Equations Flashcards
Learn renewable energy system equations
(General) Convert 1kWh to Joules
1kWh = 3,600 seconds x 1000W
= 3.6 x 10^6 J
= 3.6 MJ
(General) Give the equation for work with respect to power and time.
W = Pt
(General) State the equation for capacity factor
Capacity Factor = Actual output energy / Maximum possible output
(Hydropower) State the relationship between mass, density and volume
M = ρV
(Hydropower) Give the equation for stored potential energy
PE = ρVgH
where g = 9.81 m/s^2
(Hydropower) Give the equation for input power
P = ρQgH
where g = 9.81 m/s^2
(Hydropower) Give the equation for output power
P = ηρQgH
(Hydropower) State the equation for Volume Flow Rate in a Pelton turbine
Q = A x √(2gH)
(Hydropower) Derive the equation for Power in a Pelton turbine
PE = MgH and KE = 1/2 Mv^2
1/2 Mv^2 = MgH
For Potential Energy:
M = ρV so:
E = ρVgH
E = P/t so:
P = (ρVgH)/t
But Q = V/t so:
P = ρQgH
For Kinetic Energy:
KE = 1/2 Mv^2 so:
v = √(2gH)
Therefore:
Q = Av becomes Q = A x √(2gH)
P = ρQgH and Q = A x √(2gH)
So Final Equation is:
P = A x √(2gH) x ρgH
(Hydropower) Give the equation for efficiency in pumps
η = (ρQgH) / (Input Power)
Note: Output energy and power are the same for pumps as the input power and energy of the hydropower i.e P = ρQgH and E = ρVgH
(PV Cells) What is the value of 1eV in joules?
1eV = 1.602 x 10^-19
(PV Cells) Derive the equation for an ideal current source with a parallel diode
Diode current:
I(Diode) = I(Output) x (e^(qV/KT) - 1)
Total Current:
I = I(Short Circuit) - I(Diode)
so:
I = I(Short Circuit) - I(Output) x (e^(qV/KT) - 1)
where:
q is electric charge (1.602 x 10^-19 Coulombs)
K is Boltzmann’s constant (1.38 x 10^-23)
T is temperature in Kelvin
(PV Cells) Convert Celsius to Kelvin
Add 273 to the Celsius value to convert it to Kelvin
(PV Cells) Give the equations for the following:
1) Ideal current source with parallel diode and parallel resistor
2) Ideal current source with parallel diode and series resistor
3) Ideal current source with parallel diode, series resistor and parallel resistor
4) Source voltage
1) I = (I(Short Circuit) - I(Diode)) - (V/R(Parallel))
2) I = I(Short Circuit) - I(Output)(e^(q(IR(Series)+V)/KT) - 1)
3) I = I(Short Circuit) - I(Output)(e^38.9V(Diode) - 1) - (V(Diode)/R(Parallel))
4) V = V(Diode) - IR(Source)
(PV Cells) Give the equation for theoretical output power
P(Theoretical) = V(Open Circuit) x I(Short Circuit)
(PV Cells) State the equation for Fill Factor
FF = P(Max) / P(Theoretical) = I(Max Point)V(Max Point) / V(Open Circuit)I(Short Circuit) x 100%
(PV Cells) Give equations for the following:
1) Number of cells in module
2) Number of parallel branches
3) Number of cells in series
1) Z = P(Theoretical) / P(Cell)
where Z is the number of cells and P(Cell) is calculatedby multiplying the cell’s Short Circuit Current by Open Circuit Voltage
2) N(Parallel) = I(Short Circuit) / Cell Short Circuit Current
3) N(Series) = V(Open Circuit) / Cell Open Circuit Voltage
(PV Cells) Give the equation for input power of a module
P(In) = aG
where a is the area of the module (Metres squared)
G is the insolation (Watts per metre squared)
(PV Cells) State the equation for module efficiency
η = P(Max) / P(In) x 100%