Steam Calculator - Superheated Steam

This page contains a steam table calculator and information on reaction rates, reaction equilibrium constants and chemical process productivity analysis.

Calculation Units

For technical information on superheated steam generators, visit MHI's superheated steam page.

FAQ's

How to enhance Productivity of a chemical process?

Let us assume that the productivity (tons/year) is determined by the amount of product made by a chemical reaction. The direction of any reaction is influenced by the Equilibrium Constant. (See definition below).  For Commercial Chemical Process Development the productivity can be considerably enhanced by Temperature (T), and somewhat by Pressure(P) and the use of a Catalyst. The symbols R and E below represent the Universal Gas Constant and a genereic activation barrier.  R=8.314kJ/kmol.K (0.0831 bar dm³ mol^-1 K^-1) and E is a generic energy barrier with units kJ/kmol.

 

As a rule of thumb an increase in the productivity of an endothermic but favorable reaction (measured in throuput increase per unit time) can be enabled by:

• Increasing the Temperature (Impacts Reaction Rate, Energy Efficiency and the Equilibrium Constant - see below).  The impact of temperature on productivity (amount produced) is exponential ~ (T. e^RT/E) based on rate analysis..
• Increasing the Pressure (somewhat Increases Reaction Rate but not the Equilibrium Constant – see below). The impact of Pressure is linear ~(P) based on rate analysis.
• Increasing the Throughput (e.g. prefer continuous processing over batch processing or choose open system processing over closed system. See www.mhi-inc.com).  Note that temperature increase is relatively easy and cheaper to implement compared to certifying and creating pressure containment methods.  One may select and scale the feed rate for continuous reactions a bit more easily than for batch reactors.
• Improve uniformity and optimize the type of energy transfer (convection, radiation or e-ion plasma type of energy efficiency enhancers). Use hybrid processing.

 

Reaction rates increase when the possibility of making reactant molecules encounter each other is enhanced.  As discussed pictorially below in the reaction rate section, such an increase can be enabled in several ways. Both exothermic and endothermic reactions rates are increased when the temperature of the reaction is increased. Reaction rates increase exponentially with Temperature. Pressure also impacts the reaction rate somewhat but is a costly addition to a process because of pressure vessel certification, specialized welding and thickness requirements and safety. On the other hand a temperature increase is now available for high temperature steam and other gases (see www.mhi-inc.com for devices, and for safe insulation www.fiberfree.com). The best available catalysts should always be used as they impact activation energy (see below).  Open processes i.e. steady state flow is a continuous process, are almost always more energy efficient than a batch process – e.g. high pressure steam batch sterilization is slower than continuous sterilization. The use of high pressures for open processes is very difficult and expensive.  The temperature can be easily increased for open processes with modern thermal devices.

 

 

Equilibrium Constant of a Reaction.

 

Assume a hypothetical reaction: aA + bB +mM--> yY + zZ+mM.  Here Z and Y are the product species.  A and B are the reactant species.  M is a species inert to the reaction.  A chemical reaction consevers mass and the total number of atoms of all elements involved in the reaction.  If all species are gaseous we write the reaction with the (g) in parantheses to indicate gaseous species e.g. aA(g) + bB(g)+mM(g)--> yY(g) + zZ(g)+mM(g). The lower case letters x, y, a, b, m are moles. The arrow signals the direction of reaction that is of interest. 

 

This same reactionin its stoichiometric form may be written as a₀A(g) + b₀B(g)+ m₀M ⇔ y₀Y(g) + z₀Z(g)+ m₀M, the two way arrow indicates that a forward and backward reaction are feasible. The stoichiometric form is not necessarily the equilibrium mixture.  Let us designate the equilibrium mix in its most general form as a_0eA(g) + b_0eB(g) ⇔ y_0eY(g) + z_0eZ(g).  Regardless of how we represent the reaction, the chemical reaction consevers the initial mass and the total number of atoms of all elements involved in the reaction.

 

When at equilibrium i.e. a_0e, b_0e, y_0e, z_0eare present of each species respectively. These are not known apriori. Only m₀ is known and the initial amounts of A and B can be knoiwn.  The heat of reaction, DHr of compounds can be measured or calculated from bond energies or by assigning standard state thermodynamic properties to all elements. The entropies of each elelemt and compounds is also known.  At constant Temperature and Pressure, the equilibrium for a reaction is reached when the Gibbs free energy change between products and reactants is zero.  Based on this concept, an equilibrium constant, K_ec for every reaction reaction is known.  The equilibrium constant can be calculated from a knowledge of such free energies of the reactants and products at the reaction temperature.  Once known, it can be used to calculate the extent of a reaction at equilibrium conditions.  The equilibrium constant thus is a very useful number because it quickly indicates whether a reaction e.g the hypothetical forward one above is possible or not and how much product may be expected from a chosen reactant mix at equilibrium or under steady state conditions. The equilibrium constant allows for the determination of y_0e, z_0e in the reaction above when m₀ is known and the initial amounts of A and B are knoiwn. 

