- #1
MNZ
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Hello,
I have been reading this forum for quite some time now and I find it to be very helpful. Today, I am going to need the help from all of you generous people in the forum to assist me in the following question:
My colleagues and I designed a downdraft gasifier that is able to deliver 50kW power at 70% efficiency. In order to achieve this efficiency, assuming that all 30% loss (~22kw) is due to heat loss, an insulating layer is therefore needed to maintain this losses. Our insulating material was chosen to be refractory cement, and assuming that refractory cement has similar properties with that of fire brick, then we have its thermal conductivity, k to be 0.69 W/m.K.
To ease the calculation, the arrangement of the gasifier is assumed to be in this manner: a reactor box encased in an insulation box encased in an external box (to hold the refractory cement insulation box intact in between it and the reactor box). The dimension of the insulation box is as the following: 1.0m x 0.44m x 0.44m. Assume that the top and the bottom of the box is not covered but only the sides.
From here, the thickness of the insulator lining can be determined using this following equation: q = kAT/d, where A=area,sq.m; T=temperature difference across the insulator surfaces, Kelvin; d= thickness, m; k = thermal conductivity, W/mK; and q = heat loss, W.
Therefore the required thickness of the insulator, assuming that gasification temperature is 800 degree Celsius and ambient temperature is 27 degree Celsius:-
d = kAT/q = (0.69 x (1.0 x 0.44 x 4 sides) x (1073-300))/22,000 = 0.043m
The thing is, due to insufficient fund and whatnots, we only managed to get the refractory lining manufactured at 0.012m. And if this thickness is taken into consideration, the heat loss will be:-
q = kAT/d = (0.69 x (1.0 x 0.44 x 4 sides) x (1073 - 300))/0.012 = 78,227.6W = 78.23kW
This means we have a total loss for all the generated heat inside the gasifier, since at 100% efficiency the gasifier will produce a total of 71.43kW. Which means the gasifier now is more or less a very efficient fireplace instead of what it was supposed to be.
My question is, what went wrong? I am pretty sure my calculation was wrong somewhere but I couldn't quite pointed it out. My colleagues and I (we are both doing MSc in ME) have been very nervous in the previous days since this error we found was not justifiable, and our supervisor was constantly asking about the real insulation capability of the refractory material. Hope those who might have the knowledge in this field could assist me on this one. And with that, I thank you very much in advance.
I have been reading this forum for quite some time now and I find it to be very helpful. Today, I am going to need the help from all of you generous people in the forum to assist me in the following question:
My colleagues and I designed a downdraft gasifier that is able to deliver 50kW power at 70% efficiency. In order to achieve this efficiency, assuming that all 30% loss (~22kw) is due to heat loss, an insulating layer is therefore needed to maintain this losses. Our insulating material was chosen to be refractory cement, and assuming that refractory cement has similar properties with that of fire brick, then we have its thermal conductivity, k to be 0.69 W/m.K.
To ease the calculation, the arrangement of the gasifier is assumed to be in this manner: a reactor box encased in an insulation box encased in an external box (to hold the refractory cement insulation box intact in between it and the reactor box). The dimension of the insulation box is as the following: 1.0m x 0.44m x 0.44m. Assume that the top and the bottom of the box is not covered but only the sides.
From here, the thickness of the insulator lining can be determined using this following equation: q = kAT/d, where A=area,sq.m; T=temperature difference across the insulator surfaces, Kelvin; d= thickness, m; k = thermal conductivity, W/mK; and q = heat loss, W.
Therefore the required thickness of the insulator, assuming that gasification temperature is 800 degree Celsius and ambient temperature is 27 degree Celsius:-
d = kAT/q = (0.69 x (1.0 x 0.44 x 4 sides) x (1073-300))/22,000 = 0.043m
The thing is, due to insufficient fund and whatnots, we only managed to get the refractory lining manufactured at 0.012m. And if this thickness is taken into consideration, the heat loss will be:-
q = kAT/d = (0.69 x (1.0 x 0.44 x 4 sides) x (1073 - 300))/0.012 = 78,227.6W = 78.23kW
This means we have a total loss for all the generated heat inside the gasifier, since at 100% efficiency the gasifier will produce a total of 71.43kW. Which means the gasifier now is more or less a very efficient fireplace instead of what it was supposed to be.
My question is, what went wrong? I am pretty sure my calculation was wrong somewhere but I couldn't quite pointed it out. My colleagues and I (we are both doing MSc in ME) have been very nervous in the previous days since this error we found was not justifiable, and our supervisor was constantly asking about the real insulation capability of the refractory material. Hope those who might have the knowledge in this field could assist me on this one. And with that, I thank you very much in advance.