Space Shuttle Main Engine (SSME) perfomance model

This is a simple model of the Space Shuttle Main Engine (SSME) performance. It is based on the data from the Dr. Rodney Bowersox:

Combustion Chamber Data
Fuel	Hydrogen
Oxidizer	$O_{2}$
Fuel/Oxidizer Ratio $\phi$	0.166
Hydrogen Injection Temperature	850 K
Oxygen Injection Temperature	530 K
Combustion Chamber Pressure	204 atm

Nozzle Geometry
$R_{t}$	5.15 in
$R_{e}$	45.35 in
$\theta_{p}$	32.00 deg

Oxygen First Enthalpy Exchanger

We begin with computing the equivalence ratio $\phi$ for the oxygen first enthalpy exchanger. The equivalence ratio is defined as the ratio of the mass of fuel to the mass of oxidizer. The mass of fuel is the mass of hydrogen, and the mass of oxidizer is the mass of oxygen. The mass of hydrogen is $\phi \frac{mol_{O_{2}}}{mol_{H_{2}}} = \Phi$

From here we are to calculate the standard heat of formation by:

$\Delta \hat{h_{i}} = \hat{h_{H_{2}}}(850K) - \hat{h_{H_{2}}}(298K) + \hat{R}T^2 \int_{p_{f}}^{p_{i}} \frac{1}{p} \frac{-\hat{b} p}{\hat{R}T^2} dp|_{T_i}$

Where $\hat{b}$ is read from a table developed in exchanger.py. $\Delta \hat{h_{i}}$ is calculated for both hydrogen and oxygen. Once we calculate $\Delta \hat{h_{i}}$ for each, we then calculate the enthalpy of the mixture:

$\Delta \hat{h_{mix}} = \Phi \Delta \hat{h_{H_{2}}} + \Delta \hat{h_{O_{2}}}$

Which is then applied to our chemical equation:

$O_{2} + \Phi H_{2} \rightarrow (2-\Phi)H_{2} + 2H_{2}O$

Note: notice the $(2-\Phi)H_{2}$ portion of the equation. This is used since we calculated the enthalpy of formation for hydrogen (H) and not hydrogen gas $H_{2}$.

From here we can sum the molar enthalpies of formation to get the first estimate of the first reaction enthalpy:

$\Delta \hat{h_{r}} = \Phi \Delta \hat{h_{H_{2}}} + 2( \Delta \hat{h_{H_{2}O}} )$

We are to then interpolate this value from the given turn tables to get the actual value of the combustion temperature for Oxygen First Enthalpy Exchanger. This is done in exchanger.py. At the end of this process we have the combustion temperature for the Oxygen First Enthalpy Exchanger, $T_{c}$.

Oxygen - Combustion Chamber

For the Combustion Chamber we wil using the two sub-reactions:

$H_{2}O \rightarrow OH+H$

$H_{2}\rightarrow 2H$

From here we are to calculate the Gibbs Free Energy of each species.

For $H_{2}\rightarrow 2H$

$\Delta G_{H_{2}} = 2(\bar{g_{H}} -\bar{g_{H_{2}}})$

For $H_{2}O \rightarrow OH+H$

$\Delta G_{H_{2}O} = \bar{g_{H}} + \bar{g_{OH}} - \bar{g_{H_{2}O}}$

Where $\bar{g_{fh}}$ is the Gibbs Free Energy of formation of $H_{2}$, $\bar{g_{foh}}$ is the Gibbs Free Energy of formation of OH, and $\bar{g_{fh_{2}o}}$ is the Gibbs Free Energy of formation of $H_{2}O$. These values are read from the Iterater.py file.

