calculating for Reynalds Number

[Smokey2007]

Did I do this right?? all places where i put "**" are the places i did calculations or made assumptions. the rest i found online.


here are my specs for a heat exchanger:
surface area: 2.4sq ft
capacity: 240k Btu/hr
flow capacity: 24 gpm
shell pipe ID: 1"
tube pipe ID: 3/4"
overall size: 4.5"*10 7/8"

** the shell's thickness is not given, so i will assume 1/4"
** and that the pipe's thickness is 1/8"

steam pressure 50PSI which equates to a temp of 281 F or 411.48K

density: 927.64 kg/m^3
specific heat: 4284.82 J/kg K
dynamic viscosity: .7168 kg/m hr

**calculate for velocity of steam
**v=4*flow rate/(pi*pipe's diameter^2)
**v=4*24GPM/(3.14*1"^2)
**v=10578 m/hr

**calculate hydraulic diameter of tube
**D=(r{outside}-r{inside})/2
**D=2(.875/2"-.75/2")=.125"=.003175m

therefore reynalds # = Dvp/u
**Re = .003175m*10578(m/hr)*927.64(kg/m^3)/.7168(kg/m hr)
**Re = 43463.9

now once i have this value and calculate the Prandtl number, I can determine the Nusselt number. i know these all deal with "convective heat transfer." but how does use this knowledge in determining the overall heat transfer properties of my overall system?????

i am aware of the following forumla and which each mean
dQ=kA(delta T)/dx

i am trying to solve for the "delta T" for both outside the shell, and inside the tube.

i realize i still need to calculate the resistances for 1. outside air against shell(convective) 2. shell's material(conductive), 3. steam against shell's material(convective) 4. steam against tube's material(convective) 5. tube's material(conductive), and 6. inside fluid against tube(convective).

what i just solved references #4(steam against tube's material).

 

[Smokey2007]

to clarify, i am trying to figure out how the Nusselt number fits into the overall equation for equating the Q for a system.

is it as simple as this?
Nusselt number = hD/k

so once i have the Prandtl and Reynalds number, i can plug in the "k" and "D" and then solve for "h."

and this "h" will be the only data that plays a greater role and can be used as when determining the "overall heat transfer coefficient" of the overall system???

so once i have this i move to the next resistance?

 

[Transient1]

I don't have much experience with heat exchanger design/selection. I can provide the following information though:

(1) The Prandtl number for a good number of fluids can be found tabulated versus temperature in most heat transfer textbooks.

(2) For a given set of flow conditions and thermal boundary conditions, Nusseult number is a constant and is related to Pr and Re. However, this relationship is just a good approximation. It is not an exact solution (sometimes its real close to exact). Yet, for engineering purposes, its fine.

(3) Yes, once you have your heat transfer coefficient, the next step is to move on to the next resistance. Q is continuous at each interface. Each section serves to provide the boundary conditions for the next section.

(4) It sounds like your trying to solve the system in 1-D, but don't forget it is a 3-D heat transfer problem. That may be okay for your purposes.

 

[Smokey2007]

excellent, so the "h" determined is what will allow me to get the overall heat transfer coeffiecient.

not sure how it sounds 1D. i am aware i have to count for each wall. just keeping it simple here.

 

[Transient1]

The h, is the heat transfer coefficient between a wall and the fluid. It wouldn't feasible to solve this problem by hand, but you have conduction and convection along the length of a heat exchanger. Therefore, the rate changes dependent on position (in 3-Dimensions), and you'll have hot spots at certain points due to pipe elbows, fittings, etc.

 

[25362]

Just rememeber that the overall U would be lower than the lowest "h" you may find.

 

[pmover]

ok, for heat exchanger

Q = UA LMTD

so if process conditions and properties are known (i.e. fluid, P, T, etc.), then Q and LMTD are known/determined.

find the UA.

using the GPSA data book &/or other resources, one can reasonably determine a U for the fluid @ process conditions and exchanger type. now the A can be determined.

however, since the surface area is known (? why, i do not know) the U can be determine. however, the U that is calculated should be compared to known values typical for the fluid, process conditions, and exchanger type.

fyi, the heat exchanger mfg should be able to provide you with the information you are seeking. also, you should obtain the U calc from the exchanger mfg as well.

good luck!
-pmover

 

[Smokey2007]

i do not understand why this doesnt work. all things are a known except for the temperatures of the plate wall, as well as the fluid on the other side. i want to determine the temperature of the fluid. the process is:
warm steam heats up wall, which heats liquid.


