How many small strands needed to equal pipe pressure drop?

[DaBit]

I can obtain a large batch of 0.8mm I.D., 1.2mm O.D. (.031" I.D. / .047" O.D.) brass pipe. Now, this seems quite ideal stuff to construct low power shell-tube heat exchangers with. Many thin strands provide a very large surface area.

But, how many of these pipes do I need to put parallel to equal the pressure drop of an 1/4" pipe?

One could simply say: divide the cross-section surface area of both pipes. This would equal to about 40 parallel strands needed to equal the cross-section surface area of a 1/4" pipe.

But with multiple parallel pipes the wall surface area is greatly increased, increasing friction.

I ran a few calculations with the Gas Pipes module of Coolpack, and if Coolpack is right, I would need 80-90 strands in parallel to equal 1/4" pipe pressure drop.

But, I cannot use Coolpack for ethane (R170) at real temperatures, so I used propane (R290) at -30C evaporation instead, to provide Coolpack with density and viscosity information matching ethane as close as possible within the limits of the software.

I don't really thrust Coolpack with these small pipes, since I expect these small pipes to behave different from larger ones.

Can anyone verify this information?

If anyone needs design parameters, use R170 vapour at -100C, mass flow 1.8kg/hr.

[Prof Sporlan]

DaBit, pressure drop across a tube or pipe can be estimated using the Darcy-Weisbach equation:

h_f = f * L / D * v² / 2g
where:

h_f = head loss, ft
f =friction factor
L = equivalent length, ft
D = diameter, ft
v = velocity, ft/sec
g = acceleration due to gravity, 32.174 ft/sec²

By equating the cross sectional areas of your 0.047” capillary tubes to your 1/4" OD tube, you are keeping velocity constant. However, you will have slightly greater pressure drop with the 0.047” tubes since their L / D will be greater than that of the 1/4" tube. Friction factor for copper tubing is generally figured to be 5e-6 ft. If you have brass tubing, this friction factor should be the same.

[DaBit]

Velocity is reduced when the number of strands is larger than (d1*0.5)^2 / (d2*0.5)^2 where d1 is the I.D. of the 1/4" pipe and d2 is the I.D. of the strands.

Thus, I solved the equation accounting for the velocity decrease when more strands are used.

The result is that I would need 98 strands in parallel to equal the pressure drop of the 1/4" pipe. Velocity is then reduced to 40% of the 1/4" pipe velocity.

When using 39 strands, thus keeping cross sectional area constant, pressure drop would increase with a factor 6.3

But I am missing some variables in the equation. Wouldn't viscosity of the fluid influence the result, for example?

And just curious: how to interpret head loss in feet?

[Prof Sporlan]

Arrggghhhhh! the Prof made a small error. Friction factor is a function of Reynold's Number, which is a function of viscosity:

R = v * D * rho / (u * g_c)
where:
R = Reynold's number
v = velocity, ft/sec
D = diameter, ft
rho = density, lb_m/ft³
u = absolute viscosity, lb_f-sec/ft²
g_c = gravitational conversion, 32.174 lb_m-ft/lb_f-sec²

Friction factor is either read from a Moody chart (simplest) or calculated using the Colebrook equation (a bit difficult as it requires an iterative procedure). The 5e-6 ft value is the effective roughness of the tube required with the Moody chart or Colebrook equation. So the value of friction factor does move around a bit

[Prof Sporlan]

DaBit, you treat head loss like any column of liquid. Multiply it by density and you will get pressure loss.

[frank]

Surely Prof you would also need to consider the velocity profile in the smaller tubes - is it laminar or turbulent? this will have an effect on the Reynolds number. Also, the heat transfer coefficients on the "inner" tubes would be totally different to the coefficient on the "outer" tubes as any heat transfered from the "inner" tubes would be absorbed by the "outer" tubes before being transfered to the secondary fluid.

[Prof Sporlan]

Surely Prof you would also need to consider the velocity profile in the smaller tubes - is it laminar or turbulent? this will have an effect on the Reynolds number.

Actually, it is the other way around. Reynold's number will predict if the flow is laminar or turbulent. In a refrigeration system, however, one can expect all refrigerant flow to have Reynold's numbers well into the turbulent range.

Also, the heat transfer coefficients on the "inner" tubes would be totally different to the coefficient on the "outer" tubes

Fluid flow analysis and heat transfer are really two different subjects. Using the Darcy-Weisbach equation, Reynold's numbers, Colebrook equation, and the Moody chart, one is assuming no change in fluid temperature is taking place. If DaBit increases his heat exchange surface effectiveness by going to smaller tubes, resulting in increased refrigerant flow requirements, then that will have to be accounted for.

[DaBit]

Originally posted by frank
Also, the heat transfer coefficients on the "inner" tubes would be totally different to the coefficient on the "outer" tubes as any heat transfered from the "inner" tubes would be absorbed by the "outer" tubes before being transfered to the secondary fluid.

This can be prevented with enough spacing between the small tubes.

But if you have any tips and tricks regarding the design and construction of such a heat exchanger, I would love to know.

Originally posted by Prof Sporlan
Actually, it is the other way around. Reynold's number will predict if the flow is laminar or turbulent. In a refrigeration system, however, one can expect all refrigerant flow to have Reynold's numbers well into the turbulent range.

As I recall the Reynolds number is defined as:

R =(4*dens*flow)/ (diam*pi*visc)

when this number stays below 2100, the flow can be considered laminar. With numbers >2100, flow can be considered turbulent.

