PEMFC mass and energy model Stephan Strahl

From WikIRI

1 Introduction
2 PEMFC mass and energy model
3 Measured parameters
4 Mass balance
- 4.1 1. Calculating the mass fracitons of the individual gases in mixture using partial pressures
- 4.2 2. Determination of water mass flow rates
5 Energy balance
- 5.1 1. Temperature based
- 5.2 2. Fuel cell efficiency based
6 Dynamic model
- 6.1 1. Short circuit unit
  - 6.1.1 Heat transport
- 6.2 2. Purge valve
7 Calculation of single channel velocity
- 7.1 1. Cathode
- 7.2 2. Anode
8 Calculation of water concentration
9 Experiments for mass and energy balances
10 Velocity sensor experiments

Introduction

For the past 20 years astonishing progress in terms of PEM fuel cell materials, component design, production, and system power density improvements have been achieved. However, there is still a lot to be done in the field of fuel cell system controls, which makes it essential to understand the different physical phenomena within a fuel cell and how they need to be controlled in order to improve efficiency, operating range and durability. The hypothesis is that if the water movement within a PEM fuel cell could be controlled quickly to maintain optimal membrane water content and minimal liquid water would thus improving efficiency. As shown in the experiments of Springer et al., membrane proton conductivity is a strong function of water content. Thus, the performance of PEM fuel cells is very sensitive to membrane hydration. Although water is produced during the reaction, the anode catalyst layer is often dehydrated because water is dragged from the anode to the cathode by protons moving through the membrane, which is called electro-osmotic drag (EOD).

Besides the EOD, the main water transport mechanism in a PEM fuel cell is diffusion through the membrane due to concentration differences between anode and cathode. The third transport mechanism is hydraulic permeation, which is caused by pressure difference. EOD always transports water from the anode to the cathode whereas diffusion can occur in both directions.

Water is needed to maintain good proton conductivity and therefore has to be kept in the membrane, however liquid water on the catalyst reduces the active area, and in the GDL it hinders the reactant gases from diffusing to the catalyst surface and thus reduces performance. The goal is to maintain an optimal water concentration in the membrane electrode assembly (MEA) by keeping a balance between the two conflicting requirements. Thus, to control water transport within a fuel cell system and thereby optimize the membrane hydration at any operation point, proper dynamic water management strategies have to be developed. This has recently been analyzed by Hussaini \& Wang.

In order to characterize, understand and manipulate the water transport mechanisms, experimental work is needed as well as a mathematical model that describes the physical phenomena.

A recently published review of water balance in the MEA by Dai et al. states that further work is needed to better understand the fundamentals of water transport in the MEA, not only to improve performance, but also to develop new materials for better water management and to improve durability. In order to develop and simulate dynamic water management strategies that match the application load requirements and the operating conditions, new models need to be based on a broad understanding of water transport in the MEA.

This wiki page describes parts of the developed dynamic model, as well as the performed experimental work and the model validation of an open cathode, self-humidified PEM fuel cell.

PEMFC mass and energy model

The water balance across the fuel cell is needed to determine the transfer rate which is required for proper water management in the fuel cell.

The nomenclature can be found here.

The full water balance equation is:

$\dot m_{H_{2}O,ca,in}+\dot m_{H_{2}O,an,in}+\dot m_{H_{2}O,gen}=\dot m_{H_{2}O,ca,out}+\dot m_{H_{2}O,an,out}\qquad (1)$

This equates the water that enters the fuel cell and water that leaves the fuel cell. In this system on the anode side the hydrogen enters the fuel cell dry so the term for water coming into the system from the anode can be neglected.

Another assumption is that water that enters and exits the cell will be assumed to be in the vapor form. This is a good assumption for the following reasons; on the cathode inlet the water vapor that enters the cell is the humidity in the environment, on the cathode outlet there is always a stoichiometry in the order of ten that does not allow the gas to reach 100% relative humidity. The only controversial mass flow would be the anode outlet where the water vapor does condense. This is solved by placing a gas line heater at the exit of the anode and heating the gas up to about 70ºC before measuring the dew point temperature. So, consequently the mass flow of water can be assumed to be in the vapor form in a gas streams where the dew point temperatures are measured.

