Overall Heat Transfer Coefficient for a Cylindrical Pipe

[@vincenciussufijanhadi1287]

I just hope that there should be a retest on the calorific value of the heat change on the cylinder wall. which is closer to the test results than the existing formula or the formula that I found.

[@vincenciussufijanhadi1287]

The problems faced by solving cylindrical heat transfer using natural logarithms are: 1. When R1 = 0, then the heating value is ~ + 2. The result is that the temperature at the center of the cylinder also has a value of ~ + 3. You cannot find out the change in the value of the circumference of the cylinder (2πR). This is because you don't know the R value. 4. The R to dR ratio value becomes inconsistent. I have explained this in a previous open letter. 5. A serious problem occurs when the radius of the cylinder is large and the thickness approaches the radius of the cylinder, then the ratio of R to dR has the value (R >> dR), R is very large compared to dR. It seems that if viewed from T1 the value is reasonable. However, the T1 value actually has a fatal error. I just hope that you read very carefully the meaning of my open letter.

[@vincenciussufijanhadi1287]

For ball shape I only focus on the ratio of the radius of the ball casing to the thickness of the ball. If R/ dR = 1 is found, then the formula must be reviewed. The general equation is: Q = - k. A. (dT)/ (dR ), Fourrier law Q = k. 4π R2. (∆ T)/ ( R2 - R1 ). ………. 1 Q = k. 4π R1. R2. (∆ T)/ ( R2 - R1 ). ………. 2 Q = k. 4π R1 . R2.(∆ T)/ ( R2 - R1 ) = k. 4π R2.(∆ T)/ ( R2 - R1 ). This formula emphasizes the ratio of the radius of the ball casing to the thickness of the ball. So the equation is: k. 4π R1 . R2.(∆ T)/ ( R2 - R1 ) = k. 4π R2. (∆ T)/ ( R2 - R1 ), So the eliminated part is k .4. π. R1 . R2. (∆ T)/ ( R2 - R1 ) = R2.(∆ T)/ ( R2 - R1 ). ( R1 . R2.)0.5 (∆ T)/ ( R2 - R1 ) = ( R2 )0.5.(∆ T)/ ( R2 - R1 ). ( R1 . R2.)0.5 . (∆ T)/ ( R2 - R1 ) = R.(∆ T)/ ( R2 - R1 ). So we get the ratio of the radius of the ball casing to the thickness of the ball. ( R1 . R2.)0.5 / ( R2 - R1 ) = R/ ( R2 - R1 ). So the ratio of the radius of the ball casing to the thickness of the ball. ( R1 . R2.)0.5 : ( R2 - R1 ), ), is the ratio of the radius of the ball casing to the thickness of the ball. ………. 3 dR : Instantaneous thickness. (R2 - R1) or (outer -inner) dT : Instantaneous temperature K : Thermal conductivity of the material Q : Calorific value R : Length of the radius of the ball casing ∆T : Integral result of dT ∆R : Result (R2 - R1) = (outer - inner) = thickness 4πR2 : Area of the sphere's shell R/ dR : Proportional to R / (R2 - R1) The ratio of the length of the radius of the ball casing to the thickness of the ball. My objections are: R2 : R2 .- R1 R2 : R2 .- R1 ~ R2. R1 : R2 .- R1, R2 = R2. R1 My comment is that R2 ≠ R1 x R2. The R in question has the same value, namely R x R. Description of reasoning: 1. R2 : R2 - . R1 R2 : R2 .- R1, for R2 = R x R. R2 = R x R R. R = R1 x R2 , …??? R . R = R1 x R2, because R has the same value, then R1 = R2. This writing will have problems with the equation: Q = k. 4π R2.L (∆ T)/ ( R2 - R1 ), where R2 = R1. Q = k. 4π R2.L (∆ T)/ ( R2 - R2 ). Q = k. 4π R2.L (∆ T)/ ( 0 ). Q = + ∞ … ???. The alternatives are: R2 : R2 - . R1 , R2 : R2 .- R1, untuk R2 = R x R. (R . R}0.5 = ( R2 . R1 )0.5 R = ( R2 . R1 )0.5 … ???. Is the radius of the spherical casing (R) (R2. R1)0.5...??? If that's the case. What can be explained from ( R2 .R1)0.5 in visualization ... ??? My comment is that the conclusion is R2 ≠ R1. R2 2. I will insist that R2 = R1. R2. to prove my objection. Next, by substituting into the initial equation, when R1 = 0. (Solid). Q = - k. A(dT)/ (dR ). Fourrier law Q = k. 4π R2.L (∆T)/ (∆R) Q = k. 4π R2.L (∆T)/ ( R2 - R1 ) R2. Is outer radius dan R1 is inner radius R2 = R1.R2 Q = k. 4π R2.L (∆T)/ ( R2 - R1 ). I focus on the equations in bold. What if R1 is 0? I substituted the basic formula. Q = k. 4π R2.L (∆T)/ ( R2 - R1 ). Q = k. 4π R1. R2.L (∆T)/ ( R2 - R1 ). Q = k. 4π 0. R2.L (∆T)/ ( R2 - 0 ). Q = k. 4π 0. .L (∆T)/ ( R2 ).. I focus on the equations in bold What if R1 is 0? I substituted the basic formula. In this equation, 2 answers appear. In this case, 2 alternative answers will appear. Q = k. 4π 0. L (∆T)/ ( R2 ), 0/R2 = 0, Then, Q = 0 ( alternatif 1 ). My comment is whether when the ball is in a solid state, then Q = 0???. this is very unreasonable. Or second opinions is : Q = k. 4π R2.L (∆T)/ ( R2 - R1 ), R2 = R1. R2 Q = k. 4π R1. R2.L (∆T)/ ( R2 - R1 ), 0 = R1 Q = k. 4π 0. R2.L (∆T)/ ( R2 - 0 ), R2 - 0 = R2 Q = k. 4π 0. L (∆T)/ ( R2 ). Q = k. 4π 0. .L (∆T)/ ( R2 ), 0/R2 = 0, turn left side Q / 0 = k. 4π .L (∆T) +∞ = k. 4π .L (∆T ), k. 4π.L is constant( C ), and then +∞ = C ((∆T ), and then +∞ = ( ∆T ), ∆T = T1 - T2 +∞ = T1 - T2, , T1 = +∞ dan T2 known, and then +∞ = ( +∞ - T2 ), left side is watt and right side is temperature. so it cannot be reduced +∞ = T1 +∞ = (∆T ), ∆T = ( +∞ - T2 ) (second opinions ) +∞ = T1, ∆T = ( +∞ - T2 ) +∞ = T1, My comment is whether when the ball is in a solid state, then (∆ T ) = (+∞ ) or (T1 ) = (+∞ ) ???, this doesn't really make sense.

