Ali Yuksel said:
Using numerous parabolic mirrors, i need to focus the reflected light onto a mutual point, and achieve a temperature of about 3000*C. Why 3000? That is when thermal decomposition of H2O occurs. What i want to calculate first, is the amount of energy i will need to make this happen. Am i supposed to go with Q=MCdT?
That's too simple because,
1. C changes with Temperature
2. you also need energy for the thermal decomposition and
3. you would neglect the loss of energy due radiation and thermal conduction
Let's start with the energy collected by the mirrors:
\dot Q_{in} = A_m \cdot E_0
where Am is the total area of the mirrors and Eo the local solar constant.
The heat loss by radiation is
\dot Q_{rad} = A_c \cdot \sigma \cdot T^4
where Ac is the area of the collector.
The heat loss by thermal conduction is
\dot Q_{cond} = k \cdot \left( {T - T_0 } \right)
Now it's going to be interesting. The heat required for heating and thermal decomposition of water is the difference of the formation enthalpies. If water with the Temperature To flows into the reactor and a mixture of water vapor, hydrogen and oxygen with the temperature T comes out the corresponding change of enthalpy is
\dot Q_H = \dot n_{H_2 O} \cdot \Delta _f H_{H_2 O} \left( T \right) + \dot n_{H_2 } \cdot \Delta _f H_{H_2 } \left( T \right) + \dot n_{O_2 } \cdot \Delta _f H_{O_2 } \left( T \right) - \dot n \cdot \Delta _f H_{H_2 O} \left( {T_0 } \right)
The stoichiometry of the reaction
H_2 O \Leftrightarrow H_2 + {\textstyle{1 \over 2}}O_2
results in the following amounts of substances
\dot n_{H_2 O} = \left( {1 - x} \right) \cdot \dot n
\dot n_{H_2 } = x \cdot \dot n
\dot n_{O_2 } = {\textstyle{1 \over 2}}x \cdot \dot n
Where
\dot n = \frac{{\dot m}}{{M_{H_2 O} }}
is the total flow into the reactor.
The degree of conversion x can be obtained from the equilibrium constant
K = \frac{{x_{H_2 } \cdot \sqrt {x_{O_2 } } }}{{x_{H_2 O} }}
The amounts of substances give the mole fractions
x_{H_2 O} = \frac{{2 - 2 \cdot x}}{{2 + x}}
x_{H_2 } = \frac{{2 \cdot x}}{{2 + x}}
x_{O_2 } = \frac{x}{{2 + x}}
resulting in
K^2 = \frac{{x^3 }}{{\left( {2 + x} \right) \cdot \left( {1 - x} \right)^2 }}
The last step is the calculation of the equilibrium constant
\ln K = - \frac{{\Delta _r G}}{{R \cdot T}}
from the change of Gibbs free energy:
\Delta _r G\left( T \right) = \Delta _r H\left( T \right) - T \cdot \Delta _r S\left( T \right)
Now T, To, Am, Ac and dm/dt need to be adjusted in order to balance the sum of all heat flows out:
\dot Q = \dot Q_{in} \left( {A_m } \right) - \dot Q_{rad} \left( {A_c ,T} \right) - \dot Q_{cond} \left( {k,T,T_0 } \right) - \dot Q_H \left( {\dot m,T,T_0 } \right) = 0The required thermodynamic data can be obtained here:
http://webbook.nist.gov/cgi/cbook.cgi?ID=C7732185&Units=SI&Mask=2#Thermo-Condensed
http://webbook.nist.gov/cgi/cbook.cgi?ID=C7732185&Units=SI&Mask=1#Thermo-Gas
http://webbook.nist.gov/cgi/cbook.cgi?ID=C1333740&Units=SI&Mask=1#Thermo-Gas
http://webbook.nist.gov/cgi/cbook.cgi?ID=C7782447&Units=SI&Mask=1#Thermo-Gas
PS: How can I limit the formulas to a reasonable size?
