Heat Transfer - Chapter 3 - Cylindrical Systems - Temperature profile, Thermal Resistance, U-Value

[@chibigon01]

Nice and simple analysis of the HT equation in cylindrical systems. Thank you.

[@evokeer1929]

What happens to the equation of total heat rate q_r if we add an outer wall "T_surrounding" and attribute an emissivity to the surface?

[@jpmarquez49]

Once you have the heat rate q_r you add the resistance of the new wall to the Sum of thermal resistances, then you use the emissivity to calculate the new rate of heat that is coming out of the new wall (this one should be minor than the original q_r). Then you have the net value of the heat rate that is going to be in direct contact to the fluid. That way you have q_{new r}= Delta(T)/Sum(Thermal Resistance), beacuse Delta(T) is T_{fluid in the middle} - T_surrounding, you can find now T_surrounding. T_{surrounding} = T_{fluid in the middle} - Sum(Th. Res.) * q_{new r}

[@ivanaristover]

Great explaination. How'd you calculate for radial heat flux though solid cylinder? Thank you.

[@vincenciussufijanhadi1287]

maaf, sebelumnya, saya ingin bertanya mengenai kasus silinder berongga. berapakah nilai x ( jari jari selubung silinder )

[@samMe-vs1cx]

Great work.thanks

[@vincenciussufijanhadi1287]

hello sir, how to define r (casing radiant )? sorry distrub you

[@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.

[@patrickobogo7779]

Thanks alot.

[@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. continue.....

[@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 )], 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...