Question 2 of 6: Manure Storage, Pumping and Land Application System

Question 2: Manure Storage, Pumping and Land Application System (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. 1,500 grower-finisher pigs; dry-solids production 1.2 kg/pig/day; liquid manure at 95% moisture content stored 1 year; open circular concrete tank(s), $D \le 10\ \text{m}$; net precipitation minus evaporation on the open tank surface = 250 mm/yr; pumped 500 m through 8-in PVC pipe at $Q = 0.02\ \text{m}^3/\text{s}$; pipe friction coefficient $f = 0.07$; manure specific gravity $SG = 1.01$; tank surface elevation 320.00 m, field elevation 330.00 m, pivot nozzle 2 m above the field, discharge pressure 550 kPa; pump and engine efficiency both 75%.

Find. The number and dimensions of storage tanks, the pump's total dynamic head (TDH), and the required engine size.

Approach. Size the tank volume from a dry-solids/moisture-content mass balance plus the collected precipitation, cap the tank diameter at 10 m to find the number of tanks; find TDH as the sum of static lift, discharge-pressure head, and pipe friction head; size the engine from the hydraulic power divided by the combined pump/engine efficiency.

1. Daily and annual manure volume from the dry-solids/moisture balance. Total dry solids $= 1{,}500 \times 1.2 = 1{,}800\ \text{kg/day}$. At 95% moisture content the dry solids are 5% of the wet mass, so wet manure mass $= 1{,}800/0.05 = 36{,}000\ \text{kg/day}$. With $\rho = SG \times 1000 = 1010\ \text{kg/m}^3$, the daily volume is $$V_{day} = \dfrac{36{,}000}{1010} = 35.64\ \text{m}^3/\text{day}$$ and the annual manure volume to be stored is $V_{manure} = 35.64 \times 365 = \boxed{13{,}010\ \text{m}^3/\text{yr}}$.
2. Number of tanks. An open tank of diameter $D = 10\ \text{m}$ has plan area $A = \pi D^2/4 = 78.54\ \text{m}^2$. Check: the paper gives no target tank depth, so a working liquid depth of $h = 4.5\ \text{m}$ is assumed — a typical operating depth for a 10 m precast/cast-in-place circular concrete manure storage tank (MWPS-18); a further 0.3 m of freeboard would be added above this in the final structural design. Each tank of area $A$ also collects the net 250 mm/yr of precipitation over its own open surface, adding $0.25A$ m³/yr per tank that must also be stored. For $N$ tanks, total capacity $N\!A h$ must hold both the manure and the collected precipitation on that same $N$ tanks: $$N A h = V_{manure} + N A (0.25) \ \Rightarrow\ N = \dfrac{V_{manure}}{A(h-0.25)} = \dfrac{13{,}010}{78.54 \times 4.25} = 38.98$$ Rounding up, $\boxed{N = 39\ \text{tanks}}$, each $D = 10.0\ \text{m}$ diameter by $h \approx 4.5\ \text{m}$ working depth (total capacity 39 × 78.54 × 4.5 = 13,779 m³, comfortably covering the 13,010 m³ of manure plus the 39 × 78.54 × 0.25 = 766 m³ of collected precipitation).
3. Static lift. The pump must lift manure from the tank surface (El. 320.00 m) to the pivot nozzle, which sits 2 m above the field surface (El. 330.00 m): $$h_{static} = (330.00 + 2) - 320.00 = \boxed{12.00\ \text{m}}$$
4. Discharge-pressure head. The required 550 kPa at the nozzle converts to a head using the manure's own density: $$h_{p} = \dfrac{P}{\rho g} = \dfrac{550{,}000}{1010 \times 9.81} = \boxed{55.51\ \text{m}}$$
5. Friction head in the 500 m pipe run. Pipe diameter $D_p = 8 \times 2.54 = 20.32\ \text{cm} = 0.2032\ \text{m}$, giving cross-section $A_p = \pi D_p^2/4 = 0.03243\ \text{m}^2$ and velocity $V = Q/A_p = 0.02/0.03243 = 0.617\ \text{m/s}$. With the Darcy–Weisbach relation, $$h_f = f\dfrac{L}{D_p}\dfrac{V^2}{2g} = 0.07 \times \dfrac{500}{0.2032} \times \dfrac{0.617^2}{2(9.81)} = \boxed{3.34\ \text{m}}$$
6. Total dynamic head. Summing the three components: $$TDH = h_{static} + h_p + h_f = 12.00 + 55.51 + 3.34 = \boxed{70.85\ \text{m}}$$
7. Engine size. The hydraulic power delivered to the fluid is $$P_{hyd} = \rho g Q\, TDH = 1010 \times 9.81 \times 0.02 \times 70.85 = 14{,}037\ \text{W} = 14.04\ \text{kW}$$ Dividing by the pump efficiency gives the shaft power the pump needs, and dividing again by the engine efficiency gives the required engine output: $$P_{engine} = \dfrac{P_{hyd}}{\eta_{pump}\,\eta_{engine}} = \dfrac{14.04}{0.75 \times 0.75} = \boxed{24.96\ \text{kW}\ (33.5\ \text{hp})}$$