The K_ec (most books use the symbol Kc) values are tabulated and well known for almost all chemical and biological reactions. The K_ec changes with temperature. The Equilibrium Constant is the power ratio of activities (e.g. a_Y) (shown below) of products over reactants for a reaction that is at equilibrium.  The equilibrium constant expression is written as K_ec:

K_ec= {a_Y}^y0{a_z}^z0/{a_A}^a0{a_B}^b0

 

The activities can be written in terms of mole fractions or partial pressure ratios.  The gaseous reaction equilibrium constant Kp, is expressed by partial pressure ratios.

 

K_P= {P_Y/P0}^y0{Pz/P0}^z0/{P_A/P0}^a0{P_B/P0}^b0

 

Here P0 is a the standard state pressure (normally 25C, 1 atmosphere is chosen for chemical reactions)

 

where P_Y is the partial pressure of gas Y defined as P_Y=(y/ N_T)P_T where P_T is the total pressure including any inert constituents in the gas mixture and total number of moles is N_T including any inert constituents in the gas mixture.  The partial pressure P_z, P_A, P_B are similarly defined.

 

For ideal gasses a_Y= P_Y/P_T.  For pure solid or liquids a_Y~1.  For solutions a_Y=g_Yx_Y.  Where g_Y is called the activity coefficient (ranges from 0 to 1), and x_Y is the mole fraction including the fullaccounting for any inert species.

 

K_ec = {[P_Y/P_T]^y0[P_Z/P_T]^z0}/{[P_A/P_T]^a0[P_B/P_T]^b0}= Kp{(P0) ^∆n0/(P_T) ^∆n0}

 

where ∆n0 = (y^_o + z^_o) – (a^_o + b^_o) i.e. (number of moles of gaseous products – number of moles of gaseous reactants for the stoichiometric form of the reaction) i.e. in the balanced chemical reaction.

 

The relation between K_p and K_ec (same as the symbol Kc, commonly employed in other chemical texts) is given as:  Kp(P0) ^∆n0 = Kc (P_T) ^∆n0=Kc (N_TRT/V) ^∆n0  for a volume V (remember the molar volume of an ideal gas is same regardless of the gas and also to use the correct form of R e.g if P is in bar then a form of R could be 0.0831 bar dm³ mol^-1 K^-1) (if P is in Pa the R= 8.314J/mol.K).  Note that 1 bar ~100kPa.  For gases, especially for ideal gasses (that is described by a Equation of State PV=NRT) the Equilibrium Constant Kp is not a function of the total Pressure.  The equilibrium constant will remain the same for each reaction, independent of initial concentrations.

 

For any given temperature, there is only one value for the equilibrium constant, K_ec (or more commonly referred to as Kc when we express the equilibrium constant as a ratio of concentrations) only changes if the temperature of the reaction changes.
If K_ec > 1 then it favors product formation.   In fact if K_ec>1000 then you find mostly products i.e. if you mix A and B, they react almost completely to become product when K_ec>1000.
If K_ec < 1 then it favors the reverse reaction.

 

The dependence of the equilibrium constant with temperature is given by the van't Hoff equation: d(ln(K_ec)/dT = DHr/RT²  or that ln(K_ec)=-DHr/RT + (a constant of integration).

Assume again aA + bB -->yY + zZ; with Heat of Reaction> 0 (endothermic i.e. DHr>0).   If net heat is required to inspire this reaction (endothermic), then an increase in Temperature causes an increase in the products conversion for the reaction because the K_ec increases.   Almost always we have to enable processes whether they are chemical, melting, sintering, rolling, forging or ionization processes.   In other words, they are endothermic for almost all commercial objectives.  Even exothermic processes may present an activation energy barrier (discussed below) which needs to be overcome by a higher reactant temperature before the reaction can go forward.

Example: Consider the reaction:

C(s) + H₂O(g) ⇔ H₂(g) + CO(g). The thermodynamic equilibrium constant for the reaction is ~ 6.5 x 10^-19 at 298K.  The value of the equilibrium constant (K) at two higher temperatures is given below for this reaction. Note that K increases dramatically from less than 1 to above 1 with an increase in the temperature for this endothermic reaction.

