2022-03-07, 11:22
  #241
Medlem
Citat:
Ursprungligen postat av pollenper
Hur funkar din uträkning?
En kropp, en timme.
Två kroppar, två timmar
Det är din tes, antar jag.

Två kroppar som bränns samtidigt, under en timme.
Blir genomsnittstiden en halv timme per styck eller fortfarande två timmar.?

Jag skulle påstå att den ugnen har kapaciteten två kroppar per timme,vad kallar du det?

Visa det matematiskt ('1+1=2 nivån' räcker inte till förstår du).
Citera
2022-03-07, 11:32
  #242
Medlem
pollenpers avatar
Citat:
Ursprungligen postat av Stor-Men-Snabb
Visa det matematiskt ('1+1=2 nivån' räcker inte till förstår du).
Var exemplet för komplicerat?

Antalet kroppar är integralen av brända kroppar per timme.

Edit
Du är fascinerande med din vurm för molekylbindningar.
Erfarenhet är en mycket god egenskap i ämnet.
Tekniken att återvinna värmen är viktigare än molekyler.

Räkna dina molekyler på en bra ugn och jämför siffrorna med en dålig ugn, blir det någon skillnad då?

Hur många kroppar per timme fixade mitt exempel?
__________________
Senast redigerad av pollenper 2022-03-07 kl. 11:38.
Citera
2022-03-07, 11:38
  #243
Medlem
Citat:
Ursprungligen postat av pollenper
Var exemplet för komplicerat?

Antalet kroppar är integralen av brända kroppar per timme.

Visa det då i en uträkning.
Citera
2022-03-07, 11:39
  #244
Medlem
pollenpers avatar
Citat:
Ursprungligen postat av Stor-Men-Snabb
Visa det då i en uträkning.
Jag editerade mitt inlägg.
Läs en gång till är du snäll.
Citera
2022-03-07, 18:01
  #245
Medlem
Skogsvattes avatar
Citat:
Ursprungligen postat av Stor-Men-Snabb
Visa det då i en uträkning.

Så här fungerar deras "uträkning": släng in fler kroppar i ugnen och dividera kremeringstiden med antalet kroppar. De anser att en kropp kunde kremeras på 20 minuter, vilket innebär att 5 kroppar i samma ugnsöppning blir 20/5 = 4 minuter per lik.

Denna flagrant antivetenskapliga metodologi är så klart fullständigt löjeväckande. Logistiska parametrar tas inte ens in i ekvationen, t.ex. det faktum att man omöjligen kunde introducera 5 lik staplade på varandra i ugnen, än mindre 8 eller 12.
Citera
2022-03-07, 21:41
  #246
Medlem
pollenpers avatar
Citat:
Ursprungligen postat av pollenper
Var exemplet för komplicerat?

Antalet kroppar är integralen av brända kroppar per timme.

Edit
Du är fascinerande med din vurm för molekylbindningar.
Erfarenhet är en mycket god egenskap i ämnet.
Tekniken att återvinna värmen är viktigare än molekyler.

Räkna dina molekyler på en bra ugn och jämför siffrorna med en dålig ugn, blir det någon skillnad då?

Hur många kroppar per timme fixade mitt exempel?

Citat:
Ursprungligen postat av Skogsvatte
Så här fungerar deras "uträkning": släng in fler kroppar i ugnen och dividera kremeringstiden med antalet kroppar. De anser att en kropp kunde kremeras på 20 minuter, vilket innebär att 5 kroppar i samma ugnsöppning blir 20/5 = 4 minuter per lik.

Denna flagrant antivetenskapliga metodologi är så klart fullständigt löjeväckande. Logistiska parametrar tas inte ens in i ekvationen, t.ex. det faktum att man omöjligen kunde introducera 5 lik staplade på varandra i ugnen, än mindre 8 eller 12.
visa oss en berökning av molekyler i en effekriv ugn jämfört med en usel ugn.

Är det samma värden på en grill utomhus som en effektiv ugn i en konternuerlig process?
Citera
2022-03-07, 22:13
  #247
Medlem
Skogsvattes avatar
Citat:
Ursprungligen postat av pollenper
visa oss en berökning av molekyler i en effekriv ugn jämfört med en usel ugn.

Är det samma värden på en grill utomhus som en effektiv ugn i en konternuerlig process?

