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Aswin A.
Tutoring for last 7 years.
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Physics (Heat Transfer)
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Question:

A concrete wall, which has a surface area of 20 m2 and is 0.30 m thick, separates conditioned room air from ambient air. The temperature of the inner surface of the wall is maintained at 25C, and the thermal conductivity of the concrete is 1 W/m  K. (a) Determine the heat loss through the wall for outer surface temperatures ranging from 15C to 38C, which correspond to winter and summer extremes, respectively.

Aswin A.
Answer:

Assumptions used here are: 1. One-dimensional heat conduction in the x-direction 2. Steady state conditions 3. Constant properties 4. Outside wall temperature is that of the ambient air Now using the Fourier's law $-\frac{dT}{dx} = -\frac{q_{x}''}{k}$ , is a constant if $q_{x}''$ and k are constants. k is approximately constant if it depends only weakly on temperature. The heat flux and heat rate when the outside wall temperature is $T_{2} = -15^{0}C$ are $q_{x}'' = -k\frac{dT}{dx} = k(\frac{(T_{1}-T_{2})}{L}) = 1.\frac{(25-(-15))}{0.30} = 133.3 W/m^{2}$ $ q_{x}=q_{x}''\times A = 133.33 \times 20 = 2667W $.

Trigonometry
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Question:

Find the minimum value of 5cosA+12sinA+12?

Aswin A.
Answer:

This can be written as 13(5/13cosA+12/13sinA)+12. But we can find a value of B such that sinB=5/13 and cosB=12/13. Thus the question can now be written as 13(sinBcosA+cosBsinA)+12. That is, 13sin(A+B)+12. Now it's easy the minimum value of sin is -1. So the minimum value of the expression is -13+12 = -1

Aerospace Engineering
TutorMe
Question:

For isotropic materials show that the principal axes of strain coincide with the principal axes of stress. Further, show that the principal stresses can be expressed in terms of the principal strains as $ \sigma_{i} = 2\mu e_{i} + \lambda e_{kk} $.

Aswin A.
Answer:

If $\sigma_{i}$ and $n_{i}^{\sigma}$ are the principal values and directions for the stress, then $(\sigma_{ij}-\sigma_{i}\delta_{ij})n_{i}^{\sigma} = 0$. But for isotropic materials, $\sigma_{ij} = \lambda e_{kk} \delta_{ij} + 2\mu e_{ij}$ and thus $(\sigma_{ij}-\sigma_{i}\delta_{ij})n_{i}^{\sigma} = (\lambda e_{kk}\delta _{ij} + 2\mu e_{ij} - \sigma _{i}\delta _{ij})n_{i}^{\sigma } = 0. which implies that $(e_{ij} - \frac{1}{2\mu }(\sigma _{i} - \lambda e_{kk})\delta _{ij})n_{i}^{\sigma } = 0$. Now this can be written as $e_{ij}-e_{i}\delta_{ij})n_{i}^{\sigma} = 0 $ where $e_{i} = \frac{1}{2\mu }(\sigma _{i} - \lambda e_{kk})$. However the principal value problem for the strain can be expressed as $e_{ij}-e_{i}\delta_{ij})n_{i}^{e} = 0 $ where e_{i} and n_{i}^{e} are the principal values and the directions for the strain which shows that the principal axes of strain coincides with that of stress. That is $n_{i}^{\sigma} = n_{i}^{e}$. These results for the isotropic materials also imply that $ \sigma_{i} = 2\mu e_{i} + \lambda e_{kk} $.

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