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BAE 103 Energy in Biosystems
Spring 2012
Lecture 1 (Jan. 11): THINKING LIKE AN ENGINEER
The System Concept…….
Control
Control
Rules
Component 2
Outputs
Input
OUTPUTS
Output
Component 3
Input
Component 1
Rules
Input
Input
Rules
Control
A very important concept…….
Amount
Component Processing Rate 
time
Other examples of time rates….
Distance
Velocity 
time
Area
Field Capacity 
time
Volume
Fluid Flow 
time
Work
Power 
time
Basic Quantities, Properties,
Parameters…….
Four basic Dimensions:
Displacement (L)
Mass (M)
Time (T)
Temperature (D)
Table 1.1 Some fundamental parameters and dimensions
Quantity
Dimensions
Units
Distance
L
meters (m), feet (ft)
Time
T
minutes, hours, seconds
Mass
M
kg, g, lbm
Temperature
D
oC, oF
Area
L2
m2, ft2
Volume
L3
m3, ft3
Velocity
L/T
m/s, km/hr, ft/s, mi/hr
Acceleration
L/T2
m/s2, ft/s2
Mass density
M/L3
kg/m3, g/cm3, lbm/ft3
Force
ML/T2
N = kg-m/s2, lbf = lbm-ft/s2
Energy
ML2/T2
J = N-m, cal, ft-lbf, BTU
Power
ML2/T3
W = J/s, HP = 550 ft-lbf/s
Heat capacity
L2/(T2-D)
cal/(oC-g), BTU/(oF-lbm)
Fundamental Laws and
Relationships
•Geometric equations
•Area, Volumes, Perimeters
•Physical Laws
•Newton’s Laws of Motion
•Conservation of Mass, Energy
Table 1.2 Some fundamental laws and equations which govern systems
Law or Equation
F=m∙a
Description
Newton’s 2nd law
W=F∙d
Definition of work/energy
H = W/t
Definition of power
T=F∙r
Definition of torque
H = 2π∙T∙n
Definition of rotational power
v = d/t
Definition of velocity
a = ∆v/t
Definition of acceleration
p = F/A
Definition of pressure
Q=v∙A
Definition of volume flow rate
Inputs/Output
Force (F), mass (m),
acceleration (a),
Work/energy (W), Force (F),
distance (d)
Power (H), Work/energy (W),
time (t)
Torque (T), Force (F),
radius (r)
Power (H), radius (r),
Torque (T),
rotational speed (n)
velocity (v), distance (d),
time (t)
acceleration (a), change in
velocity (∆v), time (t)
pressure (p), Force (F),
area (A)
Volume flow rate (Q),
velocity (v),
cross-section area (A)
Table 1.3 Geometric formulas
Shapes
Formula
Rectangle:
Area = Length X Width
A = l∙w
Perimeter = 2 X Length + 2 X Width
P = 2l + 2w
Parallelogram
Area = Base X Height
A = b∙h
Triangle
Area = 1/2 of the base X the height
A = 1/2 b∙h
Perimeter = a + b + c
(add the length of the three sides)
Table 1.3 Geometric formulas
Trapezoid
Area, A = (b1 + b2)/2 ∙ h
Perimeter, P = a + b1 + b2 + c
Circle
d = 2r
c = pd = 2 pr
A = pr2
Rectangular Solid
Volume = Length X Width X Height
V = l∙w∙h
Surface = 2∙l∙w + 2∙l∙h + 2∙w∙h
Table 1.3 Geometric formulas
Prisms
Volume = Base X Height
v=b∙h
Surface = 2b + Ph (b is the area of the base P is the perimeter of the
base)
Cylinder
Volume = pr2 x height
V = pr2 h
Surface = 2p radius x height
S = 2prh + 2pr2
Pyramid
V = 1/3 b∙h
b is the area of the base
Surface Area: Add the area of the base to the sum of the areas of all
of the triangular faces. The areas of the triangular faces will have
different formulas for different shaped bases.