[@vincenciussufijanhadi1287]

execuse me, where are you from. am indonesia

[@vincenciussufijanhadi1287]

Untuk bentuk bola Saya hanya fokus pada perbandingan jari jari selubung bola terhadap ketebalan bola. Jika ditemukan R/ dR = 1, maka rumus tersebut haru dikaji ulang. Dengan persamaan umumnya adalah : Q = - k. A. (dT)/ (dR ), Hukum fourrier Q = k. 4π R2. (∆ T)/ ( R2 - R1 ). ………. 1 Q = k. 4π R1. R2. (∆ T)/ ( R2 - R1 ). ………. 2 Q = k. 4π R1 . R2.(∆ T)/ ( R2 - R1 ) = k. 4π R2.(∆ T)/ ( R2 - R1 ). Pada rumus ini menekankan pada perbandingan jari jari selubung bola terhadap ketebalan bola. Sehingga persamaannya adalah : k. 4π R1 . R2.(∆ T)/ ( R2 - R1 ) = k. 4π R2. (∆ T)/ ( R2 - R1 ), Sehingga bagian yang tereliminasi adalah k .4. π. R1 . R2. (∆ T)/ ( R2 - R1 ) = R2.(∆ T)/ ( R2 - R1 ). ( R1 . R2.)0.5 (∆ T)/ ( R2 - R1 ) = ( R2 )0.5.(∆ T)/ ( R2 - R1 ). ( R1 . R2.)0.5 . (∆ T)/ ( R2 - R1 ) = R.(∆ T)/ ( R2 - R1 ). Sehingga didapatkan perbandingan jari jari selubung bola terhadap ketebalan bola. ( R1 . R2.)0.5 / ( R2 - R1 ) = R/ ( R2 - R1 ). Sehingga perbandingan jari jari selubung bola terhadap ketebalan bola ( R1 . R2.)0.5 : ( R2 - R1 ), adalah perbandingan jari jari selubung bola terhadap ketebalan bola. …….… 3 Dimana, dR : Ketebalan sesaat. (R2 - R1 ) atau ( outer -inner ) dT : Temperature sesaat K : Konduktifitas thermal materi Q : Nilai kalor R : Panjang jari jari selubung bola ∆T : Hasil integral dari dT ∆R : Hasil ( R2 - R1 ) = ( outer - inner ) = tebal 4πR2 : Luas kulit bola R/ dR : Sebanding dengan R / ( R2 - R1 ) Perbandingan panjang jari jari selubung bola terhadap ketebalan bola. Keberatan saya adalah : 1. R2 : R2 .- R1 R2 : R2 .- R1 ~ R2. R1 : R2 .- R1, R2 = R2. R1 Komentar saya adalah bahwa R2 ≠ R1 x R2. R yang dimaksud adalah memiliki nilai yang sama, yaitu R x R. Uraian penalaran : R2 : R2 - . R1 R2 : R2 .- R1, untuk R2 = R x R. R2 = R x R R. R = R1 x R2 …??? R . R = R1 x R2, oleh karena R bernilai sama, maka R1 = R2. Penulisan ini akan bermasalah dengan persamaan : Q = k. 4π R2.L (∆ T)/ ( R2 - R1 ), dimana R2 = R1. Q = k. 4π R2.L (∆ T)/ ( R2 - R2 ). Q = k. 4π R2.L (∆ T)/ ( 0 ). Q = + ∞ … ???. Alternatifnya adalah : R2 : R2 - . R1 , R2 : R2 .- R1, untuk R2 = R x R. (R . R}0.5 = ( R2 . R1 )0.5 R = ( R2 . R 1 )0.5 … ???. Apakah jari jari selubung bola ( R ) adalah ( R2 . R1 )0.5 … ??? Jika demikian halnya. Apa yang bisa dijelaskan dari ( R2 .R1)0.5 secara visualisasi … ???. Komentar saya adalah kesimpulannya R2 ≠ R1. R2. continue.....

[@vincenciussufijanhadi1287]

Can I discuss with you? I feel there is something wrong with the derivation of the formula. Maybe I'm one of those who doesn't agree with the existing formula. please respond

[@FooPiew]

@@vincenciussufijanhadi1287 Hi, thank you for your willingness to point out the potential problem in the derivation. Could you please specify the specific equation that you do not think is correct? I will strive to respond to you here when I’m available.

[@vincenciussufijanhadi1287]

@@FooPiew In short, my objections are:. 1. For the cylinder shape a. The ratio of the sheath radius to the thickness of the cylinder is never less than 1. This condition occurs when the thickness is (e - x), where e (natural logarithm) is the outer radius and x is the inner radius which ranges from 0 < x < e ^ 0. The reason for my rejection is that the sheath radius always moves larger following changes in area. While the thickness moves linearly. so that there is never an event where the sheath radius is shorter than the thickness. Thus the sheath radius is always greater than the thickness under any conditions. b. When the cylinder is solid (has no holes), the temperature will be + ∞ (the temperature increases very large without limits at the center of the cylinder). This condition occurs when the thickness is (e - 0). e (natural logarithm) is the outer radius is a natural number and 0 is the inner radius. The reason for my rejection is that it is very unreasonable for a material in a solid condition to be able to withstand heat with a temperature of + ∞. My opinion is that the temperature can still be known even though it is difficult to practice in solid conditions. c. The ratio of the sheath radius to the thickness is greater than 1 (has no limit) and ends at 0 (NULL) when (e - e) is outer = inner. The reason for my rejection is that the sheath radius must be greater than the radius. But it has a certain value limit !!! d. In general, the reason for my rejection of points a, b and c is that the use of natural logarithms causes uncertainty in the ratio of the sheath radius to the radius. a. When the thickness ( e - x ), 0 < x < e ^ 0, then the ratio of the sheath radius < thickness. b. When the thickness ( e - x ), x = 0, then the ratio of the sheath radius to the thickness + ∞. c. When the thickness ( e - x ), x = e ^ 0, then the ratio of the sheath radius = thickness. d. When the thickness ( e - x ), e^0 < x < e, then the ratio of the sheath radius > thickness e. My question is: How to calculate the temperature at the center of the cylinder when the heat comes from outside the cylinder by conduction with steady state flow ??? f. The predicted calculation error is very serious. When the outer becomes very large and the inner approaches the outer (relatively thin). Then the calculation error in temperature becomes very large. Why is the existing formula reduction considered as the truth?, this is because the calorific value must always (constant) remain the same in each change in thickness that experiences an increase in temperature when approaching the center of the cylinder along with the change in thickness where the initial temperature comes from the inner radius. With the note that the heat source comes from inside the cylinder. However, for the case of steady state flow conductor heat transfer in the form of a cylinder. There are other conditions that must be followed. Namely, the value of the sheath radius to the thickness must always be more than 1. Why must it always be more than 1? this is due to the continuous change in cross-sectional area as explained in point C. PHILOSOPHICAL VIEWPOINT ON THE THEORY OF TRUTH Based on the theory of truth in the formula, it is a consensus theory of truth. Which is mutually agreed upon. Does not yet cover universal truth in terms of thickness. This allows for propositions that contradict objects and results when confronted. Understand the Types of Truth Theory: 1. Coherence Theory (The Consistence/Coherence Theory of Truth) 2. Correspondence (The Corespondency Theory of Truth) 3. Pragmatism (The Pragmatic Theory of Truth) 4. Performative Theory 5. Consensus theory. The approach to the theory of truth in this case is still an agreement between scientists Sources of Knowledge: 1. Rationalism 2. Empiricism (in this section I have not conducted physical testing on material temperature) 3. Criticism 4. Intuitionism Quote philosophy: It is with logic that one proves, it is with intuition that one invents (Henri Poincare). Vincent Sufijan Hadi

