# Determine Heat Transfer and Mass Flow Rate in a Ramjet Combustion Chamber

This example shows how to use Aerospace Toolbox functions to determine heat transfer and mass flow rate in a ramjet combustion chamber.

### The Ramjet Engine

When calculating the thrust of a ramjet engine, optimize the amount of heat added and the mass flow rate through an air breathing engine. This optimization is important because the thrust generated by the engine is governed by these parameters. The ramjet thrust equation is the following:

$$Thrust={\underset{}{\overset{\dot{}}{m}}}_{exit}{V}_{exit}-{\underset{}{\overset{\dot{}}{m}}}_{enter}{V}_{enter}+({p}_{exit}-{p}_{enter}){A}_{exit}$$

where

$$\underset{}{\overset{\dot{}}{m}}=Mass\phantom{\rule{0.2777777777777778em}{0ex}}flow\phantom{\rule{0.2777777777777778em}{0ex}}rate\phantom{\rule{0.2777777777777778em}{0ex}}[kg/s]$$

$$V=Velocity\phantom{\rule{0.2777777777777778em}{0ex}}[m/s]$$

$$p=pressure\phantom{\rule{0.2777777777777778em}{0ex}}[kPa]$$

$$A=Cross-sectional\phantom{\rule{0.2777777777777778em}{0ex}}area\phantom{\rule{0.2777777777777778em}{0ex}}[{m}^{2}]$$

In the ramjet thrust equation, the subscripts denote the location of the parameter.

enter - Denotes the entrance of the entire ramjet.

exit - Denotes the exit of the ramjet engine.

inlet - Used for the beginning of the combustion chamber.

outlet - Used for the end of the combustion chamber.

This difference is illustrated in the following figure, (RJ) stands for the entire ramjet engine and (CC) refers to the combustion chamber.

ramjetPicture = plotRamjetSchematic;

Note, the thrust equation directly takes into account the mass flow rate. Heat addition correlates to higher exit velocity from the energy equation; higher exit velocity means more thrust. Modeling the ramjet combustion chamber as a constant area duct where heat addition is the main driver for the change in the flow variables enables the use of Rayleigh line flow principles.

### Problem Definition

This section describes the problem to be solved. It also provides necessary equations and known values.

After a series of shock waves, flow enters the combustion at a velocity of 100 m/s and static temperature of 400K. We want to:

Maximize the amount of heat added in the combustion chamber without decreasing the mass flow rate.

Calculate the fuel-air ratio associated with the maximum allowable heat added.

The heating value of the fuel is 40 megajoules per kilogram and the mass of the fuel is negligible compared to the mass of the air. We assume that the working fluid behaves like a perfect gas with constant specific heat ratio and specific heat at constant pressure given as:

$$\gamma =k=Specific\phantom{\rule{0.2777777777777778em}{0ex}}heat\phantom{\rule{0.2777777777777778em}{0ex}}ratio=1.4\phantom{\rule{0.2777777777777778em}{0ex}}[dimensionless]$$

$${c}_{p}=Specific\phantom{\rule{0.2777777777777778em}{0ex}}heat\phantom{\rule{0.2777777777777778em}{0ex}}at\phantom{\rule{0.2777777777777778em}{0ex}}constant\phantom{\rule{0.2777777777777778em}{0ex}}pressure=1.004\phantom{\rule{0.2777777777777778em}{0ex}}[kJ/(kgK)]$$

Given data for problem is listed below.

inletVelocity = 100; % Velocity of fluid at combustor intake [m/s] inletTemperature = 400; % Temperature of fluid at combustor intake [K] heatingValue = 40e+03; % Heating value of the fuel [kJ/kg] k = 1.4; % Specific heat ratio [dimensionless] cp = 1.004; % Specific heat at constant pressure [kJ/(kg*K)]

Because the fluid is air, it also has the following gas constant:

`R = 287; % Gas constant of air [J/(kg*K)]`

Therefore, the speed of sound is:

`speedOfSound = sqrt(k * R * inletTemperature); % [m/s]`

The inlet Mach number is

`inletMach = inletVelocity/speedOfSound; % [dimensionless]`

### Solving for the Stagnation Quantities and Reference Values

To apply the energy equation to find the heat transfer rate, calculate the stagnation temperature at the inlet. Use isentropic flow ratios and the static temperature at that point to calculate this temperature. The `flowisentropic`

function calculates the ratio of the static temperature to the total (stagnation) temperature.

$$\frac{{T}_{inlet}}{{T}_{{t}_{inlet}}}=inletTempRatio$$

where

$${T}_{{t}_{inlet}}=Total\phantom{\rule{0.2777777777777778em}{0ex}}temperature\phantom{\rule{0.2777777777777778em}{0ex}}(at\phantom{\rule{0.2777777777777778em}{0ex}}the\phantom{\rule{0.2777777777777778em}{0ex}}inlet)$$

[~, inletTempRatio, inletPresRatio] = flowisentropic(k, inletMach);

With the temperature ratio at the inlet, calculate the total temperature at the inlet. Be careful. Note that the form in which we need the temperature ratio is inverted from the form as given in the `flowisentropic`

function.

$${T}_{{t}_{inlet}}={T}_{inlet}\frac{{T}_{{t}_{inlet}}}{{T}_{inlet}}$$

inletTotalTemp = inletTemperature / inletTempRatio;

Use the energy equation to describe the flow in the combustion chamber:

$$\underset{}{\overset{\dot{}}{q}}={\underset{}{\overset{\dot{}}{m}}}_{air}{c}_{p}({T}_{{t}_{outlet}}-{T}_{{t}_{inlet}})={\underset{}{\overset{\dot{}}{m}}}_{fuel}*heatingValue$$

where

$$\underset{}{\overset{\dot{}}{q}}=rate\phantom{\rule{0.2777777777777778em}{0ex}}of\phantom{\rule{0.2777777777777778em}{0ex}}heat\phantom{\rule{0.2777777777777778em}{0ex}}transfer\phantom{\rule{0.2777777777777778em}{0ex}}[kW]$$

To maximize the rate of heat transfer, the stagnation temperature at the outlet station must be the reference stagnation temperature:

$${T}_{{t}_{outlet}}={T}_{t}^{*}$$

Use the `flowrayleigh`

function to calculate the total temperature ratio at the inlet. After this, you can calculate the reference total temperature.

[~,~,~,~,~,totalTempRatio] = flowrayleigh(k, inletMach);

In this equation, note that this ratio is found by the function as the local value over the reference value.

$$\frac{{T}_{{t}_{inlet}}}{{T}_{t}^{*}}=totalTempRatio$$

Now calculate the reference total temperature. Note that the total temperature ratio has been inverted to allow the proper cancellation of terms.

$${T}_{t}^{*}={T}_{{t}_{inlet}}\frac{{T}_{t}^{*}}{{T}_{{t}_{inlet}}}$$

inletTotalTempRef = inletTotalTemp / totalTempRatio;

### Calculating the Fuel-to-Air Ratio and Maximum Heat Added

Calculate the fuel-to-air ratio by rearranging the energy equation.

$$\frac{{\underset{}{\overset{\dot{}}{m}}}_{fuel}}{{\underset{}{\overset{\dot{}}{m}}}_{air}}=\frac{{c}_{p}({T}_{t}^{*}-{T}_{{t}_{inlet}})}{heatingValue}$$

fuelAirRatio = cp * (inletTotalTempRef - inletTotalTemp) / heatingValue

fuelAirRatio = 0.0296

The maximum heat added is:

$${q}_{max}={c}_{p}*({T}_{t}^{*}-{T}_{{t}_{inlet}})$$

heatMax = cp * (inletTotalTempRef - inletTotalTemp)

heatMax = 1.1826e+03

### Accounting for an Increase in the Fuel-Air Ratio

Consider the case where there is a 10% increase in the fuel-air ratio. Calculate how much the mass flow rate decreases with a 10% increase in fuel-to-air ratio, holding the stagnation temperature and pressure constant. The new fuel-air ratio is:

fuelAirRatio10 = 1.1 * fuelAirRatio;

