Exercises — Efficiency

Humein kya chahiye: $70\text{W}$ mein se kitna fraction useful air-motion ban gaya. Divide kyun karein: definition ke anusaar $\eta =P_{\text{useful}}/P_{\text{in}}$, aur dono numbers already watts mein hain, isliye koi conversion ki zaroorat nahi. $\eta =\frac{42}{70}=0.6=60\%$ Baaki $70-42=28\text{W}$ heat ke roop mein nikal gaya — isliye fan ke motors warm lagte hain (dekho Heat and Internal Energy).

$1-\eta$ kyun use karein: parent identity $\eta =1-E_{\text{wasted}}/E_{\text{in}}$ ko rearrange karne par milta hai $E_{\text{wasted}}/E_{\text{in}}=1-\eta$. $\frac{E_{\text{wasted}}}{E_{\text{in}}}=1-0.25=0.75=75\%$ Is machine mein jo bhi daalte ho uska teen-chauthai hissa faink diya jaata hai.

Step 1 — Useful energy nikalo. Useful kaam crate ko uthaaana hai, isliye useful energy = gravitational PE gain (dekho Gravitational Potential Energy): $E_{\text{useful}}=mgh=120\times 10\times 5=6000\text{J}$ Yeh useful kyun hai: jo height gain hui hai wahi woh kaam hai jiske liye crane ko paisa diya gaya. Step 2 — Input se divide karo. $\eta =\frac{6000}{8000}=0.75=75\%$ Baaki $2000\text{J}$ cables aur gears mein heat ban gaya (Friction).

Rearrange kyun karein: hume $\eta$ aur output pata hai, lekin input chahiye. $\eta =E_{\text{useful}}/E_{\text{in}}$ se shuru karke palat do: $E_{\text{in}}=\frac{E_{\text{useful}}}{\eta }=\frac{9\text{kJ}}{0.6}=15\text{kJ}$ Sanity check: input ($15$) output ($9$) se bada hai — sahi hai, ek real machine ko hamesha useful return se zyada feed karna padta hai.

Step 1 — Ek unit mein convert karo. $4\text{kW}=4000\text{W}$ (kilo matlab $\times 1000$). Power form kyun: question mein power diya hai, energy nahi. Kyunki input aur output ka samay ek hi hai, $t$ cancel ho jaata hai aur $\eta =P_{\text{useful}}/P_{\text{in}}$ milta hai (dekho Work and Power). Step 2 — Multiply karo. $P_{\text{useful}}=\eta P_{\text{in}}=0.80\times 4000=3200\text{W}$

(a) $\eta =\frac{30}{50}=0.6=60\%$. (b) Conservation ke anusaar, input useful + wasted mein split hota hai: $P_{\text{wasted}}=P_{\text{in}}-P_{\text{useful}}=50-30=20\text{kW}$ (c) Energy = power × time. $2$ ghante mein: $E_{\text{wasted}}=20\text{kW}\times 2\text{h}=40\text{kWh}$ kWh seedha kyun kaam karta hai: ek kilowatt-hour literally "kilowatts multiplied by hours" hota hai, isliye koi unit gymnastics ki zaroorat nahi.

Step 1 — Pehle useful energy nikalo. Input ka sirf $70\%$ load tak pahunchta hai: $E_{\text{useful}}=\eta E_{\text{in}}=0.70\times 2100=1470\text{J}$ Step 2 — Woh PE height mein convert karo. Useful energy $mgh$ hai, isliye $h$ ke liye solve karo: $h=\frac{E_{\text{useful}}}{mg}=\frac{1470}{30\times 10}=\frac{1470}{300}=4.9\text{m}$ $E_{\text{in}}$ ki jagah $E_{\text{useful}}$ kyun use kiya: waste joules ne mass ko kabhi nahi uthaya, isliye $2100$ daalna height over-predict karta.

Raw joules kyun mislead karte hain: Y zyada useful energy produce karta hai ($630>150$), lekin khaata bhi bahut zyada hai. Efficiency size ko normalise kar deti hai (ratio ka yahi toh point hai), aur fair comparison possible hota hai. ${\eta }_X=\frac{150}{200}=0.75=75\%,{\eta }_Y=\frac{630}{900}=0.70=70\%$ X zyada efficient hai, haalaanki Y zyada total kaam karta hai.