[@vincenciussufijanhadi1287]

@@FooPiew In short, my objections are:. 1. For the cylinder shape a. The ratio of the sheath radius to the thickness of the cylinder is never less than 1. This condition occurs when the thickness is (e - x), where e (natural logarithm) is the outer radius and x is the inner radius which ranges from 0 < x < e ^ 0. The reason for my rejection is that the sheath radius always moves larger following changes in area. While the thickness moves linearly. so that there is never an event where the sheath radius is shorter than the thickness. Thus the sheath radius is always greater than the thickness under any conditions. b. When the cylinder is solid (has no holes), the temperature will be + ∞ (the temperature increases very large without limits at the center of the cylinder). This condition occurs when the thickness is (e - 0). e (natural logarithm) is the outer radius is a natural number and 0 is the inner radius. The reason for my rejection is that it is very unreasonable for a material in a solid condition to be able to withstand heat with a temperature of + ∞. My opinion is that the temperature can still be known even though it is difficult to practice in solid conditions. c. The ratio of the sheath radius to the thickness is greater than 1 (has no limit) and ends at 0 (NULL) when (e - e) is outer = inner. The reason for my rejection is that the sheath radius must be greater than the radius. But it has a certain value limit !!! d. In general, the reason for my rejection of points a, b and c is that the use of natural logarithms causes uncertainty in the ratio of the sheath radius to the radius. a. When the thickness ( e - x ), 0 < x < e ^ 0, then the ratio of the sheath radius < thickness. b. When the thickness ( e - x ), x = 0, then the ratio of the sheath radius to the thickness + ∞. c. When the thickness ( e - x ), x = e ^ 0, then the ratio of the sheath radius = thickness. d. When the thickness ( e - x ), e^0 < x < e, then the ratio of the sheath radius > thickness e. My question is: How to calculate the temperature at the center of the cylinder when the heat comes from outside the cylinder by conduction with steady state flow ??? f. The predicted calculation error is very serious. When the outer becomes very large and the inner approaches the outer (relatively thin). Then the calculation error in temperature becomes very large. Why is the existing formula reduction considered as the truth?, this is because the calorific value must always (constant) remain the same in each change in thickness that experiences an increase in temperature when approaching the center of the cylinder along with the change in thickness where the initial temperature comes from the inner radius. With the note that the heat source comes from inside the cylinder. However, for the case of steady state flow conductor heat transfer in the form of a cylinder. There are other conditions that must be followed. Namely, the value of the sheath radius to the thickness must always be more than 1. Why must it always be more than 1? this is due to the continuous change in cross-sectional area as explained in point C. PHILOSOPHICAL VIEWPOINT ON THE THEORY OF TRUTH Based on the theory of truth in the formula, it is a consensus theory of truth. Which is mutually agreed upon. Does not yet cover universal truth in terms of thickness. This allows for propositions that contradict objects and results when confronted. Understand the Types of Truth Theory: 1. Coherence Theory (The Consistence/Coherence Theory of Truth) 2. Correspondence (The Corespondency Theory of Truth) 3. Pragmatism (The Pragmatic Theory of Truth) 4. Performative Theory 5. Consensus theory. The approach to the theory of truth in this case is still an agreement between scientists Sources of Knowledge: 1. Rationalism 2. Empiricism (in this section I have not conducted physical testing on material temperature) 3. Criticism 4. Intuitionism Quote philosophy: It is with logic that one proves, it is with intuition that one invents (Henri Poincare). Vincent Sufijan Hadi

[@FooPiew]

@@vincenciussufijanhadi1287 I prefer answering from here to make conversation traceable. Can you provide a link that shows the formula that you agree with?

[@vincenciussufijanhadi1287]

@@FooPiew OPEN LETTER UNDERSTANDING THE CONDUCTION HEAT TRANSFER EQUATION OF STEADY STATE FLOW IN CYLINDER SHAPE To the honorable, Anyone studying heat transfer courses (mechanical engineering, chemical engineering and nuclear engineering). In Place, Together with this open letter, with humility, we send an open letter to anyone, which may be read by anyone who is willing to read this letter. The purpose of this open letter is to re-discuss the explanation of steady state heat transfer on hollow cylinder walls I am the writer of the letter refuting the formula that has been written in books and literature. Especially on conductor heat transfer on steady state cylinder walls. Briefly my objections are: 1. For cylindrical shapes a. The comparison of the radius of the sheath to the thickness of the cylinder is never less than 1. This condition occurs when the thickness is ( e - x ), where e (natural logarithm) is the outer radius and x is the inner radius which ranges from 0 < x < e ^ 0 . The reason for my objection is that the radius of the sheath always moves larger following changes in area. While the thickness moves linearly. so that there is never an event where the radius of the sheath is shorter than the thickness. Thus the radius of the sheath is always greater than the thickness under any conditions. b. When the cylinder is solid (has no holes), the temperature will be + ∞ (the temperature increases very large without limits at the center of the cylinder). This condition occurs when the thickness is ( e - 0). e (natural logarithm) is the outer radius is a natural number and 0 is the inner radius. The reason for my objection is that it is very unreasonable for a material in a solid condition to withstand heat with a temperature of + ∞. My opinion is that the temperature can still be known even though it is difficult to practice in solid conditions. c. The ratio of the sheath radius to the thickness is greater than 1 (has no limit) and ends at 0 (NULL) when (e - e) is outer = inner. The reason for my objection is that the sheath radius must indeed be greater than the radius. However, it has a certain value limit !!! d. In general, the reason for my objection from points a, b and c is that the use of natural logarithms causes uncertainty in the ratio of the sheath radius to the radius. a. When the thickness is (e - x), 0 < x < e ^ 0, then the ratio of the sheath radius < thickness. b. When the thickness is (e - x), x = 0, then the ratio of the sheath radius to the thickness is + ∞. c. When the thickness is (e - x), x = e ^ 0, then the ratio of the sheath radius = thickness. d. When the thickness is (e - x), e ^ 0 < x < e, then the ratio of the sheath radius > thickness e. My question is: How to calculate the temperature at the center of the cylinder when the heat comes from outside the cylinder by conduction with steady state flow ??? f. The prediction of the calculation error is very serious. When the outer becomes very large and the inner approaches the outer (relatively thin). Then the calculation error in temperature becomes very large. Why is the existing formula considered as the truth? This is because the calorific value must always be (constant) in each change in thickness that experiences a change in temperature increasing when approaching the center of the cylinder along with the change in thickness where the initial temperature comes from the inner radius. With a note that the heat source comes from inside the cylinder. However, for the case of steady state flow conductor heat transfer in the form of a cylinder. There are other conditions that must be followed. Namely, the value of the radius of the sheath against the thickness must always be more than 1. Why must it always be more than 1? This is due to changes in the cross-sectional area that do not move linearly against the radius of the cylinder as explained in point C. PHILOSOPHICAL VIEWPOINT ON THE THEORY OF TRUTH Based on the theory of truth in the formula, it is a consensus theory of truth. Which is mutually agreed upon. Does not yet cover universal truth in thickness conditions. This allows for propositions that contradict the object and results when confronted. Understand the Types of Truth Theory: 1. The Consistency Theory of Truth 2. Correspondence Theory of Truth 3. Pragmatism 4. Performative Theory 5. Consensus Theory. The approach to the theory of truth in this case is still based on the agreement of scientists Sources of Knowledge: 1. Rationalism 2. Empiricism (in this section I have not conducted physical testing on the temperature of the material) 3. Criticism 4. Intuitionism Quote philosophy: It is with logic that one proves, it is with intuition that one invents (Henri Poincare). My quote philosophy: Mathematics is able to predict errors, but has not fully proven absolute truth.