Table 1.3 Geometric formulas
Cones
Volume = 1/3 pr2 x height
V= 1/3 pr2h
Surface = pr2 + prs
S = pr2 + prs
=pr2 + pr
Sphere
Volume = 4/3 pr3
V = 4/3 pr3
Surface = 4pr2
S = 4pr2
Table 1.4 Unit conversions
_____________________________________________________________________________
Mass and Weight
1 ounce = 437.5 grains = 28.35 grams
1 pound = 16 ounces = 7,000 grains = 453.6 grams
1 ton = 2,000 pounds
1 kilogram = 1000 grams = 2.2046 pounds
1 metric ton = 1000 kilograms
Length
1 mile = 5,280 feet = 1,760 yards = 320 rods = 80 chains = 1.609 km
1 chain = 66 feet = 22 yards = 4 rods = 100 links
1 rod = 16.5 feet = 5.5 yards
1 meter = 39.37 inch = 3.28 feet
1 foot = 12 inches = 30.48 centimeters
1 inch = 2.54 centimeters = 25.4 millimeters
Table 1.4 Unit conversions
Area
1 hectare = 10,000 square meters = 2.47 acres
1 acre = 160 square rods = 43,560 square feet = 0.405 hectares
1 square mile = 640 acres
Volume
1 cubic inch =16.39 cubic centimeters
1 cubic foot = 1,728 cubic inches = 7.48 gallons
1 cubic yard = 27 cubic feet
1 gallon = 4 quarts = 8 pints = 231 cubic inches = 128 fluid ounces = 3.785 liters
1 bushel = 1.244 cubic feet
1 cubic meter = 1000 liters
1 liter = 1000 cubic centimeters = 1000 milliliters
Time
1 hour = 60 minutes = 3,600 seconds
1 minute = 60 seconds
Table 1.4 Unit conversions
Water Volume/Mass Properties
1 gallon = 8.34 pounds
1 cubic inch = 0.03611 pounds
1 cubic foot = 62.4 pounds
1 liter = 1 kilogram
1 cubic centimeter = 1 gram
1 cubic meter = 1,000 kilograms = 1 metric ton
Specific and Latent Heat
Specific Heat of Water – 1.0 Btu/pound/oF, 1.0 calorie/gram/oC
Specific Heat of Ice – 0.48 Btu/pound/oF, 0.48 calorie/gram/oC
Latent Heat of Fusion for Ice – 144 Btu/pound, 80 calories/gram
Latent Heat of Vaporization of Water – 970 Btu/pound, 540 calories/gram
Table 1.4 Unit conversions
Force, Work, Power and Energy
1 calorie = 4.186 J
1 Calorie = 1000 calories = 3.9683 Btu
1 pound-force = 4.448 Newtons
1 Newton = 1 kg-m/s2
1 Joule = 1 Newton-meter
1 Btu = 1054 Joules = 252 calories = 780 feet-pounds-force
1 kilowatt = 1000 Joules/second = 1 kilojoules/second
1 kilowatt.hours = 3414 Btu
1 horsepower = 550 feet-pounds-force/second = 33,000 foot-pounds-force/minute
1 horsepower = 0.746 kilowatts
Pressure
1 Pascal = 1 Newton/m2
1 atm = 101.325 kilo-Pascal = 14.7 pounds/square inch = 34.0 feet of H2O = 29.92 in Hg
= 760 mm Hg
Example 1……
A runner covers a distance of 5.8 miles in 63
minutes. What is the runner’s average velocity
(mi/hr)? If the runner can maintain this velocity
for 2.5 hours, what distance will be travelled?
Example 2……
A tractor is mowing a 15.4 ha hay field. If the length
of the cutter bar is 2.2 m and the average speed
of the tractor is 10.3 km/hr, how long is required
to mow the field (hr)?
Example 3……
A pump is filling a tank with dimensions: h = 15 ft, D
= 25 ft. If 30 hours is required to fill the tank,
what is the pumping rate (gal/min)? If the hose
delivering water from the pump to the tank has an
inside diameter of 4 in., what is the flow velocity
of water in the hose (ft/sec)?
Example 4……
A tractor is pulling a wagon up a hill of 10% slope at
a speed of 4 km/hr. If the load on the wagon has
a mass of 5000 kg, what is the required power
(kW)?
Importance and Power of Units
and Dimensions…….
•Necessary to quantify systems parameters
•Indicates “reasonableness” of calculations
•Indicates validity of relationships between
parameters
•Can help determine relationships between
parameters
BE Problem-Solving Procedure
Given:
1. Always draw a picture of the system.
• Establishes relationships between
parameters.
2. State all assumptions.
3. Identify all factors/parameters and their units.
Required:
4. Label unknown quantities with a question
mark.
BE Problem-Solving Procedure
Relationships:
5. Write or derive the main equation which
contains the unknown required quantity.
6. Algebraically manipulate the main equation to
solve for the required quantity.
7. Write subordinate equations needed to
determine quantities in the main equation.
• Indent subordinate equations.
BE Problem-Solving Procedure
Solution:
8. Insert numerical quantities and their units into
equations.
9. Ensure that units cancel correctly and check
for correct sign.
10. Compute the answer.
11. Mark the final answer, with its units, by
enclosing it in a box.
12. Make sure the final anwer is physically
reasonable.
13. Ensure that all questions have been answered.
Example 5…..
A combine grain harvester is operating in a field
that yields 200 bu/ac. The combine has a swath
width of 15 ft and the grain tank holds 250 bu.
The harvesting speed is 5 mi/hr. How long does
it take to fill the tank (hr)?
Grain combine harvester……
Example 6…..
An adult has approximately 1.6 x 105 km of blood
vessels and a total blood volume of 4.3 L. What is the
average diameter (mm) of the blood vessels? If the
heart displaces approximately 65 ml per beat and beats
75 times per minute, how long (min.) is require to
circulate the blood volume? What is the average velocity
of blood flow in the vessels (m/sec)?
Example 7…..
A student late for class ran up the stairs in the FPAT
stairway to the 2nd floor in 5 sec. The 2nd floor is 18 ft
higher than the 1st floor and the student weighs 125 lb.
How much power (hp) was required?
What volume of gasoline (gal) would be required if the
energy content of gasoline is 125,000 Btu/gal and the
typical efficiency of an engine is 40%.
Example 8…..
How many acres of corn yielding 180 bu/ac will be
required to fill a bin which has a diameter of 45 ft
and a height of 25 ft?
If the bulk density of corn is 55 lb per bushel and corn
is planted in rows spaced 30 in. apart at a spacing of
10 in. between plants, what is the average mass of corn
(lb) produced by each plant?