Energy balance over a methanol reformer

[Mads_DK]

TL;DR Summary

I am currently trying to determine the heat which has to be supplied to a methanol reformer. However, I am unsure how the energy balance should be set up. I know the process is supposed to be endothermic, whereas my small model tells me the process releases energy.

In terms of the information I have been given, I know the following:
The mass flow of the methanol/water mixture at the inlet of the reformer is 30 kg/hr (which therefore must be the same mass flow at the outlet). I likewise know the mole fractions to be 60 vol% methanol and 40 vol% water.
I know the process to be a complete reformation of methanol, so the following species are assumed to be present at the outlet: H2, CO2, CO and H2O. Their respective molefractions are: 0.646, 0.215, 0.007 and 0.132.
The inlet and outlet temperatures are 433.15 K and 483.15 K.

I know that in terms conservation of mass, what comes into the reformer goes out. However, from what I understand, the number of moles going into the reformer does not have to be equal to the number of moles leaving the reformer.
I have determined the mole flows of each species from the equations:
n_total = sum(MW_i * y_i) / m_mixture
n_i = n_total * y_i
Where n = mole flow [mol/s], MW = molar weight [g/mol], y = mole fraction [-], m = mass flow [g/s], i = a given species, mixture = methanol/water mixture.

To determine the enthalpy of each species, I have used the software EES (Engineering Equation Solver) and I have assumed the species to be ideal gases.
I have set up the energy balance as:
Q_out = Q_in + Heat
Where the unit is [W] and 'Heat' refers to the heat which has to be supplied to the reformer

Written out:
Heat = Q_out - Q_in = (h_H2(483.15 K) * n_out,H2 + h_CO2(483.15 K) * n_out,CO2 + h_CO(483.15 K) * n_out,CO + h_H2O(483.15 K) * n_out,H2O) - (h_CH3OH(433.15 K) * n_in,CH3OH + h_H2O(433.15 K) * n_in,H2O)
From the equation above I get -2873 W, which I know to be wrong.

My confusion in terms of this problem also lies in the fact that I do not know whether or not you are supposed to account for the heat of reaction for the three main reactions (methanol steam reforming (MSR), water gas shift (WGS), methanol decomposition(MD)) in the energy balance in order to get the correct heat requirement?

For good measure, the three reactions assumed to be the main reactions in the methanol reformer is given below:
MSR: CH3OH + H2O <-> CO2 + 3H2 (DeltaH_rxn = 49.7 kJ/mol)
WGS: CO + H2O <-> CO2 + H2 (DeltaH_rxn = -41.2 kJ/mol)
MD: CH3OH <-> CO + 2H2 (DeltaH_rxn = 90.7 kJ/mol)

I have been trying to find an example where the energy balance accounts for three reactions in terms of their heat of reaction. However, I have only been able to find examples where they account for one reaction. I therefore also tried to setup a single global reaction based on the three equation above, though without much luck.
The reaction I have found, where they account for the heat of reaction is given by:
Q_heat_added = xi * DeltaH_rxn + sum(h_i(T_out) * n_out,i) - sum(h_i(T_in) * n_in,i)
Where xi is the extend of reaction given by:
xi = (n_A,out - n_A,in) / v_A
Where A denotes a given species in a reaction and v is the stoichiometric coefficient of species A given in [mol/s].

So to sum up, I hope you can help to point me in the right direction, as I clearly do not get the correct answer to this problem. The problem should be straight forward as I have been given all the information, but the role of the heat of reaction in the energy balance has completely confused me. Hope to hear from you!

 

[Chestermiller]

Step 1: What is the average molecular weight of the stream entering the reformer?
Step 2: What is the average molecular weight of the stream exiting the reformer?
Step 3: What is the number of kg-moles per hour entering the reformer?
Step 4: What is the number of kg-moles per hour exiting the reformer?
Step 5: What is the number of kg-moles per hour of the two species entering?
Step 6: What is the number of kg-moles per hour of the 4 species exiting?
Step 7: What are the numbers of moles of C, H, and O entering and exiting per hour?

Are the molar balances on C, H, and O individually satisfied (entering vs exiting)?

Are you familiar with the concept of heat of formation of a compound?

 

[Chestermiller]

I will try doing it and see what I get.

MW of feed = (0.6)(32)+(0.4)(18)= 26.4
MW of exit stream = (0.646)(2)+(0.215)(44)+(0.007)(28)+(0.132)(18)=13.324
$n_{in}$=30000/26.4=1136.4 moles/hr
$n_{out}$=30000/13.324=2251.6 moles/hr

SPECIES FLOW RATES

$n_{CH_3OH}$=(0.6)(1136.4)=681.8 moles/hr
$n_{H_2O,in}$=(0.4)(1136.4)=454.5 moles/hr
$n_{H_2}$=(0.646)(2251.6)=1454.5 moles/hr
$n_{CO_2}$=(0.215)(2251.6)=484.1 moles/hr
$n_{CO}$=(0.007)(2251.6)=15.8 moles/hr
$n_{H_2O,out}$=(0.132)(2251.6)=297.2

CHECK ON MASS BALANCES:

Carbon in = 681.8 moles/hr
Carbon out = 484.1 +15.8 = 499.9 moles/hr

Hydrogen in = (4)(681.8)+(2)(454.5)=3636.2 moles/hr
Hydrogen out = (2)(1454.5)+(2)(297.2)=3503.4 moles/hr

Oxygen in = 681.8+454.5 = 1136.4 moles/hr
Oxygen out = (2)(484.1)+15.8+297.2= 1281.2 moles/hr

None of these checks balance. The balance would match almost perfectly if the feed were 40 % methanol and 60% water, rather than 60 % methanol and 40% water.

If we are willing to accept the chemical analysis of the exit stream, then it is possible to uniquely determine all the key mass flow parameters for the inlet stream, and determine the precise split between methanol and water. This is because (1) each molecule of the feed contains exactly one oxygen atom, so that the outlet molar flow of oxygen atoms determines to total molar flow of the feed stream and (2) only a single carbon atom is present in the methanol, and none in the water feed, and only a single carbon atom is present in the CO and CO2, and none in the H2 and water of the exit stream, so that the outlet molar flow of carbon atoms determines the molar flow rate of methanol in the feed. Thus, methanol in the feed must be equal to 499.9 moles/hr (see above) and the total molar flow in the feed must be equal to 1281.2 moles per hour. This means that the mole fraction of methanol in the feed, rather than being equal to 0.6, must be equal to $$y_{methanol}=\frac{499.9}{1281.2}=0.3902$$and the mole fraction water in the feed, rather than being equal to 0.4, must be equal to 0.6098. From this, it follows that

$$n_{in}=1281.3$$
$$n_{in,methanol}=499.9\ mole/hr$$
$$n_{in,water}=781.3\ moles/hr$$

Carbon in = 499.9 moles/hr
Hydrogen in = 4(499.9)+2(781.3)=3562.2 moles/hr
Oxygen in = 499.9 + 781.3 = 1281.2
Comparing these with the values for the outlet streams (above) shows an exact match of C and O; the slight mismatch on H is due to roundoff error.

So with this comparison, we now see that we have closure on the mass balance, and we can then proceed with confidence on the heat balance.

 

[Chestermiller]

Assuming ideal gas behavior, the molar enthalpy of each of the 5 chemical species in this problem can be expressed as $$h=h^0+\bar{C}_p(T-298.15)$$where $h^0$ is the heat of formation of the species from the elements at 298.15 K, and $\bar{C}_p$ is the average molar heat capacity of the species between 298.15 K and T.

 

[Chestermiller]

I get +6768 W.