12.9 An Energy Balance Calculation

In making the calculations that follow, we assume fleece thickness of 0.1 m, which is the depth of wool usually attained by Cheviot sheep (Brockway et al., 1965). Then skin diameter for our “standard sheep” is 0.3 m. The layer of subcutaneous fat is about 0.01 m thick (Porter and Gates, 1969). We take thermal conductivity of sheep’s wool as $5.91\times10^{-2}W m^{-2}{}^\circ C^{-1}$ (Blaxter, Graham, and Weinman, 1959) and that of fat as $20.51\times10^{-2} W m^{-2}{}^\circ C^{-1}$ (Porter and Gates, 1969). Body temperature of the sheep is taken as 39°C (Priestley, 1957). We assume a body weight of 70 kg.

To consolidate our calculation, we return to our general heat balance relation (Eqn. (12.1)): $$M+R_+=R_-+C+H+E$$ Our objective at this point is to calculate $T_h$, the fleece-tip temperature. We now define net radiation received by the animal at the fleece tips as $$\begin{align*} R_{Nh}=R_+-R_-&=a_1A_1S+a_2A_2s+a_3A_s(S+s) \\ &+a_4A_4R_a+a_5A_5R_g-\varepsilon A_h\sigma T_h^4 \tag{12.17} \end{align*}$$ and $R_{Na}$ as the net radiation which would be received in the same environment if the fleece tips were at air temperature, so that $$R_{Na}=a_1A_1S+a_2A_2s+a_3A_3(S+s)+a_4A_4R_a+a_5A_5\sigma T_a^4-\varepsilon A_h\sigma T_a^4$$ Then $$R_{Nh}-R_{Na}=a_5A_5\sigma T_h^4-\varepsilon A_h\sigma T_h^4-(a_5A_5\sigma T_a^4-\varepsilon A_h\sigma T_a^4)$$ and, since $a_5=1$, $A_h=\pi dL$ and $A_5=1/2A_h$, $$\begin{equation} R_{Nh}-R_{Na}=-\pi dL\sigma(T_h^4-T_a^4)(1/2-\varepsilon) \tag{12.18} \end{equation}$$

We now write $$\begin{equation} R_{Nh}\stackrel{.}{=}R_{Na}=2T_a^3(T_h-T_a)A_h \tag{12.19} \end{equation}$$ To obtain this relationship, we have assumed $\varepsilon\stackrel{.}{=}1$ and have used an approximation based on the Taylor Series expansion of $T_h^4 = g(T_h)$, say, near the point $T_h = T_a$: $$\begin{align*} T_h^4=g(T_h)&=g(T_a)+g^,(T_a)(T_h-T_a)+... \\ &=T_a^4+4T_a^3(T_h-T_a)+... \end{align*}$$ Since from Eqns. (12.1) and $$M-E+R_{Nh}=H+C$$ If we ignore heat loss through sweat (which can be relatively low in cool air temperatures), we can use Eqn. (12.14) to obtain, $$\begin{equation} \frac{T_b-T_h}{Z_h+Z_f}+R_{Nh}=H \tag{12.20} \end{equation}$$ We have dropped the conductive term $C$, since heat conduction from the animal to the ground is small in comparison to other heat loss if the sheep is not lying down. We now substitute Eqn. (12.19) into Eqn. (12.20) and, remembering that $H$ is given by Eqn. (12.15), we obtain an expression which we will use to determine $T_h$: $$\begin{equation} \frac{T_b-T_h}{Z_h+Z_f}+R_{Na}-2T_a^3\sigma A_h(T_h-T_a)=\frac{Nu}{d}k_aA_h(T_h-T_a) \tag{12.21} \end{equation}$$ Rearranging, we obtain $$\begin{equation} T_h=\frac{\alpha T_a+\beta T_b}{\alpha+\beta}+\frac{R_{Na}} {\alpha+\beta} \tag{12.22} \end{equation}$$ where $$\begin{equation} \alpha=A_h(Nuk_a/d+2\sigma T_a^3) \tag{12.23} \end{equation}$$ and $$\begin{equation} \beta=1/(Z_h+Z_f) \tag{12.24} \end{equation}$$ Formula (12.22) states that the temperature at the fleece tips $T_h$ is a weighted average of air and body temperatures $T_a$ and $T_b$, increased by a contribution from all sources of radiation. The weights $\alpha$ and $\beta$ have the units of “conductances,” that is, of conductivities multiplied by the areas across which heat flows. The conductance $\alpha$ has components associated with convection and radiation, $\beta$ with conduction in fleece and fat.

