12.12 Answers to the Problem Set
- We need to find the total and evaporative heat losses and compare them with the results of Brockway et al. Since there are three air temperatures and three fleece thicknesses, there will be nine separate calculations. We need to calculate \(T_h\) using equation (12.22). To do this, we need the \(\alpha\)’s, \(\beta\)’s and the Nusselt numbers.
Nusselt numbers can be calculated from section 12.5 by first obtaining Reynolds number and using the right coefficients \(a\) and \(b\).
Nusselt Numbers as a function of air temperature (°C) and diameter (m)
| \(T_a\)(°C) | 0.32 | 0.36 | 0.50 |
|---|---|---|---|
| 10 | 38.40 | 41.31 | 50.64 |
| 20 | 36.96 | 39.76 | 48.75 |
| 30 | 35.66 | 38.36 | 47.03 |
Once we have \(Nu\), \(\alpha\) can be obtained from \(\alpha=A_h(Nuk_a/d+2\sigma T_a^3)\) (12.23). The calculations for \(\beta\) is \(\beta=1/(Z_h+Z_f)\) (12.24). \(Z_f\) and \(Z_h\) to get \(\beta\) are calculated from (12.9) and (12.10).
\(\alpha\) and \(\beta\) values (W °C-1) as a function of air temperature (°C) and diameter (m)
| 0.32 | 0.36 | 0.50 | |
|---|---|---|---|
| 10 | 5.644 | 6.200 | 8.079 |
| 20 | 5.898 | 6.447 | 8.484 |
| 30 | 6.178 | 6.803 | 8.927 |
| \(\beta\)(W °C-1) | 4.43 | 1.850 | 0.705 |
Since we are only taking long-wave radiation into consideration, \(R_{Na} = \frac{1}{2}A_h(1.04B−σT_a^4)\).
\(R_{Na}\) (W) values as a function of air temperature (°C) and diameter (m)
| 0.32 | 0.36 | 0.50 | |
|---|---|---|---|
| 10 | -51.45 | -57.86 | -80.36 |
| 20 | -59.09 | -66.48 | -92.33 |
| 30 | -67.58 | -76.03 | -105.60 |
Using \(\alpha\), \(\beta\), \(T_a\), \(R_{Na}\) and \(T_b = 39^\circ C\), \(T_h\) can be calculated from \(T_h=\frac{\alpha T_a+\beta T_b}{\alpha+\beta}+\frac{R_{Na}} {\alpha+\beta}\) (12.22).
\(T_h\) (°C) values as a function of air temperature (°C) and diameter (m)
| 0.32 | 0.36 | 0.50 | |
|---|---|---|---|
| 10 | 17.65 | 9.48 | 3.18 |
| 20 | 22.43 | 16.24 | 11.41 |
| 30 | 27.39 | 23.14 | 19.69 |
From equation (12.14), the total heat loss can be estimated at \(M\) since \(T_b\) is always greater than \(T_h\). \(M=\beta(T_b-T_h) + E\), and \(M\) = 8800 (kJ/day) = 101.8 (J/s = W) for \(T_a\) = 15 ~ 30 °C and \(E\) = 20.4 (W) for 10 °C.
Total Heat Loss (W) as a function of air temperature (°C) and diameter (m)
| 0.32 | 0.36 | 0.50 | |
|---|---|---|---|
| 10 | 114.3 | 75.0 | 45.7 |
| 20 | 101.8 | 101.8 | 101.8 |
| 30 | 101.8 | 101.8 | 101.8 |
Evaporative Heat Loss (W) as a function of air temperature (°C) and diameter (m)
| 0.32 | 0.36 | 0.50 | |
|---|---|---|---|
| 10 | 20.4 | 20.4 | 20.4 |
| 20 | 28.4 | 59.7 | 82.4 |
| 30 | 50.4 | 72.5 | 88.2 |
Brockway, McDonald and Puller found that metabolic and evaporative heat losses were fairly insensitive to fleece thickness. Exceptions were for a sheep with (a) 1 cm fleece where the percentage of evaporative water loss was generally lower and these animals had a higher metabolic rate especially at lower air temperatures, and (b) with 10 cm fleece metabolic rate was higher at higher air temperatures. At 10°C, they measured metabolic rate to be 140 W for a sheep with 32 cm fleece and 109 W for other sheep. This is higher than the model predicts (114.3 to 45.7 W). At 20 and 30°C, evaporative water loss was measured to be 40.7 and 91.5 W; then values should be compared with model predictions of 28.4 to 82.4 (W) and of 50.4 to 88.2 (W). The comparisons suggest that the model is excessively sensitive to changes in fleece thickness.
