1. Thermodynamics Solar Hydrothermal Energy By: Aaron Vanderpool

2. Problem: To find a cheap and easy way to heat a home. Question: How much heat can be obtained using the Sun and a water mass collector? Hypothesis: Water can be heated and transfer a significant amount of energy into a building with little to no maintenance if designed correctly using our understanding of physics.

3. My Collector Glass donated by Incline Tahoe Glass

4. Heat is like work and is a form of energy that can be transferred by having a difference in temperature. Ways of transferring heat: Radiation: Happens through empty space in the form of electromagnetic waves. Conduction: Usually through a solid as molecules collide with one another. Convection: Usually in a liquid of gas as lots of molecules are moving about mixing with one another.

5. We have about 5 kWh/m^2/day average (NREL) 1366W/m^2 strikes Earth’s atmosphere. 1000 Watts / m^2 My collector panel 0.7558 m^2 = 755.8 watts Watts * Seconds = Joules 755.8 watts * 60 seconds * 60 minutes = 2721 kj in 1 hour 2721 kj = 2579 BTU’s 2579 BTUs / 125.2 lb (15 Gal/56.78 kg Tank) = 20.60 F (11.45 C) in 1 hour @ 100% efficiency 4/24/2015 5:410.6 4/24/2015 6:414.4 4/24/2015 7:4118.1 4/24/2015 8:4130.6 4/24/2015 9:41144.4 4/24/2015 10:41196.9 4/24/2015 11:41460.6 4/24/2015 12:411104.4 4/24/2015 13:41291.9 4/24/2015 14:41429.4 4/24/2015 15:41100.6 4/24/2015 16:41245.6 4/24/2015 17:41109.4 4/24/2015 18:4110.6 4/24/2015 19:410.6 4/24/2015 20:410.6 3148.7 2/5/2015 6:460.6 2/5/2015 7:4630.6 2/5/2015 8:4631.9 2/5/2015 9:4651.9 2/5/2015 10:46598.1 2/5/2015 11:46641.9 2/5/2015 12:46639.4 2/5/2015 13:46576.9 2/5/2015 14:4646.9 2/5/2015 15:4634.4 2/5/2015 16:4633.1 2/5/2015 17:460.6 2686.3 Reverse Engineer Q = mc ∆T Q = 56.78 kg * 4186 J/kg C (specific heat of H2O @ 15C) * 11.45 C = 2721 kj Radiation

6. Conductivity A = Area, d = thickness, k = smC (thermal conductivity of glass) ∆T = change in temperature CPVCCPVC (.95 BTU/in/hr/ft.2/°F) Q/t = 6 smC * 0.2444 m^2 * 11.45 C /.002 m = 8395 J/s Radiation = 1000W/m^2 *.2444 m^2 = 244.4 Watts * 1 second = 244.4 J/s The back of the collector if just wood Q/t =.1 smC * 0.7558 m^2 * 11.45 C /.018 m = 48 J/s The glass glazing of the collector Q/t =.84 smC * 0.7558 m^2 * 11.45 C /.0032 m = 2271 J/s (8178 kj/s) Air Q/t =.023 smC * 0.7558 m^2 * 11.45 C /.06 m = 3.317 J/s Water Q/t =.56 smC * 0.7558 m^2 * C /.06 m = J/s Limit convection of air. Limiting factor is air.

7. Convection 4 °C Accel = 9.8 * (1-( ϱ w / ϱ )) 15C 9.8*(1-(1/999.1)) = 9.790191172 26C 9.8*(1-(1/999.1)) = 9.790168539 9.790191172 - 9.790168539 =.00002263267415730337 m/s^2 = 1.4 mm/min^2 =.053 in/min^2 (need a drag force) buoyancy

8. Flow.0001131 *.762 = 0.0000861822 Boyles law Length 15.34m of CPVC pipe Area 1.131 x 10 -4 m^2 Volume 1.735 x 10 ^-3 m^3 inside piping Viscosity η =.001 Pa*s (water at 20C) F = η A v/l Idea = Find velocity  Find pressure (Bernoulli’s equation)  Find flow (Poiseuille’s equation) V =.007m/s P1 = 1 atm (Bernoulli’s equation) P2 = P1 + ϱ g(y1-y2) + 1/2 ϱ (v1^2-v2^2) P2 = (100000 N/m^2) + (996.8 kg/m^3 * 9.8 * -.5m) + ½ (996.8*.007^2 m/s) = 95115.704 Q = (100000-95115.7) * pi *.00597^4 / (8 *.001 * 1) = 0.00243646 m^3/s = 2436 cm^3/s Poiseuille's equation Non-laminar

9. Convection 4 °C = 1.4 mm/min^2 * 5 min =.007mm/min * 1.131 x 10 -4 m^2 = 7.917×10^-7 m^3 =.7917 cm^3/min (need a drag force of pipe and viscosity) buoyancy

10. Putting heat back into the room Thermal Radiation, conduction and convection Stefan-Boltzman equation Tank Q = 1 * 5.67 x 10^-8 W/m^2*K *.759 m^2 * 299.2^4 K = 345W Watts * Seconds = Joules 345 * Seconds = 2.721×10^6 joules Seconds = 7887 = 2.19 hours in a vacuum