Radiant Heating

8.9.4.1 Radiant Temperature

The idea of radiant heating in an industrial hall is to install local heating panels with a high surface temperature. Because of the high temperature, the proportion of thermal radiation in the heating power from the panel will in­crease drastically. But we must remember that the heating panels will also ra­diate directly on outer surfaces of the buildings, thus raising their surface temperature and increasing the heat flow through the walls.

The heating panels can be heated with electricity, burning gas flames, or hot water.

The temperature that a thermometer with an almost black bulb indicates when it is placed near the heating panel (Fig. 8.51) is practically the same tem­perature that a human being will feel in the neighborhood of the panel.

The heat balance for the bulb in Fig. 8.51 is

Fcre(T*-Tt) = kc(Tb-Ta) + (l-F)(re(Tt-T*) . (8.30)

Ts is the surface temperature of the panel, Tb the thermometer bulb tempera­ture, Ta the air temperature, and Tw the temperature of the walls of the build­ing. F is the view factor from the bulb to the heating panel, e is the emissivity of the thermometer bulb at temperature Th, a is the Stefan-Boltzmann con­stant (5.67 x 10 “ 8 W m “ 2 K ~ 4), and hc is the convective heat transfer coeffi­cient from bulb to air.

The unknown temperature Tb can be calculated from Eq. (8.30) when the other temperatures Ts, Ta, and Tw are known and also F, bc, and e. The tem­perature of the wall and air is practically the same, so we can set T, = TH, We also use for the heat transfer coefficient hc = 4 W m ~2 K “l.

The emissivity e of the bulb is quite insensitive to the material of the bulb, as long as it is not a polished bright metal. Because the bulb radiates long­wave heat radiation, we can use the value e = 0.95.

Radiant panel

„, Bulb of

^ a thermometer

FIGURE 8.51 Heat balance for a black bulb in front of a radiant panel.

The view factor F can be calculated with the aid of Fig. 8.52. With the notations in Fig. 8.52, the view factor for a position for the bulb like that in Fig. 8.52 is

F = — r-arcsin ■■■■■. . (8.31)

4 77 L „2 r, l r>2 r-,l

‘vl+Jj + C +J5 G

If the thermometer is situated symmetrically relative to the hearing panel, then the view factor will be four times higher than calculated from Eq. (8.31).

Example

Figure 8.53 gives measured values of the “radiant temperature,” which means Tb — Ta, caused by an electrical heating panel. The “effective surface” temperature of the panel can be estimated from the curves, then used to calcu­late the temperature of a thermometer bulb at a few other places. These results can be compared to measured results.

Solution The first step is choosing a reference point. At the point 1 m right below the panel, the radiant temperature is 29 °C, and if we assume that the temperature of air is 20 °C, then Th = 49 °C.

0.5 m

1.0 m

1.5 m

2.0 m

■ 2.5 m

1.0 m

1.0 m

FIGURE 8.53 Measured values of radiant temperature near an electrical heating panel. Length of panel: 1.53 m; width: 0.28 m.

The view factor at that point is

0.14×0.765

— 0.02685 .

F = 4-p-arcsin 4tr

Inserting this value into Eq. (8.30) and also Ta = Tw = 20 °C = 293 K, we get (by iteration) the temperature Ts = 740 K = 467 °C. We can use this value for calculating the radiant temperature at other locations. For each location a new view factor has to be calculated. The results are shown in Fig. 8.54.

We see that the agreement between measured and calculated temper­atures is fairly good. Only in the right corner near the heating panel is there a big difference between the measured and calculated temperature. However, the measured results cannot be reliable here either, because it is not possible that the radiant temperature is 80° just near a surface of 467 °C.

8.9.4.2 Radiant Heating Panels Heated by Water

Radiant heating can be realized also by heating the panels by water. The principal construction of such a device is shown in Fig. 8.55.

The temperature of the water in the heating panel is limited by the boiling of the water. The boiling temperature of water increases with increasing pres­sure. If the pressure of the water is about normal atmospheric pressure, then the temperature of the water can be 90-95 °C. If we want to raise the temper­ature of water to 120 °C, the absolute pressure of water must be above 2 bar.