Jag har redan visat relevanta värmetekniska uträkningar. De har emellertid mötts av tystnad.
The cremation of four to five corpses in one muffle within 20-25 minutes, or half an hour (or a little more than half an hour) is absurd on two counts: first of all because it took one hour to burn a single corpse and secondly because the time needed to burn multiple corpses at once would have extended the time necessary for each corpse well beyond one hour. In practice, however, such a procedure would have brought along insurmountable problems of thermal technology.

The necessary condition for carrying out a cremation is that the temperature of the muffle never drops below 600°C; otherwise there is no longer any incineration, but only carbonization of the corpse. A body of 70 kg contains some 45.5 kg of water. The heat of vaporization at 600°C of the water contained in three corpses is 3×45.5×[640+0.477× (500–20)] ≈ 118,500 kcal. It is known from experience that the process of evaporation took about half an hour. The loading rate of the grate of the triple-muffle furnace was about 70 kg/h of coke (two hearths with grate loadings of 35 kg/h each), hence the theoretical availability of heat over half an hour was (6,470×35=) 226,450 kcal. The effective availability was much lower because a large part of the heat generated in the gasifiers was lost. During evaporation, the major heat losses came from radiation and conduction, some 62,500 kcal/h at 800°C; at 600°C we may assume them to be 46,900 kcal/h or 23,450 kcal in half an hour, i.e. (23,450÷226,450×100=) 10.3%. To this we must add the heat loss through the smoke at 600°C: about 31.3% according to calculations; of uncombusted gases from the hearth: 4%; of uncombusted solids from the hearth: 3.1%. The efficiency of the furnace was thus (100–[10.3+31.3+4+3.1]=) 51.3%, the effective specific heat of combustion of the coke (6,470×0.513) ≈ 3,320 kcal/kg, which brings the effective heat supplied to the furnace over half an hour to (35×3,320) ≈ 116,200 kcal. To keep the furnace at 600°C, an additional heat contribution of (118,500–116,200=) 2,300 kcal was thus needed during that time: it could easily be supplied by the radiation from the muffle walls.

Let us now look at the case of the evaporation of the water contained in four corpses in each of the three muffles, 12 corpses altogether. The water content of the corpses is (45.5×12=) 546 kg: the heat of vaporization at 600°C is 546×[640+0.477×(500–20)] ≈ 474,500 kcal. The available heat input stays at 116,200 kcal in 30 minutes, hence the additional heat needed is (474,500–116,200=) 358,300 kcal or some 119,400 kcal per muffle.

We must now examine whether the radiation from the muffle walls could possibly supply this amount of heat. It is difficult to calculate the heat radiated by these walls and absorbed by the corpses, both for reasons of geometry and because of the continual cooling of the wall temperature. However, in a specific technical article, Professor Schläpfer, one of the major experts in cremation in Europe in the 1930s, does give us a reliable estimate of the heat radiated to a single corpse from the muffle walls at various temperatures. He published a chart, from which we may derive the data:

Kod:
Wall temp. [°C]              Heat flow, kcal/min
800                          1,400
700                          930
600                          600

The geometry changes somewhat when a hypothetical load of three corpses in one muffle is irradiated, but the surface-to-volume ratio of such a load is less favorable than that of a single corpse, because the corpses partly cover one another. Even if we leave this consideration aside, the amount of heat required for the evaporation of the water contained in three normal corpses, about 119,400 kcal, would require over three hours at a constant wall temperature of 600°C according to Schläpfer’s data. The wall temperature, however, would certainly not stay constant over such a long period of time, and conditions would quickly become very unfavorable, because, as shown by Schläpfer’s chart, the heat radiated by the walls drops sharply with a decrease in wall temperature.

In his discussion of a similar thermal problem, Kori writes:

"If the inner wall of the cremation chamber has a surface area of about 4 m², with a specific gravity of 2.1, a layer 5 cm thick would weigh about 420 kilograms. The specific heat of the fire clay is about 0.2. Hence, if this layer could supply its total heat content sufficiently fast, only 200×0.2×420 = 16,800 kcal would have become available for an internal temperature dropping from 1,000 to 800°C. Actually, not even this would have been possible, because the brickwork does not release its accumulated heat as quickly as the [muffle] temperature drops."

