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igor_vitrenko [27]
3 years ago
6

A solar cell, 3.0 cm square, has an output of 350 mA at 0.80 V when exposed to full sunlight. A solar panel that delivers close

to 1.0 A of current at an emf of 120 V to an external load is needed. How many cells will you need to create the panel? How big a panel will you need, and how should you connect the cells to one another? How can you optimize the output of your solar panel?

Physics
1 answer:
Vesna [10]3 years ago
8 0

Answer:

You will need 450 cells (3 cm each) to meet the voltage/current requirement.

The panel must be 3 cells in one side, by 150 cell in another side. 1350 cm^2 or 0.135 m^2. They must be connected 3 in row in parallel (to add current), then each of the former group must be connected in series to meet the voltage, so it would be 150 rows of connected in series.

The panel can be optimized using a voltage inverter, to convert current to voltage. In this way, less cells can be used achieving the same output specs.

Explanation:

To meet the voltage:

120 [v] required voltage

0.8 [v] voltage of each cell

\frac{120}{0.8} =150[v]\\

So we need 150 cells in series for the voltage.

To meet the current

1.0 [A]      Required current

350[mA]=0.35[A] cell current

1/0.35=3 cell So we need 3 cells in parallel to add the currents and meet the requirement.

See the attached figure

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An ideal Otto cycle has a compression ratio of 8. At the beginning of the compression process, air is at 95 kPa and 278C, and 75
Lina20 [59]

Answer:

a) T_3 = 1539 K , P_3 = 3898 KPa

b) w_net,out = 392.4 KJ/kg

c) n_th = 0.523 - 52.3 %

d) mep = 495 KPa

Explanation:

Given:

- Sate 1: P_1 = 95 KPa, T_1 = 27 C

- q_in = 750 KJ/kg  (constant volume process)

- R_air = 0.287 KJ/kgK

- r = 8  (compression ratio)

Find:

a) Pressure and Temperature (P_3 & T_3) at the end of Heat addition process.

Analysis:

Path 1 to 2 (isentropic compression):

T_1 = 300 K   ----------> u_1 = 214.07 KJ/kg  , v_r1 = 621.2

v_r2 = v_r1 * (V_2 / V_1) = 621.2 *(1 / 8) = 77.65

v_r2 = 77.65 -----------> u_2 = 491.2 KJ/kg , T_2 = 673.1 K

Use Ideal Gas Law for states 1 and 2:

P_2 * v_2 / T_2 = P_1 * v_1 / T_1

P_2 = P_1 * (v_1 /v_2) * ( T_2 / T_1 )

P_2 =  95 KPa * (8)*(673.1/300) = 1705 KPa

Path 2 to 3 (constant volume Heat - Addition):

q_in = (u_3 - u_2 )

u_3 = q_in + u_2

u_3 = 750 +491.2 = 1241.2 KJ/kg

u_3 = 1241.2 KJ/kg  ----------> T_3 = 1539 K , v_r3 = 6.588

Use Ideal Gas Law for states 2 and 3:

P_3 * v_3 / T_3 = P_2 * v_2 / T_2

P_3 = P_2 * ( T_3 / T_2 )

P_3 =  1705 KPa*(1539/673.1) = 3898 KPa

Answer: T_3 = 1539 K and P_3 = 3898 KPa

Path 3 to 4 (isentropic expansion):

v_r4 = v_r3 * (V_1 / V_2) = 6.588 *(8) = 52.7

v_r4 = 52.70 -----------> u_4 = 571.69 KJ/kg , T_4 = 774.5 K

Path 4 to 1 (constant volume Heat - Rejection):

q_out = (u_4 - u_1 )

q_out = 571.69 - 214.07

q_out = 357.62 KJ/kg

b) The net work output w_net,out

Analysis:

w_net, out = q_in - q_out

w_net, out = 750 - 357.62

w_net, out = 392.4 KJ/kg

Answer: The net work output w_net,out = 392.4KJ/kg

c) Thermal Efficiency n_th

Analysis:

n_th = w_net, out / q_in

n_th = 392.4 / 750

n_th = 0.523 ~ 52.3%

Answer: = Thermal Efficiency n_th = 52.3%

d) mean effective pressure for the cycle.

Use ideal gas Law at state 1:

v_1 = R*T_1 / P_1 = 0.287 * 300 / 95

v_1 = v_max= 0.906 m^3 /kg

v_min = v_max / r

mep = w_net,out / (v_1-v_2) = w_net,out / v_1*(1 - 1 / r)

mep = 392.4 / 0.906 * (1 - 1/8)

mep = 495.0 KPa

Answer: mean effective pressure for the cycle mep = 495.0 KPa

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If vector C is added to vector D, the result is a third vector that is perpendicular to D and has a magnitude equal to 3D. What
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Answer:

(e) 3.2

Explanation:

We are given that vector C and D.

Let R be the magnitude of C+D.

According to question

R=3D

We have to find the ratio of the magnitude of C to that of D.

By using right triangle property

C^2=R^2+D^2

C^2=(3D)^2+D^2

C^2=9D^2+D^2

C^2=10D^2

C=\sqrt{10D^2}=3.2D

\frac{C}{D}=3.2

Hence, the ratio of the magnitude of C to that of D=3.2

(e) 3.2

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