droop

$$\mathrm{Droop}\ \delta=-\frac{\frac{f_{final}-f_{initial}}{f_N}}{\frac{P_{final}-P_{initial}}{P_N}}=-\frac{\frac{\Delta f}{f_N}}{\frac{\Delta P}{P_N}}=-\frac{\varDelta f}{\varDelta P}\frac{P_N}{f_N}$$ $$-\frac{\varDelta f}{\varDelta P}=\frac{\delta f_N}{P_N},\quad-\frac{\varDelta P}{\varDelta f}=\frac{P_N}{\delta f_N}$$

$$-\frac{\varDelta P}{\varDelta f}=\frac{P_N}{\delta f_N}$$

$$-\frac{\Delta P_1}{\Delta f}=\frac{P_{1N}}{\delta_1f_N},\quad-\frac{\Delta P_2}{\Delta f}=\frac{P_{2N}}{\delta_2f_N},\quad-\frac{\Delta P_3}{\Delta f}=\frac{P_{3N}}{\delta_3f_N},\quad\cdots$$ $$-\frac{(\Delta P_1+\Delta P_2+\Delta P_3+\cdots)}{\Delta f}=\frac1{f_N}(\frac{P_{1N}}{\delta_1}+\frac{P_{2N}}{\delta_2}+\frac{P_{3N}}{\delta_3}+\cdots)=\frac1{f_N}\sum_{k=1}^n\frac{P_{kN}}{\delta_k}$$ speed at no load $f_0,\quad\delta=-\frac{f_N-f_0}{P_N-0}\frac{P_N}{f_N}=\frac{f_0-f_N}{f_N},\quad f_0=f_N+\delta f_N$

Ex 1. $P_N=P_{initial}=180MW,\,f_N=f_{initial}=60Hz,\,\delta=0.06=6\%\to f_{final}=60.3Hz$ then $P_{final}=?$

Solution 1.

$$-\frac{\Delta P}{\Delta f}=\frac{P_N}{\delta f_N}$$ $$-\frac{P_{final}-180}{0.3}=\frac{180}{0.06\times60}=\frac{18}6\frac{100}6=50$$ $$P_{final}=180-50\frac3{10}=165MW$$

Solution 2.

$$\frac x{3.6-0.3}=\frac{180}{3.6}$$ $$x=\frac{180}{3.6}3.3=165MW$$

Ex 2. $P_N=240MW,\,P_{initial}=240\frac34=180MW,\,f_N=f_{initial}=60Hz,\,\delta=0.06=6\%\to f_{final}=60.3Hz$ then $P_{final}=?$

Solution 1.

$$-\frac{\Delta P}{\Delta f}=\frac{P_N}{\delta f_N}$$ $$-\frac{P_{final}-180}{0.3}=\frac{240}{0.06\times60}=\frac{24}6\frac{100}6=\frac{200}3$$ $$P_{final}=180-\frac{200}3\frac3{10}=160MW$$

Solution 2.

$$\frac x{2.7-0.3}=\frac{180}{2.7}$$ $$x=\frac{180}{2.7}2.4=160MW$$

Ex 3. $P_{1N}=P_{1initial}=180MW,\,\delta_1=0.06$
$P_{2N}=P_{2initial}=1080MW,\,\delta_2=0.05$
$P_{3N}=P_{3initial}=480MW,\,\delta_3=0.04$
$P_{4N}=P_{4initial}=270MW,\,\delta=0.03$
$f_N=f_{initial}=60Hz$

$f_{final}=60.3Hz$ then $P_{1final},\,P_{2final},\,P_{3final},\,P_{4final}$ $$\Delta P=-\frac{P_N}{\delta f_N}\Delta f$$ $$P_{1final}-180=-\frac{180}{0.06\times60}\frac3{10}=-15,\quad P_{1final}=165MW$$ $$P_{2final}-1080=-\frac{1080}{0.05\times60}\frac3{10}=-108,\quad P_{1final}=972MW$$ $$P_{3final}-480=-\frac{480}{0.04\times60}\frac3{10}=-60,\quad P_{3final}=420MW$$ $$P_{4final}-270=-\frac{270}{0.03\times60}\frac3{10}=-45,\quad P_{4final}=225MW$$

