Q Ground source heat pumps operate by using a liquid rather

Q: Ground source heat pumps operate by using a liquid, rather than ambient air, as the heat source (or sink) for winter heating (or summer cooling). The liquid flows in a closed loop through plastic tubing that is buried at a depth for which annual variations in the temperature of the soil are much less than those of the ambient air. For example, at a location such as West Lafayette, Indiana, deep-ground temperatures may remain at approximately 11C, while annual excursions in the ambient air temperature may range from 25C to +35C. Consider winter conditions for which the liquid is discharged from the heat pump into high- density polyethylene tubing of thickness t = 8 mm and thermal conductivity k = 0.47 W m1 K1. The tubing is routed through soil that maintains a uniform temperature of approximately 10C at the tube outer surface. The properties of the fluid may be approximated as those of water (i.e. = 1000kgm3, = 0.001Pas, k = 0.6Wm1 K1, and Cp = 4.2kJkg1 K1).

(a) For a tube inner diameter Di =25mm,flowratem =0.03kgs1,and a fluid inlet temperature of Tm,i = 0C, determine the tube outlet temperature (heat pump inlet temperature), Tm,o as a function of the tube length L for 10 L 50m. For laminar flow, you can use Nu = 3.66, while for turbulent flow you can use the Dittus- Boelter equation. 1 (b) Recommend an appropriate length for the system. How would your recommendation be affected by variations in the liquid flow rate?

Solution

a)

Area A = 3.14/4*25^2 = 490.625 mm^2

Velocity V = m/(rho*A) = 0.03 / (1000*490.625*10^-6) = 0.0611 m/s

Reynolds number Re = rho*V*D/u = 1000*0.0611*0.025 / 0.001 = 1528.7

Since Re < 2300, flow is laminar and hence Nusselt number Nu = 3.66

Nu = hD/k = 3.66

h*0.025 / 0.6 = 3.66

h = 87.8 W/m^2-K