When running these calculations for refrigerant speeds high enough to carry oil, the number is well above the 20000.

If DaBit increases his heat exchange surface effectiveness by going to smaller tubes, resulting in increased refrigerant flow requirements, then that will have to be accounted for.

Wherever i plan to use these heat exchangers flow is dictated by other factors such as refrigerant mass flow.

[frank]

I assumed that the heat exchanger would be constructed with the smaller tubes bunched together and soldered into some sort of distributor then put inside a larger tube. With such small tubes being used I did not see any other option.

That is why I said that the "inner" tubes would give their heat up to the adjacent "outer" tubes before the secondary refrigerant.

Perhaps you could give an insight as to how you envisage the construction?

Frank

[DaBit]

Originally posted by frank
I assumed that the heat exchanger would be constructed with the smaller tubes bunched together and soldered into some sort of distributor then put inside a larger tube. With such small tubes being used I did not see any other option.

Either this, or using the double amount of small tubes, soldering them tightly together, and interleave the flow between the two sets of tubes.

That is why I said that the "inner" tubes would give their heat up to the adjacent "outer" tubes before the secondary refrigerant.

This assumes that no secondary refrigerant or suction gas can enter between the inner and outer tubes.

If I keep some spacing between the inner and outer tubes (say: tube, 1mm void, tube, 1mm void), this problem is still present, but not as much, right?

[frank]

Thats right. I assumed that with such a small diameter then you would solder them all together to make a commom heat exchanger. In this instance the heat absorbed by the central tubes would naturally be transfered to the adjacent tubes and so on until you get to the outer tubes in contact with secondary refrigerant. Not the best way to go. Spacing the tubes so that the secondary refrigerant is in contact with all the surface area will obviously lead to a greater heat transfer capacity.

With such a small heat load, i.e. overclocking, i still struggle to see the advantage of using lots of smaller tubes in lieu of a single heat transfer medium (evaporator).

Frank

[DaBit]

Originally posted by frank
Thats right. I assumed that with such a small diameter then you would solder them all together to make a commom heat exchanger.

Then there would be no advantage over using a single pipe.

With such a small heat load, i.e. overclocking, i still struggle to see the advantage of using lots of smaller tubes in lieu of a single heat transfer medium (evaporator).

Using multiple small strands allows me to pack more heat transfer surface in a smaller package.

Actually, I plan to use these heat exchangers for the SG<->LL HX. And I am not certain yet if my next system is going to be a classic cascade or autocascade. The latter one might prove a bit too difficult for an inexperienced DIY-guy. If I choose cascade, these heat exchangers are also very useful as cascade interstage heat exchanger.

Are there cases where using many small strands instead of a regular coaxial HX (with 1/4" and 1/2" for example) would have an adverse effect?

[Gary]

With that many tubes, I would suggest making a flat plate heat exchanger, using sheet copper and laying the tubes in flat rows for easy soldering. I would have the inlets and outlets at opposite corners, so that the flow path through each tube is equal length.

[DaBit]

Originally posted by Gary
With that many tubes, I would suggest making a flat plate heat exchanger, using sheet copper and laying the tubes in flat rows for easy soldering. I would have the inlets and outlets at opposite corners, so that the flow path through each tube is equal length.

Then I would have more thermal resistance from the solder and additional metal, in addition to more surface area to insulate.

Though, I could stack the tubes like you would stack pipes:

o o o o o o o o        <- circuit A
 o o o o o o o o       <- circuit B
o o o o o o o o        <- circuit A
 o o o o o o o o       <- circuit B

without using extra sheet metal.

1/4" pipe with numerous 1.2mm holes in it, capped on one side, could be used to connect the parallel strands.

In case of, for example, an interstage heat exchanger, what would be the advantage of plate-HX like construction versus evaporating the high stage refrigerant in the small strands and condensing the low stage refrigerant in the shell?

[Gary]

Easier to split the tube for a header than to drill all those holes.

[DaBit]

Bad ASCII drawing:

          A out
         #   # 
         #   #
       **#|||#**
       * / | \ *
 ******* | | | *
 B in    | | | *
 ******* | | | *
       * | | | *
       * | | | *
       * | | | *
       * | | | *
       * | | | *
       * | | | *
       * | | | ****
       * | | |   B out
       * | | | ****
       * \ | / *
       **#|||#**
         #   #
         #   #
         A in

Thus, the small strands carrying fluid A are all bundled and brazed into larger ID piping. Then, the bundle of strands expands to create space between the strands and allow good penetration of fluid B between the strands.

I don't see why this would be difficult. Spacing the strands is easy; I would use pieces of pre-drilled PCB with 100 mil hole spacing. Bundling and brazing all the strands into a larger ID pipe sounds easy enough too. Cutting all the strands is probably the toughest job.

The only disadvantage of this construction is that the strands in the middle of the shell are somewhat shorter than the strands near the walls of the shell.

[frank]

Actually, I plan to use these heat exchangers for the SG<->LL HX.



I would assume that "A" would be your liquid and "B" would be your suction gas on what is basically a counter flow HX. If this was the case then all of your small diameter tubes would act as metering devices due to their small diameter and high resistance to flow. Control would be difficult to determine. With a single capillary you would charge accordingly but this arrangement of multiple capillaries is something I have not experienced.

[DaBit]

The many 'capillary tubes' parallel would lower flow resistance. When ~90 parallel 'captubes' are used, flow resistance is equal to that of an 1/4" pipe.