Note, any variable that ends with "10" indicates that the given value is related to the 10% increase in fuel-to-air ratio. Rearrange the energy equation to calculate the difference in total temperatures from the inlet to the outlet of the combustion chamber:

$${T}_{{t}_{outlet}}-{T}_{{t}_{inlet}}=\frac{\frac{{\underset{}{\overset{\dot{}}{m}}}_{fuel}}{{\underset{}{\overset{\dot{}}{m}}}_{air}}*heatingValue}{{c}_{p}}$$

totalTempDiff = fuelAirRatio10 * heatingValue / cp;

The maximum heating condition is where the flow is choked at the outlet:

$${T}_{{t}_{outlet}}={T}_{t}^{*}$$

Therefore, inlet reference total temperature and the ratio of total temperature to the reference value are:

$${T}_{{t}_{inlet}}^{*}={T}_{{t}_{outlet}}-{T}_{{t}_{inlet}}+{T}_{t}^{*}$$

$$\frac{{T}_{{t}_{inlet}}}{{T}_{{t}_{inlet}}^{*}}=totalTempRatio$$

inletTotalTempRef10 = totalTempDiff + inletTotalTemp; totalTempRatio10 = inletTotalTemp / inletTotalTempRef10;

### Calculating the Decrease in Mass Flow Rate

Given the total temperature ratio, `flowrayleigh`

calculates the Mach number at the inlet of the combustion chamber:

`inletMach10 = flowrayleigh(k, totalTempRatio10, 'totaltsub');`

In this equation, the string input causes the function to use the subsonic total temperature ratio input mode. We know that the flow will be subsonic entering the combustion chamber because the flow will have gone through several shock waves leading up to the combustion chamber. With this Mach number at the inlet, use `flowisentropic`

to find the isentropic temperature ratio and pressure ratio at the inlet:

[~, inletTempRatio10, inletPresRatio10] = flowisentropic(k, inletMach10);

The static temperature at the inlet is:

inletTemperature10 = inletTotalTemp * inletTempRatio10;

From the equation of state, the mass flow rate can be written as:

$$\underset{}{\overset{\dot{}}{m}}=\frac{p}{RT}A(M\sqrt{\gamma RT})$$

With a 10% fuel-air ratio increase, relate a ratio showing a decrease in mass flow from a 10% increase in the mass flow rate to the ratio of decreasing Mach number. The increase in pressure ratio contributes to increasing the mass flow rate, but not as much as the decrease in Mach number decreases the mass flow rate. All other variables are constant between the two cases.

$$\frac{{\underset{}{\overset{\dot{}}{m}}}_{10}}{\underset{}{\overset{\dot{}}{m}}}=\frac{{M}_{10}}{M}\frac{{\left(\phantom{\rule{-0.16666666666666666em}{0ex}}\frac{{p}_{inlet}}{{p}_{{t}_{inlet}}}\phantom{\rule{-0.16666666666666666em}{0ex}}\right)}_{\phantom{\rule{-0.16666666666666666em}{0ex}}\phantom{\rule{-0.16666666666666666em}{0ex}}10}}{\frac{{p}_{inlet}}{{p}_{{t}_{inlet}}}}$$

massFlowRateRatio = inletMach10 / inletMach * inletPresRatio10 / inletPresRatio;

This ratio represents the percentage of the mass flow rate of the case with a 10% increase in fuel-to-air ratio. It uses the original mass flow rate as a whole. The percentage decrease in mass flow rate is just one minus the above ratio (times 100):

`percentageDecrease = ( 1 - massFlowRateRatio ) * 100 % [percent]`

percentageDecrease = 3.7665

These results show that adding fuel to the fuel-air mixture decreases the mass flow rate, making the thrust decrease. As a result, once a certain amount of fuel is added to the combustion chamber, adding more produces an inefficient result. Therefore, preemptive calculations such as these help you maximize fuel efficiency around the design conditions of an engine.

### Reference

[1] James, J. E. A., "Gas Dynamics, Second Edition", Allyn and Bacon, Inc, Boston, 1984.