Multiply kyun karein: har stage ka output agli stage ka input banta hai, isliye energy ka surviving fraction stage by stage multiply hota hai (parent note: chained efficiencies multiply). Neeche ke figure mein $100$ units fuel chain se trace karo.

${\eta }_{\text{total}}=0.40\times 0.90\times 0.85=0.306=30.6\%$ Lesson: total ($30.6\%$) sabse chhoti single stage se bhi neeche hai ($40\%$). Kamzor links poori chain ko neeche kheenchte hain.

Step 1 — Useful energy. $E_{\text{useful}}=0.65\times 5000=3250\text{J}$. Step 2 — Kinetic energy ke barabar set karo. Saari useful energy motion ban jaati hai, isliye $\frac{1}{2}m{v}^2=3250$. Step 3 — $v$ ke liye solve karo. Square root kyun lein: $v$ ek square ke andar chhupa hai, isliye square undo karna (inverse operation) use free karta hai. $v=\sqrt{\frac{2E_{\text{useful}}}{m}}=\sqrt{\frac{2\times 3250}{10}}=\sqrt{650}\approx 25.5\text{m/s}$

(a) Output → input jaane ke liye, $\eta$ se divide karo: $P_{\text{in}}=\frac{P_{\text{useful}}}{\eta }=\frac{4.5}{0.75}=6\text{kW}$ (b) Wasted = input − useful: $P_{\text{wasted}}=6-4.5=1.5\text{kW}$ Check: $1.5/6=0.25=1-\eta$. $1-\eta$ waste-fraction identity ke saath consistent hai.

(a) Har second release hone wali PE = input power. Har second, $2000\text{kg}$ $40\text{m}$ girta hai: $P_{\text{in}}=\frac{mgh}{t}=\frac{2000\times 10\times 40}{1}=800,000\text{W}=800\text{kW}$ "Per second" ise power kyun banata hai: energy ko ek-second ke interval se divide karna exactly power hai (Work and Power). (b) Turbine output. $P_{\text{turbine}}=0.90\times 800=720\text{kW}$ (c) Cables ke baad (ek aur stage $90\%$ par). $P_{\text{town}}=0.90\times 720=648\text{kW}$ ${\eta }_{\text{total}}=0.90\times 0.90=0.81=81\%$ Do identical $90\%$ stages combine hokar $81\%$ dete hain, $90\%$ nahi — chaining hamesha kisi bhi single stage se zyada cost karta hai.

Step 1 — Wasted power. $P_{\text{wasted}}=(1-\eta )P_{\text{in}}=0.40\times 500=200\text{W}$, yaani har second $200\text{J}$ heat. Step 2 — Heat → temperature. Temperature raise karne ke liye energy $Q=mc\Delta T$ hai (dekho Heat and Internal Energy). Ek second mein $Q=200\text{J}$ use karke, rise per second ke liye rearrange karo: $\Delta T=\frac{Q}{mc}=\frac{200}{0.2\times 500}=\frac{200}{100}=2{}^{\circ }\text{C per second}$ Efficiency kyun matter ki: temperature rise purely wasted fraction se drive hoti hai — zyada efficient drill zyada slowly garm hoti.

Step 1 — Required useful energy. $E_{\text{useful}}=mgh=4\times 10\times 8=320\text{J}$. Step 2 — Implied efficiency calculate karo. $\eta =\frac{320}{300}\approx 1.07=107\%$ Step 3 — Judge karo. $\eta >100\%$ impossible hai — iska matlab hoga ki total input se zyada useful energy nikal rahi hai, jo Conservation of Energy tod deta hai. Isliye claim bakwaas hai: ya toh input understate kiya gaya hai ya lift overstate ki gayi hai. Real machines mein hamesha $\eta <100\%$ hoti hai; heat-based engines ke liye asli limit aur bhi strict hai (dekho Heat Engines & Carnot Efficiency).

L1 sirf "useful ÷ in" spot karta hai. L2 ise rearrange karke ek missing quantity dhundhta hai. L3 tumhe raw joules ki jagah fractions compare karne par lagata hai. L4 machines chain karta hai — energy baar baar kati jaati hai, isliye tum multiply karte ho, kabhi average nahi. L5 efficiency ko motion, heat, aur us iron rule ke saath fuse karta hai ki kuch bhi $100\%$ nahi beat kar sakta. Poore raaste ek hi formula — sirf surrounding physics barhti jaati hai.