The weight of the refractory brickwork of one muffle was about (5×1.5×200=) 1,500 kilograms. To compensate for the heat lost due to the evaporation of the water content of the corpses, each muffle would have had to contribute 119,400 kcal, corresponding to a decrease in the average temperature of the refractory brickwork of the muffle of about (119,400 kcal ÷ [0.2 kcal/kg/°C × 1,500 kg] ≈ 400°C. The effective amount of heat supplied to each muffle is therefore:

(3‚320 kcal/kg × 70 kg/h)(/3 × 60 min/h) ≈ 1,290 kcal/min

This corresponds to the supply of 119,400 kcal in (119,400 kcal ÷ 1,290 kcal/min) ≈ 92 minutes. I have only sketched the evaporation process, which is actually more complex, depending on further factors. But these factors apply in the same way both to single cremations and to the hypothetical cremation of several bodies at the same time. The enormous difference between the two set out above still applies. It proves not only that the simultaneous cremation of four bodies in half an hour was impossible, but also that not even the evaporation of the water they contained could have been brought about during that span of time. If assuming an average weight of 60 kg per body, the figures of the above calculations drop by a mere 15%, and the conclusions are basically the same.
Citera
2022-03-08, 05:22
  #248
Medlem
Citat:
Ursprungligen postat av Skogsvatte
Jag har redan visat relevanta värmetekniska uträkningar. De har emellertid mötts av tystnad.
The cremation of four to five corpses in one muffle within 20-25 minutes, or half an hour (or a little more than half an hour) is absurd on two counts: first of all because it took one hour to burn a single corpse and secondly because the time needed to burn multiple corpses at once would have extended the time necessary for each corpse well beyond one hour. In practice, however, such a procedure would have brought along insurmountable problems of thermal technology.

The necessary condition for carrying out a cremation is that the temperature of the muffle never drops below 600°C; otherwise there is no longer any incineration, but only carbonization of the corpse. A body of 70 kg contains some 45.5 kg of water. The heat of vaporization at 600°C of the water contained in three corpses is 3×45.5×[640+0.477× (500–20)] ≈ 118,500 kcal. It is known from experience that the process of evaporation took about half an hour. The loading rate of the grate of the triple-muffle furnace was about 70 kg/h of coke (two hearths with grate loadings of 35 kg/h each), hence the theoretical availability of heat over half an hour was (6,470×35=) 226,450 kcal. The effective availability was much lower because a large part of the heat generated in the gasifiers was lost. During evaporation, the major heat losses came from radiation and conduction, some 62,500 kcal/h at 800°C; at 600°C we may assume them to be 46,900 kcal/h or 23,450 kcal in half an hour, i.e. (23,450÷226,450×100=) 10.3%. To this we must add the heat loss through the smoke at 600°C: about 31.3% according to calculations; of uncombusted gases from the hearth: 4%; of uncombusted solids from the hearth: 3.1%. The efficiency of the furnace was thus (100–[10.3+31.3+4+3.1]=) 51.3%, the effective specific heat of combustion of the coke (6,470×0.513) ≈ 3,320 kcal/kg, which brings the effective heat supplied to the furnace over half an hour to (35×3,320) ≈ 116,200 kcal. To keep the furnace at 600°C, an additional heat contribution of (118,500–116,200=) 2,300 kcal was thus needed during that time: it could easily be supplied by the radiation from the muffle walls.

Let us now look at the case of the evaporation of the water contained in four corpses in each of the three muffles, 12 corpses altogether. The water content of the corpses is (45.5×12=) 546 kg: the heat of vaporization at 600°C is 546×[640+0.477×(500–20)] ≈ 474,500 kcal. The available heat input stays at 116,200 kcal in 30 minutes, hence the additional heat needed is (474,500–116,200=) 358,300 kcal or some 119,400 kcal per muffle.