One note on the anode line heater is that if the dew point does reach 100% then the data after that reading have to be discarded and the experiment needs to be rerun with a higher anode outlet line heater temperature.

Measured parameters

With our test station we can measure the input and output parameters of the fuel cell, which are needed for calculating mass flow.

Anode: Inlet

-Gas temperature $T_{t,an,in}\!$ [K]

-Dew point temperature $T_{dp,an,in}\!$ [K]

-Mass flow rate of hydrogen $\dot m_{H_{2},an,in}$ [SLPM]

-Pressure of hydrogen $P_{an,in}\!$ [Pa]

Anode: Outlet

-Gas temperature $T_{t,an,out}\!$ [K]

-Dew point temperature $T_{dp,an,out}\!$ [K]

Cathode: Inlet

-Gas temperature $T_{t,ca,in}\!$ [K]

-Dew point temperature $T_{dp,ca,in}\!$ [K]

-Air velocity $v_{ca,in}\!$ [m/s]

-Molar fraction of oxygen in air $\chi_{O_{2_{ca,in}}}\!$

Cathode: Outlet

-Gas temperature $T_{t,ca,out}\!$ [K]

-Dew point temperature $T_{dp,ca,out}\!$ [K]

Stack parameters:

-Voltage $V_{stack}\!$ [V]

-Current $I\!$ [A]

With these known parameters the equations for describing the mass and energy balance of the fuel cell can be set up.

Mass balance

1. Calculating the mass fracitons of the individual gases in mixture using partial pressures

The total pressure of a given gas mixture is the sum of all the individual components of the gas mixture(Dalton´s law)

So therefore knowing the total pressure and just one of the two partial pressures in a gas mixture the second partial pressure can be determined. In the case of water vapor the partial pressure can be determined knowing the dew point temperature of the gas.

The equation for the water vapor partial pressure is defined by a curve fit of the flowing empirical equation 2005 Barbir pg 125 for temperatures between 0ºC and 100ºC.

$p_{v}=e^{aT_{dp}^{-1}+b+cT_{dp}+dT_{dp}^2+eT_{dp}^3+f\ln (T_{dp})}\qquad [Pa]\qquad (2)$

where:

$T_{dp}=\!$ the dew point or saturation temperature [K]

$a=-5800.2206\!$

$b=1.3914993\!$

$c=-0.048640239\!$

$d=0.41764768\times10^{-4}\!$

$e=-0.14452093\times10^{-7}\!$

$f=6.5459673\!$

To calculate the amount of water vapor in the gas mixture we first need to determine the humidity ratio, which is the ratio of mass of water vapor and the mass of the dry gas. In this case it is:

$x=\frac {m_v}{m_g}=\frac {M_v}{M_g}\cdot \frac{p_{v}}{(P-p_{v})}\Rightarrow$ e.g. $x_{H_2O,an,in}=\frac {M_{H_2O}}{M_{H_2}}\cdot \frac {p_{v,an,in}}{P_{an,in}-p_{v,an,in}}\qquad (3)$

The ratio of partial pressures in equation (3) is also known as the humidity molar ratio:

$\chi_{H_2O}= \frac{p_{v}}{(P-p_{v})}\qquad (4)$

The advantage of using dew point temperatures instead of relative humidities to calculate the amount of water in the gas is that the dew point of a gas does not change with temperature, contrary to the relative humidity.

2. Determination of water mass flow rates

2.1 Anode inlet $H 2 O$ mass flow rate

First of all the $H 2$ inlet mass flow rate unit [SLPM] (standard liters per minute) has to be transformed into [kg/s]. The standard conditions for the used mass flow controller (Bronkhorst EL-FLOW F-201C) are:

T 0 = 273.15 K

P 0 = 101325 P a

The measured Flowrate in [SLPM] has to multiplied by the hydrogen density, which has to be calculated with the standard values of pressure and temperature, and afterwards devided by 60000 to obtain SI-units.

