def find_power_larminie(current_density, fuel_cell, sign=1.0):
'''
Function that determines the power output per cell, based on in
input current density
Assumptions:
None(calls other functions)
Inputs:
current_density [Amps/m**2]
fuel cell.
interface area [m**2]
Outputs:
power_out [W]
'''
# sign variable is used so that you can maximize the power, by minimizing the -power
i1 = current_density
A = fuel_cell.interface_area
v = find_voltage_larminie(fuel_cell,current_density) #useful voltage vector
power_out = sign* np.multiply(v,i1)*A #obtain power output in W/cell
#want to minimize
return power_out
def find_power_larminie(current_density, fuel_cell, sign=1.0):
'''
Function that determines the power output per cell, based on in
input current density
Assumptions:
None(calls other functions)
Inputs:
current_density [Amps/m**2]
fuel cell.
interface area [m**2]
Outputs:
power_out [W]
'''
# sign variable is used so that you can maximize the power, by minimizing the -power
i1 = current_density
A = fuel_cell.interface_area
v = find_voltage_larminie(fuel_cell,
current_density) #useful voltage vector
power_out = sign * np.multiply(v, i1) * A #obtain power output in W/cell
#want to minimize
return power_out
def find_power_larminie(current_density, fuel_cell, sign=1.0): #adds a battery that is optimized based on power and energy requirements and technology
#sign variable is used so that you can maximize the power, by minimizing the -power
i1 = current_density/(Units.mA/(Units.cm**2.))
A = fuel_cell.interface_area/(Units.cm**2.)
v = find_voltage_larminie(fuel_cell,current_density) #useful voltage vector
power_out = sign* np.divide(np.multiply(v,i1),1000.0)*A #obtain power output in W/cell
#want to minimize
return power_out
def find_power_larminie(current_density, fuel_cell, sign=1.0):
# sign variable is used so that you can maximize the power, by minimizing the -power
i1 = current_density / (Units.mA / (Units.cm**2.))
A = fuel_cell.interface_area / (Units.cm**2.)
v = find_voltage_larminie(fuel_cell,
current_density) #useful voltage vector
power_out = sign * np.divide(np.multiply(v, i1),
1000.0) * A #obtain power output in W/cell
#want to minimize
return power_out
def larminie(
fuel_cell, conditions, numerics
): #adds a battery that is optimized based on power and energy requirements and technology
power = fuel_cell.inputs.power_in
lb = .1 * Units.mA / (Units.cm**2.
) #lower bound on fuel cell current density
ub = 1200.0 * Units.mA / (Units.cm**2.)
current_density = np.zeros_like(power)
for i in range(len(power)):
current_density[i] = sp.optimize.fminbound(find_power_diff_larminie,
lb,
ub,
args=(fuel_cell, power[i]))
v = find_voltage_larminie(fuel_cell, current_density)
efficiency = np.divide(v, fuel_cell.ideal_voltage)
print 'efficiency=', efficiency
print 'current_density=', current_density
print 'v=', v
mdot = np.divide(
power, np.multiply(fuel_cell.propellant.specific_energy, efficiency))
'''
A = fuel_cell.interface_area/(Units.cm**2.)
r = fuel_cell.r/(Units.kohm/(Units.cm**2))
Eoc = fuel_cell.Eoc
A1 = fuel_cell.A1
m = fuel_cell.m
n = fuel_cell.n
power = fuel_cell.inputs.power_in
i1 = np.linspace(.1,1200.0,200.0)*Units.mA/Units.cm**2 #current density(mA cm^-2): use vector of these with interpolation to find values
v = Eoc-r*i1-A1*np.log(i1)-m*np.exp(n*i1) #useful voltage vector
efficiency = np.divide(v,1.48) #efficiency of the cell vs voltage
p = fuel_cell.number_of_cells* np.divide(np.multiply(v,i1),1000.0)*A #obtain power output in W
imax = np.argmax(p)
if power.any()>p[imax]:
print "Warning, maximum power output of fuel cell exceeded"
p = np.resize(p,imax+1) #resize vector such that it only goes to max power to prevent ambiguity
print len(power)
print len(efficiency)
mdot_vec = np.divide(power,np.multiply(fuel_cell.propellant.specific_energy,efficiency)) #mass flow rate of the fuel based on fuel cell efficiency
ip = np.argmin(np.abs(p-power)) #find index such that that operating power is equal to required power
v1 = v[ip] #operating voltage of a single stack
efficiency_out= efficiency[ip]
mdot = np.divide(power,np.multiply(fuel_cell.propellant.specific_energy,efficiency_out)) #mass flow rate of hydrogen
'''
return mdot
def larminie(fuel_cell,conditions,numerics): #adds a battery that is optimized based on power and energy requirements and technology
power = fuel_cell.inputs.power_in
lb =.1*Units.mA/(Units.cm**2.) #lower bound on fuel cell current density
ub =1200.0*Units.mA/(Units.cm**2.)
