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    Building Physics - 1: Thermal Balance Equation

    Uploaded by

    Nidhi Chadda Malik
    The document discusses the thermal balance equation used to calculate the heating and cooling load of a building. It provides the equation that sets internal heat gain, solar heat gain, conductive heat gain, ventilative heat gain, and mechanical heat gain equal to heat loss through evaporation. As an example, it calculates the various heat gain and loss terms for an office space and determines that the mechanical cooling capacity needed is 5,672.859 watts, or 5.5-6.0 kilowatts, to maintain thermal balance.

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    Building Physics - 1: Thermal Balance Equation

    Uploaded by

    Nidhi Chadda Malik
    443 views5 pages
    The document discusses the thermal balance equation used to calculate the heating and cooling load of a building. It provides the equation that sets internal heat gain, solar heat gain, conductive heat gain, ventilative heat gain, and mechanical heat gain equal to heat loss through evaporation. As an example, it calculates the various heat gain and loss terms for an office space and determines that the mechanical cooling capacity needed is 5,672.859 watts, or 5.5-6.0 kilowatts, to maintain thermal balance.

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    Thermal Balance Equation

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    The document discusses the thermal balance equation used to calculate the heating and cooling load of a building. It provides the equation that sets internal heat gain, solar heat gain, conductive heat gain, ventilative heat gain, and mechanical heat gain equal to heat loss through evaporation. As an example, it calculates the various heat gain and loss terms for an office space and determines that the mechanical cooling capacity needed is 5,672.859 watts, or 5.5-6.0 kilowatts, to maintain thermal balance.

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    443 views5 pages

    Building Physics - 1: Thermal Balance Equation

    Uploaded by

    Nidhi Chadda Malik
    The document discusses the thermal balance equation used to calculate the heating and cooling load of a building. It provides the equation that sets internal heat gain, solar heat gain, conductive heat gain, ventilative heat gain, and mechanical heat gain equal to heat loss through evaporation. As an example, it calculates the various heat gain and loss terms for an office space and determines that the mechanical cooling capacity needed is 5,672.859 watts, or 5.5-6.0 kilowatts, to maintain thermal balance.

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
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    Building Physics -1

    Thermal Balance Equation

    NIDHI CHADDA
    M.ARCH SEM 1
    JNEC
    2

    Problem
    Heat Gain Equation :

    Qi + Qs± Qc ±Qv ±Qm – Qe = 0

    Qi : Internal Heat Gain.


    Qs : Heat gain due to solar (transparent)
    Qc : Heat gain due to Conduction (thru walls)
    Qv : Heat gain due to Ventilation
    Qm : Mechanical Controls.
    Qe : Heat loss due to Evaporation

    DOOR
    5.0
    2.0 3.0

    5.0

    1.5
    WINDOW
    3.0 WINDOW

    3.0
    1.5
    N
    PLAN WESTERN FACADE

    BP-Thermal Balance Equation, By Nidhi Chadda (M.Arch -Environ, Sem 1)


    3

    Solution
    Heat Gain Equation :

    Qi + Qs± Qc ±Qv ±Qm – Qe = 0

    Qi : Artificial lighting consumption + Heat dissipated by occupants (Std. values provided activity wise) Assuming the office is working continuously
    for 24 hr.s while 12 hr.s is artificial light n 12
    = (4 x 80 x 12)Watt + ( 4x 140)Watts hr.s is catered ny natural sunlight.)
    = 3840 + 560
    = 880 Watts

    Qs : We always get heat from sun , thus this value shall always be positive. In this case , for the window –

    Qs for Window : A (area of window) x I (incident solar radiation for West wall) x Solar gain factor of window
    =[(1.5m x 3.0 m) x 155.2 ](m2 x W/m2 )x 0.86
    = 4.5 x 155.2 x 0.86
    = 600.624 Watts

    Qc for Wall: A (area of wall ) x U (U value) x ΔT ( Ts – Ti) (for Opaque surface)


    Calculating Ts for wall: Ts = To +( I x Wall absorbant (a)) / Surface Conductance (f)
    = 33 + ( 155.2 x 0.6) / 22.7
    = 37.1
    ΔT = 37.1 – 24 = 13.1  C
    Qc for Window:
    A (area of window) x U (U value) x ΔT ( To – Ti) To – Temperature outside
    Ti – Temperature inside
    Calculating Ts for window: ΔT = To - Ti Ts – Sole air temperature.
    = 33 - 24
    =9C
    BP-Thermal Balance Equation, By Nidhi Chadda (M.Arch -Environ, Sem 1)
    4

    Solution
    Qc : A x U x ΔT
    = (13.5 x 3 x 13.1 ) + ( 4.5 x 5.77 x 9 ) (m2 x W/m2K x C)
    = 530.55 + 233.685
    = 764.235 Watts4

    Qv : 1300 x V x ΔT V = [Room volume x N (no. of air change) ] / 3600


    = 1300 x 0.04 x 9 =[(5 x 5 x 3) x 2 ] / 3600
    = 468 Watt = [ 75 x 2 ]/ 3600
    = 0.04

    Qe : shall not be considered here as not provisions are mentioned .

    Thus,
    Qi + Qs± Qc ±Qv ±Qm – Qe = 0

    3840 + 600.624 + 764.235 + 468 + Qm = 0


    Thermal balance occurs when the sum of all the different types of
    5672.859 + Qm = 0 heat flow into and out of a building is zero. That is, the building is
    losing as much heat as it gains so it can be said to be in. equilibrium.
    Qm = -5672.859 Watts The heat loos or heat gain takes place via:-
    Conduction
    Therefor, we need to cool the space with the capacity of 5.5-6.0 KW. Convection
    Radiation

    BP-Thermal Balance Equation, By Nidhi Chadda (M.Arch -Environ, Sem 1)


    Thank You !

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