# Question about hybrid propulsion :

#### mael

##### Well-Known Member
Hello everyone I would love to gather your point of view on a the following question about hybrid propulsion:

How to resolve the first mathematical problem ?

We know that the expression of the thrust of the rocket equals : F= m° * Ve

Now to get both of those variable we need, I guess, the regression rate but I don't know where and how to use it.
Some elements of the regression rate must (so have been) experimentaly determined if I got it well.
I think that by knowing the ratio oxydizer/fuel needed, we can manipulate the equation of the regression rate and obtain the mass flow rate ?

Once we'll have m⁰ we'll be able to calculate Ve necessary to lift the weigh we will have choosed.

This will help me to design of the nozzle I think ?

As you can see I just discovered the mathematical environment of hybrid propulsion and I'm sure about nothing haha, but I'm really here to learn,
So every feedbacks are welcome !

Great day !

#### LayNLow

##### Member
Find a copy of Rocket Propulsion by George Sutton, and it’ll have everything you need in it.
You’re missing quite a few steps there. But you can determine the nozzle area ratio by either (well there are many ways, these two are likely what you’ll do) a pressure ratio, or a Mach number for the mass flow rate exiting the motor.

#### rocket_troy

##### Well-Known Member
Thrust only equals ve * mdot if the the nozzle is optimally expanded to the ambient conditions. It's more correct to say that F = c * Mdot (where c is the *effective* exhaust velocity) ie. it allows for a mismatching of Pe and Pa. So, it's the same as ve * mdot + (Pe-Pa)*Ae

You sound like you're doing things backwards. You generally start with Ve and design for an Mdot to achieve your mission objectives.

Your Mdot will obviously be from the combination of your oxidizer flow rate and fuel flow rate (resulting from grain regression)
There are countless equations for fuel regression for hybrids, but the one most people use is based on the common regression equations used for solids ie. where in solids r = a p^n or r = c + a p^n, for hybrids, r = a Go^n where Go is the oxidizer flux rate ie. oxidizer flow rate divided by the port area and where "a" and "n" are internal ballistic parameters given to each specific fuel typically "a" being the burn rate coefficient and "n" being the exponent.

TP

#### mael

##### Well-Known Member
Find a copy of Rocket Propulsion by George Sutton, and it’ll have everything you need in it.
You’re missing quite a few steps there. But you can determine the nozzle area ratio by either (well there are many ways, these two are likely what you’ll do) a pressure ratio, or a Mach number for the mass flow rate exiting the motor.
Done, indeed it seems pretty interesting !

#### mael

##### Well-Known Member
Thrust only equals ve * mdot if the the nozzle is optimally expanded to the ambient conditions. It's more correct to say that F = c * Mdot (where c is the *effective* exhaust velocity) ie. it allows for a mismatching of Pe and Pa. So, it's the same as ve * mdot + (Pe-Pa)*Ae

You sound like you're doing things backwards. You generally start with Ve and design for an Mdot to achieve your mission objectives.

Your Mdot will obviously be from the combination of your oxidizer flow rate and fuel flow rate (resulting from grain regression)
There are countless equations for fuel regression for hybrids, but the one most people use is based on the common regression equations used for solids ie. where in solids r = a p^n or r = c + a p^n, for hybrids, r = a Go^n where Go is the oxidizer flux rate ie. oxidizer flow rate divided by the port area and where "a" and "n" are internal ballistic parameters given to each specific fuel typically "a" being the burn rate coefficient and "n" being the exponent.

TP

#### mael

##### Well-Known Member
Thrust only equals ve * mdot if the the nozzle is optimally expanded to the ambient conditions. It's more correct to say that F = c * Mdot (where c is the *effective* exhaust velocity) ie. it allows for a mismatching of Pe and Pa. So, it's the same as ve * mdot + (Pe-Pa)*Ae

You sound like you're doing things backwards. You generally start with Ve and design for an Mdot to achieve your mission objectives.

Your Mdot will obviously be from the combination of your oxidizer flow rate and fuel flow rate (resulting from grain regression)
There are countless equations for fuel regression for hybrids, but the one most people use is based on the common regression equations used for solids ie. where in solids r = a p^n or r = c + a p^n, for hybrids, r = a Go^n where Go is the oxidizer flux rate ie. oxidizer flow rate divided by the port area and where "a" and "n" are internal ballistic parameters given to each specific fuel typically "a" being the burn rate coefficient and "n" being the exponent.

TP
With the risk of looking stupid do we have : mdottotal kg/s= mdotoxy kg/s + (regressionrate m/s * ro kg/m³ * A m²)

#### rocket_troy

##### Well-Known Member
Well, the units you choose depend on a few things (obviously metric/imperial) – if you want to minimise conversion coefficients and whatnot and keep the equations clean, then you might be using some weird units in places.

Below is a test case borrowed from Paul Kelly (from >2 decades years ago) with various descriptive, value and unit changes in places by myself:

Let’s start with a simple N2O hybrid for our design example. N2O hybrids will provide much of their energy from the decomposition of the N2O molecule and as a result will provide a nice performance plateau from O:F ratios between 5:1 to 10:1 or thereabouts. That’s where I was coming from with my comment regarding settling on a Ve or Isp 1st and designing the Mdot to satisfy your propulsion objectives. True oxidizers like LOX won't provide such a plateau, so significant more care is required designing for the Isp or or c or Ve whichever units you're basing your propulsive performance off.

