What is the best resource to derive the Heisler charts in xy, cylindrical, and spherical? I want to calc the answers exactly instead of using a graph. Thanks

why am i seeing this now (;´༎ຶД༎ຶ`)

great video thank you

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. continue.....

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 )], such x : 0 ≤ x ≤ e ..….…. Equation 3 R : dR = 1: [ Ln ( e/ x )], such x : 0 ≤ x ≤ e …….… Equation 4 R = dR : [ Ln ( e/ x )], such x : 0 ≤ x ≤ e ….…... Equation 5 R = ( e - x ) : [ Ln ( e/ x )], such 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. Such, dR : Instantaneous thickness (R2 - R1 ) atau ( outer - inner ). dT : Instantaneous Temperature e : Euller logaritm 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 Figure apply logaritma natural. Ratio R : dR is inconsistant 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. continue....

In the stage 1-2 shouldn’t the hot piston be in the BDC and about to move upward?

Thanks a lot

Thank you, professor.

Bro I'm gonna touch you

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 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. continue...

SURAT TERBUKA KESALAHPAHAMAN PERSAMAAN PERPINDAHAN PANAS KONDUKSI ALIRAN STEADY STATE PADA KASUS BENTUK SILINDER DAN BENTUK BOLA Kepada yang terhormat, Siapa saja mempelajari mata kuliah perpindahan panas (teknik mesin, teknik kimia dan lain lain). Di Tempat, Bersama dengan surat terbuka ini, dengan kerendahan hati, kami mengirimkan surat terbuka kepada siapapun , yang boleh dibaca oleh siapa saja yang berkenan membaca surat ini. Maksud dari surat terbuka ini adalah berdiskusi ulang mengenai penjelasan perpindahan panas ( Heat transfer ) aliran tunak ( Steady state ) pada kasus dinding silinder berongga dan bentuk bola Saya penulis surat keberatan dengan rumus yang sudah ditulis di buku dan literatur. Khususnya pada perpindahan panas konduktor pada dinding silinder dan bentuk bola aliran steady state. Secara singkat keberatan saya adalah : Uraian singkat keberatan saya dengan rumus perpindahan panas ( Heat transfer ) yang sudah ada adalah : Untuk bentuk silinder Persamaan umumnya adalah Q = -k. A. (dT)/ (dR ). Hukum fourrier Sehingga persamaan umum diturunkan untuk kasus silinder menjadi, Q = k. 2π.R.L ( dT )/ ( dR ), ……… Persamaan 1 Pindah ruas kiri di Integral dR/R. Dengan batas atas outer ( R2 ) dan batas bawah inner ( R1 ), dT dintegralkan menjadi ∆T. sehingga persamaan berubah menjadi, Q. Ln ( R2/R1 ) = k. 2π.L ( ∆T ) Q = k. 2π.L ( ∆T) / [ Ln ( R2/ R1 )]. ………. Persamaan 2 Persamaan 1 equivalen dengan persamaan 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 adalah konstan ( C ), sehingga C = ( dT ).R / ( dR ) = ( ∆T) / [ Ln ( R2/ R1], C tidak dilibatkan lagi. ( 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 = tebal ( ∆T ).R / tebal = ( ∆T) / [ Ln ( R2/ R1 )], ∆T di eliminasi. R / tebal = 1/ [ Ln ( R2/ R1 )]. 1: [ Ln ( R2/ R1 )]. = R : tebal. 1: [ Ln ( R2/ R1 )]. = R : ( R2 - R1 ) R : ( R2 - R1 ) = 1: [ Ln ( R2/ R1 )], diubah untuk R1 adalah x dan R2 adalah e ( bilangan euller ) sehingga, R : ( e - x ) = 1: [ Ln ( e/ x )], dimana x : 0 ≤ x ≤ e ..……. Persamaan 3 R : dR = 1: [ Ln ( e/ x )], dimana x : 0 ≤ x ≤ e ….…… Persamaan 4 R = dR : [ Ln ( e/ x )], dimana x : 0 ≤ x ≤ e ….…… Persamaan 5 R = ( e - x ) : [ Ln ( e/ x )], dimana x : 0 ≤ x ≤ e ….…… Persamaan 6 Persamaan diatas adalah perbandingan jari jari selubung silinder terhadap jari jari silinder Dimana, dR : Ketebalan sesaat. (R2 - R1 ) atau ( outer - inner ) dT : Temperature sesaat e : Bilangan euller atau R2 atau outer K : Konduktifitas thermal materi L : Panjang silinder Q : Nilai kalor R : Panjang jari jari selubung silinder X : Inner atau R1 ∆T : Hasil integral dari dT ∆R : Hasil ( R2 - R1 ) = ( outer - inner ) = tebal 2πR : Keliling selubung silinder 2πR.L : Luas selubung silinder Uraian keberatan saya adalah : a. Dari persamaan 3 dan 4. Perbandingan jari jari selubung terhadap ketebalan silinder tidak pernah kurang dari 1. Kondisi ini terjadi saat ketebalannya ( e - x ) r , yang mana e ( logaitma natural ) adalah jari jari outer dan x adalah jari jari inner yang berkisar 0 < x < e^0 . Alasan keberatan saya adalah jari jari selubung selalu bergerak membesar mengikuti perubahan luas penampang silinder. Sedangkan ketebalan bergerak secara linier. sehingga tidak pernah terjadi suatu peristiwa jari jari selubung lebih pendek dari pada ketebalan. Dengan demikian jari jari selubung selalu lebih besar terhadap ketebalan dalam kondisi apapun. b. Dari persamaan 3 dan 4. Saat silinder pejal ( tidak memiliki lubang ), maka suhu akan menjadi +∞ (suhu meningkat sangat besar tak terbatas pada pusat silinder). Kondisi ini terjadi saat ketebalannya ( e - 0). e ( logaritma natural ) adalah jari jari outer merupakan bilangan natural dan 0 adalah jari jari inner. Alasan keberatan saya adalah sangat tidak masuk akal suatu materi dalam kondisi pejal dapat menahan panas dengan suhu + ∞. Pendapat saya adalah suhu masih bisa diketahui meskipun sulit di praktekkan dalam kondisi pejal . c. Dari persamaan 3 dan 4. Perbandingan jari jari selubung terhadap ketebalan lebih besar dari 1 ( tidak memiliki batas ) dan berakhir pada 0 saat ( e - e ) yaitu outer = inner . Alasan keberatan saya adalah jari jari selubung memang harus lebih besar daripada jari jari. Namun memiliki batas nilai tertentu !!! d. Dari persamaan 3 dan 4. Secara umum alasan keberatan saya dari point a, b dan c adalah penggunaan logaritma natural menyebabkan ketidakpastian perbandingan jari jari selubung terhadap jari jari. Yang ditampilkan pada grafik 1 dibawah. a. Saat ketebalan ( e - x ) , 0 < x < e^0, maka perbandingan jari jari selubung < ketebalan. b. Saat ketebalan ( e - x ) , x = 0 , maka perbandingan jari jari selubung terhadap ketebalan + ∞. c. Saat ketebalan ( e - x ) , x = e^0, maka perbandingan jari jari selubung = ketebalan. d. Saat ketebalan ( e - x ) , e^0 < x < e, maka perbandingan jari jari selubung > ketebalan. e. Dari persamaan 3 dan 4. Pertanyaan saya adalah : Bagaimana menghitung suhu di pusat silinder saat panas berasal dari luar silinder secara konduksi dengan aliran steady state ???. mungkinkah jawabannya adalah ( - o C )?. f. Dari persamaan 3 dan 4. Penggunaan logarotma natural adalah mengitung satuan luas tak berdimensi tidak menghitung jari jari selubung silinder ( R ). g. Dari persamaan 3 dan 4. Ramalan kesalahan hitung sangat serius. Saat outer menjadi sangat besar dan inner mendekati outer ( relative tipis ). Maka kesalahan hitung pada suhu menjadi sangat besar. Mengapa rumus yang sudah ada dianggap sebagai kebenaran ?, hal ini disebabkan oleh nilai kalor harus selalu ( konstan ) di tiap tiap perubahan ketebalan yang mengalami perubahan suhu meningkat saat mendekati pusat silinder seiring dengan perubahan ketebalan, yang mana suhu awal berasal dari inner jari jari. Dengan catatan sumber panas berasal dari dalam silinder. continue...

