1 ec

• Chris Nkinthorn

• 20190610

Table of Contents

[TOC]

The convective heat transfer is speed dependent and can be represented as:

$$h=20\left[\frac{\mathrm{W}}{\mathrm{m}^{2}-\mathrm{K}}\right]+1500\left[\frac{\mathrm{W}}{\mathrm{m}^{2}-\mathrm{K}}\right]\left(\frac{\omega}{100[\mathrm{rad} / \mathrm{s}]}\right)^{1.25}.$$

This is a correlation to the fractional power of the speed. It is unknown where what speeds this approximation is valid for so its credibility is treated as an assumption.

Part A: Derivation

There are two regions in this one dimensional problem, each with a two heat transfer modes acting simultaneously. In region 2, where $r\leq R_p$, thermal energy is moved by both conduction and convection. In region 1, heat transfers from the break pad/disk interface to both the pad and the disk because of frictional heating.

Region 1: Brake Disk $r \in [R_p,R_D]$

The brake region, has heat flux applied because of the frictional force is converting energy into heat. To define the differential control region inside of the brake disk, we must define the energy balance for each region. Energy by frictional heating is written below, then expanding rdar terms produces:

$$\cancel{\dot{q}_{r}}+\dot{q}_{f h}=\dot{q}_{r+d r} = \cancel{\dot{q}_{r}}+ \frac{d \dot{q}}{d r} d r \rightarrow \quad \dot{q}_{f h}=\frac{d \dot{q}}{d r} d r.$$

Now, the terms the energy equation: $\dot{q} $, rate equation for conduction; $F$, the force generated by the clamping force; $P$, is the rate of frictional heating in terms of power and $T_1$ is the temperature in region:

$$\dot{q}=-b 2 \pi r k \frac{d T_{1}}{d r},\quad F=4 \pi r d r P \mu, \quad\dot{q}_{f h}=4 \pi r^{2} d r P \mu \omega.$$

Substitution of these terms into the differential energy balance equation produces an equation to be simplified:

$$4 \pi r^{2} d r P \mu \omega=\frac{d}{d r}\left[-b 2 \pi r k \frac{d T_{1}}{d r}\right] d r\\ 2\cdot r^{2} P \mu \omega \cancel{\left( 2 \pi dr\right)} = \frac{d}{d r}\left[-b r k \frac{d T_{1}}{d r}\right] \cancel{\left( 2 \pi dr\right)}\\ \frac{d}{d r}\left[r \frac{d T_{1}}{d r}\right]=-\beta r^{2}; \beta=\frac{2 P \mu \omega}{b k}$$

The parameter $\beta$ collects all of the coefficients terms into a generation parameter, simplifying the differential equation. This now is an ordinary differential equation (ODE), in terms of $r$, so that direct integration is straightforward:

$$\begin{align} \int d\left[r \frac{d T_{1}}{d r}\right]&=-\beta \int r^{2} d r\rightarrow r \frac{d T_{1}}{d r}=-\beta \frac{r^{3}}{3}+C_{1}\\ \int d T_{1}&=\int\left(-\beta \frac{r^{2}}{3}+\frac{C_{1}}{r}\right) d r \end{align} r\\$$

Then the generic equation

$$\boxed{T_{1}=-\beta \frac{r^{3}}{9}+C_{1} \ln (r)+C_{2}}$$

This is the general solution for temperature in region 1, but constants of integration need to selected to satisfy boundary conditions.

