Natural Convective Flow Inside an Air Space Consider double-glazed window plates with spacing of 12.7 mm and height of 0.6 m kept at temperatures of 17°C and 0°C. Calculate the natural convective heat transfer coefficient. Given: Spacing L = 0.0127 m, height H = 0.60 m Find: [latex]h_{con}[/latex] using Equation 12.25 for natural internal flow over plates Lookup values: From property tables of air (Table A1) at a mean temperature of (0°C + 17°C)/2 = 8.5°C: [latex]\rho = 1.271 kg/m^{3}, \mu = 1.75 × 10^{−5} Pa ⋅ s, c_{p} = 1004 J/(kg ⋅ K), k = 0.0246 W/(m · K)[/latex], [latex]\beta = \frac{1}{(273.15 + 8.5)} = 0.00355 K^{-1} [/latex] [latex]u = 1.40 \times 10^{-5} m^{2}/s, \alpha = 1.95 \times 10^{-5} m^{2}/s [/latex] and Pr = 0.717

First, the Rayleigh number is determined from Equation 12.18 as [latex]Ra = \frac{g \beta (\Delta T) L^{3}}{v \alpha} [/latex] [latex] = \frac{9.81 m/s^{2} \times 0.0355 K^{-1} \times (17°C - 0°C) \times 0.0127^{3} m^{3}}{1.40 \times 10^{-5} m^{2}/s \times 1.95 \times 10^{-5} m^{2}/s} [/latex] = 4442 Next, we will determine the Nusselt number using Equation 12.27: [latex]N u_{L} = 0.42 {Ra_{L}}^{1/4} Pr^{0.012} \left\lgroup \frac{L}{H} \right\rgroup^{0.3} [/latex] [latex] = 0.42 \times 4442^{1/4} \times 0.717^{0.012} \times \left\lgroup \frac{0.0127}{0.6} \right\rgroup^{0.3} = 1.083 [/latex] from which the convective heat transfer coefficient is deduced from Equation 12.16 as [latex]Nu = \frac{h_{con} D}{k} [/latex] (12.16) [latex]h_{con} = \frac{0.0246 W/(m \cdot K) \times 1.083}{0.0127 m} = 2.10 W/(m^{2} \cdot K) [/latex] Comments The Nusselt number is close to unity, suggesting that the interpane heat transfer (neglecting the radiative component) is mostly by conduction. The spacing is small enough to have largely suppressed the onset of a convection loop.

Question 12.10: Natural Convective Flow Inside an Air Space Consider double-......

Heating and Cooling of Buildings Principles and Practice of Energy Efficient Design [1026599]

Natural Convective Flow Inside an Air Space

Consider double-glazed window plates with spacing of 12.7 mm and height of 0.6 m kept at temperatures of 17°C and 0°C. Calculate the natural convective heat transfer coefficient.

Given: Spacing L = 0.0127 m, height H = 0.60 m

Find: $h_{con}$ using Equation 12.25 for natural internal flow over plates
Lookup values: From property tables of air (Table A1) at a mean temperature of (0°C + 17°C)/2 = 8.5°C: $\rho = 1.271 kg/m^{3}, \mu = 1.75 × 10^{−5} Pa ⋅ s, c_{p} = 1004 J/(kg ⋅ K), k = 0.0246 W/(m · K)$ ,

\beta = \frac{1}{(273.15 + 8.5)} = 0.00355  K^{-1}

$u = 1.40 \times 10^{-5} m^{2}/s, \alpha = 1.95 \times 10^{-5} m^{2}/s$ and Pr = 0.717

