I have seen this Channel and now I am here again for mathematical rewiew. This is brilliant because normally exchanger effectiveness curvrs are available as a function and capacity ratios of both streams, for perfectly mixed or plug-flow conditions. Yours purpose to give us knowledge is valuable and give the result of a problem that is addressed starting from basic equations and using compatible specific geometry. In the case of heat transfer we can use Paclet numbers from mass transfer experiments but than we have approximation. I think that is new and better path but I must incurage to input in "Tube and tubeside". 😮
Awesome explanation. Just one question how to change external ho and internal heat transfer rate hi ? Do we have to change to material of tube to do so ? Please let us know the standard to get the heat transfer rate of selected material, eg SA105 316L.
The specific heat capacity of water at the temperature of 80 degrees and 20 degrees does not differ much and both are almost equal to 4180 J/kg*k . As a result Cc=Ch=4180, not 1500
These are assumed values. For hi, there are correlations, and I should make a video on them. Unfortunately, the exact values for ho are difficult to calculate, and I’m not aware of any correlations. You can probably get some approximate values by assuming the tube is a flat plate.
These coefficients depend on the flow velocities (or on throughput) on both sides. They can be obtained by interpolating through the data found in the literature like HEAT EXCHANGER DESIGN GUIDE M. NITSCHE AND R.O. GBADAMOSI for ethanol and water. I interpolated the coefficient and obtained 6th degree polinomials for them using Mathematica. I used Fit[] comment for interpolation. This is a great video, congrats
A double pipe exchanger was oversized because no data were available on the rate at which dirt accumulated. The exchanger was originally designed to heat 13,000 lb/hr of 100 percent 5 acetic acid from 150 to 250 °F by cooling 19,000 lb/hr of butyl alcohol from 157 to 100 °F. A design coefficient UD 85 was employed, but during initial operation a hot-liquid outlet temperature of 117 °F was obtained. It rose during operation at the average rate of 3 °F per month. What dirt factor should have been specified for a 6-month cleaning cycle?
I’m looking for very simple exclamation to show people with very simple limited minds who cannot comprehend math that is no more difficult than pluses and minuses and simple surface area using whole round numbers. I remember having this in HVAC School and thermal dynamics and heat heat exchanger design class. I even had a little bit of a hard time so there’s definitely no way I can describe using this method to people who just can’t comprehend anything more complicated than simple analogies. Thank you for your video.
"the quantity of heat transmitted through a UNIT THICKNESS of a material - in a direction normal to a surface of unit area - due to a unit temperature gradient under steady state conditions" I don't know why thickness of fluids is lost in your equations when you calculating total resistance. When you zoom in pipe wall to calculate overall resistance it looks like you just consider molecules of fluids( both inside and outside pipe) very close or just attached to pipe inner and outer walls, so why do you bring hi and ho when fluids are not there in your zoomed illustration
Typically, film coefficients incorporate all the details of fluid flow, including the fluid thickness. Having a larger outer diameter (and thus more fluid thickness) will affect the overall heat transfer, but this is reflected in the h values. If the fluid were perfectly still, we would be able to calculate the heat transfer through conduction, which directly incorporates the thickness of the fluid.
@@robertbrown7812 Thanks. I perform your calculation very similar to my case and get strange results when I want to calculate surface needed for the exchanger metal plates separate 2 equal flows of air. As flow rates are equal ( 1 cu.meter/ second), temperature rise ( for cold air) must be equal to temperature drop in hot air. My exchanger then will have 3000 watts power transmission between 2 fluids ( 1 kilo of air with 1.0 cp and temp rise/drop 3 degrees). Tcold(in) =19, Tcold(out) =22, Thot(in) 36 and therefore Thot(out) 33. putting temp numbers in mean log formula I get delta T= 15. then I calculate Rtot like you did (mean deltaT/ power= 15/3000= 0.005 T/W) Air thermal conductivity is 0.025 (W/m.K) . hi and ho are the same ( both fluid air) ,and I consider plates clean with no rust ,( maybe I'm applying wrong number(0.025 w/k here?!,). So Rtot=1/A( 1/h + 1/h ) = 2/ A.h = 2 / 0.025 A= 80/A . Rtot being calclated 0.005 before, 80/A=0.005 ==> A= 80/ 0.005 = 16000 m2. Do I really need 16k sq.meter of heat heat exchanging surface? where am I wrong. I'm very confused about it.
@@AM-lu5hm Thermal conductivity (k) is only one piece of what makes up a film coefficient (h). A film coefficient has units of W/(m^2 K), and is usually calculated by finding the Nusselt number (Nu = hL/k). After a bit of Googling, I found a paper: "Convective Heat Transfer Coefficients in Concentric Annuli" by Dirker and Meyer. Hopefully that's enough to get you started.
HI, I'm from Brazil and I have a project to submit to my professor tomorrow. Your video helped a lot, but I still have some doubts. Can you help me please? By email or something?
The temperature of methanol steam leaving the synthes is equipment is 450℃, which is cooled by waste heat boiler.The waste heat boiler produces 0.45MPa saturated steam.It is known that the in let water temperature is 20℃ and the pressure is 0.45MPa.The molar ratio of feed water to methanol is 0.2.Assuming that the boiler is adiabatic operation, calculate the out let temperature of methanol.The given data are as follows: The molar constant pressure heat capacity of CH3OH (vapor) at 450~300℃ can be calculated as follows: Cp,m=(19.05+9.15× 10-2T)(J/(mol•K)) Enthalpy of liquid water at 20C:H1= 83.74kJ/kg= 1507kJ/k mol； Enthalpy of 0.45MPa saturated steam:H2 = 2747.8kJ/kg = 49460 kJ/kmol。
I have seen this Channel and now I am here again for mathematical rewiew. This is brilliant because normally exchanger effectiveness curvrs are available as a function and capacity ratios of both streams, for perfectly mixed or plug-flow conditions. Yours purpose to give us knowledge is valuable and give the result of a problem that is addressed starting from basic equations and using compatible specific geometry. In the case of heat transfer we can use Paclet numbers from mass transfer experiments but than we have approximation. I think that is new and better path but I must incurage to input in "Tube and tubeside". 😮
I must say through this video, I could recollect the concepts I learned ages ago!! It helped a lot. Keep up the good work Professor! :)
Bro can u explain calculation about car radiator I'm designing radiator for F1 style car having ktm 390 engine can u help me in this
tike
This was very hopeful ... Many thanks, my good sir.
hey i am from ethiopian and this video help me a lot and i got a lot understand about it, thank u so much
Ante. Mn tiseraleh? 👏
With mH and mC data in kg/h I'm able to get same results for Q. Thanks for the video.
Excellent explanation
Awesome explanation. Just one question how to change external ho and internal heat transfer rate hi ? Do we have to change to material of tube to do so ? Please let us know the standard to get the heat transfer rate of selected material, eg SA105 316L.
Thank you sir, this calculation is very interested
hi i would like to know where did we obtain the convection heat transfer from the hot side and the two other values that follow?
Same question
@@hassanismail6120 he used tabulated values but you can calculate it using
Correlations for Nusselt number
Really great explanation.
at the 5.30 minute, how did you consider Celsius values to Kelvin?
Nice video. congrats 😄
Beautiful. Thank you sir
very nice my friend :)
The specific heat capacity of water at the temperature of 80 degrees and 20 degrees does not differ much and both are almost equal to 4180 J/kg*k . As a result Cc=Ch=4180, not 1500
how did you transform from celsius to kelvin temperature?
Can you please show how you got the ho coefficients of 700 and 1300 W/m^2K?
I would also like to know this please
Yes please explain
These are assumed values. For hi, there are correlations, and I should make a video on them. Unfortunately, the exact values for ho are difficult to calculate, and I’m not aware of any correlations. You can probably get some approximate values by assuming the tube is a flat plate.
@@PostcardProfessor Without knowing how to get these coefficients, this video is useless. we cannot calculate the area required.
These coefficients depend on the flow velocities (or on throughput) on both sides. They can be obtained by interpolating through the data found in the literature like HEAT EXCHANGER
DESIGN GUIDE M. NITSCHE AND R.O. GBADAMOSI for ethanol and water. I interpolated the coefficient and obtained 6th degree polinomials for them using Mathematica. I used Fit[] comment for interpolation. This is a great video, congrats
OMG,! You saved my life. Thank you 🙏
Thankyou
But please explain that how to find the valu of hi &ho.
A double pipe exchanger was oversized because no data were available on the rate at which dirt accumulated. The exchanger was originally designed to heat 13,000 lb/hr of 100 percent 5 acetic acid from 150 to 250 °F by cooling 19,000 lb/hr of butyl alcohol from 157 to 100 °F. A design coefficient UD 85 was employed, but during initial operation a hot-liquid outlet temperature of 117 °F was obtained. It rose during operation at the average rate of 3 °F per month. What dirt factor should have been specified for a 6-month cleaning cycle?
R = L /KA
As L thickness of tube
Why you repeat it as one time meaning heat transfer to the surface so no need to repeat
Is that true ??
Can you make one for finned tube condenser in an airconditioning? And how to determine the number of fins to be used
can u show us how u calculate those ho and hi
plzz
The problem i have is i dont know the diameter. How should i proceed?
How you take ho and hi values
I’m looking for very simple exclamation to show people with very simple limited minds who cannot comprehend math that is no more difficult than pluses and minuses and simple surface area using whole round numbers. I remember having this in HVAC School and thermal dynamics and heat heat exchanger design class. I even had a little bit of a hard time so there’s definitely no way I can describe using this method to people who just can’t comprehend anything more complicated than simple analogies.