$\dot m_{H_{2},an,in} [kg/s] = \dot m_{H_{2},an,in} [SLPM] \cdot \rho_{H_2} \cdot \frac {1}{60000} = \dot m_{H_{2},an,in}[SLPM] \cdot \frac {P_0 \cdot M_{H_2}}{R \cdot T_0 \cdot 60000}\qquad (5)$

The total anode inlet mass flow rate is:

$\dot m_{an,in}=\dot m_{H_{2},an,in} + \dot m_{H_{2}O,an,in}\qquad [kg/s]\qquad (6)$

Knowing the amount of water vapor in the entering hydrogen by measuring the dew point temperature, the anode inlet water mass flow rate is determined by:

$\dot m_{H_{2}O,an,in}=\dot m_{H_{2},an,in}\cdot x_{H_2O,an,in} \qquad [kg/s]\qquad (7)$

So the total anode inlet mass flow rate can be calculated by the following equation:

$\dot m_{an,in}= \dot m_{H_{2},an,in}\cdot (1 + x_{H_2O,an,in})\qquad [kg/s]\qquad (8)$

2.2 Cathode inlet $H 2 O$ mass flow rate

To determine this mass flow rate, first the dry cathode inlet mass flow rate is calculated using the measured inlet average velocity, the area of the cross section where the velocity is measured, and the density of dry air:

$\dot m_{ca,in}=\bar{v}_{ca,in}\cdot A_{ca}\cdot \rho_{ca,in}\qquad [kg/s]\qquad (9)$

$\dot m_{H_{2}O,ca,in}=\dot m_{air,in,d}\cdot x_{H_2O,ca,in} = \bar{v}_{ca,in}\cdot A_{ca}\cdot \rho_{air,d} \cdot x_{H_2O,ca,in} \qquad [kg/s]\qquad (10)$

To determine the total cathode mass flow rate, the density of the gas mixture has to be calculated, because the addition of water vapor to air reduces the density of the air, which may at first appear counter intuitive. This occurs because the molecular mass of water (0.018 kg/mol) is less than the molecular mass of air (around 0.029 kg/mol). For any gas, at a given temperature and pressure, the number of molecules present is constant for a particular volume. So when water vapor molecules are introduced to the air, the number of air molecules must reduce by the same number in a given volume, without the pressure or temperature increasing. Hence the mass per unit volume of the gas (its density) decreases.

The density of humid air may be calculated as a mixture of ideal gases.

$\rho_{air,w} = \frac{p\cdot M_{air,w}}{R\cdot T}\qquad [kg/m^3]\qquad (11)$

where the molecular mass of air is calculated using the three different mass fraction for the individual gases in the mixture:

$\dot m_{ca,in}=\dot m_{O_{2},ca,in} + \dot m_{N_{2},ca,in} + \dot m_{H_{2}O,ca,in}\qquad [kg/s]\qquad (12)$

Using the molar fractions of the respective gases in the gas mixture, the molar mass of wet air can be calculated.

$\chi_{H_2O,ca,in}=\frac {p_{vs,ca,in}}{P_{ca}-p_{vs,ca,in}}\qquad (13)$

$\chi_{O_{2_{ca,in}}}= \mathrm{measured}$

$\chi_{N_2,ca,in}=1-\chi_{O_{2_{ca,in}}}-\chi_{H_2O,ca,in}\qquad (14)$

The molecuar mass of humid air is then found by:

$M_{air,w}=(\chi_{N_2,ca,in}\cdot M_{N_2})+(\chi_{O_{2_{ca,in}}}\cdot M_{O_2})+(\chi_{H_2O,ca,in}\cdot M_{H_2O})\qquad [kg/mol]\qquad (15)$

2.3 Anode outlet $H 2 O$ mass flow rate

This mass flow rate can be calculated in the same way than the inlet flow rate, but adding the fact that hydrogen is consumed inside the fuel cell.

$\dot m_{an,out}=\dot m_{H_{2},an,out} + \dot m_{H_{2}O,an,out}\qquad [kg/s]\qquad (16)$

where:

$\dot m_{H_{2}O,an,out} = \dot m_{H_{2},an,out}\cdot x_{H_2O,an,out} \qquad [kg/s]\qquad (17)$

and:

$\dot m_{H_{2},an,out} =\dot m_{H_{2},an,in} - \dot m_{H_{2},an,cons} \qquad [kg/s]\qquad (18)$

Equation (16) can be rewritten as:

$\dot m_{an,out} = (\dot m_{H_{2},an,in} - \dot m_{H_{2},an,cons})\cdot (1 + x_{H_2O,an,out}) \qquad [kg/s]\qquad (19)$

Using Faraday's law, the consumed hydrogen flow rate is found by:

$\dot m_{H_{2},an,cons} = \frac {M_{H_2}}{2F}\cdot I\cdot n_{cell}\qquad [kg/s]\qquad (20)$

2.4 Cathode outlet $H 2 O$ mass flow rate

Finally the cathode outlet water mass flow rate has to be determined.