Region 2: Convection Disk $r \in (0,R_p]$

The process for the inner region is similar, starting with an energy balance

$$\dot{q}_{r}=\dot{q}_{r+d r}+\dot{q}_{c o m v}\rightarrow 0=\frac{d \dot{q}}{d r} d r+\dot{q}_{c o n v}$$

Then, find the terms, which is same as previous but convection is removing heat, to cool the disk down:

$$\dot{q}=-b 2 \pi r k \frac{d T_{2}}{d r},\quad \dot{q}_{c o m v}=4 \pi r d r \cdot h\left(T_{2}-T_{a}\right).$$

Substitution of these into the energy balance produces:

$$\frac{d}{d r}\left[-b 2 \pi r k \frac{d T_{2}}{d r}\right] d r+4 \pi r d r h\left(T_{2}-T_{a}\right)=0\\ \frac{d}{d r}\left[r \frac{d T_{2}}{d r}\right]-m^{2} r T_{2}=-m^{2} r T_{a},\quad m=\sqrt{\frac{2 h}{b k}}$$

This is a inhomogenous ODE, so it’s solutions are separated into its homogenous and particular solutions

$$T_{2}=u_{2}+v_{2}$$

So that the particular equation has solution:

$$\cancel{\frac{d}{d r}\left[r \frac{d v_{2}}{d r}\right]}-m^{2} r v_{2}=-m^{2} r T_{a}\rightarrow v_{2}=T_{a}.$$

This is justified because of the adiabatic boundary condition, which forces the first term $d/dr$ to be zero. This allows makes clear the offset of the time independent solution. The inhomogeneous component gives the useful transient temperature change. The homogenous equation is a Bessel equation, because of the bijective relationship between the problem and traditional parameters:

$$\frac{d}{d r}\left[r \frac{d u_{2}}{d r}\right]-m^{2} r u_{2}=0\leftrightarrow \frac{d}{d x}\left(x^{p} \frac{d \theta}{d x}\right) \pm c^{2} x^{s} \theta=0.$$

The substitution relationship here is ${{x=r}, {\theta=u_{2}}, {p=1}, {c=m}, {s=1}}$ and last term is negative. The Bessel equation has known functions which have similar properties of the $\sin(\cdot)$ and $\cos{(\cdot)}$ in that they are orthogonal on the same period, dictated by the order:

$$u_{2}(r)=C_{3} I_{0}(m r)+\cancel{C_{4} K_{0}(m r)}.$$

The Bessel function of the second kind $K_0$, which in a sense relates better to the $\sin{(\cdot)}$ than it would $\cos{(\cdot)}$, is singular at the origin. The temperature is finite, so this term must zero. The general region 2 distribution the sum of the homogenous and particular solutions:

$$\boxed{T_{2}(r)=C_{3} I_{0}(m r)+T_{a}. }$$

Again, the constants remain which must be solved to satisfy the boundary conditions.

Boundary Conditions

Both the temperature and gradient are continuous at the surface so that they are equal and their slopes are also equal, to produce a smooth solution:

$$T_{2, r=R_{p}}=T_{1, r=R_{p}}, \left.\frac{d T_{2}}{d r}\right|_{r=R_{p}}=\left.\frac{d T_{1}}{d r}\right|_{r=R_{p}}$$

with adiabatic ends

$$\left.\frac{d T_{1}}{d r}\right|_{r=0}=0, \left.\frac{d T_{1}}{d r}\right|_{r=R_{p}}=0.$$

Solution Construction

This step creates a system of equations to be analyzed so that the boundary conditions are fulfilled. Though this could be solved simultaneously in Maple or EES, if available, the interface conditions as well as the boundary conditions allows for solving sequentially. First, the adiabatic end condition tells us that the slope at the interface is 0:

\left.\frac{d T_{1}}{d r}\right|_{r=R_{p}}= 0 = \left.\frac{d}{d r}T_{1}(r)\right|_{r=R_{p}} = \left.\frac{d}{d r} \left[-\beta \frac{r^{3}}{9}+C_{1} \ln (r)+\cancelto{0}{C_{2}}\right]\right|_{r=R_{p}}\\ \texttt{Evaluation at end point: } \quad \quad-\beta \frac{R_{d}^{2}}{3}+\frac{C_{1}}{R_{d}}=0\\ \texttt{Solving for boundary: } \quad \quad \boxed{C_1 =\frac{3}{\beta R_d^3}.}\\