We must now examine whether the radiation from the muffle walls could possibly supply this amount of heat. It is difficult to calculate the heat radiated by these walls and absorbed by the corpses, both for reasons of geometry and because of the continual cooling of the wall temperature. However, in a specific technical article, Professor Schläpfer, one of the major experts in cremation in Europe in the 1930s, does give us a reliable estimate of the heat radiated to a single corpse from the muffle walls at various temperatures. He published a chart, from which we may derive the data:

Kod:
Wall temp. [°C]              Heat flow, kcal/min
800                          1,400
700                          930
600                          600

The geometry changes somewhat when a hypothetical load of three corpses in one muffle is irradiated, but the surface-to-volume ratio of such a load is less favorable than that of a single corpse, because the corpses partly cover one another. Even if we leave this consideration aside, the amount of heat required for the evaporation of the water contained in three normal corpses, about 119,400 kcal, would require over three hours at a constant wall temperature of 600°C according to Schläpfer’s data. The wall temperature, however, would certainly not stay constant over such a long period of time, and conditions would quickly become very unfavorable, because, as shown by Schläpfer’s chart, the heat radiated by the walls drops sharply with a decrease in wall temperature.

In his discussion of a similar thermal problem, Kori writes:

"If the inner wall of the cremation chamber has a surface area of about 4 m², with a specific gravity of 2.1, a layer 5 cm thick would weigh about 420 kilograms. The specific heat of the fire clay is about 0.2. Hence, if this layer could supply its total heat content sufficiently fast, only 200×0.2×420 = 16,800 kcal would have become available for an internal temperature dropping from 1,000 to 800°C. Actually, not even this would have been possible, because the brickwork does not release its accumulated heat as quickly as the [muffle] temperature drops."

The weight of the refractory brickwork of one muffle was about (5×1.5×200=) 1,500 kilograms. To compensate for the heat lost due to the evaporation of the water content of the corpses, each muffle would have had to contribute 119,400 kcal, corresponding to a decrease in the average temperature of the refractory brickwork of the muffle of about (119,400 kcal ÷ [0.2 kcal/kg/°C × 1,500 kg] ≈ 400°C. The effective amount of heat supplied to each muffle is therefore:

(3‚320 kcal/kg × 70 kg/h)(/3 × 60 min/h) ≈ 1,290 kcal/min

This corresponds to the supply of 119,400 kcal in (119,400 kcal ÷ 1,290 kcal/min) ≈ 92 minutes. I have only sketched the evaporation process, which is actually more complex, depending on further factors. But these factors apply in the same way both to single cremations and to the hypothetical cremation of several bodies at the same time. The enormous difference between the two set out above still applies. It proves not only that the simultaneous cremation of four bodies in half an hour was impossible, but also that not even the evaporation of the water they contained could have been brought about during that span of time. If assuming an average weight of 60 kg per body, the figures of the above calculations drop by a mere 15%, and the conclusions are basically the same.

Bra inlägg!!

Pollenper ville se matematik och hintade själv om kunskap i det..
Citat:
Ursprungligen postat av pollenper
..
Antalet kroppar är integralen av brända kroppar per timme.
..

Kan vi hoppas på att pollenper nu efter att ha gått ner på specifika, detalj och hint om matematiskt kunnande håller sig kvar där efter fått svar på detaljnivå? ...eeeellleeeer kommer pollenper som vanligt gå upp flera abstraktionsnivåer och sidsteppa med en annan tramsig 'motfråga'?

..förintelseivrarna här är inte direkt en källa av kunskap och intellektuell hederlighet.., det har man ju fått visat för sig gång på gång på gång.
Citera
2022-03-08, 07:24
  #249
Medlem
pollenpers avatar
Citat:
Ursprungligen postat av Skogsvatte
Jag har redan visat relevanta värmetekniska uträkningar. De har emellertid mötts av tystnad.
The cremation of four to five corpses in one muffle within 20-25 minutes, or half an hour (or a little more than half an hour) is absurd on two counts: first of all because it took one hour to burn a single corpse and secondly because the time needed to burn multiple corpses at once would have extended the time necessary for each corpse well beyond one hour. In practice, however, such a procedure would have brought along insurmountable problems of thermal technology.