$\dot m_{H_{2}O,ca,out} = \dot m_{air,out,d}\cdot x_{H_2O,ca,out} \qquad [kg/s]\qquad (21)$

where:

$\dot m_{air,out,d} = \dot m_{air,in,d} - \dot m_{O_{2},ca,cons} \qquad [kg/s]\qquad (22)$

Using Faraday's law, the consumed oxygen flow rate is found by:

$\dot m_{O_{2},ca,cons} = \frac {M_{O_2}}{4F}\cdot I\cdot n_{cell}\qquad [kg/s]\qquad (23)$

Energy balance

Fuel cell energy balance requires that the sum of all energy inputs must be equal to the sum of all energy outputs:

$\sum (H_i)_{in} = \sum (H_i)_{out} + W_{el} + Q \qquad (23)$

The inputs are the enthalpies of all the flows into the fuel cell plus the enthalpy of water vapor present in those gases. The outputs are the enthalpies of all the flows out of the fuel cell plus the enthalpy of water vapor present in those gases, as well as any liquid water, the electric power produced, and the heat flux due to heat dissipation from the fuel cell surface (radiation and convection).

In our case we assume that no liquid water leaves the fuel cell. Moreover, the heat dissipation due to radiation and convection is considered to be negligible. The resulting equation that describes the Energy balance is then:

$H_{H_2,an,in} + H_{Air,ca,in} + H_{H_2O,an,in} + H_{H_2O,ca,in} = H_{H_2,an,out} + H_{Air,ca,out} + H_{H_2O,an,out} + H_{H_2O,ca,out} + W_{el} \qquad (24)$

For each dry gas or a mixture of dry gases, the enthalpy is:

$H = \dot m\cdot c_p\cdot T \qquad[J/s] \qquad (25)$

Because hydrogen is a combustible, its higher heating value has to be included in equation (25):

$H_{H_2} = \dot m_{H_2}\cdot (c_{p,{H_2}}\cdot T + h_{HHV}^0) \qquad [J/s] \qquad (26)$

Because water leaves the cell as vapor the heat of evaporation has to be included in the enthalpy of water in the gases:

$H_{H_2O,v} = \dot m_{H_2O,v}\cdot (c_{p,{H_2O,v}}\cdot T + h_{fg}^0) \qquad [J/s] \qquad (27)$

Note that all the enthalpies have to be calculated using ºC as the unit for temperature because the reference zero state is at 0ºC.

The following constants are used for determing the energy balance:

$c_{p,{H_2}} = 14.2 \cdot 10^3~\frac {J}{kg\cdot K}$

$c_{p,{H_2O,v}} = 1.87 \cdot 10^3~\frac {J}{kg\cdot K}$

$c_{p,{O_2}} = 0.913 \cdot 10^3~\frac {J}{kg\cdot K}$

$c_{p,{N_2}} = 1.04 \cdot 10^3~\frac {J}{kg\cdot K}$

$h_{HHV}^0 = 142.3 \cdot 10^6~\frac {J}{kg}$

$h_{fg}^0 = 2.5 \cdot 10^6~\frac {J}{kg}$

The energy balance equation allows us to check mass flow model, because now there are two independet ways to calculate power dissipation of the fuel cell:

1. Temperature based

To calculate the power dissipation the generated heat energy in the fuel cell can be determined by separating the energy balance equation (23) using equation (25) to (27):

$\sum (H_i)_{in} = \sum (H_i)_{out} + W_{el} \qquad \rightarrow \qquad (W_{H_2} + W_{evap} + Q(T))_{in} = (W_{H_2} + W_{evap} + Q(T))_{out} + W_{el}\qquad (28)$

where:

$W_{H_2} = \dot m_{H_2}\cdot h_{HHV}^0 \qquad [J/s] \qquad (29)$

$W_{evap} = \dot m_{H_2O,v}\cdot h_{fg}^0 \qquad [J/s] \qquad (30)$

So the total heat energy, which equals the power dissipation, is determined by:

$Q_{tot} = Q_{out} - Q_{in} \qquad [J/s] \qquad (31)$

with:

$Q_{out} = (\dot m_{H_2,an,out}\cdot c_{p,{H_2}}\cdot T_{an,out}) + (\dot m_{H_2O,an,out}\cdot c_{p,{H_2O,v}}\cdot T_{an,out}) + (\dot m_{O_2,ca,out}\cdot c_{p,{O_2}}\cdot T_{ca,out}) + (\dot m_{N_2,ca,out}\cdot c_{p,{N_2}}\cdot T_{ca,out}) + (\dot m_{H_2O,ca,out}\cdot c_{p,{H_2O,v}}\cdot T_{ca,out})\qquad [J/s] \qquad (32)$

$Q_{in} = (\dot m_{H_2,an,in}\cdot c_{p,{H_2}}\cdot T_{an,in}) + (\dot m_{H_2O,an,in}\cdot c_{p,{H_2O,v}}\cdot T_{an,in}) + (\dot m_{O_2,ca,in}\cdot c_{p,{O_2}}\cdot T_{ca,in}) + (\dot m_{N_2,ca,in}\cdot c_{p,{N_2}}\cdot T_{ca,in}) + (\dot m_{H_2O,ca,in}\cdot c_{p,{H_2O,v}}\cdot T_{ca,in})\qquad [J/s] \qquad (33)$

2. Fuel cell efficiency based

The total generated heat energy inside the fuel cell can also be calculated using the fuel cell efficiency, which is defined by the ratio between the electricity produced and hydrogen consumed:

$\eta = \frac {W_{el}}{W_{H_2}} \qquad (34)$

with:

$W_{el} = I\cdot V \qquad [W] \qquad (35)$

$W_{H_2} = \frac {h_{HHV}^0\cdot M_{H_2}}{2\cdot F} \cdot I \qquad [W]\qquad (36)$

By combining equation (34) and (35), the fuel cell efficiency is directly proportional to the cell potential:

$\eta = \frac {V}{1.482\,V} \qquad (37)$

If $η$ is the fuel cell efficiency, fuel cell power dissipation is then:

$Q_{tot} = W_{H_2} - W_{el} = W_{el}\cdot (\frac {1}{\eta} - 1)\qquad [W]\qquad (38)$

Assuming that water is generated at the catalyst in its liquid state but leaves the fuel cell as vapor, the evaporation energy has to be considered in the total heat loss:

$Q_{tot} = W_{H_2} - W_{el} - W_{evap}\qquad [W]$

where $W e v a p$ is determined by the mass flow rate of generated water multiplied by the enthalpy of evaporation:

$W_{evap} = (\dot m_{H_2O,gen}-\dot m_{H_2O,an,out,l})\cdot h_{fg}^0 \qquad [W]$

The mass flow rate of liquid water that leaves the anode has to be subtracted. As the anode outlet is heated to 70ºC in order to measure dewpoint, this liquid water mass flow can be determined by comparing the anode outlet dewpoint temperature to the stack temperature. If the dewpoint is higher than the stack temperature, liquid water has left the anode. As experiments have shown, the total heat loss calculated both ways equal each other if the evaporation energy is considered, which justifies our assumption.

Using the measured gas temperatures to determine the total heat loss does not take into account that water is generated in its liquid state, because only inlet and outlet are regarded, where water enters and leaves as vapor. This means, if water would be generated as vapor, the outlet temperatures of the gases should be higher, because no energy is needed for evaporation.

Dynamic model

1. Short circuit unit

The system periodically performs a short circuit every 10 seconds, which creates heat and water at the cathode catalyst layer.

Heat transport

During a short circuit the useful electrical energy is zero. All energy is transformed into heat, which increases temperature inside the fuel cell stack. Nevertheless some energy is lost due to circuit and contact resistance, but is neglected at this point.

