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Health Clinic – Las Mercedes, Honduras Janelle Barth, Stephanie Chang, Greer Mackebee, Walter Li. Structural Design. Current Footprint – 60’x80’. Notes – Footprint/Foundation. Cut and fill enough to fill hole without extra soil left over Strip footing Soil – mostly clay
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Health Clinic – Las Mercedes, HondurasJanelle Barth, Stephanie Chang, Greer Mackebee, Walter Li
Notes – Footprint/Foundation • Cut and fill enough to fill hole without extra soil left over • Strip footing • Soil – mostly clay • Minimum 5’ from boundaries • Allow space for latrine
Notes - Plan • Use spaces on exterior – optimize sunlight • Kitchen on right – minimize distance to latrine • Sink near shower – minimize piping • Delivery near kitchen for access to hot water (as opposed to being connected to Recovery) • 8’ hallways for easy maneuverability (i.e. moving beds, etc.) • 2 entrances – main (to waiting area) and kitchen (to load supplies/easy access to latrine)
Other Notes • Roof – corrugated tin and corrugated translucent material (optimize sunlight) • 10’ walls • Materials: CMU (8”x8”x16”), rebar, poured concrete base, wood beams for roof support system • Sketch-up Model
Next steps • Determine optimal base depth for cut and fill • Design roof structure • Determine loads • Find North for solar panel/roof design • Convert to metric • Door/Window Design
Our System 6 x Isofoton 75 Watt solar panel WyckomarUV-250 UV water filters Xantrex 40 amp + 30 amp charge controller 600 watt power inverter 12 V 115 amp hr Nautlius deep cycle battery 3 x Low powered laptops Sunfrost Vaccine Refrigerator Thin-Lite DC fluorescent lights
Location of Load componets 30 watt lights 13 watt lights Battery /Charge Controller/Inverter Storage 13 watt lights Vaccine Fridge 13 watt lights/Laptops
Calculations for Total Power Consumption Total power consumption = Σ Appliance Wattage rating * Hours used/day *Total number of Appliance
Sizing the solar panel Monthly Averaged Insolation Incident On A Horizontal Surface (kWh/m2/day) Solar panels are tested at 1000 W/m at 25 ⁰ C The solar panel should be able provide the required wattage at both the summer and winter hours For temperature ranges between 25⁰ and 40⁰ C, the power output is close to linear http://www.reuk.co.uk/Effect-of-Temperature-on-Solar-Panels.htm] 75 watt panels = 75/1000 = 0.075 % efficiency At lowest solar isolation of 4.13 kWh/m2/day produces 4.13 * 0.075 = 310 watts per panel Total number of panels needed = Total power consumption/watts per panel = 1760/310 = 5.6 panels ≈ 6 panels minimum To compensate for cloudy days and system losses, 7-8 solar panels may be necessary
Sizing the battery system Nautilius Deep cycle are rated at 115 amp hrs and 12 volts for a total of 1380 watt hrs Although deep cycles can be discharged to 80 %, they have a much larger cycles if discharged at lower percentages This is a graph comparing cycles of discharge vs. discharge %
We want both a long lifespan for our system and also the power the solar fridge for several days if there are consecutive cloudy or rainy days/solar panel malfunctions Aiming for optimally 20 % discharge rate, each battery would be able to provide 0.2* 1380 = 276 Watt hours Number of batteries needed = Total wattage/ Watt hour per battery = 1760 /276 = 6.3 ≈ 7 batteries After 2 days, total watt hours available = (0.8)2 * 7 * 1380 = 6048 watt hrs This can still provide 16 days of the vaccine fridge running by itself At 20% discharge rate, the batteries will last for 2500 cycles. Assuming 1 cycle per day for each battery = 2500/365 = 6.8 years before replacement. This will probably be 6 years, due to some over discharging.
Sizing the Charge controller The charge controller prevents the batteries from overcharging/discharging, maintains the rates of charging/discharging, keeps power from batteries from going back into solar panels, and also converts the variable voltage from the solar panels into a steady voltage For 75 watt solar panel, should be at 75 watts/12 volts = 6 amps, however this current can spike up to 8 amps. For our solar system, 7 * 8 amps = 56 amps. Thus we could use two charger controllers, a 40 amp and a 30 amp charge controller.
Sizing the Inverter A 600 watt system will have enough capacity to power 5 amp tools, fans, and also our needs for a water filter, and laptops. The water filter = 30 watts 3 laptops total = 24 watts The 600 watt system will exceed our daily needs but will allow greater expansion of AC appliances.
Choosing the laptop Netbooks are very power efficient The Dell mini 9 makes it easy to upgrade RAM and hard drive The solid-state hard drive is less likely to fail Linux drive Operating System can be re-installed and rebooted from a USB drive, and thus making system recovery very easy
Choosing the Vaccine Fridge Sunfrost energy specifications Energy Consumption (12 volt) making 2.2 kg ice/day .38 kWh/day @ 32° C (90° F) Running Current: 4.5 amps for 12 volt system, 2.3 amps for 24 volt system Starting Current: 15 amps for 12 volt system, 7.5 amps for 24 volt system
The Sunfrost Fridge: • has been approved by the World Health Organization • is very power efficient • can maintain a fridge temperature of 3° C at a large range of temperatures
Manning's Eqn for Sewage System: Determination of Pipe Angle V=(k/n)*R^(2/3)*S^(1/2)
Calculations • Assumptions: • pan required 2 L/flush • latrine will serve an average of 15 people/year • 2 L/flush is used for anal cleansing • soil infiltration rate is 30 L/(m2*day) – corresponds to “sandy loam, loams” soil type • 1 urination/(person*day) • deep water table • 2 flushes/(person*day) • bricks are 4 in by 4 in by 8 in (101.6 mm by 101.6 mm by 203.2 mm)
Calculations q = wastewater flow per person q = Nf ( vw + vc ) + vf + ( a Nu vw ) + vu q = (2 flush/day)(2 L/flush + 2 L/flush) + (0.3 L/day + 1.2 L/day) + [(1)(1 flush/day)(2 L/flush)] q = 11.5 L/day Q = total wastewater flow Q = p * q Q = (15 people)(11.5 L/day) = 172.5 L/day
Calculations vs = solids storage volume vs = (25 L/(person*year))(10-3 m3/L)(2 year)(15 people) = 0.75 m3 Infiltration area required = Q / soil infiltration rate = 172.5 L/day / 30 L/(m2*day) = 5.75 m2 0.75 m3 = (π)(d2/4)(h), where d is the internal pit diameter and h is the effective depth h = [(4)(0.75)] / [(π)(d2)]
Calculations Outside pit diameter = D = d + 2(block width) = d + 2(0.1016) = d + 0.2032 Infiltration area = Ai = (D)(h)(π) = (d + 0.2032)*[(4*0.75)/(d2*π)]*(π) = (3d + 0.6096)/d2 Setting Ai equal to 5.75 m2, d = 0.67 m. Therefore, effective depth (h) = 2.127 m Adding the 0.5 m free space necessary at the top of the cylinder, the actual height (H) ≈ 2.6 m
PRELIMINARY DIMENSIONS: d = 0.67 m D = 0.875 m H = 2.6 m
Slab Brick lining V = 1.56 m2 2.6 m Pit 0.1016 m 0.1016 m 0.2032 m 0.875 m
SUPERSTRUCTURE 2.5 m PAN 1.2 m 1.2 m 20 mm 15° PIT TRAP VALVE