The necessary condition for carrying out a cremation is that the temperature of the muffle never drops below 600°C; otherwise there is no longer any incineration, but only carbonization of the corpse. A body of 70 kg contains some 45.5 kg of water. The heat of vaporization at 600°C of the water contained in three corpses is 3×45.5×[640+0.477× (500–20)] ≈ 118,500 kcal. It is known from experience that the process of evaporation took about half an hour. The loading rate of the grate of the triple-muffle furnace was about 70 kg/h of coke (two hearths with grate loadings of 35 kg/h each), hence the theoretical availability of heat over half an hour was (6,470×35=) 226,450 kcal. The effective availability was much lower because a large part of the heat generated in the gasifiers was lost. During evaporation, the major heat losses came from radiation and conduction, some 62,500 kcal/h at 800°C; at 600°C we may assume them to be 46,900 kcal/h or 23,450 kcal in half an hour, i.e. (23,450÷226,450×100=) 10.3%. To this we must add the heat loss through the smoke at 600°C: about 31.3% according to calculations; of uncombusted gases from the hearth: 4%; of uncombusted solids from the hearth: 3.1%. The efficiency of the furnace was thus (100–[10.3+31.3+4+3.1]=) 51.3%, the effective specific heat of combustion of the coke (6,470×0.513) ≈ 3,320 kcal/kg, which brings the effective heat supplied to the furnace over half an hour to (35×3,320) ≈ 116,200 kcal. To keep the furnace at 600°C, an additional heat contribution of (118,500–116,200=) 2,300 kcal was thus needed during that time: it could easily be supplied by the radiation from the muffle walls.

Let us now look at the case of the evaporation of the water contained in four corpses in each of the three muffles, 12 corpses altogether. The water content of the corpses is (45.5×12=) 546 kg: the heat of vaporization at 600°C is 546×[640+0.477×(500–20)] ≈ 474,500 kcal. The available heat input stays at 116,200 kcal in 30 minutes, hence the additional heat needed is (474,500–116,200=) 358,300 kcal or some 119,400 kcal per muffle.

We must now examine whether the radiation from the muffle walls could possibly supply this amount of heat. It is difficult to calculate the heat radiated by these walls and absorbed by the corpses, both for reasons of geometry and because of the continual cooling of the wall temperature. However, in a specific technical article, Professor Schläpfer, one of the major experts in cremation in Europe in the 1930s, does give us a reliable estimate of the heat radiated to a single corpse from the muffle walls at various temperatures. He published a chart, from which we may derive the data:

Kod:
Wall temp. [°C]              Heat flow, kcal/min
800                          1,400
700                          930
600                          600

The geometry changes somewhat when a hypothetical load of three corpses in one muffle is irradiated, but the surface-to-volume ratio of such a load is less favorable than that of a single corpse, because the corpses partly cover one another. Even if we leave this consideration aside, the amount of heat required for the evaporation of the water contained in three normal corpses, about 119,400 kcal, would require over three hours at a constant wall temperature of 600°C according to Schläpfer’s data. The wall temperature, however, would certainly not stay constant over such a long period of time, and conditions would quickly become very unfavorable, because, as shown by Schläpfer’s chart, the heat radiated by the walls drops sharply with a decrease in wall temperature.

In his discussion of a similar thermal problem, Kori writes:

"If the inner wall of the cremation chamber has a surface area of about 4 m², with a specific gravity of 2.1, a layer 5 cm thick would weigh about 420 kilograms. The specific heat of the fire clay is about 0.2. Hence, if this layer could supply its total heat content sufficiently fast, only 200×0.2×420 = 16,800 kcal would have become available for an internal temperature dropping from 1,000 to 800°C. Actually, not even this would have been possible, because the brickwork does not release its accumulated heat as quickly as the [muffle] temperature drops."

The weight of the refractory brickwork of one muffle was about (5×1.5×200=) 1,500 kilograms. To compensate for the heat lost due to the evaporation of the water content of the corpses, each muffle would have had to contribute 119,400 kcal, corresponding to a decrease in the average temperature of the refractory brickwork of the muffle of about (119,400 kcal ÷ [0.2 kcal/kg/°C × 1,500 kg] ≈ 400°C. The effective amount of heat supplied to each muffle is therefore:

(3‚320 kcal/kg × 70 kg/h)(/3 × 60 min/h) ≈ 1,290 kcal/min

This corresponds to the supply of 119,400 kcal in (119,400 kcal ÷ 1,290 kcal/min) ≈ 92 minutes. I have only sketched the evaporation process, which is actually more complex, depending on further factors. But these factors apply in the same way both to single cremations and to the hypothetical cremation of several bodies at the same time. The enormous difference between the two set out above still applies. It proves not only that the simultaneous cremation of four bodies in half an hour was impossible, but also that not even the evaporation of the water they contained could have been brought about during that span of time. If assuming an average weight of 60 kg per body, the figures of the above calculations drop by a mere 15%, and the conclusions are basically the same.
vad är det för snömos.
Det liknar beräkningen om myten att en humla inte kan flyga
Citera
2022-03-08, 07:41
  #250
Medlem
pollenpers avatar
Citat:
Ursprungligen postat av pollenper
Hur funkar din uträkning?
En kropp, en timme.
Två kroppar, två timmar
Det är din tes, antar jag.

