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Sohel et al. UT =UST(mB/mS,B)0.5 The mass flow of pentane vapor depends on the amount of heat given to the fluid from the brine. From energy balance, the following equation can be derived: vaporizer and vaporizer. • m 2 b is the pentane vapor flow from the •T •T •TT (4) Here, m B is the brine mass flow, U S •T transfer coefficient and m S , B is the flow. is the standard brine mass At first, the total heat transfer is calculated, then it is divided between the separator and the vaporizer in the same ratio as observed, QV=87.2% and QS=12.8% of QT. Equation 4 is an approximated relationship devised from experimental data from a shell and plate heat exchanger. However, in the ORC unit, the vaporizer and separator are shell and tube type. Therefore, to increase the accuracy of the calculation, iteration was carried out. At first using equation (4), UT was calculated. Using equation (2), QT was calculated. Now, brine side heat transfer, • QT=mBcp(Ti,B−To,B) (5) From the above equation, keeping Ti,B constant, the value of To,B can be calculated. Using the new value of To,B, the value of ∆tmT , can be calculated and so forth. From conservation of mass in the vaporizer we get, Where, Lv is the latent heat of vaporization. The outlet temperature from the separator is calculated depending on the state of the pentane vapor. If h2b= hsat then: The combined space inside the vaporizer and the separator is constant. As the holdup mass of the upper half is ignored, the following relationships can be expressed for the vaporizer: standard heat V• • Q −m2(h −h) m2b= 2a 2 (11) Lv T3 =Tsat + If h2b < hsat QS • (12) cpm2b then S T =T +Q −m2b(hsat −h2b) (13) 3sat • cpm2b If h2b > hsat T3 =Tsat + then QS +m2b(h2b −hsat) • cpm2b (14) d•• dt(Ml +Mv)=m2−m2b (6) Here, the suffix sat stands for saturation. All the physical properties needed for equations 9 - 14 are taken from (REFPROP 2007). Where, Ml and Mv are holdup masses of pentane liquid and vapor in the vaporizer and lower half of the separator, • respectively. m2 is the pentane liquid mass flow to the Figure 6 represents a simplified view of Figure 4 with an explanation of various parameters. The line AC represents the level of pentane in the vaporizer. The volume of vapor above this line can be calculated by multiplying area bounded by ABC with the length of the vaporizer. This area is equal to the area bounded by ABCO minus the area of the triangle ACO. Figure 6: Simplified cross sectional view of the vaporizer It can be easily shown that for a pentane level higher than or equal to the top of the tube channel, the height of the pentane level from the bottom of the shell, hl, can be calculated from the following equation: h=h +QS 3 2b • m2b (15) (7) VT =Vv +Vl (8) VT =VV +0.5VS (9) Vv =Mv /ρv (10) Here, VT is the total volume of the space inside the vaporizer and the lower half of the separator, VV is the volume of the space inside the vaporizer and VS is the volume of the space inside the separator. Vl is the volume of the liquid pentane inside the vaporizer and Vv is the volume of the vapor on top of the pentane liquid level and the space of the lower half of the separator. ρl and ρv are the density of saturated liquid and saturated vapor. Vl =Ml /ρl 4PDF Image | Dynamic Modelling and Simulation of an Organic Rankine Cycle Unit of a Geothermal Power Plant
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