2
$\begingroup$

I've found many documents that leave out the exit diameter of the LE-5B hydrolox engine. If anyone has found the detail in any public doxuments, I would appreciate it. If you can still provide the diameter without a source, it'd still be very much appreciated.

$\endgroup$
4
$\begingroup$

I haven't tried to find any official sources, but from the data available on the Wikipedia article we can estimate it quite precisely. The biggest mistake we make should be that this is an expander bleed cycle engine, and we don't know how much actually bleeds but it shouldn't be very much so we estimate it to be 0%.

We first start off by calculating the mass flow rate.

m_fuel = F_rated / (Isp * g) = 137.2 kN / (447s * 9.81 m/s^2) = 31.3 kg / s

Now we need to calculate the chemical equilibrium and the conditions at the throat and the exit to calculate the throat area using the mixture ratio of 5 and the expansion ratio of 110. Luckily this can be done rather easy using the tool cpropep which is even available as an online tool. We could also use NASA's chemical equilibrium's with applications which is bit more major but I find cpropep easier to use.

Propellant composition
Code  Name                                mol    Mass (g)  Composition
457   HYDROGEN (CRYOGENIC)                0.4961 1.0000   2H  
686   OXYGEN (LIQUID)                     0.1563 5.0000   2O  
Density :  0.328 g/cm^3
2 different elements
H  O  
Total mass:  6.000000 g
Enthalpy  : -1083.68 kJ/kg

9 possible gazeous species
2 possible condensed species

                      CHAMBER      THROAT        EXIT
Pressure (atm)   :      35.332      20.236       0.017
Temperature (K)  :    3236.859    3029.977     822.915
H (kJ/kg)        :   -1083.683   -2312.272  -11140.960
U (kJ/kg)        :   -3371.140   -4435.131  -11706.647
G (kJ/kg)        :  -65370.681  -62490.407  -27484.812
S (kJ/(kg)(K)    :      19.861      19.861      19.861
M (g/mol)        :      11.765      11.867      12.095
(dLnV/dLnP)t     :    -1.01444    -1.00984    -1.00000
(dLnV/dLnT)p     :     1.27093     1.19591     1.00000
Cp (kJ/(kg)(K))  :     7.77987     6.86742     2.94009
Cv (kJ/(kg)(K))  :     6.65463     5.87515     2.25267
Cp/Cv            :     1.16909     1.16889     1.30516
Gamma            :     1.15245     1.15750     1.30516
Vson (m/s)       :  1623.62822  1567.55251   859.24941

Ae/At            :                 1.00000   110.00001
A/dotm (m/s/atm) :                66.92186  7361.40529
C* (m/s)         :              2364.48308  2364.48308
Cf               :                 0.66296     1.89679
Ivac (m/s)       :              2921.80378  4611.05594
Isp (m/s)        :              1567.55251  4484.92531
Isp/g (s)        :               159.84587   457.33511

Molar fractions

H                     3.1008e-02  2.2736e-02  0.0000e+00
HO2                   5.0658e-06  1.8655e-06  0.0000e+00
H2                    3.5686e-01  3.5620e-01  3.5992e-01
H2O                   5.9030e-01  5.9868e-01  6.1281e-01
H2O2                  2.2258e-06  9.0876e-07  0.0000e+00
O                     1.0407e-03  5.0344e-04  0.0000e+00
OH                    2.0106e-02  1.2949e-02  0.0000e+00
O2                    6.7123e-04  3.3398e-04  0.0000e+00

Now the thing we are interested in is the A/dotm which is in the rather strange unit 1 m/s/atm = 9.869×10^-6 m^2 s/kg. Now to get the throat or exit area we simply multiply by the total mass flow.

A_throat = A_dotm * m_fuel =   66.9 m/s/atm * 31.3 kg / s = 206.6 cm^2
A_exit   = A_dotm * m_fuel = 7361.4 m/s/atm * 31.3 kg / s = 2.27 m^2

Now simply solve for the radius to get our final result:

r_throat = sqrt(A_throat / pi) =   8.11 cm
r_exit   = sqrt(A_exit   / pi) = *85.06 cm*

As you might have noticed the expansion ratio 2.2303 m^2 / 206.6483 cm^2 isn't exactly 110. This is actually a seemingly unknown bug in this ancient software. As I found out by looking at the source the problem is that the expansion ratio calculation ignores that in the shifting performance calculation the molar mass of the combustion products shifts between the throat and the exit. I'm gonna fix this and will look into making this change available as unmodified as possible (I adapted the version I used quite heavily in order to compile under c++).

As soon as I got better data I will update this answer.


Update: I fixed the bug and corrected the data.

The fix is rather easy:

ex->performance.ae_at =
    (ex->properties.T * t->properties.P * t->performance.Isp) /
    (t->properties.T * ex->properties.P * ex->performance.Isp);

has to be replaced with

ex->performance.ae_at =
    (ex->properties.T * ex->itn.n * t->properties.P * t->performance.Isp) /
    (t->properties.T * t->itn.n * ex->properties.P * ex->performance.Isp);

twice in shifting_performance() (file: performance.c). Not sure were to publish this best as the software is mostly unmaintained.

| improve this answer | |
$\endgroup$
  • $\begingroup$ Great analysis! $\endgroup$ – Organic Marble Sep 17 '19 at 16:17
1
$\begingroup$

There's a very good picture of the engine on the Mitsubishi web site:

enter image description here

and conveniently the length is given on that page as 2.765 m. Taking the proportions of the image and the known length, I estimate the diameter is 1.5 m, which is in fair agreement with Christoph's estimate of 1.7 m.

| improve this answer | |
$\endgroup$
1
$\begingroup$

Quite great answers about reverse engineering data from available data resources. I found this paper after hours of searching for data on LE-5B engines and Christoph, your calculation is spot on. The max diameter which i assume is of the nozzle end is 1700 mm which matches your calculations correctly. This is just validation for the various estimates put out as answers, and supplements them

Source: https://www.sciencedirect.com/science/article/abs/pii/S0094576501001655

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.