Temperature Expectations and Other Thermal Considerations
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We still haven’t run the furnace in anger (this is not too far away) but now is a good time to consider what upper temperature we are likely to achieve. The maximum temperature we can expect in an air/LPG flame is about 1980°C at the tip of the combustion zone. So this is the upper temperature bound for our furnace which is entirely determined by the nature of the exothermic reaction between our fuel and the oxidizer, and the ambient conditions.
From a thermodynamic system perspective we are burning the fuel in air. The heat released from the reaction increases the temperature of the combustion products and the inert Nitrogen in the air. The heat content of the hot gases is now transferred to the furnace cavity and its charge. In this latter step heat flow is always from a higher to lower temperature unless we do work on the system (the First Law of Thermodynamics). The heat transfer in the furnace does not involve work. So the flame temperature is determined entirely by the combustion chemistry and ambient temperature of the fuel and air. We can estimate the maximum flame temperature.
For simplicity let’s assume that our LPG is entirely Propane (and not a mixture of Propane and Butane). The ideal combustion reactions is:
The Oxygen comes from air at a concentration of 21% v/v (23% w/w). The other major constituent of air is Nitrogen, with less than 5% of other inert gases, Carbon Dioxide and water vapour.
Let’s assume that air is composed of just Nitrogen and Oxygen and add the inert Nitrogen to our reaction:
The reaction is exothermic generating 48.6 MJ for every kilogram of Propane burnt with all reactants and products at the same temperature (nominally 20°C) and pressure (nominally 101.3 kPa).
Using molecular weights we can calculate the mass ratios of the reactants and products.
The heat produced is 48.6 MJ/kg x 0.044 kg = 2,138 kJ. A portion of this heat is radiated away from the flame. The exact quantity radiated depends on the gas velocity but we will estimate the radiant fraction as 20% so the convective fraction which heats the gases is 80% of 2,138 kJ = 1,711 kJ
Using the specific heat of the reaction products, and remembering that water is a liquid at STP and undergoes a phase change at 100°C, we can calculate the temperature rise of the reaction products.
The resulting calculated gas temperature is 1,944°C assuming an initial temperature of 20°C. Most of the heat produced by the reaction (almost 60%) is being used to heat the nitrogen. So the hottest part of the flame at the point of complete combustion will be just below 2,000°C. Other parts of the flame will be cooler due to incomplete combustion or entrainment of surrounding air.
This leads to three methods of increasing the furnace temperature. The first is to preheat the combustion air, the second is to dry it, and the third is to reduce or eliminate the Nitrogen by using enriched air (readily available as a SCUBA diving mix) or pure Oxygen. An Oxygen Propane flame will burn at a maximum temperature of approximately 2,800°C!
Other furnaces using LPG/air burner have reportedly been used to melt cast iron which suggests an achievable upper temperature of about 1,250°C. And we can reasonably expect to achieve this as demonstrated by the initial firing. This will be quite satisfactory for melting the intended range of non-ferrous metals and alloys.
We can estimate the thermal performance of the furnace by considering the conductivity of our refractory (about 0.3 W/m.K). The temperature profile through the furnace walls can be approximated by considering cylindrical 2 dimensional heat flow, ignoring the cavity lining the stainless steel form, and the convective heat transfer from the form to the surrounding air.
The temperature profile through the lid and the base (ignoring the exhaust port in the lid) are approximated as linear (ignoring the transition from the walls to the base and the lid). The temperature profiles through the furnace walls, lid and base are shown in Figure 1.
Figure 1. Temperature Profile through Furnace Insulation
We can also approximate the steady-state conduction heat loss. At an internal temperature of 1,250°C and an ambient temperature of 10°C this is about 120 W through the furnace walls and about 59 W through the lid and base. So provided that we apply at least 190 W from the burner, the furnace cavity (and its charge) should eventually heat up to 1,250°C.
If we apply more heat the upper temperature will not increase but the rate at which the furnace heats up will.
This raises the question of how much heat our LPG burner can deliver to the furnace. Let’s assume that we can achieve ideal combustion with our burner. We can use the orifice equation and the vapour pressure of Propane to estimate our gas mass flow rate as approximately 0.02 g/s with an 0.5 mm diameter orifice and fully open needle valve.
Assuming ideal combustion we will generate about 10 kW of heat at this fuel mass flow rate. Our 8 kg LPG cylinder will provide slightly more than 100 hours of use before requiring a refill. With the 0.6 mm diameter nozzle we can increase our heat output to over 14 kW at full flow, albeit with a reduced cylinder life. As it will transpire I cannot use the burner at full gas flow due to the restricted 15 mm clearance between the crucible and the furnace walls. I figure that the maximum usable gas flow is about 0.01 g/s or less with a heat output of about 5 kW.
Interestingly the original Primus burner has a calculated heat output of about 1.6 kW. So our new burner is certainly an improvement for this application.
While we are messing with performance approximations now we can theorize about our exhaust port diameter. The burner tube ID is 15.3 mm in diameter and the fuel air mix temperature is about 20°C (~293 K). The furnace temperature is about 1,250°C (~1,523 K). Like most materials, gases expand when heated under constant (atmospheric) pressure. Note that there are are some interesting exceptions such as water between 0 and 4°C where it actually contracts as it heats!
We can use the Ideal Gas Equation, PV = nRT, to estimate the exhaust port velocity (assuming constant pressure and Ideal gases) which should be lower than the intake velocity. With our assumptions, the Ideal Gas Equation can be reduced to a ratio of squared diameters and temperature (in degrees Kelvin, K):
D2 = sqrt(D12 * T2 / T1)
= sqrt(15.32 * 1,532 / 293
= 35 mm diameter
So our 55 mm exhaust port gas velocity should be less than a half of our intake velocity:
I’m comfortable with this.
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