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aleksandrvk [35]
3 years ago
14

Based on the following passage, why might the government use the U.S. Army Corps of Engineers to undertake hydroelectric power p

rojects instead of allowing private businesses to undertake them?
Before World War I, most power projects were developed by private interests. During the Great Depression, president Franklin Roosevelt developed federal hydropower projects to give low-cost energy to consumers. The U.S. Army Corps of Engineers undertook three major hydroelectric power projects during this era: the Bonneville Dam on the Columbia River, the Fort Peck Dam on the Missouri River, and the Passamaquoddy Tidal Power Project in Maine. Congress organized the Bonneville Power Administration in 1937 to distribute the power and set usage rates for all power generated at Bonneville Dam.


Hydroelectric power is closely linked to national security.

Private businesses are by law disallowed from engaging in hydroelectric power projects.

The government would have less power to control the cost of electricity from privately funded projects.

Private businesses have less expertise in hydroelectric power than the U.S. Army Corps of Engineers.
Engineering
1 answer:
mr Goodwill [35]3 years ago
3 0

Answer:

The government would have less power to control the cost of electricity from privately funded projects.

Explanation:

According to the given excerpt, the government, the government decided to take over most of the hydroelectric power projects as opposed to privately owned corporations which previously held a monopoly. The major reason was to provide low-cost energy to consumers.

This decision taken by the government to use the Army Corps of Engineers to undertake hydroelectric power projects instead of allowing private businesses to undertake them was to allow the government have more power to control the cost of electricity.

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Thermal expansion occurs when a material is subjected to heat and changes it's shape, area and volume as a result of that heat. How much that material changes is dependent on it's coefficient of thermal expansion.

Different materials have different coefficients of thermal expansion (i.e. It is a material property and differs from one material to the next). It is important to understand how materials behave when heated, especially for engineering applications when a change in dimension might pose a problem or risk (eg. building large structures).

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2 years ago
Write the heat equation for each of the following cases:
jok3333 [9.3K]

Answer:

Explanation:

a) the steady-state, 1-D incompressible and no energy generation equation can be expressed as follows:

\dfrac{\partial^2T}{\partial x^2}=  \ 0  \  ;  \ if \  T = f(x)  \\ \\ \dfrac{\partial^2T}{\partial y^2}=  \ 0  \  ;  \ if \  T = f(y)  \\ \\ \dfrac{\partial^2T}{\partial z^2}=  \ 0  \  ;  \ if \  T = f(z)

b) For a transient, 1-D, constant with energy generation

suppose T = f(x)

Then; the equation can be expressed as:

\dfrac{\partial^2T}{\partial x^2} + \dfrac{Q_g}{k} = \dfrac{1}{\alpha} \dfrac{dT}{dC}

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c) The heat equation for a cylinder steady-state with 2-D constant and no compressible energy generation is:

\dfrac{1}{r}\times \dfrac{\partial}{\partial r }( r* \dfrac{\partial \ T }{\partial \ r}) + \dfrac{\partial^2 T}{\partial z^2 }= 0

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The radial directional term = \dfrac{1}{r}\times \dfrac{\partial}{\partial r }( r* \dfrac{\partial \ T }{\partial \ r}) and the axial directional term is \dfrac{\partial^2 T}{\partial z^2 }

d) The heat equation for a wire going through a furnace is:

\dfrac{\partial ^2 T}{\partial z^2} = \dfrac{1}{\alpha}\Big [\dfrac{\partial ^2 T}{\partial ^2 t}+ V_z \dfrac{\partial ^2T}{\partial ^2z} \Big ]

since;

the steady-state is zero, Then:

\dfrac{\partial ^2 T}{\partial z^2} = \dfrac{1}{\alpha}\Big [ V_z \dfrac{\partial ^2T}{\partial ^2z} \Big ]'

e) The heat equation for a sphere that is transient, 1-D, and incompressible with energy generation is:

\dfrac{1}{r} \times \dfrac{\partial}{\partial r} \Big ( r^2 \times \dfrac{\partial T}{\partial r} \Big ) + \dfrac{Q_q}{K} = \dfrac{1}{\alpha}\times \dfrac{\partial T}{\partial t}

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3 years ago
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You can use like a thickish paper but not to heavy and do it spiral and fall it with bubble paper and light things and then you spin it while you drop and it won’t crack
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