[@vincenciussufijanhadi1287]

SUDUT PANDANG FILSAFAT TENTANG TEORI KEBENARAN Berdasarkan teori kebenaran dalam rumus tersebut adalah merupakan teori kebenaran consensus. Yang disepakati bersama sama. Belum mencakup kebenaran secara universal dalam kondisi ketebalan. Hal ini memungkinkan adanya proposisi yang bertentangan dengan obyek dan hasil saat dilakukan konfrontir. Pahami Jenis teori kebenaran : 1. Teori Koherensi (The Consistence/Coherence Theory of Truth) 2. Korespondensi (The Corespondency Theory of Truth) 3. Pragmatisme (The Pragmatic Theory of Truth) 4. Teori Performatif 5. Teori consensus. Pendekatan teori kebenaran pada kasus ini masih bersifat kesepakatan para ilmuwan Sumber-Sumber Pengetahuan : 1. Rasionalisme 2. Empirisme ( pada bagian ini saya belum melakukan pengujian fisik pada suhu material ) 3. Kritisisme 4. Intuisisme Secara umum kebenaran bersifat : 1. Rigid 2. Universal 3. Nonkontradiktif 4. Bisa dibuktikan 5. Radic Quote philosophy : It is with logic that one proves, it is with intuition that one invents ( Henri Poincare ). My quote philosophy : Matematika sanggup memprediksi kesalahan, namun belum sepenuhnya membuktikan kebenaran mutlak. Vincencius Sufijan Hadi

[@vincenciussufijanhadi1287]