Law of Mass Action

Looking at For $H_{2}O \rightarrow OH+H$:

Species	$\nu_{(i)} '$	$\nu_{(i)} "$	$\nu_{(i)} " - \nu_{(i)} '$
$H_{2}O$	1	0	-1
$OH$	0	1	1
$H$	0	1	1

We know that:

$K_{n} = \prod_{n=1}^{N_{s}} X_{(n)}^{(\nu_{(i)} " - \nu_{(i)} ')} = (\frac{p}{p^{ * }})^{- \sigma_{v}} K_{p}$

Where;

$\sigma_{v} = \sum_{n=1}^{N_{s}} \nu_{(i)} " - \nu_{(i)} '$

$K_{p} = e^{- \frac{\Delta G}{\hat{R}T}}$

$\Delta G = \sum_{n=1}^{N_{s}} (\nu_{(i)} " -\nu_{(i)}') \bar{g}_{f(n)}$

We can then solve for $K_{n}$:

$K_{H_{2}O \rightarrow OH+H} = (\frac{204 atm}{1 atm})e^{- \frac{\Delta G_{H_{2}O}}{\hat{R}T_{c}}}$

$K_{H_{2}\rightarrow 2H} = (\frac{204 atm}{1 atm})e^{- \frac{\Delta G_{H_{2}}}{\hat{R}T_{c}}}$

Here we know that the fuel is $H_{2}$ therefore the mass of the fuel is 2.02 $\frac{g}{mol}$ and the mass of the oxidizer is $m_{oxidizer} =\frac{2.02}{\phi}$. And $l = \frac{m_{oxidizer}}{2 m_{O_2}}$. Once we have $l$ we re-visit the Chemistry Problem by solving the Atom Balance:

$H_{2} +l O_{2} \rightarrow N_{H_{2}O}H_{2}O +N_{H_{2}}H_{2} + N_{H}H +N_{OH}OH$

$H_{2}: 2 = 2N_{H_{2}O} + 2N_{H_{2}} + N_{H} + N_{OH}$

$O_{2}: 2l = N_{H_{2}O} + N_{OH}$

Where $N_{H_{2}O}$, $N_{H_{2}}$, $N_{H}$, and $N_{OH}$ are the number of moles of each species. We can then solve for $N_{H_{2}O}$, $N_{H_{2}}$, $N_{H}$, and $N_{OH}$:

$1 = N_{H_{2}O} + N_{H_{2}}$

$0.760592 = N_{H_{2}O}$

Resulting in us having 4 unknown values with only 2 equations. This a signal to us prompting the Law of Mass Action. When solving for the Moler Fraction of each species we get:

$X_{H_{2}O} = \frac{N_{H_{2}O}}{N_{Total}} = \frac{N_{H_{2}O}}{N_{H_{2}O} + N_{H_{2}} + N_{H} + N_{OH}}$

$X_{H} = \frac{N_{H}}{N_{Total}} = \frac{N_{H}}{N_{H_{2}O} + N_{H_{2}} + N_{H} + N_{OH}}$

$X_{OH} = \frac{N_{OH}}{N_{Total}} = \frac{N_{OH}}{N_{H_{2}O} + N_{H_{2}} + N_{H} + N_{OH}}$

$K_{H_{2}O \rightarrow OH+H} = \frac{\frac{N_{OH}}{N_{H_{2}O} + N_{H_{2}}} \frac{N_{H}}{N_{H_{2}O} + N_{H_{2}}}}{\frac{N_{H_{2}O}}{N_{H_{2}O} + N_{H_{2}}}} = \frac{N_{H}N_{OH}}{(N_{H_{2}O} + N_{H_{2}})N_{H_{2}O}}$

Similarly for the $H_{2}\rightarrow 2H$ Reaction:

Species	$\nu_{(i)} '$	$\nu_{(i)} "$	$\nu_{(i)} " - \nu_{(i)} '$
$H_{2}$	1	0	-1
$H$	0	2	2

$X_{H_{2}} = \frac{N_{H_{2}}}{N_{Total}} = \frac{N_{H_{2}}}{N_{H_{2}O} + N_{H_{2}} + N_{H} + N_{OH}}$

$X_{H} = \frac{N_{H}}{N_{Total}} = \frac{N_{H}}{N_{H_{2}O} + N_{H_{2}} + N_{H} + N_{OH}}$

