Thermodynamics Solar Hydrothermal Energy By: Aaron Vanderpool

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.

My Collector Glass donated by Incline Tahoe Glass

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.

Radiation We have about 5 kWh/m^2/day average (NREL) 4/24/2015 5:41 0.6 4/24/2015 6:41 4.4 4/24/2015 7:41 18.1 4/24/2015 8:41 30.6 4/24/2015 9:41 144.4 4/24/2015 10:41 196.9 4/24/2015 11:41 460.6 4/24/2015 12:41 1104.4 4/24/2015 13:41 291.9 4/24/2015 14:41 429.4 4/24/2015 15:41 100.6 4/24/2015 16:41 245.6 4/24/2015 17:41 109.4 4/24/2015 18:41 10.6 4/24/2015 19:41 4/24/2015 20:41 3148.7 Radiation 2/5/2015 6:46 0.6 2/5/2015 7:46 30.6 2/5/2015 8:46 31.9 2/5/2015 9:46 51.9 2/5/2015 10:46 598.1 2/5/2015 11:46 641.9 2/5/2015 12:46 639.4 2/5/2015 13:46 576.9 2/5/2015 14:46 46.9 2/5/2015 15:46 34.4 2/5/2015 16:46 33.1 2/5/2015 17:46 2686.3 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 Reverse Engineer Q = mc ∆T Q = 56.78 kg * 4186 J/kg C (specific heat of H2O @ 15C) * 11.45 C = 2721 kj

Conductivity A = Area, d = thickness, k = smC (thermal conductivity of glass) ∆T = change in temperature CPVC (.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.

Convection buoyancy Accel = 9.8 * (1-(ϱw/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) 4 °C

Flow Poiseuille's equation Non-laminar .0001131 * .762 = 0.0000861822 Boyles law Flow 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 + ϱg(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 P275

Convection buoyancy = 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) 4 °C

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 Wherever you are not putting radiation heat into the system, the system is losing heat and all those need to be added up.