Problem Solving Problem Solving Using Common Using Common

Engineering ConceptsEngineering Concepts

Introduction to Mechanical EngineeringIntroduction to Mechanical Engineering

The University of Texas-Pan AmericanThe University of Texas-Pan American

College of Science and EngineeringCollege of Science and Engineering

Objective

Practice how to set up engineering calculations and how to deal with common physical quantities and concepts.

Practice how to analyze and solve problems involving processes which change with time.

Practice how to analyze and solve problems in the absence of exact and complete information.

Problem Solving Ability

Problem solving ability only develops with experience gained through repetitive practice.

For a student, practice comes in the form of solving class and homework problems.

This presentation introduces simple concepts fundamental to engineering problem solving.

Rate Problems

The relation between rate, quantity, and time is

Temperature

Temperature is commonly measured using the Fahrenheit or Celsius scale.

When temperature appears as a term in a scientific or engineering equation, it is absolute temperature (measured in Kelvin or Rankine).

°F = 1.8 °C + 32°F = 1.8 °C + 32

°R = °F + 460°R = °F + 460

K = °C + 273K = °C + 273

°R = 1.8 K°R = 1.8 K

Temperature

Pressure

Pressure is the force per unit area exerted on a surface by a fluid (gas or liquid).

P = F / A Units: Pascals, psi

The Ideal Gas Law

PV = nRT (R is the Universal Gas Constant)

Density and Specific Gravity

ρ = M / V– Density is the ratio of the mass of a substance

to its volume. SG = ρS / ρR

– Specific gravity is the ratio of the density of a substance to the density of a reference substance.

– Reference substance is liquid water at 4°C which has a density of 1,000 kg/m3.

Mass and Volumetric Flow

MF = M / t

VF = V / t

If the fluid is incompressible, then

MF = ρ (VF) Incompressible implies that the density of

the fluid is constant.

Material Balance

Steady State: 0 = I - O + G - C0 = I - O + G - C No Chemical Reactions: I = OI = O Batch Process: A = G - CA = G - C

Heat, Work, and Energy

Kinetic Energy: EK = (mv2) / 2 – Energy due to velocity

Potential Energy: EP = mgh– Energy due to height

Power

Power (P) is the rate of energy production or consumption. P = E / t

Power can also be defined as the amount of work done per unit time. Power = Work / time

Estimation and Approximation

It is often necessary to make a reasonable estimate that gives an answer within 10 to 20% of what might be calculated with complete information.

In absence of complete information, you must make assumptions to simplify the problems and/or make intelligent guesses.

Open Forum

Assignment: Handout or visit website.

Analytic Method

1. Define the problem and summarize the problem statement.

2. Diagram and describe.

3. State your assumptions.

4. Apply theory and equations.

5. Perform the necessary calculations.

6. Verify your solution and comment.

Problem #1 – Statement

Robert lives 65 miles from you. You both leave home at 3:00 pm and drive toward each other on the same road. Robert travels at 40 mph, and you travel at 50 mph. At what time will you pass Robert on the road?

Problem #1 – Summary

At what time will you pass Robert?

Problem #1 – Solution

Initial time = 3:00 pm Distance = Average Velocity * time 65 miles = 40 mph * (t) + 50 mph * (t) t = 65 miles/(40 mph +50 mph) = .72 hrs 0.72 hrs (60 minutes/hr) = 43.3 minutes Hence, you should pass Robert at 3:43 pm.

Problem #2 – Statement

A rectangular tank 10 feet wide, 30 feet long, and 12 feet high is completely filled with gasoline. The gasoline is to be pumped into a spherical tank that is 6.0 meters in diameter. The pump moves gasoline at a rate of 30.0 gallons/minute. (a) How long, in hours, will it take to transfer the gasoline? and (b) what percentage of the spherical tank will be filled with gasoline?

Problem #2 – Summary

a) Time to transfer in (hrs) ?

b) % of the spherical tank filled?

Problem #2 – Solution

VR l w h 3600ft3

3600ft3 7.48052

gal

ft3

26930gal

a)t

26930gal

30gal

min

1 hr60 min

14.96hrs

Vs4

3 r

34

3 3 m

1

.3048

ft

m

3

3993.99ft3

b)

%3600

3993100 90.1%

Problem #3 – Statement

Juan and Charles are commercial painters. Juan has a job to paint a water storage tank that is a sphere supported above the ground by three legs. The sphere is 15.5 meters in diameter. Charles has a job to paint a gasoline storage tank that is a cylinder 48.0 feet in diameter and 39.0 feet high. The bottom of the tank sits on a concrete pad, and therefore, will not be painted. Juan has bet Charles that he will complete his job first. Assuming that Juan and Charles both paint at a rate of 150 feet square / hour, who will win the bet?

Problem #3 – Summary

Who will win the bet?– (who has less area since they paint at the same rate?)

Problem #3 – Solution

Juan

Area 4 r2

Area 4 15.5m

2

1

.3048

ft

m

2

8124.25ft2

Charles

Area r2 d h

Area 24 ft( )2 48 ft( ) 39 ft( ) 7690.60ft

2

Hence, Charles will win

Problem #4 – Statement

A liquid mixture of benzene and toluene contains 50% benzene by mass. The mixture is fed continuously to a distillation apparatus that produces vapor containing 60% benzene and a residual mixture containing 37.5% benzene, each by mass. The still operates at a steady state. If the mixture feed rate is 100 kg/hr, determine the mass flow rates for each stream.

Problem #4 – Summary

Determine the mass flow rate for the vapor and residual mixture.

x

y

100 kg/hr50 % benzene50 % toluene

60 % benzene 40 % toluene

37.5 % benzene 62.5 % toluene

Problem #4 – Solution

100kg

hr x y 1( )

benzene:

50kg

hr

60

100

x37.5

100

y 2( )

toluene:

50kg

hr

40

100

x67.5

100

y 3( )

Problem #4 – Solution cont.Use two of the three equations.

for example,

from (1)

y 100kg

hr x

substituting in (2)

50kg

hr .6 x .375 100

kg

hr x

50kg

hr.6x 37.5

kg

hr .375x

now solve for x

x

12.5kg

hr

.22555.55

kg

hr

solving for y:

y 100kg

hr x 100

kg

hr 55.55

kg

hr 44.45

kg

hr

Problem #5 – Statement

Estimate the total amount of coolant in gallons wasted by a radiator that drips continually at the rate of one drop every two minutes for six months. State all your assumptions clearly.

Problem #5 – Summary

Total amount of coolant wasted in six months (in gallons)?

Problem #5 – Solution

.5drops

minute 60

minutes

hour

24hours

day

365.25

2days

sixmonths

131490drops

sixmonths

We need to know how many drops are in a gallon?

Assumption:

1 dropsx mililiters

1000mililitersliters

3.785liters

gallon

ydrops

gallon

Answer:

131490drops

sixmonths

ydrops

gallon

Xgallons

sixmonths