Unit 1: Dimensional Analysis

Fundamentals of Dimensional Analysis

  • Dimensional Analysis: A mathematical problem-solving method used to convert a quantity from one unit of measurement to another unit of measurement.
  • Core Function: It explicitly demonstrates the proportional relationships between different physical units and ensures unit consistency across mathematical operations.

Handwritten notes page on dimensional analysis definitions and practice problems

Standard Unit Conversion Factors

  • Time Conversions:

    • Minutes to seconds: 1 min=60 sec1\,\text{min} = 60\,\text{sec}
    • Hours to minutes: 1 hr=60 min1\,\text{hr} = 60\,\text{min}
    • Years to days: 1 yr=365 days1\,\text{yr} = 365\,\text{days}

  • Length Conversions:

    • Feet to inches: 1 ft=12 in1\,\text{ft} = 12\,\text{in}
    • Meters to centimeters: 1 m=100 cm1\,\text{m} = 100\,\text{cm}

  • Mass and Volume Conversions:

    • Kilograms to pounds: 1 kg=2.2 lbs1\,\text{kg} = 2.2\,\text{lbs}
    • Gallons to quarts: 1 gal=4 quarts1\,\text{gal} = 4\,\text{quarts}

Practice Problem 1: Mass Conversion

  • Target Conversion: Convert 2.8 kg2.8\,\text{kg} to pounds (lbs\text{lbs}).
  • Conversion Factor Applied: 1 kg=2.2 lbs1\,\text{kg} = 2.2\,\text{lbs}
  • Calculation Setup:   (2.8 kg1)×(2.2 lbs1 kg)=6.16 lbs\left( \frac{2.8\,\text{kg}}{1} \right) \times \left( \frac{2.2\,\text{lbs}}{1\,\text{kg}} \right) = 6.16\,\text{lbs}

Practice Problem 2: Rate and Time Conversions

  • Part A - Converting Minutes per Day to Hours per Day:

    • Target: Convert 316 min/day316\,\text{min/day} into hours per day.
    • Conversion Factor: 1 hr=60 min1\,\text{hr} = 60\,\text{min}
    • Calculation Setup:     (316 min1 day)×(1 hr60 min)=5.26 hrs/day\left( \frac{316\,\text{min}}{1\,\text{day}} \right) \times \left( \frac{1\,\text{hr}}{60\,\text{min}} \right) = 5.26\,\text{hrs/day}
    • Note: The term "PER" denotes a rate per unit of time (e.g., per day).

  • Part B - Converting Minutes per Day to Seconds per Day:

    • Target: Convert 360 min/day360\,\text{min/day} into seconds per day.
    • Conversion Factor: 1 min=60 sec1\,\text{min} = 60\,\text{sec}
    • Calculation Setup:     (360 min1 day)×(60 sec1 min)=21,600 sec/day\left( \frac{360\,\text{min}}{1\,\text{day}} \right) \times \left( \frac{60\,\text{sec}}{1\,\text{min}} \right) = 21,600\,\text{sec/day}

Practice Problem 3: Multi-Step Volume and Rate Calculations

  • Part A - Daily Volume to Annual Volume:

    • Target: Convert 209,700,000 gallons209,700,000\,\text{gallons} of water per day into total gallons per year.
    • Conversion Factor: 1 yr=365 days1\,\text{yr} = 365\,\text{days}
    • Calculation Setup:     (209,700,000 gal1 day)×(365 days1 yr)=7.654×1010 gallons/year\left( \frac{209,700,000\,\text{gal}}{1\,\text{day}} \right) \times \left( \frac{365\,\text{days}}{1\,\text{yr}} \right) = 7.654 \times 10^{10}\,\text{gallons/year}

  • Part B - Annual Gallons to Annual Quarts:

    • Target: Determine how many quarts of water are used per year.
    • Conversion Factors: 1 yr=365 days1\,\text{yr} = 365\,\text{days} and 1 gal=4 quarts1\,\text{gal} = 4\,\text{quarts}
    • Calculation Setup:     (209,700,000 gal1 day)×(365 days1 yr)×(4 quarts1 gal)=306,162,000,000 quarts/year\left( \frac{209,700,000\,\text{gal}}{1\,\text{day}} \right) \times \left( \frac{365\,\text{days}}{1\,\text{yr}} \right) \times \left( \frac{4\,\text{quarts}}{1\,\text{gal}} \right) = 306,162,000,000\,\text{quarts/year}

Practice Problem 4: Rate and Unit Cost Conversions

  • Parameters: Fuel consumption rate of 26 L26\,\text{L} per week at a cost rate of $2.54\$2.54 per gallon.
  • Conversion Method: Requires multi-step conversion combining volume units between liters (L\text{L}) and gallons (gal\text{gal}) with time and cost factors.