Thus, knowing the short circuit current the heat energy that is released in one cell can be calculated using equation 36:

$W_{H_2} = \frac {h_{HHV}^0\cdot M_{H_2}}{2\cdot F} \cdot I(t) = \frac{\delta Q}{\delta t} \qquad [W]\qquad (39)$

$\frac{\delta Q}{\delta t} = c_p\cdot m\cdot \frac{\delta T}{\delta t} \qquad [W]\qquad (40)$

In order to obtain the heat flux q that flows through the catalyst surface equation 40 has to be divided by the cross-sectional area of the catalyst layer. Therefore the mass m is substituted by density times volume. The average heat flux, that is generated by one short circuit then results in:

$\bar q = c_p\cdot \rho \cdot x \cdot \frac{\Delta T}{\Delta t} \qquad [W\, m^{-2}]\qquad (41)$

where x is the location in through-plane direction relative to the membrane and $Δ t$ is the short circuit duration time.

2. Purge valve

According to Faraday's law, the mass flow rate of hydrogen that is consumed in one cell during the reaction is:

$\dot m_{H_{2},an,cons} = \frac {M_{H_2}}{2F}\cdot I\qquad [kg/s]$

This consumption of hydrogen is modeled by an outflow boundary condition. The velocity at which hydrogen therefore leaves the anode (gets consumed) through the active catalyst surface A_act results in:

$v_{H_{2},an,cons} = \frac {M_{H_2}}{2F\cdot \rho_{H2} \cdot A_{act}}\cdot I\qquad [m/s]$

Calculation of single channel velocity

1. Cathode

As the cathode inlet air velocity is measured in a housing structure outside the fuelcell stack, but has to be used to model a single cell, the velocity has to be adapted to the smaller flow channel dimensions:

Massflow through one channel:

$\dot m_{ch} = \frac{\dot m_{ca,in}}{n_{BPP} \cdot n_{channels\ per\ plate}} = \frac{\rho \cdot A_{housing} \cdot v_{sensor}}{n_{BPP} \cdot n_{channels\ per\ plate}}$

The mass flow through one channel can also be expressed as:

$\dot m_{ch} = \rho \cdot A_{ch} \cdot v_{ch}$

Combining these two equations and solving for v_ch leads to:

$v_{ch} = \frac{A_{housing}}{n_{BPP} \cdot n_{channels\ per\ plate} \cdot A_{ch}} \cdot v_{sensor} \qquad [m\, s^{-1}]$

In the current test structure the inner housing dimensions are 104.5mm x 91.5mm. The area occupied by the sensor is now constant at every sensor position because of the new design. It's value is (12mm x 91.5mm) - (3.3mm x 10.4mm). So the housing Area that has to be used in the formula above results in 0.008498m^2. With a single channel area of 1mm x 1.5mm = 1.5e-6m^2 the velocity in one single channel is:

$v_{ch} = 5.3 \cdot v_{sensor} \qquad [m\, s^{-1}]$

2. Anode

The flow velocity at the anode is used to describe the convective mass transport through the anode GDL during a purge.

Similar to the cathode channel velocity, the anode velocity in the GDL can be determined by the area ratio and the number of cells:

$v_{GDL} = \frac{A_{inlet}}{n_{cells} \cdot A_{GDL}} \cdot v_{inlet} \qquad [m\, s^{-1}]$

where $A i n l e t$ is the cross-sectional area of the stack inlet channel, which has a diameter of 5mm. Knowing this area, the inlet velocity can then be calculated using the measured mass flow rate:

$v_{inlet} = \frac{\dot m_{a,in}}{A_{inlet} \cdot \rho} \qquad [m\, s^{-1}]$

After passing the inlet fitting the flow is split into the 20 cells. The area $A G D L$ , through which each flux enters the anode of each cell, is assumed to be the thickness of the 2 anode GDLs times the lenth of the anode GDL channel, because the reactant flows from the inlet GDL channel to the outlet GDL channel uniformly.

Calculation of water concentration

The Comsol Application Mode "Convection and Diffusion", that describes water mass transport through the GDL uses absolute concentrations instead of mass fractions. Therefore these concentrations have to be calculated using standard densities for dry hydrogen and dry air at 1atm and 20ºC.