Temperature 430C (806F)   K_ec~= 5 x 10-5
Temperature 700C (1292F)  K_ec ~= 1.5

For more information on equilibrium constant expressions and tables please visit the Wikipedia site: http://en.wikipedia.org/wiki/Equilibrium_constant.

Examples where H₂O is a process chemical:

Hydrogen is used in ammonia making, acid making, petrochemicals and as fuel (including fuel cell). The United States alone produces more than 10 million tons of hydrogen per year.  The following examples are for steam reactions are possible for hydrogen manufacture.

3Fe(s) + 4H₂O(g) ⇔ Fe₃O4(s) + 4H₂(g)
From known tables Kc at 500C for this reaction is 5.218E+002. Concentrations of Fe and Fe3O4 are omitted in calculating Kp. Although the concentration of a gas can have various values depeding on partial pressure, the concentration of a pure solid or a pure liquid is a constant at a given temperature. The concentration of liquid solvent is usually omitted as well.  For the reaction of iron with steam, you would write  Kp=PH₂⁴/PH₂O⁴ and get the PH₂ value which will be very small.  Above 1000C this reaction does not occur in the forward direction.  The reaction 2Fe + 3H2O(g) = Fe2O3 + 3H2(g) does not go forward above 600C.    The reaction Fe + H2O(g) = FeO + H2(g) is very weak by 1000C.

A more commonnly employed reaction for making hydrogen is the methane steam reforming reaction (MSR), CH4 + 2H2O ⇔ CO2 + 4H2 for commercial bulk hydrogen production. Steam Reforming can be thought to occur in two steps, namely:
CH4 + H2O ⇔CO + 3H2 [ΔH = +206 kJ mol-1]
CO + H2O ⇔ CO2 + H2 [ΔH = -41 kJ mol-1]
________________________________________________
CH4 + 2H2O --> CO2 + 4H2 [ΔH = +165 kJ mol-1] Endothermic

 

For the forward reaction, the equilibrium constant has to exceed 1.  For the MSR reaction, equilibrium constant only increase above 1 when the temperature exceeds 864K.  At 1000°C K=4963 which is very favorable for complete conversion to hydrogen when that is the objective.  New superheated steam generation devices provide instant hot steam to be made available for this reaction.  MHI also offers air heating for combustible and non combustible gasses.  The reaction C+ 2H2O -->CO2 +2H2 has to be above 966K for the reaction to become feasible for the forward reaction.

 

Other reported uses uses for high temperature steam are in steam Hydrogasification.

Increasing the temperature will increase production output because of an increase in K and also from the significant reaction rate increase that occurs (e.g. with an OAB) if the highest throughput and temperature are chosen for the reactant steam.  Regardless, other considerations for a commercial process often exist, and a multi-step process may be used for this reaction product (see http://en.wikipedia.org/wiki/Equilibrium_constant).   Use hybrid thermal processing to get the best outcome.

 

Reaction Rate for Exothermic and Endothermic Processes.

 

Regardless of the equilibrium constant. reaction rates are either fast or slow depending on the reaction.  Conditions that increase the rate of a reaction include the specific species of the reaction, concentration, surface area, temperature and catalysts.

 

• Different reactants will have different reaction rates because of differences in the activation energy barrier, Ea, for every reaction (see below).
• More reactants (throughput) results in more collision giving rise to a faster production rate.
• Often reactions need a surface or include a surface in the reaction.  A high surface area gives a greater chance for reactants to form product.
• Higher temperature results in more energetic collisions.  See below for reaction rate constant and energy efficiency.
• Catalysts are metals and compounds that help a reaction rate by lowering the activation energy for a reaction.

 

For a reaction to occur at a certain temperature, an initial activation energy barrier is required to be overcome.  The Activation Energy Barrier, E_a is the minimum energy that a reactant mix must possess in order to convert the reactants to products.  This activation barrier (Ea) can determine how fast a reaction occurs.   The higher the activation barrier, the slower the reaction rate.  The lower the activation barrier, the faster the reaction.  Consider the analogous process of someone trying to roll a boulder over a hill.  The higher the hill,  the slower the task.  The lower the hill, the faster the process.  The height of the hill is analogous to the energy of activation (Ea) that has to be overcome for a reaction. Reactants have to collide to produce products and because this is a rate limiting step, chemical reactions encounter a barrier.   

