An UNglazed transpired mesh collector example

[nick pine]

Duffie and Beckman's 2006 SETP example 6.14.2 concerns an unglazed
transpired collector that's 10m wide x 3 m high with D = 1.6 mm
diameter holes on a P = 18 mm hex grid. The mass air flow rate Md =
0.05 kg/m^2-s, and the mesh has a 0.9 solar absorptance and emittance
E. The ambient temp Ta = 300 K (80 F) and the sky temp Ts = 292 K,
with S = 800 W/m^2 of sun...

At 300K, air density Rhoa = 1.176 kg/m^3, conductivity k = 0.0257 W/m-
K, specific heat Cp = 1007 J/kg-K, and viscosity Mu = 1/857E-5 kg/m-s.
With hole to mesh area porosity ratio Sp = 0.00717 and hole area
Aholes = 0.215 m^2, the air velocity through the holes Vholes = 0.05Ac/
(RhoaAh) = 5.93 m/s, and the Reynolds number Re = RhoaVholesD/Mu =
601. The heat transfer coefficient h = k/Dx2.75(P/D)-1.2Re^0.43 = 37.9
W/m^2, which makes NTU = h(Ac-Aholes)/(MdCp) = 0.747.

With a 0.5 collector-ground view factor, S - E[Tc^4-(Ta^4+Ts^4)/2] =
(Md/Ac)Cp(Tc-Ta)(1-e^-NTU). Everything but Tc is known. Solving... Tc
= 321.4 K (119 F), with a useful gain Qu = Ac[S-ESigma(Tc^4-
(Ta^4+Ts^4))/2] = 17 kW (a $50 shadecloth curtain behaving like $136K
of PVs :-) The outlet temp is Ta+Qu/(MdCp) = 38 C (100 F) and the
collector efficiency is 100x17K/(30x800) = 71%.

The approach velocity Va = 0.0425 m/s, and the pressure drop across
the plate dP = (RhoaVapp/2)(6.82((1-Sp)/Sp)^2Re^-0.236 = 30.7 Pa
(0.12" H20.)

How can we make this work well with Ta = 40 F (278 K)?

Nick

[nick pine]

> If we want to make a mesh heat a house with 40 F air, we need to know how to find the mesh temp...

> > With a 0.5 collector-ground view factor, S - E[Tc^4-(Ta^4+Ts^4)/2] = (Md/Ac)Cp(Tc-Ta)(1-e^-NTU). Everything but Tc is known. Solving... Tc = 321.4 K (119 F)... according to Duffie and Beckman.

> But it looks like the book has a typo in that equation...

Wiley Customer Support responds:

> Solar Engineering of Thermal Processes, 3e - errata, p288

> here is the response from our author...

There two problems with the numerical solution.

1) S is the absorbed solar radiation which is defined as I*alpha.
Consequently his 800 should be 720.

2) The flow rate is given as 0.05 kg/s-m^2 which already is per square
meter. Consequently, the /30 on the right hand side should not be
there.

With these changes your solution is exactly as given in the text.

You are correct that there is a sigma missing in the last equation; "S-
epsilon[..." should be "S-epsilon*sigma[..."

I did find a typo in Eqn 6.14.9. The plus sign should be a minus sign.

[7 page Errata Sheet Attached]

Nick

[WmJMorrison]

Blissfully ignorant here. My versions of SE of TP are 1974 and 1980.
Sun hasn't changed that much in 32 years. Fight. Right!