1. Anode

$c_{H_2O,an} = \frac{n_{H_2O}}{V_{H_2}} = \frac{\rho_{H_2O}}{M_{H_2O}} \cdot \frac{V_{H_2O}}{V_{H_2}} \qquad [mol\, m^{-3}]$

As defined above the humidity mass fraction is the amount of water vapor in the gas stream divided by the amount of dry gas:

$x_{H_2O,an} = \frac{m_{H_2O}}{m_{H_2}} = \frac{\rho_{H_2O}}{\rho_{H_2}} \cdot \frac{V_{H_2O}}{V_{H_2}}$

Combinig these two equations results in the correlation between mass fraction and concentration:

$c_{H_2O,an} = \frac{\rho_{H_2}}{M_{H_2O}} \cdot x_{H_2O,an} \qquad [mol\, m^{-3}]$

Inserting the density of hydrogen at 1 atm and 20 ºC, which is 0.08376 kg/m^3 results in:

$c_{H_2O,an} = 4.65 \cdot x_{H_2O,an} \qquad [mol\, m^{-3}]$

Nitrogen:

Inserting the density of nitrogen at 1 atm and 20 ºC, which is 1.1647 kg/m^3 results in:

$c_{H_2O,an} = 64.71 \cdot x_{H_2O,an} \qquad [mol\, m^{-3}]$

2. Cathode

Analog to the anode the cathode water concentration is:

$c_{H_2O,ca} = \frac{\rho_{Air}}{M_{H_2O}} \cdot x_{H_2O,ca} \qquad [mol\, m^{-3}]$

Inserting the density of dry air at 1\,atm and 20\,ºC, which is 1.1985 kg/m^3 results in:

$c_{H_2O,ca} = 66.58 \cdot x_{H_2O,an} \qquad [mol\, m^{-3}]$

3. Flow rate

To transform water mass flow rate into molar flow rate it only has to be divided by the molar mass of water:

$\dot n_{H_2O} = \frac{1}{M_{H_2O}} \cdot \dot m_{H_2O} = \frac{I}{2F}\qquad [mol\,s^{-1}]$

4. Concentration by humidity molar ratio

Humidity mole ratio is defined as mols of water vapor divided by mols of dry gas:

$\chi = \frac{n_{H_2O}}{n_{gas,dry}} = \frac{p_{v}}{P-p_{v}}$

Molar concentration of water vapor is defined as:

$c_{H_2O} = \frac{n_{H_2O}}{V_{gas,dry}} \qquad [mol\, m^{-3}]$

Combining the two equations leads to:

$c_{H_2O} = \frac{\chi \cdot n_{gas,dry}}{V_{gas,dry}} = \frac{\chi \cdot n_{gas,dry}}{\frac{m_{gas,dry}}{\rho_{gas,dry}}} = \frac{\chi \cdot n_{gas,dry}}{\frac{M_{gas,dry}\cdot n_{gas,dry}}{\rho_{gas,dry}}} = \frac{\chi \cdot \rho_{gas,dry}}{M_{gas,dry}}\qquad [mol\, m^{-3}]$

Using the ideal gas law results is the correlation between molar ratio and concentration:

$c_{H_2O} = \chi \cdot \frac{P_{gas,dry}}{R \cdot T_{gas,dry}} \qquad [mol\, m^{-3}]$

EOD

As the cathode catalyst boundary inward flux is not direction specific, only the water generation diffusive flux can be modeled by setting this boundary condition. The EOD is realized by setting a positive reaction rate of water at the cathode GDL and a a negitve reaction rate at the anode. As they are dependent on the GDL thickness they don't have to equal if the GDLs have different thicknesses.

Using Faradays law is used to calculate the molar flux of the EOD. To obtain the specific reaction rate on each side, the molar flux has to be multiplied by the GDL thickness, so the resulting unit is (mol/m^3/s).

This results in a flux discontinuity at the cathode catalzst boundary, because due to water "creation" at the cathode and water "consumption" at the andoe, the water concentration in the cathode GDL increases which decreases diffusive flux (less concentration gradient) and even changes the direction of the diffusive flux due to water accumulation. The other way round at the anode.