Untuk bentuk bola Saya hanya fokus pada perbandingan jari jari selubung bola terhadap ketebalan kulit bola. Yang mana perbandingan jari jari selubung bola terhadap ketebalan kulit bola = 1. Dengan persamaan umumnya adalah : Q = - k. A(dT)/ (dR ). Q = k. 4π R2.L (∆ T)/ ( R2 - R1 ). Q = k. 4π R2.L (∆ T)/ ( R2 - R1 ). saya hanya focus pada persamaan yang huruf bold Keberatan saya adalah : 1. R2 : R2 .- R1 R2 : R2 .- R1 ~ R2. R1 : R2 .- R1, R2 = R2. R1 Komentar saya adalah bahwa R2 ≠ R1 x R2. R yang dimaksud adalah memiliki nilai yang sama, yaitu R x R. Uraian penalaran : R2 : R2 - . R1 R2 : R2 .- R1, untuk R2 = R x R. R2 = R x R R. R = R1 x R2 …??? R . R = R1 x R2, oleh karena R bernilai sama, maka R1 = R2. Penulisan ini akan bermasalah dengan persamaan : Q = k. 4π R2.L (∆ T)/ ( R2 - R1 ), dimana R2 = R1. Q = k. 4π R2.L (∆ T)/ ( R2 - R2 ). Q = k. 4π R2.L (∆ T)/ ( 0 ). Q = + ∞ … ???. Alternatifnya adalah : R2 : R2 - . R1 , R2 : R2 .- R1, untuk R2 = R x R. (R . R}0.5 = ( R2 . R1 )0.5 R = ( R2 . R 1 )0.5 … ???. Apakah jari jari selubung bola ( R ) adalah ( R2 . R1 )0.5 … ??? Jika demikian halnya. Apa yang bisa dijelaskan dari ( R2 .R1)0.5 secara visualisasi … ???. Komentar saya adalah kesimpulannya R2 ≠ R1. R2 2. Saya akan paksakan bahwa R2 = R1. R2. untuk membuktikan keberatan saya. Selanjutnya dengan melakukan substitusi ke persamaan awalnya, saat R1 = 0. ( Solid ). Q = - k. A(dT)/ (dR ). Q = k. 4π R2.L (∆T)/ (∆R) Q = k. 4π R2.L (∆T)/ ( R2 - R1 ) R1.adalah jari jari outer dan R2 adalah jari jari inner R2 = R1.R2 Q = k. 4π R2.L (∆T)/ ( R2 - R1 ). Q = k. 4π R2.L (∆T)/ ( R2 - R1 ). saya fokus pada persamaan yang hurufnya bold. Bagaimana jika R1 bernilai 0 ?. Saya substitusikan rumus dasarnya. Q = k. 4π R2.L (∆T)/ ( R2 - R1 ). Q = k. 4π R1. R2.L (∆T)/ ( R2 - R1 ). Q = k. 4π 0. R2.L (∆T)/ ( R2 - 0 ). Q = k. 4π 0. .L (∆T)/ ( R2 ). Q = k. 4π 0. .L (∆T)/ ( R2 ). Pada persamaan ini muncul 2 jawaban. Dalam hal ini akan muncul 2 alternatif jawaban. Q = k. 4π 0. L (∆T)/ ( R2 ), 0/R2 = 0, Sehingga, Q = 0 ( alternatif 1 ). Komentar saya adalah apakah Q = 0 ??? Atau altenatif lainnya adalah Q = k. 4π R2.L (∆T)/ ( R2 - R1 ), R2 = R1. R2 Q = k. 4π R1. R2.L (∆T)/ ( R2 - R1 ), 0 = R1 Q = k. 4π 0. R2.L (∆T)/ ( R2 - 0 ), R2 - 0 = R2 Q = k. 4π 0. L (∆T)/ ( R2 ). Q = k. 4π 0. .L (∆T)/ ( R2 ), 0/R2 = 0 pindah ruas kiri Q / 0 = k. 4π .L (∆T) +∞ = k. 4π .L (∆T ), k. 4π.L adalah nilai konstan ( C ), sehingga +∞ = C ((∆T ), sehingga, +∞ = ( ∆T ), ∆T = T1 - T2 +∞ = T1 - T2, , T1 = +∞ dan T2 diketahui +∞ = ( +∞ - T2 ), satuan sisi kiri adalah watt dan satuan sisi kanan temperature sehingga tidak boleh dikurangi. +∞ = T1 +∞ = (∆T ), ∆T = ( +∞ - T2 ) ( alternatif 2 ) +∞ = T1, ∆T = ( +∞ - T2 ) +∞ = T1, Komentar saya adalah apakah (∆ T ) = (+∞ ) atau (T1 ) = (+∞ ) ??? 3. Apakah jari jari selubung bola dibanding ketebalan kulit bola pernah bernilai 1? Jika ditemukan rasio jari jari selubung bola terhadap ketebalan kulit bola memiliki nilai 1 , maka rumus yang digunakan perlu untuk dikaji ulang. Alasannya adalah jari jari selubung bola bergerak tidak linier terhadap ketebalan kulit bola. Sedangkan jari jari selubung bola bergerak berdasarkan fungsi hiperbola tertentu( tidak menyatakan garis linier. Yang bermakna R2 / ( R2 - R1 ) ~ R2 R1 ( R2 - R1 ). Sehingga, R / ( R2 - R1 ) = ( R2 R1 )0.5 / ( R2 - R1 ) adalah perbandingan jari jari selubung bola terhadap ketebalan kulit ( R2 - R1 ). (R1 . R2 )0.5 / R2 -. R1 tidak pernah bernilai 1. Ditulis dalam persamaan adalah (R1 . R2)0.5 / R2 -. R1 ≠ 1. Mari kita buktikan bahwa (R1 . R2)0.5 / R2 -. R1 = 1, jika bisa ditemukan , maka rumus sebelumnya harus dikaji ulang. Sebagai catatan bahwa saya akan paksakan bahwa R2 = R2. R1 dan R = ( R2 . R1)0.5. Sehingga perbandingan jari jari selubung bola terhadap ketebalan kulit bola ditulis dalam persamaan : ( R1 . R2 )0.5 / R2 -. R1 = 1 ……………. Persamaan 1 Q ` = k. 4π R2.L (∆ T)/ ( R2 - R1 ). Q = k. 4π R1.R2.L (∆ T)/ tebal. Fokus pada R2 / ( R2 - R1 ). R2 / ( R2 - R1 ) R2 = R2. R1 ( R2 )0.5 = ( R2. R1 )0.5 R = ( R2. R1 )0.5 Yang dicari adalah jari jari selubung bola terhadap ketebalan kulit bola. R / ( R2 - R1 ) = 1, R = ( R2. R1 )0.5 ( R2. R1 )0.5 / ( R2 - R1 ) = 1 ( R2. R1 )0.5 = 1.( R2 - R1 ) ( R2. R1 )0.5 = ( R2 - R1 ) ( R2. R1 ) = ( R2 - R1 )2 ( R2. R1 ) = ( R22 - 2 R2 . R1 + R12 ) ( R22 - 2 R2 . R1 + R12 ) / ( R2. R1 ) = 1 R2 / R1 - 2 + R1 / R2 = 1 R2 / R1 + R1 / R2 = 1 + 2 R2 / R1 + R1 / R2 = 3, x R1 . R2 R22 + R12 = 3 R1 . R2 R22 - 3. R1 . R2 + R12 = 0 (R2 - 0.381969.R1 ) (R2 - 2.618031.R1 ) = 0 R2 - 0.381969.R1 = 0, harus memenuhi R2 > R1 R2 = 0.381969.R1 R2 : R1 = 0.381969 : 1, ternyata R2 < R1. Persamaan ini tidak bisa digunakan. Dan berikutnya adalah persamaan yang ke dua R2 - 2.618031.R1 = 0 R2 = 2.618031.R1 R2 : R1 = 2.618031 : 1, persamaan yang digunakan adalah : (R2 - 2.618031.R1 ), memenuhi R2 > R1. Jadi untuk sembarang R2 ( outer ) akan selalu memenuhi persamaan ( R1 . R2 )0.5 / R2 -. R1 = 1, saat R2 : R1 = 2.618031 : 1 R2 : R1 = 2.618031 : 1, adalah rasio outer terhadap inner ( R1 . R2 )0.5 / R2 -. R1 = 1, Kembali pada persamaan ( 1 ). R2 = 2.618031 R1 = 1 ( 1. .2.618031 )0.5 / 2.618031 -. 1 = 1. Adalah suatu keadaan dimana panjang jari jari selubung bola terhadap panjang jari jari bola memilki panjang yang sama. Hal ini tidak pernah terjadi. Alasannya adalah jari jari selubung bola bergerak tidak linier terhadap jari jari bola. Sedangkan luas kulit bola bergerak berdasarkan fungsi hiperbola tertentu( tidak menyatakan garis linier). Namun demikian untuk kasus perpindahan panas konduktor aliran steady state bentuk silinder dan bentuk bola. Ada syarat lain yang harus diikutii. Yaitu nilai jari jari selubung terhadap ketebalan bola harus selalu bernilai lebih dari 1. Mengapa harus selalu bernilai lebih dari 1 ? hal ini disebabkan adanya perubahan luas penampang yang tidak bergerak linier terhadap jari jari silinder maupun bentuk bola.

