https://tex.z-dn.net/?f=%5Cmathbf%7B%20%5Csigma%20_2%20%3D%203000%20psi%7D)
The free flow body diagram showing the state of stress on a volume element located on the wall at point B is attached in the diagram below