Answer:
835,175.68W
Explanation:
Calculation to determine the required power input to the pump
First step is to calculate the power needed
Using this formula
P=V*p*g*h
Where,
P represent power
V represent Volume flow rate =0.3 m³/s
p represent brine density=1050 kg/m³
g represent gravity=9.81m/s²
h represent height=200m
Let plug in the formula
P=0.3 m³/s *1050 kg/m³*9.81m/s² *200m
P=618,030 W
Now let calculate the required power input to the pump
Using this formula
Required power input=P/μ
Where,
P represent power=618,030 W
μ represent pump efficiency=74%
Let plug in the formula
Required power input=618,030W/0.74
Required power input=835,175.68W
Therefore the required power input to the pump will be 835,175.68W
Answer:
a. 51.84Kj
b. 2808.99 W/m^2
c. 11.75%
Explanation:
Amount of heat this resistor dissipates during a 24-hour period
= amount of power dissipated * time
= 0.6 * 24 = 14.4 Watt hour
(Note 3.6Watt hour = 1Kj )
=14.4*3.6 = 51.84Kj
Heat flux = amount of power dissipated/ surface area
surface area = area of the two circular end + area of the curve surface
= 2.136 *10^-4 
Heat flux =
= 2808.99 
fraction of heat dissipated from the top and bottom surface
=11.75%
Answer:
Farm equipment
Explanation:
Most people have heard claims that the US government pays farmers not to grow crops. The Agricultural Adjustment Act is the legislation that started this program. It was the first “Farm Bill.” The current farm bill provides for the following:
Subsidies for farmers
Insurance for farmers
Price supports
Food assistance for economically challenged Americans (the largest portion of the Farm Bill)
Forestry conservation programs
Alternative energy programs.
Answer:
Stratification is the process of arranging the different elements classification in a specific manner.
In the storage tank, hot water and cold water create a specific layer and coexist between them for the process of stratification. Thermal storage and heat storage exchanger are the types of storage tank, in which the stratification process take place.
The density of the cold water is more as compared to hot water. Then, due to this the cold water sink to the bottom and hot ware rises up in the storage vessel. In this way, the water gets heated and the stratification process take place in the tank.
In a slowly cooled hypereutectoid iron-carbon steel, the pearlite colonies are normally separated from each other by a more or less continuous boundary layer of cementite done by Slower cooling reasons coarse Pearlite, even as rapid cooling reasons first-rate pearlite to form.
<h3>What levels is in Hypereutectoid metal?</h3>
Hypoeutectoid steels can, upon preliminary cooling from the austenite single segment field, exist as extraordinary levels, eutectoid ferrite and austenite, every with extraordinary carbon contents.
At room temperature, hypereutectoid steels have a pearlitic primary microstructure (ferrite grains with embedded cementite lamellae) with moreover induced cementite on the grain boundaries! The micrograph under suggests a hypereutectoid metal with 1.0 Carbon (C100).
Read more about the cementite:
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