Working our way toward the center, we then apply the interface condition that the temperature profile is continuous and smooth. In this case, we want to evaluate that the derivative is continuous because this allows us to define $C_3$ in terms of constants and $C_1$:

$$\begin{align} \left.\frac{d T_{1}}{d r}\right|_{r=R_{p}} &= \left.\frac{d T_{2}}{d r}\right|_{r=R_{p}}\\ \left.\frac{d}{d r}\left(-\beta \frac{r^{3}}{9}+C_{1} \ln (r)+C_{2}\right)\right|_{r=R_{p}} &= \left.\frac{d}{d r}\left(C_{3} I_{0}(m r)+T_{a}\right)\right|_{r=R_{p}} \\ C_{3} m I_1 ( m R_{p}) &= -\beta \frac{R_{p}^{2}}{3}+\frac{C_{1}}{R_{p}}. \end{align}$$

The expression for $C_3$ is now found:

$$\boxed{C_3 = \frac{-\beta \frac{R_{p}^{2}}{3}+\frac{C_{1}}{R_{p}}}{m I_1 ( m R_{p})} . }$$

The final boundary condition is applied. This solves for the remaining integration constant.

$$\begin{align} \left.T_{1}\right|_{r=R_{p}} &= \left.T_{2}\right|_{r=R_{p}}\\ C_{3} I_0( m R_{p})+T_{a}&=-\beta \frac{R_{p}^{3}}{9}+C_{1} \ln \left(R_{p}\right)+C_{2}\\ \end{align}$$

Solving for $C_2$ produces:

$$\boxed{C_{2} = \beta \frac{R_{p}^{3}}{9}-C_{1} \ln \left(R_{p}\right) + C_{3} I_0( m R_{p})+T_{a}}$$

Collected, the model derived is

$$\boxed{\begin{equation} \begin{split} & T_{1}(r)=-\beta \frac{r^{3}}{9}+C_{1} \ln (r)+C_{2}\\ & T_{2}(r)=C_{3} I_{0}(m r)+T_{a}\\ & \beta=\frac{2 P \mu \omega}{b k}, m=\sqrt{\frac{2 h}{b k}}\\ & C_1 =\frac{3}{\beta R_d^3}\\ & C_{2} = \beta \frac{R_{p}^{3}}{9}-C_{1} \ln \left(R_{p}\right) + C_{3} I_0( m R_{p})+T_{a}\\ & C_3 = \frac{-\beta \frac{R_{p}^{2}}{3}+\frac{C_{1}}{R_{p}}}{m I_1 ( m R_{p})} \end{split} \end{equation}}$$

Part B/C: Vary Parameters

If the disk material can withstand a maximum safe operation temperature of $750^o C$ we evaluate our model in the extreme edge of the disk brake. At the location $r = R_D$, we evaluate the temperature in Region 1 with pressure varying. The resulting:

$$T_{1}(r=R_D)=T_{Max}= -\beta \frac{R_D^{3}}{9}+C_{1} \ln (R_D)+C_{2}; \beta=\frac{2 P \mu \omega}{b k}\\ P_{Max} = \frac{bk}{2\mu\omega}\frac{9}{R_D^3} \left[C_2+C_1\ln{(R_D)}-T_{Max}\right]=0.57 \text{ MPa}$$

how gives us the corresponding maximum pressure at $N=3600\text{ rpm}$. At lower speeds, then the temperature is going to be lower: the disk isn’t rubbing against the brake as quickly. Then to find the maximum allowable clamping pressure that can be applied and plot the temperature in the disk at this maximum pressure. The torque is the integral of pressure over the acting arm or the region where the brake presses against the disk brake. Again, this can be integrated directly as an ODE:

$$F=4 \pi r d r P \mu \rightarrow \Tau q=\int_{R_{D}}^{R_{q}} 4 \pi r^{2} \mu P d r = \frac{4}{3} \pi \mu P\left[R_{d}^{3}-R_{p}^{3}\right]=13.2\text{ N-m}$$