[@vincenciussufijanhadi1287]

OPEN LETTER MISS UNDERSTANDING OF STEADY STATE FLOW CONDUCTION HEAT TRANSFER EQUATIONS IN CYLINDER SHAPED AND BALL SHAPED CASES Dear, Anyone studying heat transfer subjects (mechanical engineering, chemical engineering, etc.). In Place, Together with this open letter, with humility, we send an open letter to anyone, which can be read by anyone who is willing to read this letter. The purpose of this open letter is to re-discuss the explanation of steady flow heat transfer in the case of a hollow cylindrical wall and a spherical shape. I am the author of the letter objecting to the formula that has been written in books and literature. Especially in conductor heat transfer on cylindrical walls and spherical shape of steady state flow. In short, my objections are: A brief description of my objections to the existing heat transfer formula is: For cylindrical shape The general equation is : Q = -k. A. (dT)/ (dR ). Fourrier Low So the general equation derived for the cylindrical case becomes, Q = k. 2π.R.L ( dT )/ ( dR ), ……… Equation 1 Move the left side to Integral dR/R. With the outer upper limit (R2) and inner lower limit (R1), dT is integrated into ∆T. so the equation changes to, Q. Ln ( R2/R1 ) = k. 2π.L ( ∆T ) Q = k. 2π.L ( ∆T) / [ Ln ( R2/ R1 )]. ………. Equation 2 Equation 1 is equivalent to equation 2 Q = k. 2π.R.L ( dT )/ ( dR ) = k. 2π.L ( ∆T) / [ Ln ( R2/ R1]. Q / k. 2π.L = R. ( dT )/ ( dR ) = ( ∆T) / [ Ln ( R2/ R1]. Q / k. 2π.L is constant (C), so C = ( dT ).R / ( dR ) = ( ∆T) / [ Ln ( R2/ R1], C no longer involved ( dT ).R / ( dR ) = ( ∆T) / [ Ln ( R2/ R1], dT = ∆T. ( ∆T ).R / ( dR ) = ( ∆T) / [ Ln ( R2/ R1 )]. ( ∆T ).R / ( R2 - R1) = ( ∆T) / [ Ln ( R2/ R1 )], ( outer - inner ) = R2 - R1 = thicknes ( ∆T ).R / thicknes = ( ∆T) / [ Ln ( R2/ R1 )], ∆T eliminited R / thicknes = 1/ [ Ln ( R2/ R1 )]. 1: [ Ln ( R2/ R1 )]. = R : thicknes. 1: [ Ln ( R2/ R1 )]. = R : ( R2 - R1 ) R : ( R2 - R1 ) = 1: [ Ln ( R2/ R1 )], changed for R1 is x and R2 is e (Euller number) so, R : ( e - x ) = 1: [ Ln ( e/ x )], dimana x : 0 ≤ x ≤ e ..….…. Equation 3 R : dR = 1: [ Ln ( e/ x )], dimana x : 0 ≤ x ≤ e …….… Equation 4 R = dR : [ Ln ( e/ x )], dimana x : 0 ≤ x ≤ e ….…... Equation 5 R = ( e - x ) : [ Ln ( e/ x )], dimana x : 0 ≤ x ≤ e …....… Equation 6 The equation above is the ratio of the radius of the cylinder casing to the radius of the cylinder Where, dR : Instantaneous thickness (R2 - R1 ) atau ( outer - inner ). dT : Instantaneous Temperature e : Euller number or R2 or outer K : conduktifitas thermal of material L : length of cylinder Q : Calor vullue R : Length of radius of cylinder casing X : Inner atau R1 ∆T : The integral result of dT ∆R : the result ( R2 - R1 ) = ( outer - inner ) = thickness 2πR : Cylindrical casing circumference 2πR.L : Cylinder casing area The explanation of my objection is : a. From equations 3 and 4. The ratio of the casing radius to the cylinder thickness is never less than 1. This condition occurs when the thickness ( e- x ) where e (natural logarithm) is the outer radius and x is the inner radius of the range 0 < x < e^0 . The reason for my objection is that the radius of the casing always moves larger following changes in the cross-sectional area of the cylinder. Meanwhile, thickness moves linearly. so that there is never an incident where the sheath radius is shorter than the thickness. Thus the sheath radius is always greater than the thickness under any conditions. b. From equations 3 and 4. When the cylinder is solid (has no holes), the temperature will be +∞ (the temperature increases infinitely large at the center of the cylinder). This condition occurs when it is thick ( e- 0. e (natural logarithm) is the outer radius which is a natural number and 0 is the inner radius. The reason for my objection is that it is very implausible that a material in a solid state can withstand heat at a temperature of + ∞. My opinion is that the temperature can still be known even though it is difficult to practice in solid conditions. c. From equations 3 and 4. When the cylinder is solid (has no holes), the temperature will be +∞ (the temperature increases infinitely large at the center of the cylinder). This condition occurs when it is thick ( e- 0. e (natural logarithm) is the outer radius which is a natural number and 0 is the inner radius. The reason for my objection is that it is very implausible that a material in a solid state can withstand heat at a temperature of + ∞. My opinion is that the temperature can still be known even though it is difficult to practice in solid conditions. d. From equations 3 and 4. In general, the reason for my objection from points a, b and c is that the use of natural logarithms causes uncertainty in the ratio of the sheath radius to the radius. a. When the thickness ( e - x ), 0 < x < e^0, then the ratio of the sheath radius < thickness. b. When the thickness (e - x), x = 0, then the ratio of sheath radius to thickness is + ∞. c. When the thickness is ( e - x ), x = e^0, then the ratio of the radius of the casing = thickness d. When the thickness ( e - x ), e^0 < x < e, then the ratio of sheath radius > thickness. e. From equations 3 and 4. My question is: How to calculate the temperature at the center of the cylinder when the heat comes from outside the cylinder by conduction with steady state flow ???. could the answer be ( - o C )? f. From equations 3 and 4. The use of natural logaritms is to calculate dimensionless area units without calculating the radius of the cylindrical envelope (R). g. From equations 3 and 4. The forecast calculation error is very serious. When the outer becomes very large and the inner approaches the outer (relatively thin). So the calculation error in temperature becomes very large Why is the existing formula considered to be true? This is because the heating value must always be (constant) at each change in thickness which experiences changes in temperature which increases as it approaches the center of the cylinder along with changes in thickness, where the initial temperature comes from the inner radius. Note that the heat source comes from inside the cylinder.

[@vincenciussufijanhadi1287]

