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Calculating heat transfer coefficient
2

Calculating heat transfer coefficient

Calculating heat transfer coefficient

(OP)
Hello.

 I am attempting to design a system that involves running an external half-pipe cooling jacket around a firepot assembly. This jacket is limited in size to about a 1.5" radius and will be using ammonium hydroxide as the cooling agent. The NH4OH will only be flowing @ .5gpm. The firepot is made of cast iron. I attempted calculating the Reynolds number, Prandtl number and then the Nusselt number in order to get to the heat transfer coefficient but I'm not sure if I used the equations applicable to my piping and flow conditions. For example in calculating the Reynolds number for flow in a half pipe what value would be used for the Diameter? Any help would be much appreciated.



density NH4OH  .9g/cm3
thermal conductivity cast iron  29BTU/ft hr F
viscosity NH4OH 5cP
specific heat capacity of solution  1.12BTU/lb F


equations used:

Pr= kinematic visc/thermal diffusivity
kin. viscosity= viscosity/density
thermal diffusivity = therm conductivity/ (density*specific heat cap)

Nusselt= .023(Re^.8)*(Pr^.4)

Re=speed*density*diameter/viscosity

heat transfer coefficient= Nu* thermal conductivity/ L

(is L the thickness of boundary layer or length of piping?)

RE: Calculating heat transfer coefficient

I believe you should use the Hydraulic radius = x-sectional Area/wetted perimeter.

RE: Calculating heat transfer coefficient

(OP)
Great. now for the x-sectional area would that just be half the x-sectional area for a full pipe of this diameter? Thanx.

RE: Calculating heat transfer coefficient

yup

RE: Calculating heat transfer coefficient


Please note that the hydraulic radius is not half the hydraulic diameter, not even for a circular cross-section !

You should use the hydraulic diameter in your calculations of Re, Nu, and the friction factor.

Dh = 4(fluid cross section)/wetted perimeter

In this particular case, if I'm not mistaken,

Dh = (4)(½)(π/4) 32 ÷ [(π 1.5)+ 3] = 1.833" = 0.0466 m

Do you already have an idea of the number of windings around
the firepot ? Are you limited on the pressure drop ?

BTW, L in your approximate heat transfer formula is again Dh. Please note that the proposed empirical (Ditttus-Boelter) heat transfer expression formula is for fully developed turbulent flow.

Are you sure the flow régime is indeed fully turbulent with Re>10,000? Pls note that for Re<2,100, Nu = ƒ(Re,Pr,L/Dh) where L is the coil length.

I prefer the 0.33 to 0.4 as exponent of the Pr number, following Sieder and Tate classical measurements.

Good luck.



RE: Calculating heat transfer coefficient

(OP)
I've been getting a really low reynolds number so I'm not sure if I'm doing this right. If someone could check my work that'd be great.

1.I start with Vol. flow:

.5gpm*(3.785liter/gall)*(.8gm/liter)=1.51 gm/min mass flow.


2. Then I solve for velocity: v=M/(rho*(xArea)/2)

(1.51gm/min)/(.8gm/liter)(1/2)(pi/4)(3in^2) then

*61.02in^3/liter to get the units right= 32.6in/min


3. Then for the Re # =rho*velocity*Dh/mu

(.8gm/liter)(32.6in/min)(1.833in)/(.05Poise)(2.54cm/in)(61.02in^3/liter)(60sec/min)

and I get the ridiculously low .1 for my answer.


Thanks for the help.

RE: Calculating heat transfer coefficient

OMG.  You used x-sectional area/2.  I said "hydraulic radius".   To get that, you start with the x-sect area/2 but then divide that by the quantity (inside circumference of whole pipe/2  +  pipe inside diameter)

RE: Calculating heat transfer coefficient

I think that your density should be 0.8grams/milliliter not 0.8gm/liter.  ( I am assuming that the amonium hydroxide is in a liquid state, I could be wrong)

RE: Calculating heat transfer coefficient


The estimated Re is low. It is, in fact, in the laminar régime. Stonecold is right. I suspect the viscosity is too high; what is the ammonia % and the temperature ?
Anyway, given:

Linear velocity = 0.0138 m/s
Density: 900 kg/m3 (as originally stated)
Hyd. diameter = 0.0466 m
Abs. visc. = 5 cP = 0.005 kg/(m.s)

Re = (0.0138)(900)(0.0466)÷0.005 = 116

Now, assuming we are (conservatively) speaking of a straight pipe the values of  Nu*Pr-0.33 = (hD/k)(Cpμ/k)-0.33, where h is the heat transfer coefficient, and reading from a small (old) graph, would be:

for L/D=25, ~3; for L/D=100, ~2; for L/D=500, it drops to ~1.

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