Question 2 of 5: Manure Transport and Storage Lagoon

Question 2: Manure Transport and Storage Lagoon (25 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. Lagoon plan dimensions; pump total dynamic head (TDH); pump energy (power) demand.

Approach. Convert the dry-solids loading and liquid-manure solids content into an annual manure volume, size the lagoon so its storage volume simultaneously accommodates that manure and the net precipitation collected on its own open surface over one year, then compute the pump's total dynamic head from the pipeline's static lift and major (Darcy-Weisbach) friction loss, and finally convert flow and head into the required pump power at the stated efficiency.

1. Annual manure volume. At 2% solids content by weight, the wet liquid-manure mass flow is $\dot m_{wet} = 2{,}000\ \text{kg/d} / 0.02 = 100{,}000\ \text{kg/d}$. At the stated specific gravity (ρ = 1.01×1,000 = 1,010 kg/m³), $$Q_{manure} = \frac{100{,}000}{1{,}010} = 99.0\ \text{m}^3/\text{d} \;\Rightarrow\; V_{manure} = 99.0\times365 = \boxed{36{,}140\ \text{m}^3/\text{yr}}$$
2. Lagoon plan area and diameter. The lagoon's storage volume ($A\times h$, working depth $h$) must hold the annual manure volume PLUS the net precipitation ($P_{net}$) that falls on its own open top over the same year, so $A\,h = V_{manure} + A\,P_{net}$, giving the self-referential balance $$A(h - P_{net}) = V_{manure}$$ With the assumed working depth $h = 4$ m and $P_{net}=0.5$ m/yr, $$A = \frac{36{,}140}{4-0.5} = 10{,}326\ \text{m}^2 \;\Rightarrow\; D=\sqrt{\frac{4A}{\pi}} = \boxed{114.6\ \text{m}}$$ Adding the 1.0 m freeboard, the pond is excavated to a total depth of $4+1=\boxed{5.0\ \text{m}}$.
3. Pipe velocity and friction head loss. The 8-inch PVC pipe has an internal diameter $D_p = 8\times0.0254 = 0.2032$ m, area $A_p = \tfrac{\pi}{4}D_p^2 = 0.0324\ \text{m}^2$, so $$V = \frac{Q}{A_p} = \frac{0.04}{0.0324} = 1.234\ \text{m/s}$$ The Darcy-Weisbach major friction loss over the 500 m run is $$h_f = f\frac{L}{D_p}\frac{V^2}{2g} = 0.07\times\frac{500}{0.2032}\times\frac{1.234^2}{2(9.81)} = \boxed{13.4\ \text{m}}$$
4. Total dynamic head. Ignoring minor (fitting/entrance) losses as instructed, TDH is the sum of the static lift and the major friction loss (the exit velocity head, 0.08 m, is small enough to be absorbed within this simplification): $$TDH = 3 + 13.4 = \boxed{16.4\ \text{m}}$$
5. Pump energy demand. At the stated pump efficiency $\eta=0.4$, $$P = \frac{\rho\, g\, Q\, TDH}{\eta} = \frac{1{,}010\times9.81\times0.04\times16.4}{0.4} = \boxed{16.2\ \text{kW}}$$