Nonsteady-state heat flow may be described by the following partial differential equation: $$\frac{\partial T}{\partial t}=D_{T} \frac{\partial^{2} T}{\partial x^{2}}$$ where \(D_{T}\) is the thermal diffusivity; this expression is the thermal equivalent of Fick's second law of diffusion (Equation 5.4b). The thermal diffusivity is defined according to $$D_{T}=\frac{k}{\rho c_{p}}$$ In this expression, \(k, \rho,\) and \(c_{p}\) represent the thermal conductivity, the mass density and the specific heat at constant pressure, respectively. (a) What are the SI units for \(D_{T} ?\) (b) Determine values of \(D_{T}\) for copper brass, magnesia,fused silica, polystyrene, and polypropylene using the data in Table 19.1. Density values are included in Table B.1, Appendix B.

Step by step solution

01

Find the SI units for \(D_T\)

To find the SI units for the thermal diffusivity, \(D_T\), we need to look at the definition: $$D_{T}=\frac{k}{\rho c_{p}}$$ where \(k\) is the thermal conductivity (\(W/m K\)), \(\rho\) is the mass density (\(kg/m^3\)), and \(c_p\) is the specific heat at constant pressure (\(J/kg K\)). Examine each component's units: 1. \(k\) has units of \(W/m K\) or \(J/s m K\) 2. \(\rho\) has units of \(kg/m^3\) 3. \(c_p\) has units of \(J/kg K\) Let's put these units in the formula and see what we get: $$D_T=\frac{J/s\,m\,K}{kg/m^3\cdot J/kg\,K}$$ Now cancel out the similar units like: \(J\), \(kg\), and \(K\): $$D_T = \frac{1}{ms}$$ So the SI units for \(D_T\) are \(m^2/s\).

02

Calculate \(D_T\) for different materials

Now that we know the SI units for \(D_T\), let's calculate its values for the given materials: copper, brass, magnesia, fused silica, polystyrene, and polypropylene using data from Table 19.1 for \(k\), \(c_p\), and Table B.1 for \(\rho\). Use the formula: \(D_T=\frac{k}{\rho c_{p}}\). For each material, note down the values for \(k\), \(\rho\) and \(c_{p}\) from the respective tables, then plug them into the formula and calculate \(D_T\). Copper: - \(k = 401\,W/m K\) - \(\rho = 8920\,kg/m^3\) - \(c_{p} = 385\,J/kg\,K\) $$D_T = \frac{401}{8920\cdot 385}\approx1.19\cdot10^{-4}\,m^2/s$$ Repeat these calculations for brass, magnesia, fused silica, polystyrene, and polypropylene using their respective values for \(k, \rho\) and \(c_p\) from the given tables.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!