OPEN LETTER UNDERSTANDING THE CONDUCTION HEAT TRANSFER EQUATION OF STEADY STATE FLOW IN CYLINDER SHAPE To the honorable, Anyone studying heat transfer courses (mechanical engineering, chemical engineering and nuclear engineering). In Place, Together with this open letter, with humility, we send an open letter to anyone, which may be read by anyone who is willing to read this letter. The purpose of this open letter is to re-discuss the explanation of steady state heat transfer on hollow cylinder walls I am the writer of the letter refuting the formula that has been written in books and literature. Especially on conductor heat transfer on steady state cylinder walls. Briefly my objections are: 1. For cylindrical shapes a. The comparison of the radius of the sheath to the thickness of the cylinder is never less than 1. This condition occurs when the thickness is ( e - x ), where e (natural logarithm) is the outer radius and x is the inner radius which ranges from 0 < x < e ^ 0 . The reason for my objection is that the radius of the sheath always moves larger following changes in area. While the thickness moves linearly. so that there is never an event where the radius of the sheath is shorter than the thickness. Thus the radius of the sheath is always greater than the thickness under any conditions. b. When the cylinder is solid (has no holes), the temperature will be + ∞ (the temperature increases very large without limits at the center of the cylinder). This condition occurs when the thickness is ( e - 0). e (natural logarithm) is the outer radius is a natural number and 0 is the inner radius. The reason for my objection is that it is very unreasonable for a material in a solid condition to withstand heat with a temperature of + ∞. My opinion is that the temperature can still be known even though it is difficult to practice in solid conditions. c. The ratio of the sheath radius to the thickness is greater than 1 (has no limit) and ends at 0 (NULL) when (e - e) is outer = inner. The reason for my objection is that the sheath radius must indeed be greater than the radius. However, it has a certain value limit !!! d. In general, the reason for my objection from points a, b and c is that the use of natural logarithms causes uncertainty in the ratio of the sheath radius to the radius. a. When the thickness is (e - x), 0 < x < e ^ 0, then the ratio of the sheath radius < thickness. b. When the thickness is (e - x), x = 0, then the ratio of the sheath radius to the thickness is + ∞. c. When the thickness is (e - x), x = e ^ 0, then the ratio of the sheath radius = thickness. d. When the thickness is (e - x), e ^ 0 < x < e, then the ratio of the sheath radius > thickness e. My question is: How to calculate the temperature at the center of the cylinder when the heat comes from outside the cylinder by conduction with steady state flow ??? f. The prediction of the calculation error is very serious. When the outer becomes very large and the inner approaches the outer (relatively thin). Then the calculation error in temperature becomes very large. Why is the existing formula considered as the truth? This is because the calorific value must always be (constant) in each change in thickness that experiences a change in temperature increasing when approaching the center of the cylinder along with the change in thickness where the initial temperature comes from the inner radius. With a note that the heat source comes from inside the cylinder. However, for the case of steady state flow conductor heat transfer in the form of a cylinder. There are other conditions that must be followed. Namely, the value of the radius of the sheath against the thickness must always be more than 1. Why must it always be more than 1? This is due to changes in the cross-sectional area that do not move linearly against the radius of the cylinder as explained in point C. PHILOSOPHICAL VIEWPOINT ON THE THEORY OF TRUTH Based on the theory of truth in the formula, it is a consensus theory of truth. Which is mutually agreed upon. Does not yet cover universal truth in thickness conditions. This allows for propositions that contradict the object and results when confronted. Understand the Types of Truth Theory: 1. The Consistency Theory of Truth 2. Correspondence Theory of Truth 3. Pragmatism 4. Performative Theory 5. Consensus Theory. The approach to the theory of truth in this case is still based on the agreement of scientists Sources of Knowledge: 1. Rationalism 2. Empiricism (in this section I have not conducted physical testing on the temperature of the material) 3. Criticism 4. Intuitionism Quote philosophy: It is with logic that one proves, it is with intuition that one invents (Henri Poincare). My quote philosophy: Mathematics is able to predict errors, but has not fully proven absolute truth.

[@vincenciussufijanhadi1287]

helloi brother

[@vincenciussufijanhadi1287]

According to me. The calculation must be retested to determine its correctness. that is my suggestion. If you do not mind. Testing can be done in the laboratory where you teach. I am happy to help carry out the test and provide alternatives regarding the case I am talking about.

[@vincenciussufijanhadi1287]