TABLE A.1 (IP Units) Transport Properties of Standard, Dry Air (14.7 psia)
°R	°F	ρ (lb_m/ft3)	c_p (Btu/lb · °R)	k (c_p/c_v)	μ x 10⁴(lb_m/ft · s)
180	-280	0.2247	0.2456		0.0466
198	-262	0.2033	0.2440	1.4202	0.0513
216	-244	0.1858	0.2430	1.4166	0.0559
234	-226	0.1711	0.2423	1.4139	0.0604
252	-208	0.1586	0.2418	1.4119	0.0648
270	-190	0.1478	0.2414	1.4102	0.0691
288	-172	0.1384	0.2411	1.4089	0.0733
306	-154	0.1301	0.2408	1.4079	0.0774
324	-136	0.1228	0.2406	1.4071	0.0815
342	-118	0.1163	0.2405	1.4064	0.0854
360	-100	0.1104	0.2404	1.4057	0.0892
369	-91	0.1078	0.2403	1.4055	0.0911
378	-82	0.1051	0.2403	1.4053	0.0930
387	-73	0.1027	0.2403	1.4050	0.0949
396	-64	0.1003	0.2402	1.4048	0.0967
405	-55	0.0981	0.2402	1.4046	0.0986
414	-46	0.0959	0.2402	1.4044	0.1004
423	-37	0.0939	0.2402	1.4042	0.1022
432	-28	0.0919	0.2401	1.4040	0.1039
441	-19	0.0901	0.2401	1.4038	0.1057
450	-10	0.0882	0.2401	1.4036	0.1074
459.7	0	0.0865	0.2401	1.4034	0.1092
468	8	0.0848	0.2401	1.4032	0.1109
486	26	0.0817	0.2492	1.4029	0.1143
504	44	0.0787	0.2402	1.4024	0.1176
522	62	0.0760	0.2403	1.4020	0.1208
540	80	0.0735	0.2404	1.4017	0.1241
558	98	0.0711	0.2405	1.4013	0.1272
576	116	0.0689	0.2406	1.4008	0.1303
594	134	0.0668	0.2407	1.4004	0.1334
612	152	0.0648	0.2409	1.3999	0.1364
630	170	0.0630	0.2411	1.3993	0.1394
648	188	0.0612	0.2412	1.3987	0.1423
666	206	0.0595	0.2415	1.3981	0.1452
684	224	0.0580	0.2417	1.3975	0.1479
702	242	0.0565	0.2420	1.3968	0.1508
720	260	0.0551	0.2422	1.3961	0.1536
738	278	0.0537	0.2425	1.3953	0.1563
756	296	0.0525	0.2428	1.3946	0.1590
774	314	0.0512	0.2432	1.3938	0.1617
792	332	0.0501	0.2435	1.3929	0.1643
810	350	0.0490	0.2439	1.3920	0.1670
100	-173.15	3.598	1.028		6.929
110	-163.15	3.256	1.022	1.420	7.633
120	-153.15	2.975	1.017	1.416	8.319
130	-143.15	2.740	1.014	1.413	8.990
140	-133.15	2.540	1.012	1.411	9.646
150	-123.15	2.367	1.010	1.410	10.28
160	-113.15	2.217	1.009	1.408	10.91
170	-103.15	2.085	1.008	1.407	11.52
180	-93.15	1.968	1.007	1.407	12.12
190	-83.15	1.863	1.007	1.406	12.71
200	-73.15	1.769	1.006	1.405	13.28
210	-63.15	1.684	1.006	1.405	13.85
220	-53.15	1.607	1.006	1.404	14.40
230	-43.15	1.537	1.006	1.404	14.94
240	-33.15	1.473	1.005	1.404	15.47
250	-23.15	1.413	1.005	1.403	15.99
260	-13.15	1.359	1.005	1.403	16.50
270	-3.15	1.308	1.006	1.402	17.00
280	6.85	1.261	1.006	1.402	17.50
290	16.85	1.218	1.006	1.402	17.98
300	26.85	1.177	1.006	1.401	18.46
310	36.85	1.139	1.007	1.401	18.93
320	46.85	1.103	1.007	1.400	19.39
330	56.85	1.070	1.008	1.400	19.85
340	66.85	1.038	1.008	1.399	20.30
350	76.85	1.008	1.009	1.399	20.75
360	86.85	0.980	1.010	1.398	21.18
370	96.85	0.953	1.011	1.398	21.60
380	106.85	0.928	1.012	1.397	22.02
390	116.85	0.905	1.013	1.396	22.44
400	126.85	0.882	1.014	1.396	22.86
410	136.85	0.860	1.015	1.395	23.27
420	146.85	0.840	1.017	1.394	23.66
430	156.85	0.820	1.018	1.393	24.06
440	166.85	0.802	1.020	1.392	24.85
450	176.85	0.784	1.021	1.392	24.85
Source: CRC Press, Handbook of Tables for Applied Engineering Science, CRC Press, Inc., Boca Raton, FL, 1973.

Step-by-Step

The 'Blue Check Mark' means that this solution was answered by an expert.
Learn more on how do we answer questions.

First, the Rayleigh number is determined from Equation 12.18 as

Ra = \frac{g \beta (\Delta T) L^{3}}{v \alpha}

= \frac{9.81  m/s^{2} \times 0.0355  K^{-1} \times (17°C  –  0°C) \times 0.0127^{3}  m^{3}}{1.40 \times 10^{-5}  m^{2}/s \times 1.95 \times 10^{-5}  m^{2}/s}

= 4442

Next, we will determine the Nusselt number using Equation 12.27:

N u_{L} = 0.42  {Ra_{L}}^{1/4} Pr^{0.012} \left\lgroup \frac{L}{H} \right\rgroup^{0.3}

= 0.42 \times 4442^{1/4} \times 0.717^{0.012} \times \left\lgroup \frac{0.0127}{0.6} \right\rgroup^{0.3} = 1.083

from which the convective heat transfer coefficient is deduced from Equation 12.16 as

$Nu = \frac{h_{con} D}{k}$ (12.16)

h_{con} = \frac{0.0246  W/(m \cdot K) \times 1.083}{0.0127  m} = 2.10  W/(m^{2} \cdot K)

Comments
The Nusselt number is close to unity, suggesting that the interpane heat transfer (neglecting the radiative component) is mostly by conduction.
The spacing is small enough to have largely suppressed the onset of a convection loop.