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3 years ago

5

Identify renewable energy sources you will propose. Explain the key elements to your solution and the basic technical principles

involved.

1 answer:

MArishka [77]3 years ago

3 0

Answer:

A renewable electricity generation technology harnesses a naturally existing energy. But they have other features that a few fringe customers value.

Explanation:

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What is the composition, in atom percent, of an alloy that consists of 4.5 wt% Pb and 95.5 wt% Sn?

jeka57 [31]

Answer: Option A is correct -- 2.6 at% Pb and 97.4 at% Sn.

Explanation:

Option A is the only correct option -- 2.6 at% Pb and 97.4 at% Sn. While option B, which is 7.6 at% Pb and 92.4 at% Sn. and option C, which is 97.4 at% Pb and 2.6 at% Sn. and option D, which is 92.4 at% Pb and 7.6 at% Sn. are wrong.

6 0

3 years ago

What is the thermal efficiency of this regeneration cycle in terms of enthalpies and fractions of total flow?

irga5000 [103]

Answer:

$\eta =\dfrac{(h_3-h_4)-(h_2-h_1)}{(h_3-h_5)}$

Explanation:

generally regeneration of cycle is used in the case of gas turbine. due to regeneration efficiency of turbine is increased but there is no effect on the on the net work out put of turbine.Actually in regeneration net heta input is decreases that is why total efficiency increase.

Now from T-S diagram

$W_{net}=W_{out}-W_{in}$

$W_{net}=(h_3-h_4)-(h_2-h_1)$

$Q_{in}=h_3-h_5$

Due to generation $(h_5-h_2)$ amount of energy has been saved.

$Q_{generation}=Q_{saved}$

So efficiency of cycle $\eta =\frac{W_{net}}{Q_{in}}$

$\eta =\dfrac{(h_3-h_4)-(h_2-h_1)}{(h_3-h_5)}$

Effectiveness of re-generator

$\varepsilon =\dfrac{(h_5-h_2)}{(h_4-h_2)}$

So the efficiency of regenerative cycle

$\eta =\dfrac{(h_3-h_4)-(h_2-h_1)}{(h_3-h_5)}$

7 0

3 years ago

A well-designed product will increase?

Colt1911 [192]

Answer:

true

Explanation:

A well designed product will increase in sells and in stock.

8 0

2 years ago

The human eye, as well as the light-sensitive chemicals on color photographic film, respond differently to light sources with di

jeka57 [31]

Answer:

a) at T = 5800 k

band emission = 0.2261

at T = 2900 k

band emission = 0.0442

b) daylight (d) = 0.50 μm

Incandescent ( i ) = 1 μm

Explanation:

To Calculate the band emission fractions we will apply the Wien's displacement Law

The ban emission fraction in spectral range λ1 to λ2 at a blackbody temperature T can be expressed as

F ( λ1 - λ2, T ) = F( 0 ----> λ2,T) - F( 0 ----> λ1,T )

<em>Values are gotten from the table named: blackbody radiati</em>on functions

<u>a) Calculate the band emission fractions for the visible region</u>

at T = 5800 k

band emission = 0.2261

at T = 2900 k

band emission = 0.0442

attached below is a detailed solution to the problem

<u>b)calculate wavelength corresponding to the maximum spectral intensity</u>

For daylight ( d ) = 2898 μm *k / 5800 k = 0.50 μm

For Incandescent ( i ) = 2898 μm *k / 2900 k = 1 μm

3 0

3 years ago

If you had a match and a lantern and a candle in the dark which one would you choose to light.

PSYCHO15rus [73]

Answer:

The match

Explanation:

You can light both the lantern and the candle if you light the match first.

I don't know of this is a homework question, but I answered it anyway :)

5 0

3 years ago

Read 2 more answers