2. Saya akan paksakan bahwa R2 = R1. R2. untuk membuktikan keberatan saya. Selanjutnya dengan melakukan substitusi ke persamaan awalnya, saat R1 = 0. ( Solid ). Q = - k. A(dT)/ (dR ). Hukum fourrier Q = k. 4π R2.L (∆T)/ (∆R) Q = k. 4π R2.L (∆T)/ ( R2 - R1 ) R2.adalah jari jari outer dan R1 adalah jari jari inner R2 = R1.R2 Q = k. 4π R2.L (∆T)/ ( R2 - R1 ). Q = k. 4π R2.L (∆T)/ ( R2 - R1 ). saya fokus pada persamaan yang hurufnya bold. Bagaimana jika R1 bernilai 0 ?. Saya substitusikan rumus dasarnya. Q = k. 4π R2.L (∆T)/ ( R2 - R1 ). Q = k. 4π R1. R2.L (∆T)/ ( R2 - R1 ). Q = k. 4π 0. R2.L (∆T)/ ( R2 - 0 ). Q = k. 4π 0. .L (∆T)/ ( R2 ). Q = k. 4π 0. .L (∆T)/ ( R2 ). Pada persamaan ini muncul 2 jawaban. Dalam hal ini akan muncul 2 alternatif jawaban. Q = k. 4π 0. L (∆T)/ ( R2 ), 0/R2 = 0, Sehingga, Q = 0 ( alternatif 1 ). Komentar saya adalah apakah saat bola dalam keadaan solid, maka Q = 0 ???. hal ini sangat tidakl masuk akal. Atau altenatif lainnya adalah Q = k. 4π R2.L (∆T)/ ( R2 - R1 ), R2 = R1. R2 Q = k. 4π R1. R2.L (∆T)/ ( R2 - R1 ), 0 = R1 Q = k. 4π 0. R2.L (∆T)/ ( R2 - 0 ), R2 - 0 = R2 Q = k. 4π 0. L (∆T)/ ( R2 ). Q = k. 4π 0. .L (∆T)/ ( R2 ), 0/R2 = 0 pindah ruas kiri Q / 0 = k. 4π .L (∆T) +∞ = k. 4π .L (∆T ), k. 4π.L adalah nilai konstan ( C ), sehingga +∞ = C ((∆T ), sehingga, +∞ = ( ∆T ), ∆T = T1 - T2 +∞ = T1 - T2, , T1 = +∞ dan T2 diketahui, sehingga +∞ = ( +∞ - T2 ), satuan sisi kiri adalah watt dan satuan sisi kanan temperature sehingga tidak boleh dikurangi. +∞ = T1 +∞ = (∆T ), ∆T = ( +∞ - T2 ) ( alternatif 2 ) +∞ = T1, ∆T = ( +∞ - T2 ) +∞ = T1, Komentar saya adalah apakah saat bola dalam keadaan solid, maka (∆ T ) = (+∞ ) atau (T1 ) = (+∞ ) ???, hal ini sangat tidakl masuk akal. 3. Apakah jari jari selubung bola dibanding ketebalan kulit bola pernah bernilai 1? Jika ditemukan rasio jari jari selubung bola terhadap ketebalan kulit bola memiliki nilai 1 , maka rumus yang digunakan perlu untuk dikaji ulang. Alasannya adalah jari jari selubung bola bergerak tidak linier terhadap ketebalan kulit bola. Sedangkan jari jari selubung bola bergerak berdasarkan fungsi hiperbola tertentu( tidak menyatakan garis linier. Yang bermakna R2 / ( R2 - R1 ) ~ R2 R1 ( R2 - R1 ). Sehingga, R / ( R2 - R1 ) = ( R2 R1 )0.5 / ( R2 - R1 ) adalah perbandingan jari jari selubung bola terhadap ketebalan kulit ( R2 - R1 ). (R1 . R2 )0.5 / R2 -. R1 tidak pernah bernilai 1. Ditulis dalam persamaan adalah (R1 . R2)0.5 / R2 -. R1 ≠ 1. Mari kita buktikan bahwa (R1 . R2)0.5 / R2 -. R1 = 1, jika bisa ditemukan , maka rumus sebelumnya harus dikaji ulang. Sebagai catatan bahwa saya akan paksakan bahwa R2 = R2. R1 dan R = ( R2 . R1)0.5. Sehingga perbandingan jari jari selubung bola terhadap ketebalan kulit bola ditulis dalam persamaan : ( R1 . R2 )0.5 / R2 -. R1 = 1 .……. Persamaan 1 Q ` = k. 4π R2.L (∆ T)/ ( R2 - R1 ). Q = k. 4π R1.R2.L (∆ T)/ tebal. Fokus pada R2 / ( R2 - R1 ). R2 / ( R2 - R1 ) R2 = R2. R1 ( R2 )0.5 = ( R2. R1 )0.5 R = ( R2. R1 )0.5 Yang dicari adalah jari jari selubung bola terhadap ketebalan kulit bola. R / ( R2 - R1 ) = 1, R = ( R2. R1 )0.5 ( R2. R1 )0.5 / ( R2 - R1 ) = 1, sehingga ( R2. R1 )0.5 = 1.( R2 - R1 ) ( R2. R1 )0.5 = ( R2 - R1 ), persamaan di kuadratkan sehingga ( R2. R1 ) = ( R2 - R1 )2, diuraikan sehingga ( R2. R1 ) = ( R22 - 2 R2 . R1 + R12 ), pindah ruas ( R22 - 2 R2 . R1 + R12 ) = ( R2. R1 ), persamaan ( R2. R1 ) pindah ruas, sehingga ( R22 - 2 R2 . R1 + R12 ) / ( R2. R1 ) = 1, dilakukan pembagian, sehingga R2 / R1 - 2 + R1 / R2 = 1, R2 / R1 + R1 / R2 = 1 + 2 R2 / R1 + R1 / R2 = 3, persamaan dikalikan dengan R1 . R2, sehingga R22 + R12 = 3 R1 . R2 R22 - 3. R1 . R2 + R12 = 0 (R2 - 0.381969.R1 ) (R2 - 2.618031.R1 ) = 0 R2 - 0.381969.R1 = 0, harus memenuhi R2 > R1 R2 = 0.381969.R1 R2 : R1 = 0.381969 : 1, ternyata R2 < R1. Persamaan ini tidak bisa digunakan. Dan berikutnya adalah persamaan yang ke dua R2 - 2.618031.R1 = 0 R2 = 2.618031.R1 R2 : R1 = 2.618031 : 1, persamaan yang digunakan adalah (R2 - 2.618031.R1 ), memenuhi R2 > R1. Jadi untuk sembarang R2 ( outer ) akan selalu memenuhi persamaan ( R1 . R2 )0.5 / R2 -. R1 = 1, saat R2 : R1 = 2.618031 : 1 R2 : R1 = 2.618031 : 1, adalah rasio outer terhadap inner Kembali pada persamaan 3 ( R1 . R2.)0.5 : ( R2 - R1 ), adalah perbandingan jari jari selubung bola terhadap ketebalan bola. R2 = 2.618031 R1 = 1. ( 1. .2.618031 )0.5 / ( 2.618031 -. 1) = 1. ( 2.618031 )0.5 / ( 2.618031 -. 1) = 1 Adalah suatu keadaan dimana panjang jari jari selubung bola terhadap ketebalan bola memilki panjang yang sama. Hal ini tidak pernah terjadi. Alasannya adalah jari jari selubung bola bergerak tidak linier terhadap jari jari bola. Sedangkan luas kulit bola bergerak berdasarkan fungsi hiperbola tertentu( tidak menyatakan garis linier). Namun demikian untuk kasus perpindahan panas konduktor aliran steady state bentuk silinder dan bentuk bola. Ada syarat lain yang harus diikutii. Yaitu nilai jari jari selubung terhadap ketebalan bola harus selalu bernilai lebih dari 1. Mengapa harus selalu bernilai lebih dari 1 ? hal ini disebabkan adanya perubahan luas penampang yang tidak bergerak linier terhadap jari jari silinder maupun bentuk bola. Kesimpulan : Dengan menemukan persamaan jari jari selubung silinder terhadap ketebalan silinder = 1. maka saya merasa keberatan dan sebagai konsekuensinya adalah rumus yang sudah ada harus diuji di laboratorium Dengan menemukan persamaan jari jari kulit bola terhadap ketebalan bola = 1. maka saya merasa keberatan dan sebagai konsekuensinya adalah rumus yang sudah ada harus diuji di laboratorium. SUDUT PANDANG FILSAFAT TENTANG TEORI KEBENARAN Berdasarkan teori kebenaran dalam rumus tersebut adalah merupakan teori kebenaran consensus. Yang disepakati bersama sama. Belum mencakup kebenaran secara universal dalam kondisi ketebalan. Hal ini memungkinkan adanya proposisi yang bertentangan dengan obyek dan hasil saat dilakukan konfrontir. memahami Jenis teori kebenaran : 1. Teori Koherensi (The Consistence/Coherence Theory of Truth) 2. Korespondensi (The Corespondency Theory of Truth) 3. Pragmatisme (The Pragmatic Theory of Truth) 4. Teori Performatif 5. Teori consensus. Pendekatan teori kebenaran pada kasus ini masih bersifat kesepakatan para ilmuwan Sumber-Sumber Pengetahuan : 1. Rasionalisme 2. Empirisme ( pada bagian ini saya belum melakukan pengujian fisik pada suhu material ) 3. Kritisisme 4. Intuisisme Secara umum kebenaran bersifat : 1. Rigid 2. Universal 3. Nonkontradiktif 4. Bisa dibuktikan 5. Radic Quote philosophy : It is with logic that one proves, it is with intuition that one invents ( Henri Poincare ). My quote philosophy : Matematika sanggup memprediksi kesalahan, namun belum sepenuhnya membuktikan kebenaran mutlak. Saya tidak punya niat apapun selain memperbaiki pendapat akademik. Yang terlanjur menyebar luas sampai ke pelosok dunia ( khususnya pada materi perpindahan panas ). Dan kita bekerja sama untuk saling memperbaiki dan menguji kebenarannya. Alangkah indahnya dunia ini jika manusia saling belajar dan menjadi bijaksana untuk segala permasalahan yang dihadapi bersama di dunia ini. My favourite music : The lonely Shepherd by Andre rieu feat Gheorghe Zamfir Marriage d’amor -paul de senneville II Jacob’s piano The gael - last of the mohicans-Royal scots dragoon Vincencius Sufijan Hadi