Cooling Hot Nitrogen with an aluminum block

Question

I'm designing a cooling system to cool nitrogen gas stream at 300°C with a mass flow rate of 15.6 g/s moving through an undulating channel in a cylindrical aluminum ring block with external fins. How much mass of aluminum or what special feature is needed to cool 15.6 g of nitrogen moving through and exiting to 25°C EVERY SECOND. It's to be noted that the block is immersed in a large pool of water, and that should be factored in as a cooling advantage. I'll love a method of calculation that allows flexibility with variables like mass flow rate, temperature, and channel diameter.

Answer 1

Cooling a Gas in a Channel (Tube)

Assume we have a cylindrical channel that has gas flowing in it. The channel is in a material that itself is cylindrical. We have a single pass tube heat exchanger or alternatively the equivalent of a 1-D extended system but with gas flow. The outside of the tube is exposed to air or water at a temperature $T_\infty$. The energy balance at a given point $z$ for the flow of the gas through this heat exchanger is basically as below.

$$ \dot{m}\tilde{C}_p \frac{dT}{dz} = - U P (T - T_\infty) $$

In this, $P$ is the perimeter area of the inside of the tube, $P = 2\pi r_{i}$. The factor $U$ is the overall heat transfer coefficient (based from the inside of the tube). It is a combination of the convection of the gas in the tube, the conduction through the tube, and the convection outside the tube.

Recast this in a dimensionless form with $\Theta = (T - T_\infty)/(T_h - T_\infty)$ and $Z = z/L$ where $T_h$ is the entering hot temperature and $L$ is the tube length. Allow $\beta = UPL/\dot{m}\tilde{C}_p$.

$$ d\ln(\Theta) = - \beta\ dZ $$

With the boundary condition $\Theta(Z=0) = 1$, the answer is $\ln(\Theta) = -\beta Z$. This gives the temperature profile of the gas in the tube. See this post for a comparable question and answer.

For the system proposed, $T_h = 300$ $^o$C. The value of $T_\infty$ should be BELOW 25 $^o$C. Otherwise, cooling could eventually require an infinite area heat exchanger (in essence an infinite length tube). Assume $T_\infty = 0$ $^o$C, and the goal is $\Theta = (25 - 0)/(300 - 0) = 0.083$. Use a 1 mm diameter tube with values of $\dot{m} =$ 16 g/s, $\tilde{C}_p = 1$ J/g for nitrogen, and $U = h_a = 50$ W/m$^2$ $^o$C. This latter assumption says that gas flow in the channel is the limiting value. The result is

$\beta = UPL/\dot{m}\tilde{C}_p = (50)\pi(0.001)L/((16)(1)) = 0.010 L$

$\ln(0.083) = -0.010 x \Rightarrow x = 250$

You need a 1 mm diameter tube that is 250 m long with walls that are cooled at 0 $^o$C.

Every 10x increase in heat transfer coefficient cuts the required length by 10x. Every 10x decrease in flow rate cuts the required length by 10x. Decreasing the external temperature $T_\infty$ gives a modest decrease in required length.

Flowing Gas Thru a Tube

The velocity of the gas through the tube is $v = \dot{V}/A$. Use $A = \pi D$ where $D$ is the diameter. The volumetric flow of an ideal gas is related to its mass flow as below with $T$ as temperature, $M$ as molar mass, and $p$ as pressure.

$$ \dot{V} = \dot{m} \frac{RT}{Mp} $$

Consider 1 bar at 300 $^o$C with nitrogen to obtain

$$ \dot{V} = \dot{m} \frac{(8.314)(573)}{(28)(101325)} = 0.00170 \dot{m}$$

Use the value 16 g/s to obtain $\dot{V} = 0.027$ m$^3$/s. In a 1 mm diameter tube, the required gas velocity becomes 8.6 m/s.

$$ v = 0.027 / ((\pi)(0.001)) = 8.6$$

At this point, we are left to find the required pressure drop to sustain 8.6 m/s through a 250 m long tube with a 1 mm internal diameter. This involves a review of the friction factor for fluids in pipe flow. The gas contracts as it cools. This demands further investigation too, because the basic calculations with incompressible fluids will fail with such cases.

