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8. (a) Determine by degree of efficiency method, or by any other suitable method, the approximate size of a vertical window for a living room 5.0 metres long by 3.8 metres wide requiring an average illumination of 150 lux on the working plane - Leaving Cert Construction Studies - Question 8 - 2010

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8.-(a)-Determine-by-degree-of-efficiency-method,-or-by-any-other-suitable-method,-the-approximate-size-of-a-vertical-window-for-a-living-room-5.0-metres-long-by-3.8-metres-wide-requiring-an-average-illumination-of-150-lux-on-the-working-plane-Leaving Cert Construction Studies-Question 8-2010.png

8. (a) Determine by degree of efficiency method, or by any other suitable method, the approximate size of a vertical window for a living room 5.0 metres long by 3.8... show full transcript

Worked Solution & Example Answer:8. (a) Determine by degree of efficiency method, or by any other suitable method, the approximate size of a vertical window for a living room 5.0 metres long by 3.8 metres wide requiring an average illumination of 150 lux on the working plane - Leaving Cert Construction Studies - Question 8 - 2010

Step 1

Determine by degree of efficiency method

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Answer

To determine the window area required, we can use the following formula:

Li=Lo×Wf×E×AFloor AreaL_i = \frac{L_o \times W_f \times E \times A}{\text{Floor Area}}

Where:

  • LiL_i = Lux required
  • LoL_o = Standard Overcast Sky (C.I.E. = 5000 Lux)
  • WfW_f = Window factor (0.5 for observations of daylight)
  • EE = Efficiency coefficient (0.4 for reflections, obstructions, etc.)
  • AA = Area of the window
  1. Set parameters:

    • Required illumination (LiL_i) = 150 Lux
    • Standard Overcast Sky (LoL_o) = 5000 Lux
    • Window factor (WfW_f) = 0.5
    • Efficiency (EE) = 0.4
    • Floor area = 5.0mimes3.8m=19.0m25.0 m imes 3.8 m = 19.0 m^2
  2. Rearranging the formula for window area (WW): W=Li×Floor AreaLo×Wf×EW = \frac{L_i \times \text{Floor Area}}{L_o \times W_f \times E}

  3. Substitute values: W=150×19.05000×0.5×0.4W = \frac{150 \times 19.0}{5000 \times 0.5 \times 0.4} W=28501000W = \frac{2850}{1000} W=2.85m2W = 2.85 m^2

Thus, the approximate size of the window required is 2.85 m².

Step 2

Discuss in detail, using notes and freehand sketches, two design considerations for a contemporary window frame and glazing system

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Answer

Window Frame

  1. Thermal Break: Utilizing a wooden frame made from composite materials with good insulating properties. This will reduce heat loss due to cold bridging. A thermal break is included in the design to separate the frame from the sash, which enhances insulation.

  2. Good Seals: The frame should have good seals between the sash and frame to minimize air infiltration and reduce heat loss.

Glazing System

  1. Double or Triple Glazing: Using insulated units such as double or triple glazing will greatly enhance thermal performance. The gases used, like krypton or argon, have lower thermal conductivity and will thus reduce heat loss.

  2. Solar Control Coating: Incorporating a reflective coating on the glass will help reflect heat back into the room. Additionally, inert gas-filled cavities between panes improve insulation and reduce energy loss.

Step 3

Outline two environmental considerations

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Answer

  1. Use of Renewable Materials: Prioritize materials that are sourced from renewable resources. Renewable materials generally have a lower embodied energy footprint, contributing less to climate change.

  2. Sustainability: Select materials that are locally sourced or easily recyclable. Over time, these sustainable practices will minimize environmental harm and promote a healthier ecosystem.

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