To illustrate the calculation of $T_h$, we take $T_a=10^\circ C$, $T_b=39^\circ C$, $A_h=0.5\pi m^2$, $Nu=194$, $k_a=2.48\times10^{-2}Wm^{-1} {}^\circ C^{-1}$, $k_h=5.91\times10^{-2}Wm^{-1} {}^\circ C^{-1}$, $k_f=2.51\times10^{-1}Wm^{-1} {}^\circ C^{-1}$, $d=0.5m$, $L=1.0m$, $r_b=0.145m$, $r_s=0.15m$, and $r_h=0.25m$.

Then $\alpha=1.571(9.638+2.570)=19.19W{}^\circ C^{-1}$, $Z_h+Z_f=1.3748+0.05431=1.428$, so that $\beta = 0.700W{}^\circ C^{-1}$. Also, $$\begin{align*} R_{Na}&=a_1A_1S+a_2A_2S+a_3A_3r(S+s)+1/2A_h(1.04B-\sigma T_a^4) \\ &=387.3+16.2+104.1-80.6 \\ &=427W \end{align*}$$ Then $$\begin{align*} T_h&=\frac{(19.18)(283)+(0.7)(312)+427}{19.88} \\ &=305.5K=32.5^\circ C \end{align*}$$ Thus, in this case, the temperature of the fleece tips is approximately 22°C higher than that of the air. We make the following observations.

1. The contribution of body temperature to fleece-tip temperature is much smaller than that of the air temperature, because the weight $\alpha$ is much larger than $\beta$. In turn, $\beta$ is small because of the very large resistance to heat conduction offered by the wool.

2. The radiative contribution to fleece-tip temperature, which is appreciable (21°C), depends very largely on the size of the short-wave component.

3. The radiative contribution to the weight a is considerably smaller than the convective one.

4. Contribution of fat to total resistance to heat flow from the animal is very small compared to that of the wool.

Once we have determined $T_h$, a number of very important calculations become possible. The first of these is the total heat of metabolic origin that passes outward through the fleece tips, which is given by Eqn. (12.14): $$\frac{T_b-T_h}{Z_h+Z_f}=M-E_r-E_s/(1+\frac{Z_f}{Z_h})$$ In the present case, the quantity on the left works out to be $(39 - 32.5)/1.428 = 4.55 W = 393 kJ$ day$^{-1}$. Checking the other side, we have already noted that sweating heat loss is in the neighborhood of $16.8 kJ kg^{-1} day^{-1} = 1170 kJ day^{-1}$ for a 70 kg animal, independent of air temperature (Brockway et al., 1965). Many experiments have shown that 70 kg sheep lose between 8,800 and 10,000 kJ day$^{-1}$ heat. If we take $M = 8,800 kJ$, then $E_r = 7630 kJ day^{-1}$, which is about twice as great as the 3,660 kJ day reported by Brockway et al. (1965). In their experiment, however, there was no net absorption of radiation at the fleece tips, so that higher conductive heat losses were possible.

For comparative purposes, it is also useful to calculate the magnitudes of the terms in the energy balance equations (12.19) and (12.20). These are

1. metabolic sensible heat loss $$\frac{T_b-T_h}{Z_h+Z_f}=4.6W$$
2. radiation heat gain $$R_{Nh}=R_+-R_-=387.3+16.2+104.1+2.05+386.2-757.7=336W$$
3. convective heat loss $$H =\frac{Nu}{d}k_aA_h(T_h-T_a)=341W$$

Thus, in this case, net incoming radiation at the fleece tips is approximately equal to convective heat loss and metabolic heat loss by conduction from the body through fat and fleece is very small. This is due in part to the high “radiant” temperature at the fleece tips and the resulting low temperature gradient between body and fleece tips.

These examples illustrate only some of the types of problems we can solve with this technique. We have shown that, if the sheep’s environment is given in terms of temperature, wind speed, and incoming radiation, it is possible to determine the range of rates of body metabolism and evaporative losses due to respiration and sweating that will allow the animal to maintain thermal equilibrium. Porter and Gates (1969) have worked out the converse problem for a number of animals: given a range of metabolic rates that the animal can maintain as well as its physical characteristics (size, thickness of fur, etc.), the range of environmental conditions within which it can survive can be calculated.