found by adding the resistance attributable to the soil to that resulting from the pipe
insulation (Phetteplace and Meyer 1990). After simplification the following result is
obtained
Ro = ln[ (Do/d) (4Hp/Do)γ ]/2ki
(2-4)
where γ = ki/ks (dimensionless)
ki = insulation thermal conductivity (W/m C)
ks = soil thermal conductivity (W/m C)
Do = outer diameter of insulation (m)
Hp = burial depth to pipe centerline (m).
In this form it becomes easy to see how each factor affects this parameter. The
(4Hp/Do)γ factor represents the contribution of the soil to the overall thermal
resistance. If γ << 1, that is, if the soil conductivity is much greater than the insulation
thermal conductivity, then this factor will be close to unity and the overall thermal
resistance reduces to the thermal resistance of the insulation alone.
To obtain a simpler form for the cost of heat loss, we make the following
assumptions:
1. That the soil temperature at the pipe depth varies sinusoidally over the yearly
cycle about a mean temperature.
2. That the cost of heat is constant over the yearly cycle. This does not limit us to
fixed heat cost over the life of the system, since escalation factors may be used.
3. That the outer surface temperature of the carrier pipe is equal to the tem-
perature of the carrier medium.
The result of these assumptions is the following form for the cost of heat loss
Chl = I1/ln(A10/d)
(2-5)
∫
where I1 = PVFh L 4πki ( Ch Tavg dt At Ch Tm) ($)
A10 = Do(4Hp/Do)γ (m)
Tavg = (Ts+ Tr)/2 (C)
Tm = mean soil temperature (C)
Tr = return temperature (C)
Ts = supply temperature (C)
At = number of hours per year (8760).
COST OF PUMPING
Now let's consider the pumping costs. A cost is associated with the electrical
energy input to drive the pumps. The portion of this energy that results in frictional
heating of the fluid in the pipes is recovered as heat. In general the value of the heat
recovered will, of course, be less than the value of the electrical energy input to drive
the pumps. It can be significant, however, and therefore it has been included here.
Thus, we have the following for the pumping cost
∫
∫r ChPPf dt
Cpe = PVFe CePPa dt PVFh
(2-6)
yr
y
where PVFe = present value factor for electrical energy (dimensionless)
PPa = actual pumping power required, including pump and pump driver
inefficiencies (W)
6
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