Consider a composite wall that includes an 8-mm thick hardwood siding, 40mm by 130mm hardwood studs on .65m centers with glass insulation ( paper faced, 28 kg/m^3) and a 12mm layer of gypsum (vermiculite) wall board.
What is the thermal resistance associated with a wall that is 2.5m high by 6.5m wide ( having 10 studs, each 2.5m high)? Assume surfaces parallel to the x-direstion are adiabatic.

Answer / Explanation:

(1) First, we identify the known parameters:

Known Parameters: Dimensions and materials associated with a composite wall ( 2.5m x 6.5m, 10 studs each 2.5m high)

Assumptions: (1) Steady state condition (2) Temperature of comosite depends only on x ( surface normal to x are iso thermal ), (3) constant properties (4) Negligible contact resistance.

Properties: (T≈ 300k ): Hard wood siding, Kₐ = 0.094 W / m.k :

Hard wood, Kₙ = 0.16 W / m.k : Gypsum, Kₓ = 0.17 W / m.k :

Insulation = Fiber glass faced paper, 28 kg/m³ ), K₀ = 0.038W / m.k

(2) Analysis:

                    Using the isothermal surface assumption, the thermal circuit associated with a single unit.

(3)

     (Lₐ / Kₐ Aₐ) = 0.008m / 0.094 W / m.k . ( 0.65m x 2.5m )

                        = 0.0524 K/W

(Lₓ / Kₓ Aₓ) = 0.13m / 0.16 W / m.k . (0.04m x 2.5m )

                        = 8.125 K/W

    (Lₙ / Kₙ Aₙ)         = 0.13m / 0.038 W / m.k . ( 0.61m x 2.5m )

                     = 2.243 K/W

(L₀ / K₀ A₀)   =             0.012M / 0.17 W / m.k . ( 0.65m x 2.5m )

Therefore, the equivalent resistance of the core is :

Req = ( 1 / Rb + 1 / Rd) ⁻¹

= ( 1/8.125 + 1/2.243)

1.758 k/w

and the total unit resistance

= Rtotal = Ra + Req + Rc

=1.854 k/w

Rtotal = ( 10 x 1 / Rtotal-1) ⁻¹

0.1854 k/w

Hence, if the surface parallel to the heat flow direction are assumed adiabatic, the thermal circuit and the value Rtotal  will differ.

Therefore, the total wall resistance = 0.1854 k/w

i think i backed my dad

a true.

answer:

make you upset

explanation: