Thermodynamics

[Maqboole]

Hello

i was working on my crousework on Thermodynamics of chemical engineering i would like to ask for help as i would be vevry appriciateable & other might benifi from it too the question is .

 

Calculate the rate of hear transfer from vertical plate of side length 1m ,if its surface temperature is 500k & air flow by the plate due to natural convection,at a temperature of 300k.

Data:

 

For air viscosity ; 1.9x10^-5 kg m-1 s-1

specific heat capacity ;1.01 KJ kg-1 k-1

 

Cofficient of cubical expansion : 3.33x10^-3 k-1

Density ;1.165 kg m-3

thermal conductivity;0.0337 Wm-2 k-1

Gravity; 9.81s-2

For Lamimar flow: Nu = 0.59 [Gr.Pr]^0.25

For Turbulent Flow Nu=0.13 [Gr.Pr.]^0.33

[CaptainPanic]

Normally, you first determine the Nusselt number, then you determine the heat transfer coefficient using the Nusselt number:

[math]Nu=\frac{h\cdot{L}}{\lambda}[/math]

 

But to determine Nu, you have to find out the Grashof (Gr) and Prandtl (Pr) numbers...

 

First of all, your formula's and mine are different. My book says:

[math]Nu=0.55(Gr\times{Pr})^{\frac{1}{4}}[/math] for [math]10^4<Gr\times{Pr}<10^8[/math]

[math]Nu=0.13(Gr\times{Pr})^{\frac{1}{3}}[/math] for [math]Gr\times{Pr}>10^8[/math]

It's funny that the coefficients 0.55 and 0.13 are different. But more important, we now know also that laminar and turbulent can be expressed in terms of Gr x Pr... and this should help you choose between the two.

 

Wikipedia gives 3 formulas for the Grashof number (Gr), and this one seems good for you:

[math]\mathrm{Gr_L} = \frac{g \beta (T_s - T_\infty ) L^3}{\nu ^2}[/math], for vertical flat plates.

The Prandtl number is also straightforward (and you can sneak into wikipedia to see the formula):

[math]\mathrm{Pr} = \frac{\nu}{\alpha} = \frac{\mbox{viscous diffusion rate}}{\mbox{thermal diffusion rate}} = \frac{c_p \mu}{k} [/math]

 

So, now you have the heat transfer coefficient... which you use to calculate the overall heat transfer coefficient (I assume you know how to do that - hint: overall heat transfer coefficient often has symbol U, units [W/m2K], which is the same as the heat transfer coefficient).

 

Also take into account that the air is heating up. Here I am really not so sure, but I'd use the log mean temperature difference as value.

 

I hope that you can take if from there... and I hope that what I wrote is correct. Your problem is a tough one!

 

Good luck with your studies! Heat and mass transfer are a b*tch, but oh so important. And once you've learned it a bit, you're a hero! Unfortunately, it's all empirical, so I had little choice other than post formulas... it's a little bit against the forum rules to help this much (I'm not sure though in this case - please post again if you know the answer from your teacher!).

[Maqboole]

first of all i would like to say my thank to CaptainPanic. In helping me in that question i would like to inform you that i did manage to calculate Gr & Pr .so i multiplied them & my answer came up as 3.08x10^9.

What to do next now i am confused .

yes i have the answer of the question answer is 1953 W.

Can you help me

Thx

[CaptainPanic]

So, now you have to use this formula:

 

[math]Nu=0.13(Gr\times{Pr})^{\frac{1}{3}}[/math] because [math]Gr\times{Pr}>10^8 [/math]... ([math]Gr\times{Pr}=3.08\cdot{10^9} [/math]).

 

Then you know Nu, and you can continue the way I explained.

 

Sorry for not checking the entire answer now... I am not in the mood today for heat & mass transfer - I have my own problems today that I need to solve and they cost enough effort

[Maqboole]

Thz a lot i did get the solution & i did it right.Thz a lot if you would like i can explain that to you.