Master of Architecture (MArch)

Welcome to Master of Architecture (MArch). We will study Climate-Responsive Design: Indian Climate Zones, Solar Geometry, Shading Design and Wall U-Values, work through an example, and practise using the course's learning material. Advanced architecture, urbanism and specialist design studies. ZELVU Education provides supplementary academic support; university, semester, specialisation and current syllabus should be matched before exam-specific preparation. Sign in and use the study header to open your authorised course. Select this published lesson. Start Course Tutor with Auto audio on: it teaches continuously, answers a typed question, then resumes. Heat flow through walls. Conductive heat flow through a building element is summarised by its U-value (thermal transmittance, W/m squared K): the lower the U-value, the better the insulation. It is the reciprocal of the total thermal resistance: Example 3 — U-value of a brick wall. Layers (assumed values): 15 mm cement plaster (k = 0.72) each side; 230 mm brick (k = 0.81). R_si = 0.13, R_so = 0.04. Brick: 0.230 / 0.81 = 0.2840. Plaster: 0.015 / 0.72 = 0.0208 each; two layers = 0.0417. R_total = 0.13 + 0.2840 + 0.0417 + 0.04 = 0.4957 m squared K/W. U = 1 / 0.4957 = 2.02 W/m squared K. Check: 0.13 + 0.04 = 0.17; 0.17 + 0.2840 = 0.4540; 0.4540 + 0.0417 = 0.4957 ✓. Adding a 50 mm insulation layer with k = 0.035 adds 1.4286 to R, giving R_total = 1.9243 and U approximately 0.52 W/m squared K — roughly a fourfold improvement. Which mistake would change the reasoning in this example? Using solar altitude directly as the overhang angle. Correction: use the vertical shadow angle, which depends on HSA. Shading east and west glazing with overhangs. Correction: low sun needs fins, screens or reduced glazing. Forgetting surface resistances in U-value. Correction: always add R_si and R_so. Ask Course Tutor: please explain this step from the example. Plaster: 0.015 / 0.72 = 0.0208 each; two layers = 0.0417. Why does it follow? Add what you tried, so the explanation addresses your difficulty. Now try the source practice. A layer is 100 mm thick with k = 0.5 W/mK. What is its thermal resistance? Pause before the answer. The source answer is: 0.1 / 0.5 = 0.2 m squared K/W. Compare your reasoning as well as the result. If your response differs, return to the worked example and identify the step that needs correction. Explain the key idea in your own words, revisit any mistaken step, then continue to the next lesson. Use Stop to end Course Tutor when you finish.