×
INTELLIGENT WORK FORUMS
FOR ENGINEERING PROFESSIONALS

Are you an
Engineering professional?
Join Eng-Tips Forums!
• Talk With Other Members
• Be Notified Of Responses
• Keyword Search
Favorite Forums
• Automated Signatures
• Best Of All, It's Free!

*Eng-Tips's functionality depends on members receiving e-mail. By joining you are opting in to receive e-mail.

#### Posting Guidelines

Promoting, selling, recruiting, coursework and thesis posting is forbidden.

# Temperature distribution

## Temperature distribution

(OP)
Hi Everyone,

I am posting this for some assistance as I haven't done energy balances in a while, but have been struggling with this one problem as it might be seen as an easy problem it has become some what difficult for me.

7000 L water at ±83oC is collected in the vessel and it takes 3 hours 45 min to collect. Once the water has been collected it has to be cooled to 23oC the reason is so that the raw material doesn't form lumps as it does when it is added to hot water and the cooling process takes 3 hours 45 min, then the raw material is added after which the water has to be reheated again to 83oC which takes another ±3 hours 45 min and once the temperature has been reached it then has to be maintained at this temperature for another 45 min – hour. Once it has been visually determined that the material is completely dissolved the water has to then be cooled to 23oC which takes +3 hours 45 min.

What I want to be attempting as a method to reduce this amount of time as well as reducing energy consumption, I want to collect about 1000 litres of cold water @ 30C and then add the raw material and thereafter start collecting water @ 83C at a rate 32 L/min under mixing to make up to 7000L when this collection is taking place the process is operating semi-batch wise. I expect for the temperature to increase but I want to know how will this temperature increase with time without having to induce heat from steam and this will there after give me an indication.

If I say: Accumulate = Heat Gain - Heat Loss

The energy balance which I done is as follows:
Ek no change kinetic energy
Ep no change potential energy
Ws Work do by the stirrer is neglidgible
ṽ = volumetric flow rate
Tref = 0 Tf = Feed temperature
Cp = Constant ṽ (L/min) 31.1
ρ = constant V (L) 1000
Energy Accumulation = ∆Enthalpy Energy Tf (oC) 83
dρVCpT/dt = ρṽCp(Tf - T) Ti (oC) 30
dρVCpT/dt = ρṽCp(Tf -T)
ρVCpdT/dt = ρṽCp(Tf - T)
dT/dt = ṽ/V(Tf - T)
Therefore Final formula after intergration
T = 83 - 32.154 e^(t + 0.5)
When drawing this distribution I get values which are completely out.

Any ideas on where I could be making a mistake.

#### Red Flag This Post

Please let us know here why this post is inappropriate. Reasons such as off-topic, duplicates, flames, illegal, vulgar, or students posting their homework.

#### Red Flag Submitted

Thank you for helping keep Eng-Tips Forums free from inappropriate posts.
The Eng-Tips staff will check this out and take appropriate action.

Close Box

# Join Eng-Tips® Today!

Join your peers on the Internet's largest technical engineering professional community.
It's easy to join and it's free.

Here's Why Members Love Eng-Tips Forums:

• Talk To Other Members
• Notification Of Responses To Questions
• Favorite Forums One Click Access
• Keyword Search Of All Posts, And More...

Register now while it's still free!