$Pa = \frac{N}{m^{2}} = \frac{\text{kg}}{\text{ms}^{2}}$ bar = 10^5Pa = 0, 1MPa = 1, 01972at = 750Tr Tr = mmHg = 13, 595mmH2O = 133, 32Pa at(t) = 98066, 5Pa = 0, 00980665MPa at(f) = 101325Pa $J = Nm = \frac{\text{kgm}^{2}}{s^{2}}$ kWh = 3, 6 • 10^6J J = 0, 239cal cal = 4, 1868J ${n'}_{x} = \frac{x}{\text{l.at.x}}\ \left\lbrack \frac{\text{kmolx}}{\text{kg\ pal}} \right\rbrack$ (C=12, H2=2, O2=32, N2=28, S=32, H2O=18) Teoretyczne zapotrzebowanie tlenu do spalania:$\text{\ n}_{O_{2}t} = {n'}_{C} + {n'}_{S} + 0,5{n'}_{H_{2}} - {n'}_{O_{2}}\left\lbrack \frac{\text{kmolO}_{2}}{\text{kg\ pal}} \right\rbrack$ Teoretyczne objętościowe zapotrzebowanie tlenu do spalania: $V_{O_{2}t} = 22,42 \bullet n_{O_{2}t}\left\lbrack \frac{\text{um}^{3}\text{pow}}{\text{kg\ pal}} \right\rbrack$ $V_{\text{pow}_{t}} = \frac{22,42}{0,21} \bullet n_{O_{2}t}\left\lbrack \frac{\text{um}^{3}\text{pow}}{\text{kg\ pal}} \right\rbrack$ $V_{\text{pow}_{R}} = V_{\text{pow}_{t}} \bullet \lambda \bullet \frac{273,15 + T_{\text{pow}}}{273,15}\left\lbrack \frac{m^{3}\text{pow}}{\text{kg\ pal}} \right\rbrack$ $V_{\text{pow}}^{w} = V_{\text{pow}_{R}} \bullet \beta\left\lbrack \frac{m^{3}\text{pow}}{s} \right\rbrack$ Spaliny wilgotne: n”sw=n”ss+n”H2O ${n"}_{\text{ss}} = {n"}_{\text{CO}_{2}} + {n"}_{\text{CO}} + {n"}_{\text{SO}_{2}} + {n"}_{N_{2}} + {n"}_{O_{2}}$ ${n'}_{C} = {n"}_{\text{CO}_{2}} + {n"}_{\text{CO}}{+ n"}_{C}$ ${n"}_{\text{CO}_{2}} + {n"}_{\text{CO}} = {n'}_{C}{- n"}_{C}$ ${n'}_{\text{ss}} = \frac{{n'}_{C} - {n"}_{C}}{\left\lbrack \text{CO}_{2} \right\rbrack\% + \left\lbrack \text{CO} \right\rbrack\%} \bullet 100\%$ ${n"}_{\text{SO}_{2}} = {n'}_{S}$ ${n"}_{N_{2}} = {n'}_{N_{2}} + \frac{0,79 \bullet n_{O_{2}t} \bullet \lambda}{0,21}$ ${n"}_{O_{2}} = \left( \lambda - 1 \right)n_{O_{2}t} + 0,5{n"}_{\text{CO}} + {n"}_{C}$ ${n"}_{H_{2}O} = {n'}_{H_{2}} + {n'}_{H_{2}O} + X\frac{n_{O_{2}t}}{0,21} \bullet \lambda$ $\left\lbrack \text{CO}_{2} \right\rbrack_{\%} = \frac{{n"}_{\text{CO}_{2}}}{{n"}_{\text{ss}}} \bullet 100\%$ $\left( \text{CO}_{2} \right)\% = \frac{{n"}_{\text{CO}_{2}}}{{n"}_{\text{sw}}} \bullet 100\%$ zawartość węgla w żużlu: $C_{Z\%} = \frac{m_{CZ} \bullet 100\%}{m_{CZ} + m_{A}}$ $m_{CZ} = 12 \bullet {n"}_{C}$ $m_{A} = \frac{A\%}{100\%}$ wartość opałowa paliwa: $Q_{w} = 339C_{\%} + 1210\left( H_{\%} - \frac{O_{\%}}{8} \right) + 104S_{\%} - 25W_{\%}\left\lbrack \frac{\text{kJ}}{\text{kg\ pal}} \right\rbrack$ sprawność kotła: metoda bezpośrednia: $\eta_{k\%} = \frac{D\left( i_{0} - i_{\text{wz}} \right)}{BQ_{w}} \bullet 100\%$ metoda pośrednia: $\eta_{k\%} = 100\% - {\sum_{}^{}S}_{\%}$ ${\sum_{}^{}S}_{\%} = S_{g\%} + S_{n\%} + S_{w\%} + S_{pr\%} + S_{Z\%}$ (niezupełne, niecałkowite, wylot, promieniowanie, żużel) $S_{g\%} = \frac{12628 \bullet {n"}_{\text{CO}} \bullet 22,44}{Q_{w}} \bullet 100\%$ $S_{n\%} = \frac{33900 \bullet {n"}_{C} \bullet 12}{Q_{w}} \bullet 100\%$ $S_{w\%} = \frac{i_{\text{sp}} - i_{\text{pow}}}{Q_{w}} \bullet 100\%$ ipow = Cppow • Tpow • 22, 42 • npow $i_{\text{sp}} = \sum_{j = 1}^{n}{22,42} \bullet t_{\text{sp}} \bullet \left( {n"}_{j} - C_{\text{pj}} \right)\text{\ \ \ }$ $$jezeli\ C_{p}\left\lbrack \frac{\ }{um^{3}\deg} \right\rbrack \bullet 22,42n"$$ Obieg Rankinea: Bilans turbiny: $D \bullet i_{1} = \frac{\text{Nel}}{\eta_{m} \bullet \eta_{g}} + D \bullet {i'}_{k}$ Bilans skraplacza: $D \bullet {i'}_{k} + D_{\text{wch}} \bullet i_{\text{wchI}} = D \bullet {i"}_{k} \bullet D_{\text{wch}} \bullet i_{\text{wchII}}$ $D_{\text{wch}}\left( i_{\text{wchII}} - i_{\text{wchI}} \right) = D\left( {i'}_{k} - {i"}_{k} \right)$ $\eta_{k} = \frac{D\left( i_{0} - i_{\text{wz}} \right)}{B \bullet Q_{w}}$ $\eta_{\text{el}} = \frac{\text{Nel}}{B \bullet Q_{w}}$