For Heat Exchanger Shell (dont really think this applies here)

k thermal conductivity 46 W/m-K
x thickness 0.00635 m
A Surface Area 0.176625 m^2



For Steam
Cp specific heat capacity 2222.87 J/kg-K
u viscosity 1.36E-05 kg/m-s
k thermal conductivity 0.016 W/m-K
v velocity 10728 m/hr
p density 1.8797 kg/m^3
flow 24 GPM

For Transfer Oil
Cp specific heat capacity 1948 J/kg-K
u viscosity 0.00088 kg/m-s
k thermal conductivity 0.1131 W/m-K
v velocity 10757.69 m/hr
p density 957.8 kg/m^3
flow 12 GPM

For Heat Exchanger Tube1

k thermal conductivity 400 W/m-K
x thickness 0.003175 m (i estimated this value)
A Surface Area 0.002968 m^2

******calculations
Prandtl Number Reynalds Number
Pr=Cpu/k Re=Dvp/u (i assumed the diameter of the
pipe will be .01m)

For Steam
Pr=1.89 Re=4122.39

For Oil
Pr=15.16 Re=32524.35

1.Steam-Wall
q=h(T)

h=.023(Re^.8)*(Pr^a)*k/D
347.34


2.Wall
q=k(T)/x
q=400(T)/.003175

h=400/.003175

3.Wall-Oil
q=h(T)

h=.023(Re^.8)*(Pr^a)*k/D
31419.62497 W/m^2-K

Now I have the three resistances, but when i plug them in i get the nothing that looks right. shouldnt the "q" of each of these be the same. therefore, i should be able to solve for each temperature???

Heat Exchanger has a rating of 79.125 kW
Initial temp of steam is 411 K.


Help please

 

[IRstuff]

Plug them into what? You need to solve at least 2 equations simultaneously.

TTFN

FAQ731-376

 

[Smokey2007]

I was thinking if i had the three Resistances, "1/hA" "x/kA" and "1/hA" than i found find U with 1/(R1+R2+R3).

and then use this in the equation Q=UA(delta T).

something that doesnt make sense is i have the value of BTU/hr that my heat exchanger produces, but not sure what to do with it.

 

[pmover]

Smkoey2007,

it is good to see your desire to solve this matter. however, after reading the last statement in your posting, i suggest that you review some basics theory/knowledge of heat exchangers. perhaps the attached file and webpage link will help you.
Also, steam flow is generally given as lbs/hr (kg/hr), but GPM is given, which indicates liquid flow (condensate).
Also, are you sure about the equation Q=UAdT? Instead of dT, how about LMTD?
Also, I am a little confused as to what type exchanger your analysis is for and what you are trying to determine.

hope this info helps . . .

http://www.wlv.com/products/databook/db3/DataBookIII.pdf

good luck!
-pmover

 

[pmover]

Smokey2007,

Whoops! had this thought after previous post . . .

if the exchanger is condensing steam, then the latent heat needs to be taken into consideration (from memory, roughly 970 Btu/lbm). so, if the steam is at saturation temp, then this becomes a 2-part problem in 1) determining the latent heat transfer and then 2) heat transfer below the condensing temp (post latent heat transfer).

fyi, since the heating fluid is steam, the enthalpy values at saturation pressure/temp and at condensing temp can be determined from the steam tables. the difference between these is the dH, which is the amount energy transferred to the oil. from here, the oil dT can then be determined.

good luck!
-pmover

 

[IRstuff]

The proper formulation is that the deltaT's add, not the resistances, so Total_delta_T = Q/h1 + Q/h2 + Q/h3, assuming that your htc's can be put into closed form equations. Otherwise, you have 3 unknowns to be solved for simultaneously.

TTFN

FAQ731-376

 

[Smokey2007]

So is it as easy as this? the following are enthalpies:


Saturated Liquid (hf) 581.99 kJ/kg

Saturated Vapor (hg) 2731.6 kJ/kg

therefore dH = 2731.6-581.99=
2149.7 kJ/kg

and this dH = amount of energy going into my oil???

if so that sounds way to simple that has been so difficult to find.

i appreciate all out there trying to help me.

 

[pmover]

provided that the P & T steam data used to determine enthalpy is representative of the exchanger operating conditions, then yes. with the understanding that there are some heat losses (i.e. minor/minimal losses). steam tables are useful and have been for years.

you are welcome.

-pmover