Evidently, heat transfer through the fleece is not accurately represented by our model. An improved model might take into account factors such as change in thermal conductivity of the fleece as a function of its water content. Such refinements are always possible. A point is always reached, however, where for all practical purposes the costs involved in making the improvement may exceed the benefits gained from it. In the present case, benefits consist chiefly in improved understanding of the process of heat transfer.
\[\begin{align*} k_a&=2.37\times10^{-2}Wm^{-1}°C^{-1} \\ \nu&=1.24\times10^{-5}m^2s^{-1} \\ v&=2.778ms^{-1} \\ E&=20.4W \end{align*}\]
| 0.32 | 0.50 | |
|---|---|---|
| Re | 71690 | 112016 |
| Nu | 205.65 | 295.21 |
| \(\alpha\) (W°C-1) | 17.387 | 25.22 |
| \(\beta\) (W°C-1) | 4.43 | 0.705 |
| \(R_{Na}\) (W) | -38.08 | -59.51 |
| \(T_h\) (°C) | -0.695 | -11.09 |
| \(\beta(T_b-T_h)\) (W) | 203.2 | 35.86 |
| \(M\) (W) | 223.6 | 56.3 |
- 991 kcal day-1 = 47.98 W.
\(M = E + \beta (T_b - T_h)\) and insert the values from problem 2.
| 0.32 | 0.50 | |
|---|---|---|
| Required M (W) | 251.2 | 83.84 |
| Required grass (g day-1) | 1331.4 | 444.4 |
- Since 1 cal = 4.184 J,
\(M\) = 1900 kcal/day = 92.0 W
\(E_s\) = 4 kcal/kg\(\cdot\)day = 13.6 W
To get \(R_{Na}\), the following formula is used. \[R_{Na}=a_1A_1S+a_2A_2s+a_3A_3(S+s)+a_4A_4R_a+a_5A_5\sigma T_a^4-\varepsilon A_h\sigma T_a^4\] Since heat gain from long-wave radiation can be ignored, \[R_{Na}=a_1A_1S+a_2A_2s+a_3A_3(S+s)-\varepsilon A_h\sigma T_a^4\] Inserting the provided values, we get \[R_{Na}=0.74 \cdot dL\cdot 800+0.74\cdot\frac{\pi d L}{2}\cdot 150+0.25\cdot 0.74\cdot \frac{2}{3}\frac{\pi dL}{2}(800+150)- 0.1396 \cdot \pi dL\sigma T_a^4\]
| 0.32 | 0.50 | |
|---|---|---|
| \(R_{Na}\) (W) | 237.1 | 370.4 |
| \(T_h\) (°C) | 55.96 | 68.96 |
| \(\beta(T_b-T_h)\) | -75.13 | -21.12 |
| \(\frac{E_s}{1+(Z_f/Z_h)}\) | 12.10 | 13.39 |
| \(E_r\) (W) | 155.0 | 99.73 |
(Recall that \(M = \beta(T_b-T_h) + E_r + \frac{E_s}{1+(Z_f/Z_h)}\).)
\(E_r\) ranges from 6.78 to 61.0 W, which is far less than needed to maintain the steady body temperature. The animals need to lower their metabolic rate or seek shade if they are to prevent a large increase in body temperature.
