# Calculation for heat transfer rate / cooling capacity for air cooled solid I have a pipe which is channeling air flow from a compressor with a known mass flow rate and temperature. I can adjust the mass flow rate (power setting) and measure the delta T for the temp of air coming out the nozzle for each setting. It's for a cooling application and I want to maximise mass flow rate and minimise temperature of the flow. How can I optimise this system?

The air is channeled perpendicular to the disc solid with a gap between the nozzle and the solid to allow the air flow to dissipate. The nozzle and the disc are fixed.

• What have you considered in terms of conduction, convection and radiation? Aug 30, 2018 at 9:32
• I think the dominant heat loss of the solid disc will be through convection although I'm open to other theories.
– Don
Aug 30, 2018 at 10:32
• How conversant are you in developing, understanding, and solving second order differential equations? Alternatively, do you know the concept of lumped analysis or pseudo one-dimensional analysis? Aug 30, 2018 at 13:43
• @Don Take an object with an emissivity of $\epsilon$ at $T$ in air at 298 K. Radiation is $\epsilon \sigma T^4$. Convection is $h_a (T - 298)$. Start with a black body in stagnate air ($\epsilon = 1, h_a = 1$). Radiation will always be larger than convection. Decrease emissivity by x(1/10). Convection dominates almost immediately above 300 K. Increase convection by x10 on the black body. Convection dominates again almost immediately above 300 K. Both changes are realistic (gray body and flowing air). Short answer: Convection dominates for real objects in air. Aug 30, 2018 at 21:51

## Stagnate Gas

When the gas is stagnate, view the system as though it is an extended surface (from compressor to outlet). Take the heat transfer inside the tube as having no radial component. Develop the differential equation along the tube $z$ as

$$\frac{d}{dz} k_G A \frac{dT}{dz} - \frac{P}{R'}\left(T - T_\infty\right) = 0 \\ \frac{1}{R'} = \left(\frac{1}{h} + \frac{w}{k_w}\right)^{-1}$$

where $A$ is the cross-sectional area and $P$ is the perimeter length at a given $z$.

Allow that gas thermal conductivity $k_G$ is constant to obtain

$$\frac{d^2T}{dz^2} - m^2\left(T - T_\infty\right) = 0 \\ m^2 = \frac{P}{k_G R'A}$$

where $P$ is the (outer) perimeter of the tube, $w$ is the tube thickness, $k_w$ is the thermal conductivity of the tube, and $A$ is (inner) circular area of the gas at a position $z$ along the tube. For this geometry

$$\frac{P}{A} = \frac{2\pi r_o}{\pi r_i^2} = \frac{2r_o}{r_i^2}$$

Fix the temperature at the compressor. Three solutions are obtained for the temperature along the (center of the) tube $T(z)$ depending on the boundary conditions. One is when the end temperature is fixed, one is when the end is insulated (no heat flow), and the final is when the end allows heat flow. Examples are found at this link to MIT course notes.

The system you have is modeled as one where heat flows out of the end. To first order, the heat flow out the end is

$$\dot{q}_L = -\left.k_G\frac{dT}{dz}\right|_L = \dot{m} \tilde{C}_p\left(T - T_\infty\right)$$

This adds complexity to the simple answers and requires analysis at an advanced level. In short though, as you increase $\dot{m}$, the temperature $T_L$ will become closer to $T_o$. The relationship has nothing that lends itself to be "optimized". In short, you cannot find a relationship to obtain a minimum $T_o - T_L$ based on setting a maximum $\dot{m}$ because $T_L \rightarrow T_o$ as $\dot{m} \rightarrow \infty$.

## Gas Flow - Well Mixed Radial

When the gas flows along $z$ and is well-mixed along $r$, we replace the term $-k_G A dT/dx$ by $\dot{m}\tilde{C}_p T$ to obtain

$$\dot{m}\tilde{C}_p\frac{dT}{dz} - \frac{A}{R'}\left(T - T_\infty\right) = 0$$

This first order differential equation can be solved for $T(z)$ with the boundary condition $T(0) = T_o$. Here again, the system has no optimization between $\dot{m}$ and $T_o - T_L$. The same limits apply as above.

## Gas Flow - Non-Mixed Radial

When the gas is not well mixed in the radial direction, the temperature profile inside the tube will not be uniform along the radius. To continue with the model of an extended surface, we will have to include an internal convection coefficient. This gets ugly fast.

The simpler system in this case is to model the tube as a heat exchanger. The first order relationship is

$$\dot{q} = \dot{m}\tilde{C}_p\left(T_o - T_L\right) = \varepsilon U A \left(T_o - T_\infty\right)$$

where $\varepsilon$ is the efficiency of the exchanger, $U$ is its overall heat transfer coefficient, and $A$ is the (inner/outer) tube area. This is what is called an NTU analysis.