Heat loss through back side, appr, 10%

^ r 1 ? ? 1 1 f / ? 1 1 1 ( ?

Convection, appr. 30′

Radiation, appr. 60%

FIGURE 8.56 Heat balance of a heated panel.2

Where

Twi = inlet temperature of the water to the panel and 7wo is the outlet temperature.

T. ul. = temperature of the walls and roofs of the industrial hall; Tfl is the temperature of air in the hall, and we can put Tsur = Ta.

A = area of the heating panel and b is the convective heat transfer coefficient.

E = emissivity of the surface and, in the considered temperature range, we can put e = 0.95.

<f>[ can be estimated if we know the type and thickness of the thermal insulation of the panel. One estimation is that is 9-10% of the total heat to the pane!.

The convective heat transfer for the panel is free convection from a heated surface faced down. It can be calculated from Incropera and DeWitt:1

1/4

G(Ts-Ta)Ld

(8.34)

H = 0.27 f-

Ct

(Ts + Ta)va

Where

G — gravitational acceleration, 9.81 ms’ k = thermal conductivity of air v = kinematic viscosity of air a = thermal diffusivity of air

Numerical values for these quantities for a few temperatures are

Temperature (K)	K, W m-1 K-‘	V, m2 s-‘	A, m4 s-‘
293	0.0257	15.3 x 10	21.6 x 10
313	0.0273	17.2 x 10	24.4 x 10 “f’
333	0.0287	19.2 !0	27.4 x 10" «’
353	0.0302	21.2 x 10“*	30.4 x 10 " 6

When calculating h from Eq. (8.34), the values for k, v, and a should be taken at temperature A (Ts + Ta).

The length L in Eq. (8.34) is defined as

Lel = а, (8.35)

Where P is the perimeter of the surface area of the heating panel. In case the length of the panel is much bigger than the width, then Eq. (8.35) becomes

I _ W, o

Lcf-y.

Equation (8.36) can be written as heat per unit length of the heating panel:

<j)/L = eaW l 4 — Ts4ur + h WTf — Ta + <k//L. (8.37)

Example

The heat effect per unit length by radiation and convection from a water — heated panel can be calculated theoretically. For example, consider panels of width 0.6 m, 0.9 m, and 1.2 m. Room temperature is 3 °C and surface temper­ature 30 °C, 50 °C, and 70 °C of the panel. Let us compare the results with cal­culations for room temperature 15 °C and surface temperature 40 °C, 60 °C, and 80 °C of the panel.

Solution One of the most critical and important quantities to calculate in Eq. (8.32) is the convective heat transfer coefficient. It depends on the tem­perature conditions and also on the width of the panel. Tables 8.11 and 8.12 collect the calculated heat transfer coefficients in different conditions.

We see from the results that in these situations there is no temperature de­pendence of the heat transfer coefficient. However, the heat transfer coeffi­cient is lower the larger the width of the panel is. It is quite natural that a narrow panel has a higher heat transfer coefficient because it is easier for the air to rise upwards when the panel is narrow. Using the heat transfer coeffi­cients mentioned above, the heat effect delivered by the heating panels down­ward is

4»/L = eirW(l4 — Tt, r) + h W(TS — Ta) .

TABLE 8.11 Heat Transfer Coefficients for ATa 5 °C Heat transfer coefficient, h (W m-2 K-1) Surface temperature T, for T„ = 5 °C W = 0.6 m W = 0.9 m W = 1.2 m 30 °c 2.1 1.9 1.8 50 °C 2.4 2.2 2.0 70 °C 2.6 2.4 2.2

HH TABLE 8.12 Heat Transfer Coefficients for ATa = 15 °C

Heat transfer coefficient, h (W m-2 K’1) Surface temperature T, for T„ 1S °C W = 0.6 m W = 0.9 m W = 1.2 m 40 °c 2.1 1.9 1.8 60 °c: 2.4 2.2 2.0 80 x: 2.6 2.4 2.2

Tables 8.13 and 8.14 collect the results and compare them to results given by a manufacturer.

