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CaHeK987 [17]
3 years ago
14

At the surface of the earth, there is an approximate average solar flux of 0.75 kW/m2. A family wishes to construct a solar ener

gy conversion system to power their home. If the conversion system is 30% efficient and the family needs a maximum of 26 kW, what effective surface area is needed for perfectly absorbing collectors?
Physics
1 answer:
GenaCL600 [577]3 years ago
7 0

Answer:

S=115.6 [m^{2}]

Explanation:

If the family needs 26 kW of power, and knowing that the efficient is 30% the relation between the solar flux and the power would be:

\Phi_{solar} [kW/m^{2}]\cdot S [m^{2}] \cdot e=26 [kW] (1)

e is the efficient (in our case 0.3)

Φsolar is the solar flux

S is the effective surface

Solving (1) for S we have:

S=\frac{26 [kW]}{0.75 [kW/m^{2}]\cdot 0.3}=115.6 [m^{2}]

I hope it helps!

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Answer:

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Explanation:

This is a problem that encompasses the work and principle of energy conservation.

In this way, we establish the equation for the principle of conservation and energy.

i)

E_{k1}+W_{1-2}=E_{k2}\\where:\\E_{k1}= kinetic energy at moment 1\\W_{1-2}= work between moments 1 and 2.\\E_{k2}= kinetic energy at moment 2.

W_{1-2}= (F*d) - (m*g*h)\\W_{1-2}=(500000*2.5*10^3)-(40000*9.81*2.5*10^3)\\W_{1-2}= 269*10^6[J] or 269 [MJ]

At that point the speed 1 is equal to zero, since the maximum height achieved was 2.5 [km]. So this calculated work corresponds to the energy of the rocket.

Er = 269*10^6[J]

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E_{k2}=0.5*m*v^2\\269*10^6=0.5*40000*v^2\\v=\sqrt{\frac{269*10^6}{0.5*40000} }\\ v=116[m/s]

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