 

Regardless, high and low barriers are relative to the might and energy of the person pushing the boulder.  Again one may draw from common experience with an analogy to the hill barrier for pushin a boulder.   Any activation energy is overcome if the reactants are properly energized.  The thermal energy possessed by the reactants is related to temperature. 

 

Catalysts reduce the activation barrier. The (Ea) barrier can be lowered also by the use of a correct catalyst material surface, however catalysts may foul over time and are often expensive (many catalysts contain expensive noble metals like platinum).  Increasing the temperature is often the best bet on a cost basis to enjoy higher productivity.

 

Assume again a hypothetical reaction: aA(g) + bB(g) ⇔ yY(g) + zZ(g).  Assume we are only interested in the forward reaction i.e. aA(g) + bB(g) --> yY(g) + zZ(g) i.e. we are interested in the production of Z and Y.  Assume we are further interested in producing Z and Y as quickly as possible in order to have the highest productivity (production rate).  The rate equation is now discussed (link to example for oxides).

 

Regardless of being exothermic or endothermic there is an initial barrier Ea, to overcome for the forward reaction.  Increasing the temperature of the reaction is the same as providing higher energy reactants. The difference between an exothermic and endothermic reaction can be illustrated by the hill/boulder analogy again.  After reaching the top of the hill if the boulder rolls to a height lower than the original, it will possess a lower potential energy (the difference in energy is the net gain of energy from the process- exothermic reaction analogy).  If on the other hand the boulder rests at a higher level than the reactants, the difference in energy is the amount supplied to the process (endothermic reaction analogy) shown by the thick line in the right hand side figure.  Regardless, for both examples, Ea is the amount of energy to be overcome and needs to be initially made available to the reactants. 

 

 

 

 

 

The rate equation for a chemical process is often captured by a rate constant which is  proportional to e^-Ea/RT (the symbol most often employed for the rate constant is lower case k, note also that e^-Ea/RT = e^RT/Ea). The units for Ea are kJ/mole or kJ/volume.  R is the gas constant =8.314 kJ/kmol.K.  T is the temperature in Kelvin.  The rate of reaction in either reaction direction, is related to the initial concentration raised to a power that reflects the order of reaction, multiplied by a constant k₀ and further multiplied by e^-Ea/RT.  The pre-exponential part k₀ (units of 1/s) is only a mild function of temperature.  It increases with temperature in a close-to-linear relationship according to a theory called the activated species theory. 

 

The Ea may not be strongly influenced by temperature, but note how k is exponentially increased (very strongly influenced) by an increase in the temperature.  It is now easy to make the inference that the k is directly related to productivity.  Therefore increasing the reaction temperature (or reactant temperature) dramatically increases productivity.  The e^-Ea/RT part of the rate constant does not have dimensions but the amount made per unit time i.e. Kg/hr, is proportional to k and thus to temperature..   To learn more - choose a search engine like Google, Yahoo or Bing or any other and type in the key phrase " How is productivity influenced by high temperature, including high temperature capable MHI heating materials".  Then choose your best and most efficient thermal device.

 

How does one ensure that the forward reaction is the only one occurring? Will equilibrium conditions prevail and equalize the forward and backward reaction rates?

 

All reactions tend towards equilibrium.  When Kec > than about 1000 or less than about 0.001 the reaction is slam-dunk forward or backward respectively.  The equilibrium constant is thus clearly noted to impact the rate of reaction in a particular direction.  For reactions that are not lop-sided, there will be a mixture of reactants and products at equilibrium. 

 

For increased productivity the objective is to quickly consume the reactants to make product.  One common method is to maintain the system far from equilibrium with more reactants than what should be present at equilibrium and find ways to reduce the back reaction rate.  Note that in the boulder analogy schematic shown above, the forward and backward rates could be different because of the respective barrier to overcome to cross the 'hill'.  For the hypothetical reaction above if the products [Z] and [Y] are (a) separated quickly soon after they form ( e.g. by gravity, magnetic, electrical or distillation type processes that can separate reactants from each other) or (b) the products are made cooler than reactants by providing hotter reactants, or (c) the catalyst employed enables only the forward reaction, then the forward reaction is compelled to continuously happen even when the Ea for the backward reaction could be lower than for the forward reaction.  An open process can achieve a steady state while maintain high productivity and non-equilibrium feed-rate into a control volume.  Thus the use of higher temperature reactant(s) can again be the best way to increase productivity. Should the temperature be increased? or the plant size be increased? for a production amount rate increase. Examples from metals /chemicals industry.