Engr112 Homework Assignment 3 Spring Term For All Problems S

Engr112 Homeworkassignment3 Springtermforallproblemsshow

Engr112 Homework assignment 3 for the Spring term covers various topics in unit conversions, physics, and engineering calculations. The problems involve converting units between systems, calculating mass and velocity, determining densities, and analyzing pressure and energy transfer. The assignment emphasizes showing detailed work for full credit to demonstrate understanding of fundamental engineering principles and conversion techniques.

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Engr112 Homeworkassignment3 Springtermforallproblemsshow

This assignment encompasses a series of engineering problems requiring mastery of unit conversions, physical principles, and applied mathematics. It aims to assess students' ability to convert between different measurement systems, calculate mass from weight, determine velocities in SI units, and analyze densities and pressures within engineering contexts. Accurate work with proper unit conversions, algebraic calculations, and understanding of physical laws are vital to successfully completing these problems.

1. Conversion of force per volume units

The first problem asks to develop a conversion factor to convert a specific pressure or force per volume unit, specifically from pounds-mass per inch cubed (lbm/in^3) to kilograms per meter cubed (kg/m^3). This involves understanding the relationships among various units of mass, length, and volume. Starting with the known conversion constants: 1 lbm = 0.453592 kg, 1 inch = 0.0254 meters, and 1 ft^3 = 0.0283168 m^3, it is necessary to convert the units accordingly.

2. Mass from weight in different units

Given an object weighing 1000 lbf (pounds-force), the problem asks to derive its mass in slugs and in kilograms. Recognizing that pounds-force and mass are related through acceleration due to gravity, the conversion to slugs involves using the English unit system where 1 slug = 32.174 lbm. To find the mass in kilograms, converting the weight to mass via gravitational acceleration is essential, using the relation: mass (kg) = weight (N) / gravity (m/s^2), and converting pounds-force to newtons.

3. Velocity conversion to SI units

The third problem requires converting a velocity given in miles per hour (mph) to meters per second (m/s). It calls for multiplying the velocity in mph by the appropriate conversion factors: 1 mile = 1609.344 meters, and 1 hour = 3600 seconds.

4. Density calculations with diverse units

This problem investigates the density of a basketball in unconventional units, using given conversion units: 1 arroba = 11.5 kg, 1 peck = 9 liters, 1 batman = 3 kg, and 1 hogshead = 63 gallons. The basketball has a mass of approximately 624 grams (0.624 kg) and a volume of 0.25 ft^3, which needs to be converted into liters or cubic meters for density calculations in arroba per peck and batman per hogshead. Converting volume units from cubic feet to liters or cubic meters involves known conversion factors, and similar steps are required for calculating density ratios.

5. Weight of a power line cable

The fifth problem involves calculating the total weight of a power line segment with a given mass per unit length, spanning a certain distance. Multiplying the mass per length (kg/m) by the length yields the total mass, which can then be converted to weight in newtons by multiplying by gravity if needed, or expressed directly in mass units.

6. Velocity of a rubber ball on the Moon

The sixth question assesses kinematic calculations in lunar gravity. It involves finding the velocity of a rubber ball just before impact after being dropped from a height, considering the moon’s gravity. Using energy conservation principles or directly applying kinematic equations for free fall in gravitational acceleration of 1.62 m/s^2 gives the impact velocity.

7. Temperature increase from absorbed energy

This problem deals with thermal energy transfer and specific heat capacity. It calculates temperature rise of a human body (assumed water-like properties) soaked with a known energy flux (BTU/hr) over a specified duration. Converting BTU to Joules, then applying the specific heat formula, determines the temperature increase in Celsius.

8. Density of pressurized liquid in a tank

The final problem involves calculating the density of a liquid (tribromoethylene) in a pressurized tank, based on the pressure difference. Using the hydrostatic pressure equation, PDE = density g height, and knowing the pressure difference, allows solving for the density in kg/m^3, ensuring consistent units throughout.

