As a head-up, it is important to notice that a white dwarf only shines thanks to the stored energy and light, because a white dwarf doesn't have any hydrogen left to perform nuclear fusion.
Now the process:
First, the white dwarf accumulates all the extracted matter from its companion, onto its own surface. This extra matter increases the white dwarf's temperature and density.
After a while, the star reaches about 10 million K, so nuclear fusion can begin. The hydrogen that has been "stolen" from the other star and accumulated in the white dwarf's surface it's used for the fusion, dramatically increasing the star's brightness for a short time, causing what we know as a Nova.
As this fuel its quickly burnt out or blown into space, the star goes back to its natural white dwarf state. Since the white dwarf nor the companion star are destroyed in this process, it can happen countless of times during their lifespan.
Question:
The operations manager for a well-drilling company must recommend whether to build a new facility, expand his existing one, or do nothing. He estimates that long-run profits (in $000) will vary with the amount of precipitation (rainfall) as follows:
Alternative Precipitation
Low Normal High
Do nothing -100 100 300
Expand 350 500 200
Build new 750 300 0
If he feels the chances of low, normal, and high precipitation are 30 percent, 20 percent, and 50 percent respectively, What is EVPI (Expected value of Perfect Information)?
A. $140,000
B. $170,000
C. $285,000
D. $305,000
E. $475,000
Answer:
D. $170,000
Explanation:
The expected long run profits are for
Low Normal High
Do nothing -100*0.3 100*0.2 300*0.5 = 140
Expand 350*0.3 500*0.2 200*0.5 = 305
Build new 750*0.3 300*0.2 0*0.5 = 285
Therefore the expected long run profits are
$140,000
$305,000
$285,000
Based on his selected option being either to build new or to expand, the most profitable option is to expand
=$305,000
EVPI = EPPI-EMV =$170,000
Specific Gravity of the fluid = 1.25
Height h = 28 in
Atmospheric Pressure = 12.7 psia
Density of water = 62.4 lbm/ft^3 at 32F
Density of the Fluid = Specific Gravity of the fluid x Density of water = 1.25 x 62.4
Density of the Fluid p = 78 lbm/ft^3
Difference in pressure as we got the differential height, dP = p x g x h dP = (78 lbm/ft^3) x (32.174 ft/s^2) x (28/12 ft) [ 1 lbf / 32.174 ft/s^2] [1 ft^2 /
144in^2]
Difference in pressure = 1.26 psia
(a) Pressure in the arm that is at Higher
P = Atmospheric Pressure - Pressure difference = 12.7 - 1.26 = 11.44 psia
(b) Pressure in the tank that is at Lower
P = Atmospheric Pressure + Pressure difference = 12.7 + 1.26 = 13.96psia
There are no appropriate units for power on the list you provided