Två kroppar som bränns samtidigt, under en timme.
Blir genomsnittstiden en halv timme per styck eller fortfarande två timmar.?

Jag skulle påstå att den ugnen har kapaciteten två kroppar per timme,vad kallar du det?

Citat:
Ursprungligen postat av pollenper
visa oss en berökning av molekyler i en effekriv ugn jämfört med en usel ugn.

Är det samma värden på en grill utomhus som en effektiv ugn i en konternuerlig process?

Citat:
Ursprungligen postat av Stor-Men-Snabb
Bra inlägg!!

Pollenper ville se matematik och hintade själv om kunskap i det..


Kan vi hoppas på att pollenper nu efter att ha gått ner på specifika, detalj och hint om matematiskt kunnande håller sig kvar där efter fått svar på detaljnivå? ...eeeellleeeer kommer pollenper som vanligt gå upp flera abstraktionsnivåer och sidsteppa med en annan tramsig 'motfråga'?

..förintelseivrarna här är inte direkt en källa av kunskap och intellektuell hederlighet.., det har man ju fått visat för sig gång på gång på gång.
frågan ovan.
Hur många kroppar per timme bränns i den hypotetiska ugnen.
Pratar vi samma språk.

Vistt är det skillnad mellan en dålig och en effektiv ugn?

Du kan räkna på hur mycket detaljer som helst, som ger en fin känsla av vetenskap.
Men om data inte speglar verkligheten så blir det bara fel. och just snömos.

Matten bevisar intet.
Det finns vittnen köporder på effektiva ugnar.

Auschwitz hade kapacitet på 40k kroppar om dygnet, er kompis beräkningar motsäger inte det.
Citera
2022-03-08, 09:13
  #251
Medlem
Citat:
Ursprungligen postat av Skogsvatte
Jag har redan visat relevanta värmetekniska uträkningar. De har emellertid mötts av tystnad.
The cremation of four to five corpses in one muffle within 20-25 minutes, or half an hour (or a little more than half an hour) is absurd on two counts: first of all because it took one hour to burn a single corpse and secondly because the time needed to burn multiple corpses at once would have extended the time necessary for each corpse well beyond one hour. In practice, however, such a procedure would have brought along insurmountable problems of thermal technology.

The necessary condition for carrying out a cremation is that the temperature of the muffle never drops below 600°C; otherwise there is no longer any incineration, but only carbonization of the corpse. A body of 70 kg contains some 45.5 kg of water. The heat of vaporization at 600°C of the water contained in three corpses is 3×45.5×[640+0.477× (500–20)] ≈ 118,500 kcal. It is known from experience that the process of evaporation took about half an hour. The loading rate of the grate of the triple-muffle furnace was about 70 kg/h of coke (two hearths with grate loadings of 35 kg/h each), hence the theoretical availability of heat over half an hour was (6,470×35=) 226,450 kcal. The effective availability was much lower because a large part of the heat generated in the gasifiers was lost. During evaporation, the major heat losses came from radiation and conduction, some 62,500 kcal/h at 800°C; at 600°C we may assume them to be 46,900 kcal/h or 23,450 kcal in half an hour, i.e. (23,450÷226,450×100=) 10.3%. To this we must add the heat loss through the smoke at 600°C: about 31.3% according to calculations; of uncombusted gases from the hearth: 4%; of uncombusted solids from the hearth: 3.1%. The efficiency of the furnace was thus (100–[10.3+31.3+4+3.1]=) 51.3%, the effective specific heat of combustion of the coke (6,470×0.513) ≈ 3,320 kcal/kg, which brings the effective heat supplied to the furnace over half an hour to (35×3,320) ≈ 116,200 kcal. To keep the furnace at 600°C, an additional heat contribution of (118,500–116,200=) 2,300 kcal was thus needed during that time: it could easily be supplied by the radiation from the muffle walls.