Great! Thank you very much sir

Thanks a lot, but what if we have two pumps in this case?

Please solve vith high pressure turbine

You saved me for three years sir

That's a great way to lose a finger...jeeze... Just use a flywheel, not a fan

I didn't know Pedro Pascal taught fluid mechanics!

it's still doesn't make sense mathematically!! how can the gradient operator work on the vector field? the gradient works only on the scalar field, the divergent does.

It's Nusselt Number not new salt.

Thank you for this course! Are you planning to cover CFD also?

Thank you so much. 🙏🏼 The double pipes with dq pic ture makes me understood clearly for how the equation is derived from!

Thank you Ron for uploading this beautiful series. You have helped me design a small thermal power plant for my coursework. 11 years later and you're still a legend.

Thank you sir❤

Thanks man🎉

Thank you! Your videos are very helpful for me as I'm studying for my next exam

You look like Joel from The Last of Us series

Hope you see thus Sir. This is 2024 and ur video's are helping me

Your videos are still helping us in 2024.... Thanks a lot Sir

Have You done CPU cooling?

Hands down, the best fluid professor at UofC

There's is/was Stirling Electric Generation Plant in Victorville, California, powering thousands of homes. With an efficiency almost double that of solar photovoltaic panels, I wonder if we couldn't capture heat and store it in sand, and use that heated sand to power a Stirling engine electricity generator. I like in the Mojave Desert, so it seems like an obvious choice.

Good morning

thank u RON HUGO U SAVED MY LIFE

Gracias por el aporte

Recommending these over Alexandra Niedden's videos for Heat Transfer in 1D

thank you sir

Gracias por el aporte 🙏🏻

Excelente explicación 👍🏻

Excellent 👌🏻👏🏻

Can someone answer this. In finding V3, there are two ways that I know. One is in the process 2-3 where the pressure is constant. That gives T3/T2 = V3/V2. The other is that V3=V2+cVd. Now, the question is what is more accurate. I mean, what should I use? For example, you all had the values in both ways. Trust me, there's a discrepancy when use both. Not a small, but quite large discrepancy. Can someone explain. Thank you!

Gracias por el aporte 🙏🏻 Tengo duda en 10:10 ¿Cómo llega a nuestros oidos el sonido producido por la burbuja colapsando? ¿Se transporta por el líquido hasta las paredes del recipiente y la superficie expuesta?

Excelente 👍🏻

Excelente video 👌🏻 consumiré toda la serie. Muchas gracias🙏🏻

is the v volume or velocity?

do biot number characteristiic length is equal to diameter/6 or radius/3 then why you took radius it should be

pure gold to do it by hands, not a fancy bright new PC solver of some sort

What if "ri" is negligibly small?

not bad greg line integral f to q along curve denot3d in integral f

Pedro pascal teaching physics

Thanks grandpa for understanding our confusion learning issue :)