[Question / Bug?] in fuel_cell.use for PEM-cell

[wafab112]

In the use-function of the fuel_cell I saw that when using up a partial load, the residual_power is set to 0 when it is lower than the MinOutputPower:

MESSpy/techs/fuelcell.py

Lines 898 to 911 in 9db6970

	if residual_power >= self.MinOutputPower:
	hyd_singlemodule,power_singlemodule,FC_Heat_singlemodule,etaFC_singlemodule,water_singlemodule = fuel_cell.use1(self,step,-residual_power,residual_hydrogen)
	if hyd_singlemodule != 0: # single module operating in partial load
	self.n_modules_used[step] = full_modules + 1
	self.EFF[step] = ((full_modules*etaFC_full) + (etaFC_singlemodule))/self.n_modules_used[step] # weighted average
	self.EFF_last_module[step] = etaFC_singlemodule
	else:
	self.n_modules_used[step] = full_modules
	self.EFF[step] = etaFC_full
	else:
	residual_power = 0
	hyd_singlemodule,power_singlemodule,FC_Heat_singlemodule,water_singlemodule = [0]*4
	self.n_modules_used[step] = full_modules
	self.EFF[step] = etaFC_full

But I thought in this case the MinOutputPower should be used by the fuel cell, as it was done in the case of just one module being used:

MESSpy/techs/fuelcell.py

Lines 855 to 873 in 9db6970

	if (abs(p) <= self.Npower) or (available_hydrogen/self.max_h2_module < 1): # if required power or available hydrogen in system are lower than nominal fuel cell parameters
	if abs(p) >= self.MinOutputPower:
	hyd,power,FC_Heat,etaFC,water = fuel_cell.use1(self,step,p,available_hydrogen)
	if abs(hyd) > 0:
	self.n_modules_used[step] = 1
	else:
	self.n_modules_used[step] = 0
	self.EFF[step] = etaFC
	else:
	p = self.MinOutputPower
	hyd,power,FC_Heat,etaFC,water = fuel_cell.use1(self,step,-p,available_hydrogen)
	if abs(hyd) > 0:
	self.n_modules_used[step] = 1
	else:
	self.n_modules_used[step] = 0
	self.EFF[step] = etaFC

Is that the case or is there a mistake in my thought-process

Thanks in advance.