Thank you for your video.
i have no idea where you get rfo and rfi
Logarithmic mean temperature difference formula is different !!
Wouldn’t it be (T_h,i -T_c,o) - (T_h,o - T_c,i) for the numerator of the log mean temperature difference?
Your formula is for counter flowing exchanger i think
@@gappuma7883 No, he is right. Postcad professor made a mistake here.
how you calculate the ho and hi ??
هل من مصل على النبي محمد صل الله عليه وسلم وبارك الصلاة الإبراهيمية
How you got 700 and 1300 resistant values
sir i have make cooing volume 400m3/60min ITemp130c outt60 solv thery
how do take hi and ho values
Why you were using Celsius degrees all time except when calculating the mean global T, that in the formula suddenly you started using Kelvin degrees??
Because you use Kelvin to show changes in temperature. °C is an absolute value.
But 1K=1°C in change of temperature.
Thanks 👍
51,3?
"the quantity of heat transmitted through a UNIT THICKNESS of a material - in a direction normal to a surface of unit area - due to a unit temperature gradient under steady state conditions" I don't know why thickness of fluids is lost in your equations when you calculating total resistance. When you zoom in pipe wall to calculate overall resistance it looks like you just consider molecules of fluids( both inside and outside pipe) very close or just attached to pipe inner and outer walls, so why do you bring hi and ho when fluids are not there in your zoomed illustration
Typically, film coefficients incorporate all the details of fluid flow, including the fluid thickness. Having a larger outer diameter (and thus more fluid thickness) will affect the overall heat transfer, but this is reflected in the h values.
If the fluid were perfectly still, we would be able to calculate the heat transfer through conduction, which directly incorporates the thickness of the fluid.
@@robertbrown7812 Thanks. I perform your calculation very similar to my case and get strange results when I want to calculate surface needed for the exchanger metal plates separate 2 equal flows of air. As flow rates are equal ( 1 cu.meter/ second), temperature rise ( for cold air) must be equal to temperature drop in hot air. My exchanger then will have 3000 watts power transmission between 2 fluids ( 1 kilo of air with 1.0 cp and temp rise/drop 3 degrees). Tcold(in) =19, Tcold(out) =22, Thot(in) 36 and therefore Thot(out) 33. putting temp numbers in mean log formula I get delta T= 15. then I calculate Rtot like you did (mean deltaT/ power= 15/3000= 0.005 T/W)
Air thermal conductivity is 0.025 (W/m.K) . hi and ho are the same ( both fluid air) ,and I consider plates clean with no rust ,( maybe I'm applying wrong number(0.025 w/k here?!,).
So Rtot=1/A( 1/h + 1/h ) = 2/ A.h = 2 / 0.025 A= 80/A . Rtot being calclated 0.005 before, 80/A=0.005 ==> A= 80/ 0.005 = 16000 m2. Do I really need 16k sq.meter of heat heat exchanging surface? where am I wrong. I'm very confused about it.
@@AM-lu5hm Thermal conductivity (k) is only one piece of what makes up a film coefficient (h). A film coefficient has units of W/(m^2 K), and is usually calculated by finding the Nusselt number (Nu = hL/k). After a bit of Googling, I found a paper: "Convective Heat Transfer Coefficients in Concentric Annuli" by Dirker and Meyer. Hopefully that's enough to get you started.
Who here memorized the specific heat of water?
is it just me, or he get the wrong power from the hot side wrong? .1kg/s * 4.18 kj/kg*k * 15K = 6.27 kW, not 10.5 kW
95-70=25
HI, I'm from Brazil and I have a project to submit to my professor tomorrow. Your video helped a lot,
but I still have some doubts. Can you help me please? By email or something?
How did you get convection heat transfer
You calculation for LMTD is wrong. It will be the other way around.
I love you
The temperature of methanol steam leaving the synthes is equipment is 450℃, which is cooled by waste heat boiler.The waste heat boiler produces 0.45MPa saturated steam.It is known that the in let water temperature is 20℃ and the pressure is 0.45MPa.The molar ratio of feed water to methanol is 0.2.Assuming that the boiler is adiabatic operation, calculate the out let temperature of methanol.The given data are as follows: The molar constant pressure heat capacity of CH3OH (vapor) at 450~300℃ can be calculated as follows: Cp,m=(19.05+9.15× 10-2T)(J/(mol•K)) Enthalpy of liquid water at 20C:H1= 83.74kJ/kg= 1507kJ/k mol； Enthalpy of 0.45MPa saturated steam:H2 = 2747.8kJ/kg = 49460 kJ/kmol。
Q dot = m dot suffix H *C* delta T turns out to be 124.5Kw