$$\frac a{3.6-0.3}=\frac{180}{3.6},\quad a=\frac{180}{3.6}3.3=165MW$$ $$\frac b{3.0-0.3}=\frac{1080}{3.0},\quad b=\frac{1080}{3.0}2.7=972MW$$ $$\frac c{2.4-0.3}=\frac{480}{2.4},\quad c=\frac{480}{2.4}2.1=420MW$$ $$\frac d{1.8-0.3}=\frac{270}{1.8},\quad d=\frac{270}{1.8}1.5=225MW$$
$P_{initial}=P_{1initial}+P_{2initial}=180+1080=1260MW\to P_{final}=1137MW$ then $f_{final},\,P_{1final},\,P_{2final}$ $$\Delta P=-\frac{\Delta f}{f_N}\sum_{k=1}^n\frac{P_{kN}}{\delta_k}$$ $$1137-1260=-\frac{\Delta f}{60}(\frac{180}{0.06}+\frac{1080}{0.05})=-410\Delta f,\quad\Delta f=\frac{123}{410}=\frac3{10},\quad f_{final}=60.3Hz$$ $$P_{1final}-180=-\frac{180}{0.06\times60}\frac3{10}=-15,\quad P_{1final}=165MW$$ $$P_{2final}-1080=-\frac{1080}{0.05\times60}\frac3{10}=-108,\quad P_{1final}=972MW$$
$P_{initial}=P_{1initial}+P_{2initial}+P_{3initial}=180+1080+480=1740MW\to P_{final}=1557MW$ then $f_{final},\,P_{1final},\,P_{2final},\,P_{3final}$ $$\Delta P=-\frac{\Delta f}{f_N}\sum_{k=1}^n\frac{P_{kN}}{\delta_k}$$ $$1557-1740=-\frac{\Delta f}{60}(\frac{180}{0.06}+\frac{1080}{0.05}+\frac{480}{0.04})=-610\Delta f,\quad\Delta f=\frac{183}{610}=\frac3{10},\quad f_{final}=60.3Hz$$ $$P_{3final}-480=-\frac{480}{0.04\times60}\frac3{10}=-60,\quad P_{3final}=420MW$$
$P_{initial}=P_{1initial}+P_{2initial}+P_{3initial}+P_{4initial}=180+1080+480+270=2010MW\to P_{final}=1782MW$ then $f_{final},\,P_{1final},\,P_{2final},\,P_{3final},\,P_{4final}$ $$\Delta P=-\frac{\Delta f}{f_N}\sum_{k=1}^n\frac{P_{kN}}{\delta_k}$$ $$1782-2010=-\frac{\Delta f}{60}(\frac{180}{0.06}+\frac{1080}{0.05}+\frac{480}{0.04}+\frac{270}{0.03})=-760\Delta f,\quad\Delta f=\frac{228}{760}=\frac3{10},\quad f_{final}=60.3Hz$$ $$P_{4final}-270=-\frac{270}{0.03\times60}\frac3{10}=-45,\quad P_{4final}=225MW$$

Ex 4. $P_{1N}=240MW,\,P_{1initial}=240\frac34=180MW,\,\delta_1=0.06$
$P_{2N}=P_{2initial}=1080MW,\,\delta_2=0.05$
$P_{3N}=600MW,\,P_{3initial}=600\frac45=480MW,\,\delta_3=0.04$
$P_{4N}=540MW,\,P_{4initial}=540\frac12=270MW,\,\delta_4=0.03$
$f_N=f_{initial}=60Hz$

$f_{final}=60.3Hz$ then $P_{1final},\,P_{2final},\,P_{3final},\,P_{4final}$ $$\Delta P=-\frac{P_N}{\delta f_N}\Delta f$$ $$P_{1final}-180=-\frac{240}{0.06\times60}\frac3{10}=-20,\quad P_{1final}=160MW$$ $$P_{2final}-1080=-\frac{1080}{0.05\times60}\frac3{10}=-108,\quad P_{1final}=972MW$$ $$P_{3final}-480=-\frac{600}{0.04\times60}\frac3{10}=-75,\quad P_{3final}=405MW$$ $$P_{4final}-270=-\frac{540}{0.03\times60}\frac3{10}=-90,\quad P_{4final}=180MW$$

