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3 years ago

A photovoltaic panel of dimension 2mÃ—4m is installed on the

Engineering

1 answer:

blsea [12.9K]3 years ago

6 0

Answer:

Explanation:

Simply put, a solar panel works by allowing photons, or particles of light, to knock electrons free from atoms, generating a flow of electricity. Solar panels actually comprise many, smaller units called photovoltaic cells. (Photovoltaic simply means they convert sunlight into electricity. The attached diagram give an ilustsration of the photovotaic pannel mounted on a roof top.

Solution

To Determine the electric power generated for

a) A still summer day.

E = A * r * H * PR

E = Total Amount of Energy in kilowatt

A = Total Surface Area

r = efficiency Rating

H = global radiation value

PR = Performance Ratio

kwh = watt * Time/1000

kwh = 100 * 35/1000

3.5

kwh = watt * Time/1000

kwh = 30 *15/1000

4.5

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Which band has an average of $3.58 per hour of parking?

Minchanka [31]

C it would be c because that has more and the others have less

6 0

2 years ago

Determine the following parameters for the water having quality x=0.7 at 200 kPa:

ra1l [238]

Solution :

Given :

Water have quality x = 0.7 (dryness fraction) at around pressure of 200 kPa

The phase diagram is provided below.

a). The phase is a standard mixture.

b). At pressure, p = 200 kPa, T = $T_{saturated}$

Temperature = 120.21°C

c). Specific volume

$v_{f}= 0.001061, \ \ v_g=0.88578 \ m^3/kg$

$v_x=v_f+x(v_g-v_f)$

$=0.001061+0.7(0.88578-0.001061)$

$=0.62036 \ m^3/kg$

d). Specific energy ( $u_x$ )

$u_f=504.5 \ kJ/kg, \ \ u_{fg}=2024.6 \ kJ/kg$

$u_x=504.5 + 0.7(2024.6)$

$=1921.72 \ kJ/kg$

e). Specific enthalpy $(h_x)$

At $h_f = 504.71, \ \ h_{fg} = 2201.6$

$h_x=504.71+(0.7\times 2201.6)$

$= 2045.83 \ kJ/kg$

f). Enthalpy at m = 0.5 kg

$H=mh_x$

$= 0.5 \times 2045.83$

= 1022.91 kJ

7 0

3 years ago

A charge of +2.00 μC is at the origin and a charge of –3.00 μC is on the y axis at y = 40.0 cm . (a) What is the potential at po

Nimfa-mama [501]

a) Potential in A: -2700 V

b) Potential difference: -26,800 V

c) Work: $4.3\cdot 10^{-15} J$

Explanation:

The electric potential at a distance r from a single-point charge is given by:

$V(r)=\frac{kq}{r}$

where

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

q is the charge

r is the distance from the charge

In this problem, we have a system of two charges, so the total potential at a certain point will be given by the algebraic sum of the two potentials.

Charge 1 is

$q_1=+2.00\mu C=+2.00\cdot 10^{-6}C$

and is located at the origin (x=0, y=0)

Charge 2 is

$q_2=-3.00 \mu C=-3.00\cdot 10^{-6}C$

and is located at (x=0, y = 0.40 m)

Point A is located at (x = 0.40 m, y = 0)

The distance of point A from charge 1 is

$r_{1A}=0.40 m$

So the potential due to charge 2 is

$V_1=\frac{(8.99\cdot 10^9)(+2.00\cdot 10^{-6})}{0.40}=+4.50\cdot 10^4 V$

The distance of point A from charge 2 is

$r_{2A}=\sqrt{0.40^2+0.40^2}=0.566 m$

So the potential due to charge 1 is

$V_2=\frac{(8.99\cdot 10^9)(-3.00\cdot 10^{-6})}{0.566}=-4.77\cdot 10^4 V$

Therefore, the net potential at point A is

$V_A=V_1+V_2=+4.50\cdot 10^4 - 4.77\cdot 10^4=-2700 V$

Here we have to calculate the net potential at point B, located at

(x = 0.40 m, y = 0.30 m)

The distance of charge 1 from point B is

$r_{1B}=\sqrt{(0.40)^2+(0.30)^2}=0.50 m$

So the potential due to charge 1 at point B is

$V_1=\frac{(8.99\cdot 10^9)(+2.00\cdot 10^{-6})}{0.50}=+3.60\cdot 10^4 V$

The distance of charge 2 from point B is

$r_{2B}=\sqrt{(0.40)^2+(0.40-0.30)^2}=0.412 m$

So the potential due to charge 2 at point B is

$V_2=\frac{(8.99\cdot 10^9)(-3.00\cdot 10^{-6})}{0.412}=-6.55\cdot 10^4 V$

Therefore, the net potential at point B is

$V_B=V_1+V_2=+3.60\cdot 10^4 -6.55\cdot 10^4 = -29,500 V$

So the potential difference is

$V_B-V_A=-29,500 V-(-2700 V)=-26,800 V$

The work required to move a charged particle across a potential difference is equal to its change of electric potential energy, and it is given by

$W=q\Delta V$

where

q is the charge of the particle

$\Delta V$ is the potential difference

In this problem, we have:

$q=-1.6\cdot 10^{-19}C$ is the charge of the electron

$\Delta V=-26,800 V$ is the potential difference

Therefore, the work required on the electron is

$W=(-1.6\cdot 10^{-19})(-26,800)=4.3\cdot 10^{-15} J$

4 0

3 years ago

Fill in the blank to correctly complete the statement below.

ZanzabumX [31]

Did not engineer cables factoring wind shear

4 0

3 years ago

Steam enters an adiabatic turbine at 10MPa and 500 C and leaves at 10 kPa with a quality of 90%. Neglecting the changes in kinet

Amanda [17]

Answer:

flow ( m ) = 4.852 kg/s

Explanation:

Given:

- Inlet of Turbine

P_1 = 10 MPa

T_1 = 500 C

- Outlet of Turbine

P_2 = 10 KPa

x = 0.9

- Power output of Turbine W_out = 5 MW

Find:

Determine the mass ow rate required

Solution:

- Use steam Table A.4 to determine specific enthalpy for inlet conditions:

P_1 = 10 MPa

T_1 = 500 C ---------- > h_1 = 3375.1 KJ/kg

- Use steam Table A.6 to determine specific enthalpy for outlet conditions:

P_2 = 10 KPa -------------> h_f = 191.81 KJ/kg

x = 0.9 -------------> h_fg = 2392.1 KJ/kg

h_2 = h_f + x*h_fg

h_2 = 191.81 + 0.9*2392.1 = 2344.7 KJ/kg

- The work produced by the turbine W_out is given by first Law of thermodynamics:

W_out = flow(m) * ( h_1 - h_2 )

flow ( m ) = W_out / ( h_1 - h_2 )

- Plug in values:

flow ( m ) = 5*10^3 / ( 3375.1 - 2344.7 )

flow ( m ) = 4.852 kg/s

3 0

3 years ago