Even in this case, one does not have a condition to optimize. As $\dot{m} \rightarrow \infty$, you will find that $T_L \rightarrow T_o$.

I hope this gives you a useful starting point. In summary, there is nothing to optimize in a balance of $\dot{m}$ and $T_o - T_L$.

• Hi Jeffrey, thank you for your detailed answer. Essentially I want to cool the disc as much as possible with the restrictions of geometry and medium (air). First principles makes me assume that I want max flow rate of min temp air to pass over the disc. Is there a point of diminishing returns on increased mass flow rate as it's limited by heat transfer coeff. - air temp leaving disk only slightly above air flow temp. At which point, heat transfer coefficient is a fn of the difference in temperature, lower incoming air temp would have more impact than an increase in mass flow rate?
– Don
Aug 31, 2018 at 8:36
• Well, that is a different problem! I see how off base my answer is. I focused on the tube not the disk. How does the air flow over the disk (perpendicular or lateral flow) and where does it impact the disk (center or off-center)? Also, is the disk sitting on a solid surface (and what is that surface) or is it floating in the air? Short answer in the interim is as you guessed. Lowering the initial temperature of the incoming gas will have a linear effect on cooling rate while increasing mass flow will eventually have diminishing returns on cooling rate. Aug 31, 2018 at 13:14

## Temperature of Disk

Consider a flat, circular disk that is being cooled by convection only (ignore radiation). The configuration is illustrated below with symbols for geometry ($w$ and $R$), temperatures $T_j$, thermal conductivity $k$, density $\rho$, specific heat capacity $\tilde{C}_p$, and convection coefficient $h_a$. The problem is to find the temperature in the disk as a function of time $T(t)$. The most important number to determine first is the Biot number $Bi = h_a (V/A) / k$. For the disk, $V/A = w\pi R^2 / (2\pi R^2 + 2\pi R w)$. For a thin disk $R >> w$, this approaches $w/2$. When $Bi < 0.1$, the disk can be handled as a lumped system. Basically, heat is conducted through the disk faster than heat is pulled out of the disk. The temperature profile inside the disk can be neglected. The resultant analysis for $T(t)$ is given by example at this link. When $Bi > 1$, the analysis must include the temperature profile inside the disk. The analysis leads to $T(t,z)$ solved by what are called Heisler charts. A quick and dirty approach in this case is to pick the temperature either at the center point of the object or on its surface to follow as a function of time. With this approach, the $T(t,z=0)$ or $T(t,z=w)$ results are about the same as found by the lumped analysis. The temperature changes exponentially with time in either case.

## Convection Coefficient Correlations

Convection coefficients are correlated to gas flow rate. The starting point is to appreciate that correlations are used to find the the Nusselt number $Nu = h_a (V/A)/k$. The convection coefficient is calculated from $Nu$. Correlations in forced convection depend on the Reynolds number and the Prandtl number. See the same link as given for $Nu$ to see examples in various geometries. The flow rate of a gas over the disk appears as velocity in $Re = \rho_g v_g L_c/\mu_g$. For the system at hand with mass flow, the gas velocity can calculate at the end of the tube as $v_g = \dot{m} /(\rho_g \pi R_T^2)$. All else being equal, velocity is therefore proportional to mass flow rate. As a side note, $L_c$ is a characteristic length of flow over the disk. For flow over the disk laterally, $L_c \approx 2R$ at the opposite end(s) of the disk. For flow perpendicularly from the center line, $L_c = R$ at the edges.

## Cooling Rate

At any instance in time, the cooling rate of the plate is $\dot{q} \approx 2 h_a A (T - T_a)$ (ignore loss from the edge of the disk). As $T_a$ decreases, the cooling rate increases linearly. To first order in laminar flow, $Nu \propto \sqrt{Re}$. Therefore, to first order in laminar flow over the disk, cooling rate will depend on the square root of mass flow. The correlations are complex for turbulent flow. The convection coefficient still increases with mass flow rate, however the correlation is never directly linear.

## Summary

Decreases in temperature $T_a$ will increase cooling rate linearly. Increasing mass flow $\dot{m}$ will increase cooling rate but less effectively (less than a linear increase).

• Hi Jeffrey, thanks again for the detailed explanation. Just on the final statement, should it not be that increases in mass flow rate will increase cooling rate rather than decreases in mass flow rate?
– Don
Sep 3, 2018 at 10:58
• Yep. I fixed it. Sep 4, 2018 at 3:08