$$\frac a{2.7-0.3}=\frac{180}{2.7},\quad a=\frac{180}{2.7}2.4=160MW$$ $$\frac b{3.0-0.3}=\frac{1080}{3.0},\quad b=\frac{1080}{3.0}2.7=972MW$$ $$\frac c{1.92-0.3}=\frac{480}{1.92},\quad c=\frac{480}{1.92}1.62=405MW$$ $$\frac d{0.9-0.3}=\frac{270}{0.9},\quad d=\frac{270}{0.9}0.6=180MW$$
$P_{initial}=P_{1initial}+P_{2initial}=180+1080=1260MW\to P_{final}=1132MW$ then $f_{final},\,P_{1final},\,P_{2final}$ $$\Delta P=-\frac{\Delta f}{f_N}\sum_{k=1}^n\frac{P_{kN}}{\delta_k}$$ $$1132-1260=-\frac{\Delta f}{60}(\frac{240}{0.06}+\frac{1080}{0.05})=-\frac{1280}3\Delta f,\quad\Delta f=128\frac3{1280}=\frac3{10},\quad f_{final}=60.3Hz$$ $$P_{1final}-180=-\frac{240}{0.06\times60}\frac3{10}=-20,\quad P_{1final}=160MW$$ $$P_{2final}-1080=-\frac{1080}{0.05\times60}\frac3{10}=-108,\quad P_{1final}=972MW$$
$P_{initial}=P_{1initial}+P_{2initial}+P_{3initial}=180+1080+480=1740MW\to P_{final}=1537MW$ then $f_{final},\,P_{1final},\,P_{2final},\,P_{3final}$ $$\Delta P=-\frac{\Delta f}{f_N}\sum_{k=1}^n\frac{P_{kN}}{\delta_k}$$ $$1537-1740=-\frac{\Delta f}{60}(\frac{240}{0.06}+\frac{1080}{0.05}+\frac{600}{0.04})=-\frac{2030}3\Delta f,\quad\Delta f=203\frac3{2030}=\frac3{10},\quad f_{final}=60.3Hz$$ $$P_{3final}-480=-\frac{600}{0.04\times60}\frac3{10}=-75,\quad P_{3final}=405MW$$
$P_{initial}=P_{1initial}+P_{2initial}+P_{3initial}+P_{4initial}=180+1080+480+270=2010MW\to P_{final}=1717MW$ then $f_{final},\,P_{1final},\,P_{2final},\,P_{3final},\,P_{4final}$ $$\Delta P=-\frac{\Delta f}{f_N}\sum_{k=1}^n\frac{P_{kN}}{\delta_k}$$ $$1717-2010=-\frac{\Delta f}{60}(\frac{240}{0.06}+\frac{1080}{0.05}+\frac{600}{0.04}+\frac{540}{0.03})=-\frac{2930}3\Delta f,\quad\Delta f=293\frac3{2930}=\frac3{10},\quad f_{final}=60.3Hz$$ $$P_{4final}-270=-\frac{540}{0.03\times60}\frac3{10}=-90,\quad P_{4final}=180MW$$

Ex 1. $P_N=P_{initial}=180MW,\,f_N=f_{initial}=60Hz,\,\delta=0.06=6\%\to f_{final}=60.3Hz$ then $P_{final}=?$

Ex 2. $P_N=240MW,\,P_{initial}=240\frac34=180MW,\,f_N=f_{initial}=60Hz,\,\delta=0.06=6\%\to f_{final}=60.3Hz$ then $P_{final}=?$

Ex 3. $P_{1N}=P_{1initial}=180MW,\,\delta_1=0.06$$P_{2N}=P_{2initial}=1080MW,\,\delta_2=0.05$$P_{3N}=P_{3initial}=480MW,\,\delta_3=0.04$$P_{4N}=P_{4initial}=270MW,\,\delta=0.03$$f_N=f_{initial}=60Hz$

Ex 4. $P_{1N}=240MW,\,P_{1initial}=240\frac34=180MW,\,\delta_1=0.06$$P_{2N}=P_{2initial}=1080MW,\,\delta_2=0.05$$P_{3N}=600MW,\,P_{3initial}=600\frac45=480MW,\,\delta_3=0.04$$P_{4N}=540MW,\,P_{4initial}=540\frac12=270MW,\,\delta_4=0.03$$f_N=f_{initial}=60Hz$

Ex 3. $P_{1N}=P_{1initial}=180MW,\,\delta_1=0.06$
$P_{2N}=P_{2initial}=1080MW,\,\delta_2=0.05$
$P_{3N}=P_{3initial}=480MW,\,\delta_3=0.04$
$P_{4N}=P_{4initial}=270MW,\,\delta=0.03$
$f_N=f_{initial}=60Hz$

Ex 4. $P_{1N}=240MW,\,P_{1initial}=240\frac34=180MW,\,\delta_1=0.06$
$P_{2N}=P_{2initial}=1080MW,\,\delta_2=0.05$
$P_{3N}=600MW,\,P_{3initial}=600\frac45=480MW,\,\delta_3=0.04$
$P_{4N}=540MW,\,P_{4initial}=540\frac12=270MW,\,\delta_4=0.03$
$f_N=f_{initial}=60Hz$