LEMNISCATA
Matemàtiques, física, química…
\[T_F = \frac{q}{5} T_c + 32 \rightarrow 68 = \frac{q}{5} T_c + 32 \rightarrow T_c = 20 \, ^\circ \text{C}\]\[P_c = P_c (\text{CN}) \cdot p \cdot \frac{273}{273 + T} = 11.000 \, \text{kcal/m}^3 \cdot 3 \, \text{atm} \cdot \frac{273}{273 + 20 \, ^\circ \text{C}} = 30747 \, \text{kcal/m}^3\]La potència útil i consumida serà de:\[P_u = 200 \, \text{CV} \cdot \frac{736 \, \text{W}}{1 \, \text{CV}} = 147200 \, \text{W}\]\[\eta = \frac{P_u}{P_c} \implies P_c = \frac{147200 \, \text{W}}{0,55} = 267636 \, \text{W}\]L’energia necessària en una hora serà:\[E = P \cdot t = 267636 \, \text{W} \cdot 3600 \, \text{s} = 963489600 \, \text{J} = 963,49 \, \text{MJ} = 230,5 \, \text{Mcal} = 230,5 \cdot 10^3 \, \text{kcal}\]I el consum de gas natural:\[V = \frac{E}{P_c} = \frac{230,5 \cdot 10^3 \, \text{kcal}}{30747 \, \text{kcal/m}^3} = 7,497 \, \text{m}^3\]