A fulx discontinuty is also generated by having unequal diffusion lengths at anode and cathode due to different GDL thicknesses, because the concentration gradient in the thinner GDL decreases faster than the thicker GDL with leads to a flux dicontinuity.

Experiments for mass and energy balances

1. Heat loss to environment due conduction and external convection

a) Settings

Changes to the original measurement setup:

Thermo-couple TR3 measures endplate temperature at the gas connection side
Thermo-couple TR7 measures endplate temperature at the electrical connection side
Velocity sensor placed 55mm inside the fuel cell housing

Hydrogen flow rate set to constant flow of 0.63 SLPM which results in approximately 2.5A at a stoichiometry of 1.75 for a 21cell stack. ("ADD FORMULA")

Temperture settings:

Anode dew point temperature set point to 28ºC
Anode inlet line heater setpoint set to 70ºC
Anode outlet line heater setpoint set to 75ºC

End plate geometries exposed to environment:

H2 fluid plate: 54mm x 35mm with 4 x 5mm radii
Current plate: two boxes, 75mm x 25mm + 10mm x 49mm with 8 x 5mm radii
Side plates: 70mm x 40mm with with 4 x 5mm radii
Top plate: housing - fan, 90mm x 104mm - (((70mm)^2)/4) x $π$

b) Procedure

At a constant hydrogen flowrate the current is changed stepwise from 0 to 2A with a step size of 0.25A. Each step lasts 20 min. to guarantee a steady-state condition for the temperature measurement. Plate temperatures are measured by thermocouples and an infrared thermometer. The environmental chamber is switched off and the door is left open.
Heat loss due to natural convection at the fuel cell housing and the endplates is then calculated using the measured temperatures. At the end of this test the fan of the environmental chamber is switched on to check its influence on the cathode inlet velocity and the effect of forced convection.

c) Results

The independently calculated power dissipations as described above only match at low currents (I=0,75A). The higher the drawn current, the bigger the difference, which is, at 2A, around 30% of the theoretical power dissipation, calculated by efficiency, without including external convection.
Including external convection improves the difference to a maximum of 20%. This calculation assumes that there is only natural convection.

$Q=h\cdot A\cdot \Delta T$

However, forced convection, when the fan of the environmental chamber is switched on, has a very big influence to the energy balance and the cathode inlet air velocity, which drops due to external air flow. The reason might be the change in cathode outlet pressure.

So there is still some uncertainness, so that at the moment the energy balance does not serve to check mass balance.

2. Determination of the bulk diffusion coefficient

The diffusion coefficient of water through the membrane is dependent on temperature and water content. In order to determine this relation, water diffusion has to be seperated from the other water transport mechanisms. As the experiments of \cite[Husar et al] have shown, water transfer due to the pressure difference between the anode and cathode is at least an order of magnitude lower than those due to other two mechanisms and therefore can be neglected. To seperate diffusion from the EOD the fuel cell is not connected to an external circuit and also nitrogen is used instead of hydrogen, which also guarantees that no water can be generated due to possible leakage of the anode. For this experiment one side of the fuel cell is supplied with saturated gas while on the other side dry gas is used. Since the inlet water flow rate at the dry side is zero, its exit water flow rate must equal the water transfer rate due to diffusion. Also knowing the water concentration difference between the wet side and the dry side, the diffusion coefficient can be calculated applying Fick's law:

$J = -D \frac{\Delta c}{\Delta z} \qquad [mol\,s^{-1}\,m^{-2}]$

3. Determination of the bulk specific heat capacity

To determine a specific heat capacity for stack, a constant current is drawn for a short period of time, and the stack temperature is measured. As desrcribed in chapter \ref{heat transfer} the generated heat energy is determined by the electrical power and the stack efficiency. Thus, knowing the generated heat energy, the temperature and time difference and the mass of the stack, a specific heat capacity can be calculated:

$C_p = \frac{Q \cdot \Delta t}{m \cdot \Delta T} \qquad [J\,kg^{-1}\,K^{-1}]$

Velocity sensor experiments

1. Velocity sensor test in external pipe

Because the cathode mass flow rate is calculated using the inlet air velocity, this parameter has to be reliable. With a mass flow controler the mathematic relation (8) between mass flow rate and air velocity can be checked. For first attempt a pipe was designed and built.