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Green Building Bible, Fourth Edition
Green Building Bible, fourth edition (both books)
These two books are the perfect starting place to help you get to grips with one of the most vitally important aspects of our society - our homes and living environment.

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    • CommentAuthorjwd
    • CommentTimeApr 7th 2016
     
    Hi
    Im doing a project for a course that needs a simple calculation for the heat loss via ventilation for an office and have got myself in a muddle as the number I have found seems way too large.
    The office is 6 m X 10m by 3.5m. It has an occupancy of 10 people and a photocopier. The CIBSE standards tell me I need to allow 10l/s for each person and part F tells me I need another 20 l/s for the copier in order to keep Co2 to an acceptable level. That would make a total of 120 l/s. It is 22 deg c inside and 4 deg c out side. what would be the heat losses due to ventilation? I have assumed it is naturally ventilated. I used a formula where heat extracted is 1.2 x q (the req ventilation rate in l/s) x dt (temp differential - 18 degrees).

    Can anyone tell me if I am a)barking up the right tree or b) where I hae gone wrong ?

    Cheers
    JW
    • CommentAuthorjwd
    • CommentTimeApr 7th 2016
     
    Just answered my own question - bloody idiot. I confused my units.
    • CommentAuthorEd Davies
    • CommentTimeApr 7th 2016 edited
     
    For any-hypothetical-body else who's interested, here's how I'd calculate it:

    Specific heat capacity of air is about 1 kJ/kg·K. Density is a little less than 1.3 kg/m³ at normal sorts of pressures and temperatures. So volumetric heat capacity is 1.3 kJ/m³·K.

    120 l/s is 0.12 m³/s. Heat loss rate is therefore 0.12 m³/s × (22 - 4)K × 1.3 kJ/m³·K = 2.8 kJ/s = 2.8 kW.

    Which is indeed quite a lot (have I confused my units?) but 10 people will produce something like a 1 kW and gawd knows what a photocopier would produce. How does that compare with your answer jwd?
    • CommentAuthorCWatters
    • CommentTimeApr 8th 2016
     
    I agree with your figures. I found a reference for the volumetric specific heat capacity of air that was the same as you calculated.
    • CommentAuthorCWatters
    • CommentTimeApr 8th 2016 edited
     
    A flow rate of 0.12 m^3 per second works out at 432m^3 an hour.

    If the volume is 6 x 10 x 3.5 = 210 m^3 the that flow rate works out at..

    432/210 = 2.1 air changes an hour. That's actually on the low side according to..

    http://www.engineeringtoolbox.com/air-change-rate-room-d_867.html

    The figures you mention as coming from the CIBSE standards appear to be similar to those in Building Regs Approved Document F...

    http://webarchive.nationalarchives.gov.uk/20151113141044/http://www.planningportal.gov.uk/uploads/br/br_pdf_adf_2010.pdf

    See page 32.

    However if I'm reading it correctly you don't need to add together the rate "per machine" and "per person", instead you take the higher of the total. If I'm correct that would make a small difference.. 100L/S rather than 120L/S. Still it could be a significant percentage saving.
    •  
      CommentAuthordjh
    • CommentTimeApr 8th 2016
     
    FWIW, I don't think the problem with copiers or printers is the CO2 they produce. It's the nasty chemical vapours from fusing toner. If it was being used regularly, I'd try to put it in a separately ventilated area.

    The figure for Residences of 1-2 ACH in that engineering toolbox list is way too high if it's talking about houses. Even building regs only wants about 0.5 and that's generally too much. Don't know about the other values.
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