Question 9.3: Characteristic curve of a heat exchanger Construct the chara......
Characteristic curve of a heat exchanger Construct the characteristic curve of the heat exchanger in Fig. E.9.8.
Since the fuel of the heat exchanger is \dot{B}_{8}-\dot{B}_{9} and the product is \dot{B}_{15}-\dot{B}_{16} the unit exergy consumption of the heat exchanger is
k_{H X}=\frac{\dot{B}_{8}-\dot{B}_{9}}{\dot{B}_{15}-\dot{B}_{16}}The exergy for the i-th flow of water is
{\dot{B}}_{i}=\dot m\,c_{p}\left(T_{i}-T_{0}-T_{0}l n\frac{T_{i}}{T_{0}}\right)The independent variables (\tau_{H X}) chosen for the heat exchanger are: the primary and secondary input temperatures (T_{8},T_{16}), the mass flowrates \left(\dot{m}_{p r i m}^{},\dot{m}_{s e c}\right), the ambient temperature (T_{0}) and the global coefficient of heat transfer UA. Therefore, the exchanger outlet temperatures (T_{9},T_{15}), depend on these variables.
Since the objective is to determine the exchanger outlet temperatures of which the transfer surface is known, the method used is that of Effectiveness-Number of Transfer Units, ε-NTU. For this, we must first determine whether the maximum thermal capac-ity corresponds to the primary or secondary. If C_{p r i m}=\dot m_{p r i m}.c_{p} and C_{sec}=\dot m_{sec}.c_{p} we will call
The effectiveness of the heat exchanger depends on the geometric configuration, as well as on the configuration of the flow. For the plate heat exchanger in the installation, the expression obtained for effectiveness is
\varepsilon=\frac{1-e x p\left[-\frac{U A}{C_{m i n}}\left(1-\frac{C_{m i n}}{C_{m a x}}\right)\right]}{1-{\frac{C_{m i n}}{C_{m a x}}}e x p\bigg[-{\frac{U A}{C_{m i n}}}\bigg(1-{\frac{C_{m i n}}{C_{m a x}}}\bigg)\bigg]}From the effectiveness, the exit temperatures are determined, which are in turn the entry temperatures for M1 and T. These temperatures are
T_{9}=T_{8}-\varepsilon{\cdot}\left(\frac{C_{m i n}}{C_{p r i m}}\right)(T_{8}-T_{16}) \\ T_{15}=T_{16}-\varepsilon{\cdot}\left(\frac{C_{m i n}}{C_{sec}}\right)(T_{8}-T_{16})If the exit temperatures are known the unit exergy consumption k_{H X} can be calcu-lated. In Fig. E.9.9, its values are represented graphically, when one of the independent variables changes and the rest remain constant. The graphs in Fig. E.9.9 have been constructed for the following values: U A=133,888\,\mathrm{kJ/hK},\;\dot m_{p r i m}=\ 1.920~\mathrm{kg/h}\;\;,{\dot{m}}_{s e c}=1860~\mathrm{kg}/\mathrm{h} keeping constant two of the three temperatures T_{16}=35^{\circ}{\mathrm{C}},\ T_{0}=15^{\circ}{\mathrm{C}}, and T_{8}=75^{\circ}{\mathrm{C}}, with the third one appearing on the horizontal axis.