Other Options

The analysis used a 1 mm diameter tube. Increasing the tube diameter will decrease the required length and decrease the required velocity. The issue at some point will be that a gas does not sustain a uniform temperature profile. It will cool along the wall of the tube but less at the center.

The analysis used only one tube. An alternative is to flow the same mass (volume) of gas through a multi-channel system. With $N$ tubes, the contact area increases by $N$. This means, the required tube length decreases by $N$. This alternative says, rather having the gas flow through one longer channel at higher speed, you have the gas flow through N channels at 1/N speed because area increases by N. The contact time calculated as area per volumetric flow is the same. The appreciation of why we flow gases through mesh grids to heat or cool them rapidly strikes home. As for the pressure drop, this issue should be addressed in a separate discussion.

A gas can be cooled by adiabatic expansion. An effective design may just be to allow the hot gas at high pressure to expand rapidly to lower pressure.

Finally, analysis at this problem shows an example of how a heat exchanger design equation is used for fluid flow in a pipe surrounded by a constant temperature fluid. This may be an easier approach to take from a practical standpoint.

Cooling a Gas Flowing in a Channel through a Block

The first part of the analysis above gives you the LENGTH OF A TUBE that is required to cool a gas to a given end temperature when the tube walls are held at a constant (lower) temperature.

The second part of the analysis above gives you the GAS VELOCITY that is required to pass the gas through the tube at a specific mass flow rate (g/s).

Assume that you will make your channel with a diameter $D$. Assume the block is cooler than your desired end temperature. Determine the length of channel that you need using the first part (the equation for $\Theta$ versus $Z$). Determine the gas velocity you need to support the desired mass flow rate using the second step. Finally, follow up with pressure drop calculations for gas flow in pipes. I suspect you will find out that you likely will not be able to get the velocity that you need in a single channel because the pressure drop that you will need across such a long channel is so high as to be physically unreachable. Also, any curves in the channel only act to increase the required pressure drop. So, a "curvy" channel will be far longer but require far higher pressure to force the gas through.

In the end, you may have to realize that a better option to cool the amount of gas you want as rapidly as you want is to run the gas through a multi-channel (porous) solid that is held at a lower temperature than you need.

Heating a Mass of Solid in a Given Time (Optional)

Admittedly, the first part of the question is likely not addressed in the above. Specifically, what time is needed for a mass of a solid to absorb a specific amount of heat (from whatever source).

The approach to answer this first question depends on making one or another of many different assumptions. When the gas at constant temperature is in full contact with the solid, the problem is an unsteady state convection + conduction problem. It may require analysis comparable to the use of Heisler charts. When the gas is in full contact with the solid but also changing its temperature, the Heisler chart analysis is replaced by an incremental or numerical integration approach (solve the Heisler chart at t, increment by dt and solve again). Finally, when the gas is flowing through the solid and the solid is not maintained at a constant temperature, one goes to the need for a Heisler chart or its foundations plus time integration plus position analysis.

Answer 2

Lets say you have Q per second calories of nitrogen heat to loose to the Aluminum for every 1 cm length of your pipe.

Then we have $$Q_{1\ cm\ length} = \frac{K*A(T_{hot}-T_{cold} )}{d(R^2_{aluminium}/R^2_{pipe})}\ $$

-K= thermal conductivity of the aluminum = $o.5 (cal/sec)/(cm^2 C/cm)$

-t = time, second

-R = radius

-T = temperature Celsius

-A = area of contact per one cm length of pipe.

-d = thickness of aluminum around pipe, approximate 0.8* one side, if the aluminum section is square.

So If we have the diameter of the pipe as D, we have A=pi*D and we can calculate d the thickness of aluminum.

If we have all the dimensions by plugging in the equation we can calculate the Q heat loss by nitrogen per second per each centimeter length of the pipe and if we multiply that by t we get total heat lss after some time.

Edit

After some comments I modify my answer as follows to include the length of the pipe and the time.

$$Q_{nitrogen} = \frac{K*A*L(T_{hot}-T_{cold} )}{d(R^2_{aluminium}/R^2_{pipe})}t\ $$ -L=length of pipe

This equation gives the Q loss by nitrogen after certain time t passing through the block. So if you need to loose more heat you have to make the pipe length L longer, or change other variables.