System analysis is largely concerned with budgets. Since we exclusively focused on budgets of nutrients and suspended microorganisms so far, let’s consider something totally different (Figure 1). Let’s try to answer the question how much fish a penguin needs to ingest in order to keep its body temperature.
Figure 1: Fat penguin in the cold dreaming of a high-energy snack.
2 General concept
A rise and decrease in temperature is always a consequence of heat fluxes. At the same time, we know that heavy bodies with a large specific heat capacity (e.g. water) warm up or cool down only slowly. These relations can be expressed by the very general Equation 1
\[
\frac{d}{dt} T = \frac{Q_{in} - Q_{out}}{m \cdot hcap}
\tag{1}\]
which relates the change in temperature \(T\) (°C) to …
imported or internally produced heat, \(Q_{in}\) (kJ),
the outward-bound heat flux, \(Q_{out}\), (kJ),
the mass of the body, \(m\), (kg),
the specific heat capacity, \(hcap\), (kJ/kg/K).
For a body to keep a constant temperature, the derivative \(\frac{d}{dt} T\) must be zero. As stated by Equation 1, this requires a perfect compensation of any heat loss by a corresponding gain, i.e. \(Q_{in} = Q_{out}\) while the heat capacity (denominator of Equation 1) is irrelevant. Since heat losses (and possibly gains) happen continuously, the case of a constant body temperature represents a classic steady-state problem.
3 Simplifying assumptions
To make the question tractable, let’s make a number of simplifying assumptions about the penguin and its behavior.
The body shall be sphere-shaped such that volume and surface are given by the diameter.
The insulation is due two layers: a layer of fat plus an outer layer of air (feathers).
Heat loss happens over the body surface only. Any other losses due to, e.g., breathing or feces are neglected.
All energy taken up via food is assumed to be converted into heat rather than work. For example, we neglect that the penguin may paddle in water and energy is lost while creating turbulence.
4 Calculation
The calculation is rather straightforward. As usual, one must make sure that all units are consistent. The trickiest part is actually to account for the combined effect of the two layers of insulation. Since multi-layer insulation is common in modern construction, I found the solution in some text on energy-efficient buildings. The trick is to calculate the heat transfer coefficient \(U\) (kJ / m^2 / d) as the inverse of the sum of the heat transfer resistances of the individual layers (see detail in code below).
# assumed gradient of temperatureT.bdy <-37# body temperature (°C)T.env <--20# ambient temperature (°C)# characteristics of the penguindiam <-0.25# diameter of sphere-shaped body (m)hcap <-4.2# spec. heat capacity of body, using water (kJ / kg / K)thick.fat <-0.02# thickness of fat layer (m)lambda.fat <-12# approximate thermal conductivity of fat (kJ / m / d)# --> 0.14 W/m/K * 86400 s/d * 1e-3 kJ/Jthick.air <-0.01# thickness of air layer representing feathers (m)lambda.air <-2.25# thermal conductivity of air (kJ / m / d)# --> 0.026 W/m/K * 86400 s/d * 1e-3 kJ/J# quality of fishecont <-6370# energy content of herring (kJ / kg)# derived properties of the penguinarea <-4* pi * (diam/2)^2# body surface (m^2)mass <-4/3* pi * (diam/2)^3*1000# body mass (kg)U <-1/ (thick.fat/lambda.fat +# heat transfer coef. (kJ / m^2 / d) thick.air/lambda.air)# derivative of body temperature (K / d) depending on ingested fish (kg / d)dT_over_dt =function(ingest) { Q.in <- ingest * econt # energy gain due to food (kJ / d) Q.out <- (T.bdy - T.env) * U * area # heat loss over body surf. (kJ / d) (Q.in - Q.out) / (hcap * mass) # resulting change in temp. (K / d)}# obtain solution by root-finding using a sensible guess (interval)x <-uniroot(dT_over_dt, interval=c(0, 10))print(paste("Penguin needs",signif(x$root, 3),"kg / d to keep",T.bdy,"°C"))
[1] "Penguin needs 0.288 kg / d to keep 37 °C"
# even simpler, we could directly set Q.in = Q.out (see body of the function# dT_over_dt) and solve for the ingestion rateingest <- (T.bdy - T.env) * U * area / econtprint(paste("Direct solution:",signif(ingest, 3),"kg / d"))
