 ## Saturday, 28 October 2017

### More Heat Transfer

Here I will attempt to calculate the heat transfer from the combustion gases to the coolant of the cylindrical combustion chamber of the proposed rocket engine. This will allow further calculation of the minimum coolant flow rate required to safely remove this heat.

I am not going to give exhaustive details of the calculation method, as this was comprehensively covered in my previous post dealing with the heat transfer in the nozzle section.

In order to proceed with the heat transfer investigation, it is necessary to calculate the chamber heat transfer coefficient, or hg. RPA gives all of the relevant variables to do this apart from one, this being the velocity of the combustion gases in the chamber.

This velocity is generally defined as “low” in the literature. How low is low though, bearing in mind we are comparing it to sonic and supersonic velocities in the nozzle and divergence? RPA gives the total mass flow rate through the engine as 4.4 kg/sec. Given that the chamber area is 0.0167 metres square and the density of the gases at this point is 2.5 kg per cubic metre, this gives a velocity of about 103.5 m/sec.

So:-

Vchamber = 103.5 m/sec

Putting this and the other pertinent variables into the modified Colburn equation gives a value for hg in the chamber of:-

hg = 1220 W/m^2 K

Now we can work out what the heat transfer rate, or q is. Recall that to do this we specify a temperature that we want the hot gas wall to be at, given the materials thermal and mechanical properties. We will go with 420K as before. We now subtract this from the adiabatic wall temperature and then multiply by hg.

The adiabatic wall temperature in the chamber is derived by multiplying the stagnation temperature by a so called recovery factor, or r. Experiment shows that for practical purposes r = 0.9. RPA gives the stagnation temperature in the chamber as 2170K so the adiabatic wall is at 1953K.

Substituting these values into the relevant equation gives a figure for q:-

q = 1.87MW/m^2 (1.14 BTU/sec/in^2 deg F in old money)

Having gained a figure for q we can go on and calculate the temperature of the cooled wall, or Twc, by finding the heat conducted through the aluminium. To do this we require the thickness of the wall and the material conductivity. Recalling the wall is 3mm thick and the aluminium conductivity is 160 W/m^2 K, we arrive at a figure for Twc of:-

Twc = 385K (112 deg C)

Now we can go on to find the maximum heat transferred into the coolant or Q. This is done by multiplying q by the heat transfer area, this being the outside area of the cylindrical chamber portion. This area will be 0.005 m^2. Hence the total heat transferred, Q, becomes:-

Q = 9.35 kW

Now we can work out how much coolant flow is required to dissipate this much heat. To do this we need to know the specific heat of the coolant at it’s mean temperature within the chamber cooling jacket. This so called film temperature, or Tfilm, is evaluated by halving the sum of the coolant input temperature and Twc. Remember of course that the coolant will have already done a job of work cooling the nozzle before it enters the chamber. We know from the previous calculation that the film temperature in the nozzle was 306K or 33 deg C. If we assume a temperature rise of 30 degrees, which is perfectly within the range of coolant flow that we have, then we have a chamber inlet temperature of 63 deg C or 336K. This gives a Tfilm in the chamber of:-

Tfilm = 360K (85 deg C)

The specific heat of the fuel at this temperature is 3360 J/kgK.

Let us again allow a temperature rise of 30 degrees for the coolant in the chamber. This gives a coolant i.e. fuel temperature of 120 deg C on entering the injector. We already know that the boiling point of our fuel, which is at elevated pressure, is above this.

To calculate the minimum amount of coolant flow, or mdot, required under the above conditions we divide Q by the product of the specific heat of the coolant and the temperature increase desired.

This gives a coolant flow rate of:-

mdot = 0.093 kg/sec

As was previously mentioned, the fuel/coolant flow rate will be in excess of this figure and so it seems highly likely that the chamber will be able to survive the heat load imposed upon it by the combustion gases, given the design fuel flow rate.