Full Solutions to Selected Problems

Problem 1: Conversion factor from lbm/in^3 to kg/m^3

To convert from pounds-mass per cubic inch to kilograms per cubic meter, start with the known conversions:

• 1 lbm = 0.453592 kg
• 1 inch = 0.0254 m
• 1 ft^3 = (12 inches)^3 = 1728 in^3
• 1 m = 39.3701 inches; thus 1 in = 1 / 39.3701 m
• 1 in^3 = (0.0254 m)^3 = 1.6387 x 10^-4 m^3

Therefore, the conversion factor is:

1 lbm / in^3 = (0.453592 kg) / (1.6387 x 10^-4 m^3) ≈ 2762.9 kg/m^3

Hence,

1 lbm/in^3 ≈ 2762.9 kg/m^3

Problem 2: Mass of an object weighing 1000 lbf

Using the relation: weight = mass × acceleration due to gravity, with gravitational acceleration as 32.174 ft/s^2, the mass in slugs is:

• Mass (slugs) = Weight (lbm) / g (ft/s^2)

But since lbf is force in pounds, and in English units, 1 slug weighs 32.174 lb, then:

• Mass in slugs = 1000 lbf / 32.174 ft/s^2 ≈ 31.05 slugs

To convert to kilograms, use 1 lbm = 0.453592 kg, considering pounds-force and pounds-mass separately, but an easier approach is directly converting weight to newtons: 1 lbf ≈ 4.44822 N. Thus:

• Weight in newtons = 1000 lbf × 4.44822 N/lbf ≈ 4448.22 N
• Mass in kg = Weight / g = 4448.22 N / 9.80665 m/s^2 ≈ 454.0 kg

Problem 3: Velocity conversion from mph to m/s

Given velocity = 1530 mph:

1 mile = 1609.344 meters, 1 hour = 3600 seconds

Speed in m/s = 1530 × 1609.344 / 3600 ≈ 683.3 m/s

Problem 4: Density calculations with unusual units

Mass of basketball = 624 grams = 0.624 kg; Volume = 0.25 ft^3.

Convert volume to liters: 1 ft^3 = 28.3168 liters

Volume in liters = 0.25 × 28.3168 ≈ 7.0792 liters

Convert volume to cubic meters: 1 liter = 0.001 m^3, so:

Volume = 7.0792 liters × 0.001 = 0.0070792 m^3

Density in kg/m^3 = mass / volume = 0.624 / 0.0070792 ≈ 88.1 kg/m^3

Now, convert this to arroba per peck:

Mass in arroba: 0.624 kg / 11.5 kg/arroba ≈ 0.0543 arroba

Volume in pecks: 7.0792 liters / 9 liters/peck ≈ 0.7876 peck

Density in arroba per peck = 0.0543 / 0.7876 ≈ 0.069 arroba/peck

For batman per hogshead:

Mass in batmans = 0.624 kg / 3 kg/batman ≈ 0.208 batman

Volume in hogshead: 7.0792 liters / 63 gallons, with 1 gallon ≈ 3.785 liters

Hogshead volume in liters = 63 × 3.785 ≈ 238.4 liters

Hogshead volume in cubic meters = 0.2384 m^3

Density in batman per hogshead = 0.208 / 0.2384 ≈ 0.872 batman/hogshead

Problem 5: Weight of the power line cable

Mass per length = 0.5 kg/m, length = 75 m

Total mass = 0.5 × 75 = 37.5 kg

Weight in newtons = mass × g = 37.5 × 9.80665 ≈ 368.0 N

Problem 6: Velocity of ball on the Moon

Using conservation of energy or kinematic equations:

Velocity just before impact = √(2 × g_moon × h)

g_moon = 1.62 m/s^2, h=6 m

Velocity = √(2 × 1.62 × 6) ≈ √19.44 ≈ 4.41 m/s

Problem 7: Temperature increase from absorbed energy

Energy absorbed = 100 BTU/hr × 2 hours = 200 BTU

Convert BTU to Joules: 1 BTU ≈ 1055 Joules, so:

Energy = 200 × 1055 ≈ 211,000 Joules

Body mass = 132 lbm, convert to grams: 132 × 453.592 ≈ 59,874 grams

Specific heat of water = 4.18 J/g/°C

Temperature increase ΔT = energy / (mass × specific heat)

ΔT = 211,000 / (59,874 × 4.18) ≈ 211,000 / 250,124 ≈ 0.844°C

Problem 8: Density of tribromoethylene

Given: total pressure at bottom = 5 atm, surface pressure = 3 atm, so the pressure difference is 2 atm.

Pressure difference in Pascals: 1 atm = 101,325 Pa, thus ΔP = 2 × 101,325 ≈ 202,650 Pa

Using hydrostatic pressure relation: ΔP = ρ × g × h, rearranged as ρ = ΔP / (g × h)

g = 9.81 m/s^2, h = 25 ft = 7.62 meters

Density ρ = 202,650 / (9.81 × 7.62) ≈ 202,650 / 74.74 ≈ 2712 kg/m^3

Conclusion

This set of problems exemplifies fundamental engineering calculations involving unit conversions, physical laws, and property evaluations. Mastery over such computations ensures effective problem-solving skills in engineering contexts, bridging the gap between theoretical knowledge and practical application.

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