## 12.11 Problem Set

Assume \(T_b = 39\)°C, \(v=1\)km/hr = 0.2778ms^{-1}, \(A_h = \pi d L\)m^{2}, \(L=1\)m, \(r_b=0.29\)m, \(r_s\)=0.3, \(k_h = 5.91 \times 10^{-2}\), \(k_f = 0.251\), \(\varepsilon = 1\) and \(e = 10\)mb for the following problems unless specified.

- Brockway et al. (1965. Evaporative heat-loss mechanisms in sheep. J. Physiology 179: 554-568) compared the contributions of metabolism and evaporative water loss in sheep. Use the model in this modules to estimate the total evaporative heat losses at three temperatures (10, 20, 30 °C) and three body diameters (\(r_h\)=0.32, 0.36, 0.50) corresponding to fur thicknesses. Then roughly compare your results to heat losses measured by Brockway et al. You can estimate net radiation under the assumption that the sheep’s body temperature is steady state such that the tips of its fleece are at air temperature. The sheep is in an enclosed area so we can neglect the energy gain through direct, sky, and reflected radiations. At moderate air temperatures (15 to 30°C), metabolic rate will be high and fairly constant which from the text we will take as \(8800 kJ day^{-1}\). At 10°C, however, it is assumed that evaporative water loss is 20.4 (W). Assume the following relationships for how the kinematic viscosity of air \(\nu\) and the thermal conductivity or air \(k_a\) vary with temperature.

\(T_a\)(°C) | 10 | 20 | 30 |
---|---|---|---|

\(\nu \:(m^2s^{-1}\times10^{-5})\) | 1.42 | 1.51 | 1.60 |

\(k_a \:(W m^{-1}°C^{-1}\times10^{-2})\) | 2.50 | 2.57 | 2.64 |

- What is the total heat of metabolic origin that passes outward through the fleece tips required to maintain a \(T_b\) of 39 °C given that \(T_a\) = -10 °C for sheep with diameters of 0.32m and 0.5m? The sheep is in an enclosed area so we can neglect the energy gain through direct, sky, and reflected radiations. Assume, \(k_a= 2.37\times10^{-2}Wm^{-1\circ}C^{-1}\), \(\nu=1.24\times10^{-5}m^2s^{-1}\) and evaporative heat losses \(E = 20.4W\).

Use the values in problem 2 but now assume that the evaporative heat loss is 990 kcal per day. How much grass are sheep with diameters of 0.32m and 0.5m required to eat to compensate for the metabolic heat loss if consuming 5.30g of grass per day is required to offset 1W?

Consider a 70 kg sheep on a sunny day with \(T_a\) of 30°C experiencing direct solar radiation of 800W right from above and diffuse sky radiation of 150W with one-third of the solid angle occupied by the sheep’s shadow. Suppose your sheep is standing in a pasture in which reflectivity of the grass is 25 percent. Assume cutaneous evaporative heat loss equals 4 kcal kg

^{-1}day^{-1}, but respiratory heat loss can vary. It is known that sheep can breathe as few as 20 times or as many as 250 times per minute, and the corresponding respiratory heat loss varies from 3.0 to 18.0 kcal kg^{-1}day^{-1}. Calculate the rate of heat loss when the sheep is shorn (\(r_h = 0.32\)) and unshorn (\(r_h = 0.50\)). Assume that total metabolic heat produced is 1900 kcal day^{-1}. What does the comparison of these heat fluxes suggest about whether evaporative water loss will be sufficient for the sheep to avoid overheating in the sun? \(\varepsilon = 0.1396\), \(a = 0.74\) and heat received by long-wave radiation can be ignored. You may use the corresponding \(\alpha\) and \(\beta\) values from problem 1.Now assume that instead of a sheep your study animal is a blue whale, which you can idealize as a cylinder 25 m long and 5 m in diameter, including a layer of blubber 50 cm thick. Its body temperature is maintained at 33°C. If the whale swims at a rate of 15 km hr\(^{-1}\) through 5°C water, how much food (in kcal day\(^{-1}\)) must it consume to maintain thermal equilibrium? Take the Nusselt number \(Nu\) as \(0.332 (Re_x)^{1/2} \stackrel{.}{=} 1.6 \times 10^4\) (since \(V = 15 km \:hr^{-1}\), \(x \stackrel{.}{=} 10 m\), \(v = 1.7 \times 10^{-5} m^2 sec^{-1}\)), \(k_w=0.57Wm^{-1}{}^\circ C^{-1}\) and \(k_f = 0.2052 Wm^{-1}{}^\circ C^{-1}\).

Calculate heat loss in the blue whale when its velocity relative to the water is zero. The thermal conductivity of water is approximately \(5.7 \times 10^{-3} J cm^{-1} s^{-1} {}^\circ C^{-1}\).

Calculate \(T_h\) for a rat whose length is 18 cm, total diameter is 7 cm with a fat thickness of 0.3 cm and a hair thickness of 0.5 cm. Take \(\varepsilon = 1.0\) and \(T_b = 39^\circ C\). Assume the same environmental conditions as in problem 1, but make calculations for \(T_a\) = 10 and 30°C only. If evaporative heat loss is 40 W m\(^{-2}\), what is the metabolic heat production? Repeat your calculations for a shrew whose body length is 5 cm, hair thickness of 0.20 cm and total body diameter of 2 cm. What percentage is evaporative water loss of the total heat loss? Of the sheep, the rat and the shrew, who has the largest per-unit-surface-area heat production?

When heat transfer depends on air movements caused by differences in fluid density (i.e., buoyancy effects), the Nusselt number can be shown to depend on another dimensionless parameter, the Grashof number, \[Gr=\frac{\beta gd^3}{v^2}(T_h-T_a)\] where \(\beta\) is the coefficient of thermal expansion of the fluid, \(g\) is the acceleration due to gravity, and \(d\) is a characteristic length. For air at 20°C moving at right angles to the axis of a cylinder and \(10^4<Gr<10^9\), \(Nu = 0.48 Gr^{0.25}\) (Monteith, 1973). For the sheep, calculate rate of heat loss due to free convection, if \(T_h\) = 25 °C and \(T_a\) = 20 °C. (Take \(\beta = 3.48 \times 10^{-3} °C^{-1}\), \(g = 9.81 m \cdot sec^{-2}\), \(v = 1.45\times10^{-5} m^2\cdot sec^{-1}\), \(d = 0.5 m\).)

Suppose you did not know \(T_h\) and wished to solve for it before proceeding further with your analysis. (Recall that we linearized terms involving \(T_h^4\) by writing \(T_h^4\stackrel{.}{=}T_a^4+4T_a^3(T_h-T_a)\), supposing that \(T_h\) is sufficiently close to \(T_a\), which is taken as known.) How would you linearize the term containing \(Nu = 0.48Gr^{0.25}\)?