current_density =np.zeros_like(power)
for i in range(len(power)):
current_density[i] =sp.optimize.fminbound(find_power_diff_larminie, lb, ub, args=(fuel_cell, power[i]))
v =find_voltage_larminie(fuel_cell,current_density)
efficiency =np.divide(v, fuel_cell.ideal_voltage)
print 'efficiency=', efficiency
print 'current_density=', current_density
print 'v=', v
mdot = np.divide(power,np.multiply(fuel_cell.propellant.specific_energy,efficiency))
'''
A = fuel_cell.interface_area/(Units.cm**2.)
r = fuel_cell.r/(Units.kohm/(Units.cm**2))
Eoc = fuel_cell.Eoc
A1 = fuel_cell.A1
m = fuel_cell.m
n = fuel_cell.n
power = fuel_cell.inputs.power_in
i1 = np.linspace(.1,1200.0,200.0)*Units.mA/Units.cm**2 #current density(mA cm^-2): use vector of these with interpolation to find values
v = Eoc-r*i1-A1*np.log(i1)-m*np.exp(n*i1) #useful voltage vector
efficiency = np.divide(v,1.48) #efficiency of the cell vs voltage
p = fuel_cell.number_of_cells* np.divide(np.multiply(v,i1),1000.0)*A #obtain power output in W
imax = np.argmax(p)
if power.any()>p[imax]:
print "Warning, maximum power output of fuel cell exceeded"
p = np.resize(p,imax+1) #resize vector such that it only goes to max power to prevent ambiguity
print len(power)
print len(efficiency)
mdot_vec = np.divide(power,np.multiply(fuel_cell.propellant.specific_energy,efficiency)) #mass flow rate of the fuel based on fuel cell efficiency
ip = np.argmin(np.abs(p-power)) #find index such that that operating power is equal to required power
v1 = v[ip] #operating voltage of a single stack
efficiency_out= efficiency[ip]
mdot = np.divide(power,np.multiply(fuel_cell.propellant.specific_energy,efficiency_out)) #mass flow rate of hydrogen
'''
return mdot
def larminie(fuel_cell, conditions, numerics):
'''
function that determines the mass flow rate based on a required power input
Assumptions:
None (calls other functions)
Inputs:
fuel_cell.
inputs.
power_in [W]
ideal_voltage [V]
Outputs:
mdot [kg/s]
'''
power = fuel_cell.inputs.power_in
lb = .1 * Units.mA / (Units.cm**2.
) #lower bound on fuel cell current density
ub = 1200.0 * Units.mA / (Units.cm**2.)
current_density = np.zeros_like(power)
for i in xrange(len(power)):
current_density[i] = sp.optimize.fminbound(find_power_diff_larminie,
lb,
ub,
args=(fuel_cell, power[i]))
v = find_voltage_larminie(fuel_cell, current_density)
efficiency = np.divide(v, fuel_cell.ideal_voltage)
mdot = np.divide(
power, np.multiply(fuel_cell.propellant.specific_energy, efficiency))
print 'efficiency=', efficiency
print 'current_density=', current_density
print 'v=', v
return mdot
def larminie(fuel_cell,conditions,numerics):
power = fuel_cell.inputs.power_in
lb = .1*Units.mA/(Units.cm**2.) #lower bound on fuel cell current density
ub = 1200.0*Units.mA/(Units.cm**2.)
current_density = np.zeros_like(power)
for i in xrange(len(power)):
current_density[i] = sp.optimize.fminbound(find_power_diff_larminie, lb, ub, args=(fuel_cell, power[i]))
v = find_voltage_larminie(fuel_cell,current_density)
efficiency = np.divide(v, fuel_cell.ideal_voltage)
mdot = np.divide(power,np.multiply(fuel_cell.propellant.specific_energy,efficiency))
print 'efficiency=', efficiency
print 'current_density=', current_density
print 'v=', v
return mdot
def larminie(fuel_cell, conditions, numerics):
power = fuel_cell.inputs.power_in
lb = .1 * Units.mA / (Units.cm**2.
) #lower bound on fuel cell current density
ub = 1200.0 * Units.mA / (Units.cm**2.)
current_density = np.zeros_like(power)
for i in xrange(len(power)):
current_density[i] = sp.optimize.fminbound(find_power_diff_larminie,
lb,
ub,
args=(fuel_cell, power[i]))
v = find_voltage_larminie(fuel_cell, current_density)
efficiency = np.divide(v, fuel_cell.ideal_voltage)
mdot = np.divide(
power, np.multiply(fuel_cell.propellant.specific_energy, efficiency))
print 'efficiency=', efficiency
print 'current_density=', current_density
print 'v=', v
return mdot
