Question 4 of 5: Manure Storage, Pump Sizing and Land-Application Nitrogen Budget

Question 4: Manure Storage, Pump Sizing and Land-Application Nitrogen Budget (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.

1) Number and dimensions of the storage tanks (5 marks).

Given. Annual liquid manure production 10,000 m³/yr (4 wt% solids, not otherwise needed for tank sizing); maximum diameter 15 m (geological constraint); maximum storage duration one year (no precipitation addition given, so the tanks are sized on the manure volume alone).

Find. The number of tanks and their dimensions.

1. Single-tank capacity at the diameter cap. At $D_{max}=15\ \text{m}$ and an assumed 6 m working depth, $$V_{tank,max} = \frac{\pi}{4}(15)^2(6) = \boxed{1{,}060\ \text{m}^3}$$ — far short of the 10,000 m³ annual volume, so more than one tank is required.
2. Number of tanks. $$n = \frac{10{,}000}{1{,}060} = 9.43 \;\Rightarrow\; \boxed{n = 10\ \text{tanks}}\ \text{(rounded up to fully contain the required volume)}$$
3. Exact depth for 10 tanks at the diameter cap. Holding $D=15\ \text{m}$ and solving for the depth that exactly delivers 10,000 m³ across 10 tanks, $$h = \frac{10{,}000}{10\times(\pi/4)(15)^2} = \boxed{5.66\ \text{m}}$$

2) Total dynamic head of the pump (10 marks).

Given. Pipe length 200 m, nominal 8-inch PVC (taken as ID = 8 in = 0.2032 m); tank-surface elevation 320.00 m; barn collection-pit elevation 330.00 m; manure friction coefficient f = 0.07 (Darcy form, elevated above a clear-water value because manure slurry is a non-Newtonian fluid with higher apparent viscosity than water — used directly as given, no Colebrook iteration needed); specific gravity 1.01.

Find. Total dynamic head (TDH) of the pump.

Approach. Apply the steady-flow energy equation between the barn collection pit (source, atmospheric, negligible approach velocity) and the tank discharge (atmospheric, pipe-exit velocity): $\text{TDH} = \Delta z + h_f + V^2/2g$, with the friction loss found from the Darcy-Weisbach equation using the given friction coefficient.

1. Pipe velocity and flow at the assumed design velocity. Pipe area $A=\dfrac{\pi}{4}(0.2032)^2=0.0324\ \text{m}^2$; at $V=1.5\ \text{m/s}$, $Q=VA=\boxed{0.0486\ \text{m}^3/\text{s}}$ ($\approx$ 175 m³/h).
2. Friction head loss (Darcy-Weisbach). $$h_f = f\frac{L}{D}\frac{V^2}{2g} = (0.07)\left(\frac{200}{0.2032}\right)\frac{(1.5)^2}{2(9.81)} = \boxed{7.90\ \text{m}}$$
3. Static elevation change. The discharge (tank, El. 320.00 m) sits BELOW the source (barn pit, El. 330.00 m), so $$\Delta z = 320.00-330.00 = \boxed{-10.00\ \text{m}}\ \text{(a favourable, gravity-assisted 10 m drop)}$$
4. Total dynamic head. Including the discharge velocity head $V^2/2g=0.115\ \text{m}$, $$\text{TDH} = \Delta z + h_f + \frac{V^2}{2g} = -10.00+7.90+0.12 = \boxed{-1.98\ \text{m}}$$ The NEGATIVE result is a genuine, physically meaningful finding, not an error: the 10 m elevation drop from barn to tank exceeds the computed friction loss, so gravity alone is more than sufficient to move the design flow through the line, and the specified centrifugal pump would in practice operate at a very low (near-zero) head — functioning more as a flow-rate controller/backflow safeguard than as a true lift pump. (Minor losses from bends, valves and fittings are not given in the source data and are not included; a few metres of minor-loss head would still likely leave the line net gravity-favourable or only marginally head-positive.)

3) Annual application rate and land area requirement (10 marks).

Given. Manure fertilizer values per m³: organic N = 1.2 kg, NH₄-N = 0.3 kg, NO₃-N = 0 kg. Mineralization rate: year 1 = 40%, year 2 = 20%, year 3+ = 5%. Ammonia volatilization after application = 30%. Corn N uptake target = 120 kg/ha. Starter nitrogen = 10 kg/ha applied separately at planting.

Find. Annual manure application rate (m³/ha) and land area requirement (ha) to just satisfy the corn nitrogen requirement, using the year-2 mineralization rate as the steady-state organic-N availability (per the question's own instruction).

1. Available nitrogen per m³ of manure at steady state. Plant-available ammonium-N after volatilization loss: $$N_{NH_4,avail} = 0.3\times(1-0.30) = 0.21\ \text{kg/m}^3$$ Organic-N mineralized at the steady-state (year-2) rate: $$N_{org,avail} = 1.2\times0.20 = 0.24\ \text{kg/m}^3$$ (NO₃-N contributes nothing, as it is given as 0 kg/m³.) Total available nitrogen per m³ of manure: $$N_{avail} = 0.21+0.24 = \boxed{0.45\ \text{kg N/m}^3}$$
2. Nitrogen the manure must supply. The corn's total requirement is 120 kg/ha; 10 kg/ha of that is already supplied by the separate starter-nitrogen application, so $$N_{from\ manure} = 120-10 = \boxed{110\ \text{kg/ha}}$$
3. Annual application rate. $$\text{Rate} = \frac{N_{from\ manure}}{N_{avail}} = \frac{110}{0.45} = \boxed{244.4\ \text{m}^3/\text{ha}}$$
4. Land area requirement. Applying the full 10,000 m³/yr production at this rate, $$A = \frac{10{,}000}{244.4} = \boxed{40.9\ \text{ha}}$$