Energy and Power Concepts

Class Exercises

Class Exercise 1

  • Grading Contributions

    • The midterm exam contributes 15% to the final grade.

    • The final exam contributes 10% to the final grade.

    • The average of all quizzes contributes 25% to the final grade.

    • The average of all problem sets contributes 20% to the final grade.

    • The average of all class exercises contributes 10% to the final grade.

  • Physics Problems

    1. Work and Power Calculation

    • Given a force of 10 N applied to move a mass 100 m at constant speed with no friction over 10 minutes.

    • Work done is calculated as:
      W=Fimesd=10extNimes100extm=1000extJW = F imes d = 10 ext{ N} imes 100 ext{ m} = 1000 ext{ J}

    • Power is calculated as:
      P=racWt=rac1000extJ600extsext(10minutes)=1.67extWP = rac{W}{t} = rac{1000 ext{ J}}{600 ext{ s}} ext{ (10 minutes)} = 1.67 ext{ W}

    • Answers: 1000 J and 1.67 W

    1. Torque and Power Applied to Generator

    • Shaft radius = 10 cm (0.1 m), speed = 1800 rpm, force = 10 N. Torque is calculated as:
      au=rimesF=0.1extmimes10extN=1extNmau = r imes F = 0.1 ext{ m} imes 10 ext{ N} = 1 ext{ Nm}

    • Power is calculated as:
      P=auimesextangularvelocity;extangularvelocity=rac1800imes2extπ60extrad/sP = au imes ext{angular velocity}; ext{ angular velocity} = rac{1800 imes 2 ext{π}}{60} ext{ rad/s}
      P=1extNmimes188.5extrad/s=188.5extWP = 1 ext{ Nm} imes 188.5 ext{ rad/s} = 188.5 ext{ W}

    • Answers: 1 Nm and 188.5 W

    1. Potential Energy Calculation

    • Force of 10 N is applied to move a mass 1 m above a reference elevation.

    • Potential energy (PE) is calculated as:
      PE=mimesgimesh=10extNimes1extm=10extJPE = m imes g imes h = 10 ext{ N} imes 1 ext{ m} = 10 ext{ J}

    • Answers: 10 J

    1. Kinetic Energy Calculation

    • A mass of 10 kg at a velocity of 1 m/s.

    • Kinetic Energy (KE) is calculated as:
      KE=rac12mv2=rac12imes10extkgimes(1extm/s)2=5extJKE = rac{1}{2} m v^2 = rac{1}{2} imes 10 ext{ kg} imes (1 ext{ m/s})^2 = 5 ext{ J}

    • Answers: 5 J

    1. Power from Air Density

    • Air density = 1.225 kg/m³, velocity = 5 m/s, rotor diameter = 4 m.

    • Power exerted on the rotor:
      P=rac12<br>hoAv3P = rac{1}{2} <br>ho A v^3
      A=racextπ4d2=racextπ4imes(4extm)2=12.57extm2A = rac{ ext{π}}{4} d^2 = rac{ ext{π}}{4} imes (4 ext{ m})^2 = 12.57 ext{ m}^2
      P=rac12imes1.225extkg/m3imes12.57extm2imes(5m/s)3ext(simplifyingyieldsaround0.96kW)P = rac{1}{2} imes 1.225 ext{ kg/m³} imes 12.57 ext{ m}^2 imes (5 m/s)^3 ext{ (simplifying yields around 0.96 kW)}

    • Answers: 0.96 kW

    1. Energy Consumption by Electrical Load

    • Electrical load = 20 kW constantly for 1 day. Energy consumed:
      E=Pimest=20extkWimes24exthrs=480extkWhE = P imes t = 20 ext{ kW} imes 24 ext{ hrs} = 480 ext{ kWh}

    • Answers: 480 kWh or 1728 MJ

    1. Hydroelectric Power Calculation

    • Gross head $h=15$ m, flow $Q=90$ m³/s, head loss $5$ m due to friction.