Let us now look at the case of the evaporation of the water contained in four corpses in each of the three muffles, 12 corpses altogether. The water content of the corpses is (45.5×12=) 546 kg: the heat of vaporization at 600°C is 546×[640+0.477×(500–20)] ≈ 474,500 kcal. The available heat input stays at 116,200 kcal in 30 minutes, hence the additional heat needed is (474,500–116,200=) 358,300 kcal or some 119,400 kcal per muffle.

We must now examine whether the radiation from the muffle walls could possibly supply this amount of heat. It is difficult to calculate the heat radiated by these walls and absorbed by the corpses, both for reasons of geometry and because of the continual cooling of the wall temperature. However, in a specific technical article, Professor Schläpfer, one of the major experts in cremation in Europe in the 1930s, does give us a reliable estimate of the heat radiated to a single corpse from the muffle walls at various temperatures. He published a chart, from which we may derive the data:

Kod:
Wall temp. [°C]              Heat flow, kcal/min
800                          1,400
700                          930
600                          600

The geometry changes somewhat when a hypothetical load of three corpses in one muffle is irradiated, but the surface-to-volume ratio of such a load is less favorable than that of a single corpse, because the corpses partly cover one another. Even if we leave this consideration aside, the amount of heat required for the evaporation of the water contained in three normal corpses, about 119,400 kcal, would require over three hours at a constant wall temperature of 600°C according to Schläpfer’s data. The wall temperature, however, would certainly not stay constant over such a long period of time, and conditions would quickly become very unfavorable, because, as shown by Schläpfer’s chart, the heat radiated by the walls drops sharply with a decrease in wall temperature.

In his discussion of a similar thermal problem, Kori writes:

"If the inner wall of the cremation chamber has a surface area of about 4 m², with a specific gravity of 2.1, a layer 5 cm thick would weigh about 420 kilograms. The specific heat of the fire clay is about 0.2. Hence, if this layer could supply its total heat content sufficiently fast, only 200×0.2×420 = 16,800 kcal would have become available for an internal temperature dropping from 1,000 to 800°C. Actually, not even this would have been possible, because the brickwork does not release its accumulated heat as quickly as the [muffle] temperature drops."

The weight of the refractory brickwork of one muffle was about (5×1.5×200=) 1,500 kilograms. To compensate for the heat lost due to the evaporation of the water content of the corpses, each muffle would have had to contribute 119,400 kcal, corresponding to a decrease in the average temperature of the refractory brickwork of the muffle of about (119,400 kcal ÷ [0.2 kcal/kg/°C × 1,500 kg] ≈ 400°C. The effective amount of heat supplied to each muffle is therefore:

(3‚320 kcal/kg × 70 kg/h)(/3 × 60 min/h) ≈ 1,290 kcal/min

This corresponds to the supply of 119,400 kcal in (119,400 kcal ÷ 1,290 kcal/min) ≈ 92 minutes. I have only sketched the evaporation process, which is actually more complex, depending on further factors. But these factors apply in the same way both to single cremations and to the hypothetical cremation of several bodies at the same time. The enormous difference between the two set out above still applies. It proves not only that the simultaneous cremation of four bodies in half an hour was impossible, but also that not even the evaporation of the water they contained could have been brought about during that span of time. If assuming an average weight of 60 kg per body, the figures of the above calculations drop by a mere 15%, and the conclusions are basically the same.

Citat:
Ursprungligen postat av pollenper
vad är det för snömos.
Det liknar beräkningen om myten att en humla inte kan flyga

'Snömoset' är matematik och en tillämpning av den vid energiberäkningar. Du vet, sånt trams som vetenskapen ofta använder sig av när de belägger bevis för eller förkastar en tes.
__________________
Senast redigerad av Stor-Men-Snabb 2022-03-08 kl. 09:16.
Citera
2022-03-08, 16:58
  #252
Medlem
pollenpers avatar
Citat:
Ursprungligen postat av Stor-Men-Snabb
'Snömoset' är matematik och en tillämpning av den vid energiberäkningar. Du vet, sånt trams som vetenskapen ofta använder sig av när de belägger bevis för eller förkastar en tes.
Det var en del "assume" där och en kolförbrukning av 70 kg cole i timmen
och rökgaser som var svåra att beräkna.

Håller du med om att det är skillnad på en bra ugn och ett grillgaller.
Samma energi går åt att koka bort vattnet i en kropp.
Citera

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