Question 1 of 5: Oxygen-Transfer Capacity of a Stirred-Tank Aerator

Question 1: Oxygen-Transfer Capacity of a Stirred-Tank Aerator (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Find. (a) The maximum volumetric O₂-transfer rate (g O₂·m^-3·h^-1) deliverable to the broth at 5°C; (b) a qualitative judgement, supported by the same correlations, on whether O₂ supply is more or less likely to be limiting at 15°C.

Approach. Compute the impeller's ungassed power from the power-number definition, apply the given gassed/ungassed ratio to get P_g, feed P_g/V_L into the given k_La correlation, and combine with the saturation DO concentration (from the given solubility equation, using the maximum possible driving force C^*−0) to get the maximum oxygen-transfer rate (OTR_max = k_La·C^*). Repeat the saturation-concentration step at 15°C to quantify how the ceiling moves.

1. Confirm the turbulent regime (power number is valid as a constant). $$N_{Re}=\frac{nD_i^2\rho}{\mu}=\frac{(1)(0.936)^2(1000)}{1\times10^{-3}}=\boxed{8.76\times10^5}$$ Well above the ∼10⁴ threshold for a fully turbulent impeller, so N_p=6 is valid as a constant (no Reynolds-number correction needed).
2. Ungassed impeller power from the power-number definition. $$P=N_p\,\rho\,n^3D_i^5=(6)(1000)(1)^3(0.936)^5=\boxed{4311\ \text{W}}=4.311\ \text{kW}$$
3. Gassed power and specific power input. The given ratio P_g/P=0.6 converts ungassed to gassed (aerated) power directly: $$P_g=0.6\times4311=2586\ \text{W}=2.586\ \text{kW}\quad\Rightarrow\quad \frac{P_g}{V_L}=\frac{2.586}{10}=0.2586\ \text{kW/m}^3$$
4. Volumetric mass-transfer coefficient from the given correlation. $$k_La=9.09\times10^{-4}(0.2586)^{0.7}=3.527\times10^{-4}\ \text{s}^{-1} =\boxed{1.270\ \text{h}^{-1}}$$
5. (a) Saturation DO concentration at 5°C and maximum OTR. The O₂ partial pressure is p=y_O2·P=0.21×760=159.6 Torr, so $$DO^{*}=\frac{(760-159.6)(0.678)}{35+5}=\boxed{10.18\ \text{ppm}}\ (=10.18\ \text{g/m}^3)$$ Taking the maximum possible driving force (bulk liquid essentially depleted, C_L→0) gives the ceiling on volumetric O₂ flux: $$OTR_{max}=k_La\cdot DO^{*}=(1.270)(10.18)=\boxed{12.9\ \text{g O}_2\,\text{m}^{-3}\text{h}^{-1}}$$
6. (b) Repeat the saturation step at 15°C. k_La is unchanged (the correlation depends only on P_g/V_L, not on T), but the solubility equation's denominator (35+t) grows with t: $$DO^{*}_{15^\circ C}=\frac{(600.4)(0.678)}{35+15}=8.14\ \text{ppm}\quad\Rightarrow\quad OTR_{max,15^\circ C}=(1.270)(8.14)=\boxed{10.3\ \text{g O}_2\,\text{m}^{-3}\text{h}^{-1}}$$ The maximum deliverable rate falls by a factor DO*(15)/DO*(5) = 0.80 — about 20% lower — purely from the temperature-dependence of oxygen solubility, with the mass-transfer coefficient itself unaffected.

(b) Yes, oxygen supply is more likely to become limiting at 15°C than at 5°C. Two effects work in the same direction: the maximum deliverable O₂ flux itself falls by ∼20% because the equilibrium solubility of oxygen in water decreases as temperature rises (the given correlation's denominator grows with t, and physically warmer water simply holds less dissolved gas), while the mixing power and hence k_La are essentially unchanged (no explicit T-dependence in the given correlation, and viscosity/density changes over this range are modest). At the same time, the microorganisms' demand for oxygen (the specific oxygen-uptake rate) typically rises with temperature following Arrhenius-type kinetics, up to the organisms' thermal optimum. A plant that was already close to its 5°C transfer ceiling of ∼12.9 g·m^-3·h^-1 would therefore be squeezed from both sides at 15°C — a lower supply ceiling and a higher demand — making oxygen limitation more, not less, likely.