    • Net head = $15 - 5 = 10$ m. Power available: P=hogQhextnet=1000extkg/m3imes9.81extm/s2imes90extm3/simes10extmP = ho g Q h_{ ext{net}} = 1000 ext{ kg/m³} imes 9.81 ext{ m/s²} imes 90 ext{ m³/s} imes 10 ext{ m}

      • The final calculation gives around 880 kW.

      • Answers: 880 kW

    1. Photon Flux Power Calculation

    • Photon flux = 1 mmol/(s m²) at a wavelength of 460 nm. Power flux: Pflux=(6.626imes10−34extJs)/(460imes10−9extm)imes(6.022imes1023extphotons/mol)P_{flux} = (6.626 imes 10^{-34} ext{ J s})/(460 imes 10^{-9} ext{ m}) imes (6.022 imes 10^{23} ext{ photons/mol})

      • This results in approximately 0.259 kW/m².

      • Answers: 0.259 kW/m²

    1. Molten Salt for Heat Storage

    • Specific heat capacity = 1.54 J/(gK), temperature difference = 400 K, heat = 10 MJ.

      • Total energy calculation yields approximately amount of salt needed: 1600 kg.

      • Answers: 1600 kg

    1. Enthalpy Comparison in Reactions

      • Enthalpy of Reaction 1 = -200 kJ/mol, Reaction 2 = +250 kJ/mol.

      • Reaction to drive an engine with more energy = Reaction 2, as it yields more.

      • Answers: Reaction 2

Class Exercise 2

  • Ecosystem Terms Matching

    • Productivity ⇒ kg/(m² yr)

    • Biomass ⇒ kg/m²

    • Primary Producer ⇒ plant

    • Secondary Producer ⇒ herbivore

    • Nutrients Cycle ⇒ energy flow

  • Biomass Calculation

    • If productivity is 0.2 kg m²/day and biomass on day 30 is 2 kg m², then biomass on day 31 is 2.2 kg m².

  • Net Primary Productivity Calculation

    • For a grass plant with net primary productivity = 0.5 MJ in 7 hours, and energy conversion efficiency = 5%. Energy received: approximately 4 kW.

  • Carbon in Earth Systems

    • Most of Earth’s carbon located as fossil fuels and living biomass in the biosphere.

  • Carbon Flux Changes with Emission Cuts

    • Calculate changes assuming negligible flow from other pools. A cut of 10% emissions from fossil fuels leads to a 10% change in net carbon flux.

  • CO2 Abundance in Atmosphere

    • CO2 abundance reflects high, low, and Nitrogen represents low, high.

  • Greenhouse Gas Composition

    • Comprehensive greenhouse gas set includes N2, CH4, H2O(g);

  • Annual CO2 Fluctuation Explanation

    • Mauna Loa CO2 data fluctuations align with growing seasons of the northern hemisphere.

  • Estimating Doubling Time for CO2 Levels

    • With a coefficient for CO2 increase = 0.39% per year, doubling time is approximately 177 years.

Class Exercise 3

  • Resistance and Inductance of Copper Wire

    • For 100 m of AWG #12 wire, the diameter is 2.053 mm, cross-sectional area is 5.211 mm², and resistance is calculated as 0.311 Ω.

  • Voltage Divider Design

    • To develop 4 V from 12 V, two resistors are needed, finding various combinations leads us to:

      • V = 4V, R₁ = 50 kΩ, R₂ = 100 kΩ.

  • Power Delivered by Load

    • In off-grid scenarios, correlation between voltage, current, and wire gage impacts efficiency and energy loss.

Further Exercises

  • Carnot Efficiency and Heat Engines

    • Work extraction from a heat engine yields efficiency characteristics and necessitates understanding specific heat interactions.

  • Thermodynamic Scenarios

    • Each thermodynamic scenario yields specific energy interactions across varying conditions and materials. Evaluating these through set equations demonstrates quantifiable insights into performance and efficiency dynamics.