- For a whale, \(M=H\) since evaporative, conductive and radiative heat exchange can be ignored. From (12.15),
\[H=\frac{Nuk_w}{d}A_s(T_s-T_w)\]
Let \(\alpha =\frac{Nuk_wA_s}{d}\). Solving for \(T_s\),
\[T_s=\frac{T_b+Z_f\alpha T_w}{1+Z_f\alpha}\]
Using \(k_f=0.2052Wm^{-1}{}^\circ C^{-1}\), \(r_s=2.5m\), \(r_b=2.0m\), \(A=5\cdot\pi\cdot25=392.7m^2\), \(Z_f=\frac{1}{2\pi k_fL}ln\frac{r_s}{r_b}=6.923\times10^{-3}{}^\circ CW^{-1}\), and \(k_w=0.57Wm^{-1}{}^\circ C^{-1}\), \[\alpha =\frac{1.64\times10^4\cdot0.57\cdot392.7}{5}=7.342\times10^5W{}^\circ C^{-1}\]
\[T_s=\frac{33+6.923\times10^{-3}\cdot7.342\times10^5\cdot5}{1+6.923\times10^{-3}\cdot7.342\times10^5}=5.01^\circ C\]
Since evaporative heat loss can be neglected for whale, \(E_r = E_s = 0\). In addition, the lack of hair suggests that \(r_h = r_s\), leading to \(Z_h = 0\). The remaining equation for \(M\) is
\[\begin{align*} M&=\frac{T_b-T_s}{Z_f} = \frac{33-5.11}{0.006923}=4043W \\ &=3.49\times10^5kJ\:day^{-1} \mbox{ with perfect efficiency} \\ &=2.33\times10^6kJ\:day^{-1} \mbox{ at 15 percent conversion efficiency} \end{align*}\]
- There are several ways one might attack this problem. One could simply assume a conduction model in which the water surrounding the animal offered no resistance to heat flow (\(T_s = 5.0^\circ C\)). A Nusselt number calculation could be made using the Grashof number as in problem 9. Or finally, as is done here, one could assume that still water is equivalent to some very slow velocity (in air this is about 0.10 m s\(^{-1}\)).
\[\begin{align*} Nu&=80.5\;\;\;\;\;\;\;\alpha=3.604\times10^3 \\ T_s&=\frac{33+6.923\times10^{-3}\cdot3.604\times10^3\cdot5}{1+6.923\cdot10^{-3}\cdot3.604\times10^3}=6.08 \\ M&=\frac{33-6.08}{0.006923}=3.888\times10^3W \\ &= \mbox{ 96 percent of metabolic costs while swimming at over 4 m/s} \end{align*}\]
- See following table.
| 0.02 | 0.07 | ||
|---|---|---|---|
| \(A_n\) | (m2) | 0.003142 | 0.039584 |
| Nusselt number | 10°C | 10.25 | 18.77 |
| 30°C | 9.69 | 17.47 | |
| \(Z_f\) | 0.4544 | ||
| \(Z_h\) | 12.018 | 2.3063 | |
| \(\beta\) (W°C-1) | 0.08321 | 0.36222 | |
| \(\alpha\) (W°C-1) | 10°C | \(\frac{0.04833}{0.13154}\) | \(\frac{0.36285}{0.72507}\) |
| 30°C | \(\frac{0.05009}{0.13330}\) | \(\frac{0.38567}{0.74789}\) | |
| \(R_{Na}\) | 10°C | -0.13812 | -1.7404 |
| 30°C | -0.2096 | -2.6409 | |
| \(T_h\) (°C) | 10°C | 27.29 | 22.09 |
| 30°C | 34.05 | 30.82 | |
| Sensible heat loss (W) | 10°C | 0.9744 | 6.126 |
| 30°C | 0.4122 | 2.963 | |
| Evaporative water loss (W) | 0.1257 | 1.583 | |
| % Evaporative water loss | 10°C | 11.4 | 20.5 |
| 30°C | 23.4 | 34.8 | |
| Metabolic rate per unit area (W m-2) | 10°C | 350.1 | 194.8 |
| 30°C | 171.2 | 114.9 |
\[\begin{align*} Gr&=\frac{3.48\times10^{-3}\cdot9.81\cdot(0.5)^3}{(1.45\times10^{-5})^2}(5) \\ &=1.105\times10^8\;\;\;\;\;Nu=49.21 \\ H&= \frac{Nu\:k_a}{d}A_h(T_h-T_a)\\ &\frac{48.18\cdot0.0257\cdot\pi\cdot1.0\cdot0.5}{0.5}(5)=19.87W \end{align*}\]
\[\begin{align*} H&=\frac{Nuk_aA_h(T_h-T_a)}{d} \\ X&=\frac{k_aA_h}{d}0.48Gr^{0.25}(T_h-T_a) \\ &=\frac{k_aA_h}{d}0.48\Big(\frac{\beta gd^3}{v^2}\Big)^{0.25}(T_h-T_a)^{1.25} \\ &=k'(T_h-T_a)^{1.25} \\ &=k'[(\overline{T_h}-T_a)^{1.25}+1.25(\overline{T_h}-T_a)^{0.25}(T_h-T_a)] \end{align*}\]