$K_{H_{2}\rightarrow 2H} = \frac{(\frac{N_{H}}{N_{H_{2}}+ N_{H_{2}O}})^{2}}{\frac{N_{H_{2}}}{N_{H_{2}}+ N_{H_{2}O}}} = \frac{N_{H}^{2}}{N_{H_{2}}(N_{H_{2}}+ N_{H_{2}O})}$

After this we know have to solve a system of coupled equations:

$1 = N_{H_{2}O} + N_{H_{2}}$

$0.760592 = N_{H_{2}O}$

$K_{H_{2}O \rightarrow OH+H} = \frac{N_{H}N_{OH}}{(N_{H_{2}O} + N_{H_{2}})N_{H_{2}O}}$

$K_{H_{2}\rightarrow 2H} = \frac{N_{H}^{2}}{N_{H_{2}}(N_{H_{2}}+ N_{H_{2}O})}$

We can then solve for $N_{H_{2}O}$, $N_{H_{2}}$, $N_{H}$, and $N_{OH}$:

Species	$N_{(i)}$	$X_{(i)}$	$m_{(i)}$	$Y_{(i)}$	$\hat{c_{p}}_{(i)}$	$c_{p_{(i)}}$	$Y_{(i)}c_{p}$
$H_{2}O$	$N_{H_{2}O}$	$X_{H_{2}O} = N_{H_{2}O}$	$\hat{m} N_{H_{2}O}$	$\frac{\hat{m}}{m_{(i)}}$	$\hat{c_{p}} (T_{c})$	$\frac{\hat{c_{p}}}{\hat{m}}$	$Y_{H_{2}O}c_{p}$
$H_{2}$	$N_{H_{2}}$	$X_{H_{2}} = N_{H_{2}}$	$\hat{m} N_{H_{2}}$	$\frac{\hat{m}}{m_{(i)}}$	$\hat{c_{p}} (T_{c})$	$\frac{\hat{c_{p}}}{\hat{m}}$	$Y_{H_{2}}c_{p}$
$H$	$N_{H}$	$X_{H} = N_{H}$	$\hat{m} N_{H}$	$\frac{\hat{m}}{m_{(i)}}$	$\hat{c_{p}} (T_{c})$	$\frac{\hat{c_{p}}}{\hat{m}}$	$Y_{H}c_{p}$
$OH$	$N_{OH}$	$X_{OH} = N_{OH}$	$\hat{m} N_{OH}$	$\frac{\hat{m}}{m_{(i)}}$	$\hat{c_{p}} (T_{c})$	$\frac{\hat{c_{p}}}{\hat{m}}$	$Y_{OH}c_{p}$

Now to calculate $c_{p}$ we sum the $Y_{(i)}c_{p}$ for each species:

$c_{p} = \sum_{i} Y_{(i)}c_{p}$

$m_{(total)} = \sum_{i} X_{(i)}m_{(i)}$

We perform this step to calculate the gas constant $R$:

$R = \frac{\hat{R}}{m_{(total)}}$

$c_{v} = c_{p} - R$

Resulting in use getting a gamma of:

$\gamma = \frac{c_{p}}{c_{p} - \frac{\hat{R}}{\sum_{i} X_{(i)}m_{(i)}}} = \frac{c_{p}}{c_{p} - \frac{\hat{R}}{m_{(total)}}}=\frac{c_{p}}{c_{p} - R} =\frac{c_{p}}{c_{v}}$

Oxygen Nozzle Performance

Begin by defining our Area Ratio: $A_{r} = \frac{\pi (R_{e})^{2}}{\pi (R_{t})^{2}}$. From here we are to solve for the Mach number:

$A_{r} = (\frac{\gamma +1}{2})^{-\frac{\gamma+1}{2(\gamma-1)}} \frac{1}{M_{e}} (1+ \frac{\gamma-1}{2} (M_{e})^{2})^{\frac{\gamma+1}{2(\gamma -1)}}$

We can for Mach number $M_{e}$ by making use of scipy.optimize.fsolve, which returns the roots of a non-linear function. After solving for $M_{e}$ we can then solve for the pressure, Temperature, and velocity:

$T_{e} = T_{c} (1+ \frac{\gamma-1}{2} (M_{e})^{2})^{-1}$

$p_{e} = p_{c} (1+ \frac{\gamma-1}{2} (M_{e})^{2})^{-\frac{\gamma}{\gamma-1}}$

$v_{e} = M_{e} \sqrt{\gamma R_{sp} T_{e}}$

After Solving for these values we can then calculate the mass flow rate:

$\dot{m} = \frac{A_{t} p}{\sqrt{T_{c}}} \sqrt{\frac{\gamma}{R_{sp}}} (\frac{2}{\gamma + 1})^{\frac{\gamma + 1}{2(\gamma -1)}}$

Then once you have the mass flow rate you can calculate the thrust, specific impulse, and thrust coefficient:

$Thrust = \dot{m} v_{e}+p_{e}A_{e} - pA_{t}$

$I_{sp} = \frac{Thrust}{\dot{m} g_{0}}$

$C_{T} = \frac{Thrust}{pA_{t}}$

Code Flow

    flowchart
    A[Givens] --> B[Parameters]
    B --> SSME
    D -->|def| FCC
    FCC1 --> |req| Request
    Request --> |return| del_h_hat
    phi --> FCC2
    T_alg --> |req| Request
    Request --> |return| T_alg
    E -->|def| Comb
    Gibbs --> |req| Request
    Request --> |return| Gibbs
    cpval --> |req| Request
    Request --> |return| cpval
    N --> |def| Nozzle


    subgraph SSME[SSMEMain.py]
    C[Read in Givens] --> phi["equivalence ratio"]
    phi --> D[FCC]
    D--> E[Comb Chamber]
    E --> N[Nozzle]

  
    end
    

    subgraph FCC[FCC]
    FCC1["h_(i)(InjectionTemp)"] 
    b_hat["b_hat_table"]
    del_h_hat["del_h_hat_(i)"]
    b_hat --> del_h_hat
    FCC2["mols_(i)"]
    h_total["del_h_total"]
    T_alg["T_algorithm"]

    del_h_hat --> h_total
    FCC2 --> h_total
    h_total --> T_alg
    end 

    subgraph Iter["Iterater.py"]
    Request["Request_(i) (input,output)"] 

    Request --> |"(T,h)"|Iter1
    Request --> |"(h,T)"|Iter2
    Request --> |"(T,g)"|Iter3
    Request --> |"(T,cp)"|Iter4
    Request --> |"(T,S)"|Iter5
    Iter1["(i)_tableH = (i)_table_f"]
    Iter2["(i)_tableT"]
    Iter3["(i)_tableg"]
    Iter4["(i)_table_cp"]
    Iter5["(i)_tableS"]

    MachS["MachSolve"]
    end

    subgraph Comb["Comb Chamber"]
    Atom["Atom Balance"]
    Gibbs["Gibbs Free Energy"]
    gibbs_r["g_reaction"]
    LMA["LawOfMassAction_df"]
    cpval["cp_values_(i)"]
    Gibbs --> gibbs_r
    gibbs_r --> k_n["K_(n)"]
    T_alg --> k_n

    Atom --> eq["equations"]
    eq --> sol["solution(s)"]
    k_n --> eq
    sol --> LMA
    cpval --> LMA
    LMA --> R["R"]
    R --> cv["cv"]
    cv --> gamma["gamma"]
    end

    subgraph Nozzle["Nozzle"]
    A_r["Area Ratio"]
    gam["gamma"]
    gam --> Mach
    A_r --> Mach
    Mach -->|req| MachS
    MachS --> |return|Mach
    Mach --> mdot["mdot"]
    Mach --> T_e["T_e"]
    Mach --> p_e["p_e"]
    Mach --> v_e["v_e"]

    v_e --> Thrust["Thrust"]
    mdot --> Thrust
    p_e --> Thrust
    c_T["c_T"]
    Thrust --> c_T
    Thrust --> Isp["Isp"]

    end

    subgraph Mach["mach"]
    end

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