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integration of solid oxide fuel cells and ... - Ea Energianalyse

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3.5. Condenser - COND<br />

T [° C ]<br />

ΔT min ,r ,i<br />

T r ,o<br />

ΔT min ,r ,o<br />

ΔT min ,mp T c ,o<br />

T c ,i<br />

T r ,i<br />

˙Q[kW ]<br />

Figure 3.6: Condenser: Approach temperature differences are calculated both in the ends <strong>and</strong> at<br />

the middle to make sure that the one being given as a parameter actually is the smallest <strong>of</strong> them.<br />

The problem can be viewed as having two heat exchangers - one<br />

for the superheated region, <strong>and</strong> one for the two-phase region. Both<br />

effectiveness are calculated, <strong>and</strong> the larger <strong>of</strong> them is the one to be<br />

specified (if not the closest approach temperature difference is specified<br />

instead):<br />

ɛ SH = T r,i − T r,mp<br />

T r,i − T c,mp<br />

(3.21)<br />

ɛ 2P = T c,mp − T c,i<br />

T r,mp − T c,i<br />

(3.22)<br />

Where SH = Super Heated region, <strong>and</strong> 2P = Two-Phase region.<br />

The refrigerant midpoint temperature (T r,mp ) is calculated from the<br />

saturation temperature at the given pressure. The cooling water<br />

midpoint temperature (T c,mp ) is calculated from an energy balance<br />

between refrigerant inlet <strong>and</strong> midpoint:<br />

(h c,o − h c,mp ) · ṁ c = (h r,i − h r,mp ) · ṁ r (3.23)<br />

The amount <strong>of</strong> heat to be removed by the cooling circuit is calculated<br />

by the energy balance, which includes a loss to the surroundings ( ˙Q loss ),<br />

which should be zero (adiabatic component) or positive (heat loss to the<br />

51

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