Small hybrid motor with PVC fuel grain and liquid N2O
ISP=180s (or c = 1764m/s)
Pc=2.1 MPa
Average thrust=1000N
Burn Time=5s
Expansion ratio=4

You don’t need to worry too much about expansion ratios if you’re (a) running the Pc<2.5 Mpa and (b) keeping a low altitude. Let’s assume for this example we’re satisfying both.

Okay, so starting with the oxidizer requirements – we have our total impulse (1000N avg F x 5 sec = 5000Ns)
Now this is where the units start getting a little bit weird, for equation simplicity sake. We’ll use Newtons a lot for things like mass quantities.
If we divide our total impulse by our Isp (Isp is c/g) we get 5000Ns/180s = 27.78N of propellant required

Working on an O:F of 5 that provides us with 27.78 * (5/(5+1) = 23.15N of N2O
Designing a tank vent to maintain tank temperature of, say, 20 deg C will provide us with a target N2O density of =-2.90771182972311E-09*20^6 - 6.0312570866609E-07*20^5 - 0.0000432532975736404*20^4 - 0.00154440495997047*20^3 - 0.05881035796875*20^2 - 6.04812419669871*20 + 909.705602212855
= 743.83 Kg/M^3 or /1000 = 0.744 g/cc

So our bare minimum tank volume requirement would be 23.15/9.8/0.744 = 3.175 L tank
We should allow for a bit or ullage for expansion and perhaps ignition spikes but more importantly for the tank to be pressurised when the last bit of liquid is being squirted out the injectors so it takes:
Pt*V/Pa/22.4*M g of N2O to pressurise our tank where:
Pt=Tank Pressure (5.14 Mpa)
V=Tank Volume (3.5L)
Pa=Ambient Pressure (0.1 Mpa)
M=Molar Mass of gas (44g)
= 5.14*3.5/22.4/44 = 353g which takes up another 353/0.744 = 475cc of tank volume so I’d probably keep the 3.5L but maybe a touch more would be optimal. Obviously there’ll be boil off and whatnot too that’s also contributing to the tank pressure-volume.

Working downwards we come to the injector.
We need to flow 2.36Kg of N2O in 5 seconds that's a MOXdot of 0.472Kg/s

The formula for flow across a simple injector is:
MOXdot=Cfl*Ai*SQRT(2*Pox*DeltaP)

Where:
MOXdot=Oxidiser flow rate (Kg/s)
Cfl=Coefficient of flow for that injector(0.5 is a good guess for biphase flow)
Ai = Area of the injector (m^2)
Pox=The density of NOX in Kg/M^3 (744)
DeltaP =The pressure drop across the injector ie 5.14-2.1Mpa(3040000Pa)

Rearranging for Ai
Ai=Moxdot/(Cfl*sqrt(2*Pox*DeltaP)))

Plugging in our numbers:
Ai=0.472/(0.5*sqrt(2*744*3040000)))
=0.00001404m^2
or a diameter of sqrt(0.00001404 / pi) * 2 * 1000 = 4.23mm

Now for the fuel grain and what’s required based on the regression expected. This is where we utilise that regression equation mentioned earlier:

r= regression rate.
a= Burn rate coefficient (dimensionless – arguably - at least when you try to convert burn rate units)
n= Burn rate exponent (dimensionless - definitely)
Go= The oxidiser Flux rate (N/m^2/s) – again the units get weird

Designing the core/port around a reasonable value for Go of 4500 N/m^2/s we get a port area of:

4.63/4500=0.00103m^2 (4.63 is Moxdot in N)
or an approx port diameter of 36mm

“a” for PVC is about 0.0052 and “n” is about 0.65 (this will vary with grade and colour)
Thus r=0.0052*4500^0.65=1.23mm/s

So, we have a 3.6cm diam port, regressing at 0.123cm/sec. Given the density of PVC is 1.7, This will produce:
3.6*pi()*0.123*1.7=2.36g of fuel for every centimetre of fuel grain we have.

We know Moxdot is 472g/s and O/F=5:1 so we work out that we need 472/5=94.4g/s of Fuel

From which we derive a grain length of 94.4/2.36=40cm

Now, for most small HPR class applications, we can assume a constant regression for the purposes of determining the web thickness. Of course you can make a spreadsheet to model it more iteratively/dynamically if you wish.
So, 1.23mm/s * 5sec = 0.615cm of web required as a minimum.
That provides us with a min grain diameter of 3.6+(2*0.615) = 4.83cm

Nozzle:

F=Cf*At*Pc
F=Thrust (N)
Cf=Coefficient of thrust (worked out from isentropic equations) or assuming 1.4-1.5 for our Pc : Pa ratio
Pc=Chamber Pressure in Pascals

At = 1000/(1.4*2100000) = 0.00034m^2
a diameter of sqrt(0.00034 / pi) * 2 * 1000 = 20.8mm

So, we've nominated above using an expansion ratio of 4, which provides us with an exit diameter of sqrt(0.00034 *4 / pi) * 2 * 1000 = 41.6mm
note: at sea level, an expansion ratio of 4 would be ideally suited to a Pc of around 3.4 IIRC but by the time you factor in all the flight dynamics and avg altitude of operation, 4 is probably reasonable for 2.1mpa

TP

Ps: I’ve just typed all this out without checking much of it so verify the numbers and all

Last edited:

#### tfish

##### Well-Known Member
Well, the units you choose depend on a few things (obviously metric/imperial) – if you want to minimise conversion coefficients and whatnot and keep the equations clean, then you might be using some weird units in places.

Below is a test case borrowed from Paul Kelly (from >2 decades years ago)

Paul Kelly..sure is a name from the past..then you lost me in the math.

I'll stick with the Pitch Black calculations.

Good to see you in here..from time to time...

Tony