Heat transfer from water in a stainless steel pipe: analysis

  • Thread starter CKVEng
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  • #1
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Hi Guys, this is my first post and I'd love a bit of help here, appreciated this may seem basic but it's been a few years since I did this last.

I have water in a stainless steel pipe (ID = 0.0381m, OD = 0.0411m) network of roughly 10m length, at 120 bar, roughly 50°C (+ or - 5).
This water is having 7kW of heat transferred into it by a cylinder and piston and I am going to assume that ALL of this heat is transferred directly into the water.

I need to work out whether it is practical for me to use a passive (or active ie with a fan) radiator type design to disperse this heat input into the air, or if I have to use a tank of cool water (assumed to be 20°C), to keep it within it's operating temperatures (90°C is the cut off maximum temperature) and if so, what are the sizes required?

Currently I am thinking about using a tank of cool water and passing the piping through this to cool it, but I need to know the required length of pipe in contact with the body of water to keep the temperature stable.

My mass flow rate (q) varies like a sine wave between 0 and 687.1 litres/minute (at 2hz) and I have been told to take an RMS value of this.

Homework Equations


q RMS = √(x^2 + ∂^2)

Re=(μ.u.Dh)/μ


The Attempt at a Solution



My RMS value was determined by my assumption that x = the average value ie the max/2
∴ x=343.55
∂= the standard deviation (?) which I was unsure about and assumed to be the same ie
∂=343.55

∴ RMS of flow rate = 485.9 Litres per minute = 8.0972 x 10^-3 metres^3 per second

Re=(ρ.u.Dh)/μ

where
ρ=998.2 (water at 20°C)
μ=1x10^-3 kg/ms
u=1.274.q/d^2=7.106500505 m/s
Dh=4.a.p= 0.00054585171 m

and
d= 0.0381 m
a= [itex]\pi[/itex] .(d/2)^2 = 0.00114009183 m^2
p= [itex]\pi[/itex] .d m
q= 0.008097227 m^3/second

∴ Re= 3832 = transient flow

this is incorrect as I know the flow to turbulent. Is this because I am using an incorrect RMS value?

How do I continue this solution to find my required pipe area or length?
Let me know if more information is required for the solution,
Any help will be massively appreciated!
Thanks,
Chris

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
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The method of heating isn't clear. Is the fluid being heated before it enters the pipe? What is this thing about a piston and cylinder doing the heating? If there is a pump before the pipe, do you first want to determine the temperature coming out of the pump? Do you just have to keep the temperature less then 90 C for the water coming out of the pipe?
 
  • #3
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Yes the fluid is being heated before it enters the pipe (by losses from a cylinder and piston).

We can assume that the 'coolant' (in this case water) is being directly given 7kW of heat at a constant rate. This coolant is then passed through the cylindrical tank to cool it down.

The coolant can be assumed to be 50°C to start with (but will be constantly heated 'perfectly' at a rate of 7kW).

The coolant must be kept below 90°C, ideally it will stay stable at 50°C.

I think I need a pipe length of 12m going through the coolant tank using the following methed:
2πkL
q= Ln(R2-R1) x(T1-T2)

as R1= 0.01905m
R2 = 0.02055
T1=50
T2=20
q=7000W
k=20 (taken for stainless steel)

that gives L = 12.07m

is that correct?

I then have a tank of water to cool the coolant which needs to be kept cool.
I am thinking of having a thermometer which refills the tank with cold water when the tank T gets above 25°C.
If we assume that 7kW of heat is all going into the tank, what rate of heating do I get?
I worked it out that I'd need at least 56 litres using:
mcΔT=q
∴ m= 7000/4.182x30 = 55.79kg
but is this how much mass flow rate i'd need?
very unsure about this thanks for your help!
 
  • #4
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that equation didnt come out properly its meant to say
q= (2πkl/Ln(R2-R1))*(T1-T2)

is it correct to use k for the steel? or am i meant to use the value of water?
 
  • #5
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If the water goes into the pump at 50C, the flow rate is 486 Kg/min, and the heating rate is 7kW, what is the exit temperature of the water from the pump?

Chet
 
  • #6
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Sorry I'm not sure.
Initially the water would start at 20C but the pump is giving it 7kW of heat. We will allow this to take the coolant T up to around 50 degrees before passing the coolant through the cooling tank.
Does that help?
Sorry I cant give more info I'm still trying to wrap my head around this.
 
  • #7
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4,672
Sorry I'm not sure.
Initially the water would start at 20C but the pump is giving it 7kW of heat. We will allow this to take the coolant T up to around 50 degrees before passing the coolant through the cooling tank.
Does that help?
Sorry I cant give more info I'm still trying to wrap my head around this.
Well, for what it's worth, here's my calculation:

WCpΔT=Q

(8.1)(4.186)ΔT=7

ΔT = 0.2 C

So, according to this, if the water enters the pump at 20C, it exits at 20.2 C. Thoughts?
 
  • #8
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Where does your 8.1 come from?

I've been using mcΔT=Q
to work out how much the tank is heated up, and therefore how often I need to refill it with water.

so far ive got that the tank of 545 litres heats up by about 11C every hour (if all 7kW is transferred through the pipe wall from coolant to tank fluid).

I want to only pump new (colder at 20C) water into the tank when it reaches a cut off point, say 35 degrees.
How do i work out the mass of 35C water that needs to be replaced with 20C water to keep temperature stable or to bring it down from 35 to 25?

thanks
 
  • #9
21,156
4,672
Where does your 8.1 come from?

I've been using mcΔT=Q
to work out how much the tank is heated up, and therefore how often I need to refill it with water.

so far ive got that the tank of 545 litres heats up by about 11C every hour (if all 7kW is transferred through the pipe wall from coolant to tank fluid).

I want to only pump new (colder at 20C) water into the tank when it reaches a cut off point, say 35 degrees.
How do i work out the mass of 35C water that needs to be replaced with 20C water to keep temperature stable or to bring it down from 35 to 25?

thanks
The 8.1 came from 460 kg/min divided by 60 seconds.

Assume that the tank is well mixed, such that the outlet temperature is equal to the temperature of the water in the tank. Then from a steady state heat balance on the tank:

[tex]qρC_p(T_out -20)=Q[/tex]
where q is the volumetric flow rate of cooling water, ρ is the water density, Cp is the heat capacity of water, and Q is the heat load.

If you run the system in an on-off way, then you need to solve the transient equation:

[tex]mC_p\frac{dT}{dt}=Q-qρC_p(T -20)[/tex]

where m is the mass of water in the tank (assumed constant), t is time, and T is the temperature in the tank at any time (also equal to the exit water temperature). For this analysis to apply, the tank has to be well stirred.

Chet
 
  • #10
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The thing is, why do you have to remove heat from the water flowing through the pipe at all if the temperature coming out of the pump is only 0.2 C higher than that coming into the pump? Does it really matter whether the water temperature is 20 C or 20.2 C?

Chet
 

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