A photovoltaic panel of dimension 2m×4m is installed on the
1 answer:
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
b)
kwh = watt * Time/1000
kwh = 30 *15/1000
4.5
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Answer:
a) q' = 351.22 W/m
b) q'_total = 1845.56 W / m
c) q'_loss = 254.12 W/m
Explanation:
Given:-
- The diameter of the steam line, d = 100 mm
- The surface emissivity of steam line, ε = 0.8
- The temperature of the steam, Th = 150°C
- The ambient air temperature, T∞ = 20°C
Find:-
(a) Calculate the rate of heat loss per unit length for a calm day.
Solution:-
- Assuming a calm day the heat loss per unit length from the steam line ( q ' ) is only due to the net radiation of the heat from the steam line to the surroundings.
- We will assume that the thickness "t" of the pipe is significantly small and temperature gradients in the wall thickness are negligible. Hence, the temperature of the outside surface Ts = Th = 150°C.
- The net heat loss per unit length due to radiation is given by:
q' = ε*σ*( π*d )* [ Ts^4 - T∞^4 ]
Where,
σ: the stefan boltzmann constant = 5.6703 10-8 (W/m2K4)
Ts: The absolute pipe surface temperature = 150 + 273 = 423 K
T∞:The absolute ambient air temperature = 20 + 273 = 293 K
Therefore,
q' = 0.8*(5.6703 10-8)*( π*0.1 )* [ 423^4 - 293^4 ]
q' = (1.4251*10^-8)* [ 24645536240 ]
q' = 351.22 W / m ... Answer
Find:-
(b) Calculate the rate of heat loss on a breezy day when the wind speed is 8 m/s.
Solution:-
- We have an added heat loss due to the convection current of air with free stream velocity of U∞ = 8 m/s.
- We will first evaluate the following properties of air at T∞ = 20°C = 293 K
Kinematic viscosity ( v ) = 1.5111*10^-5 m^2/s
Thermal conductivity ( k ) = 0.025596
Prandtl number ( Pr ) = 0.71559
- Determine the flow conditions by evaluating the Reynold's number:
Re = U∞*d / v
= ( 8 ) * ( 0.1 ) / ( 1.5111*10^-5 )
= 52941.56574 ... ( Turbulent conditions )
- We will use Churchill - Bernstein equation to determine the surface averaged Nusselt number ( Nu_D ):
- The averaged heat transfer coefficient ( h ) for the flow of air would be:
- The heat loss per unit length due to convection heat transfer is given by:
q'_convec = h*( π*d )* [ Ts - T∞ ]
q'_convec = 36.58951*( π*0.1 )* [ 150 - 20 ]
q'_convec = 11.49493* 130
q'_convec = 1494.3409 W / m
- The total heat loss per unit length ( q'_total ) owes to both radiation heat loss calculated in part a and convection heat loss ( q_convec ):
q'_total = q_a + q_convec
q'_total = 351.22 + 1494.34009
q'_total = 1845.56 W / m ... Answer
Find:-
For the conditions of part (a), calculate the rate of heat loss with a 20-mm-thick layer of insulation (k = 0.08 W/m ⋅ K)
Solution:-
- To reduce the heat loss from steam line an insulation is wrapped around the line which contains a proportion of lost heat within.
- A material with thermal conductivity ( km = 0.08 W/m.K of thickness t = 20 mm ) was wrapped along the steam line.
- The heat loss through the lamination would be due to conduction " q'_t " and radiation " q_rad":
q' = ε*σ*( π*( d + 2t) )* [ Ts^4 - T∞^4 ]
Where,
T_o = T∞ = 20°C
T_s = Film temperature = ( Th + T∞ ) / 2 = ( 150 + 20 ) / 2 = 85°C
r_2 = d/2 + t = 0.1 / 2 + 0.02 = 0.07 m
r_1 = d/2 = 0.1 / 2 = 0.05 m
- The heat loss per unit length would be:
q'_loss = q'_rad - q'_cond
- Compute the individual heat losses:
Therefore,
q'_loss = 351.22 - 97.10
q'_loss = 254.12 W / m .... Answer
- If the wind speed is appreciable the heat loss ( q'_loss ) would increase and the insulation would become ineffective.