We see that there is a big difference between the calculations and the re­sults given by the manufacturer. The manufacturer’s results are about 45-50% higher. The reason for the differences is not clear. One difficulty is that we do not know in the manufacturer’s case how much heat is flowing through the in­sulation upward. We have assumed that this flow is 9% of the total heat effect of the heating panel.

TABLE 8.13 Heat Transfer from Panels. Calculated Values and Manufacturer’s Data

Heat effect per unit length (W rrr1)

W = 0.6 m W = 0.9 m W=l.2i Surface temp Ts For T„ — 5 °C Calc. Manuf. Calc. Manuf. Calc. Manuf. 30 ЬC 110 150 160 230 212 310 50 ‘ C 220 320 330 480 430 650 70 ‘ C 360 520 520 770 680 1040

TABLE 8.14 Heat Transfer from Panels. Calculated Values and Manufacturer’s Data Surface temp Ts for T„ = 15 ”C Heat effect per unit length (W m~ ‘) W = 0.6 m W = 0.9 m W = 1.2 m Calc. Manuf. Calc. Manuf. Calc. Manuf. 40 °C 120 150 170 230 230 310 60 °C 240 320 350 480 460 650 80 °C 380 520 560 770 730 1040

A common heating method in industry is using blowers that blow heated air into the room. See Fig. 8.57. An advantage is that the installation costs are low. A possible disadvantage is that it mixes the air in the room, which may not be desirable in combination with displacement ventilation.

When locating hot air blowers, we need to know the penetrarion depth for the hot air that is being discharged from the blowers. For a given discharge angle the penetration depth, 2, can be expressed as

Z~kJ

Where

Z = penetration depth for the centerline of the hot air jet (m) kz = penetration factor, see Fig. 8.58

I = characteristic length tor the discharge opening, normally the square root of the discharge area (m)

The Archimedes number is defined as

(8.39)

Ar = ATgl/Tu(i~

Where

G = acceleration of gravity (9.81 m/s2)

AT = temperature difference between the discharged warm air and the room air (K)

T = absolute temperature in the discharged air (K) m0 = discharge velocity of the warm air (m/s)

To find the penetration depth,

1. Find the Archimedes number from Eq. (8.39).

2. Read the constant from Fig. 8.58.

3. Calculate the penetration depth for the centerline of the jet from Eq. (8.38).

0.0002 0.0005 0.001 0.002 0.005 0.01 Archimedes number, Ar FIGURE 8.58 Penetration depth versus Archimedes number and discharge angle.

0.02

0.05

Example

A room is heated by hot air blowers. The indoor temperature is +10 °C. Each blower has the following data:

Discharge opening dimensions: w = h = / = 0.5 m Discharge velocity: u0 = 5 m/s Heater power: Q = 5 kW

The blowers are mounted at the walls, and tilted 45° downward. At what height should the blower be mounted so that the air jet reaches the floor? See Fig. 8.57.

Solution First, we find the discharge airflow rate from the blower:

Qv = Auq = 0.5m x 0.5m x 5m/s = 1.25(m3/s) .

The temperature increase in the blower is

AT = Q/qvpcp

= 45 kW/1.25 m3/s x 1.15(kg/m3 x l(kW s/kg K)) = 30 K.

The Archimedes number for the jet at the discharge point, Eq. (8.39), is

Ar0 = 30 Kx 9.81 m/s2 x 0.5(m/(313 Kx(5 m/s)2)) = 0.0196.

For a discharge angle of 45°. Thus, the penetration depth for the centerline of the jet becomes

H ~ Јmax = 6 x 0,5 m = 3 m.

The radius of the jet at the lowest point is approximately lira (calculated from a jet widening angle of 25° and a jet length of 4.2 m from the blower). See Section 7.4.

This means that the hot air blower should not be placed more than 4,2 m above the floor.

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