The System Concept…….
Component 1 Input
Control
Rules
Control
Outputs Component 2
Control
Component 3Input
Input
Rules
Rules
Input
OU
TP
UT
S
Output
Other examples of time rates….
time
Work Power
time
Volume Flow Fluid
time
Area Capacity Field
time
Distance Velocity
Basic Quantities, Properties, Parameters…….
Four basic Dimensions:Displacement (L)Mass (M)Time (T)Temperature (D)
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)
Table 1.1 Some fundamental parameters and dimensions
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 Description Inputs/OutputF = m a∙ Newton’s 2nd law Force (F), mass (m),
acceleration (a), W = F d∙ Definition of work/energy Work/energy (W), Force (F),
distance (d) H = W/t Definition of power Power (H), Work/energy (W),
time (t)T = F r∙ Definition of torque Torque (T), Force (F),
radius (r) H = 2π T n∙ ∙ Definition of rotational power Power (H), radius (r),
Torque (T), rotational speed (n)
v = d/t Definition of velocity velocity (v), distance (d), time (t)
a = ∆v/t Definition of acceleration acceleration (a), change in velocity (∆v), time (t)
p = F/A Definition of pressure pressure (p), Force (F),area (A)
Q = v A∙ Definition of volume flow rate 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 WidthP = 2l + 2w
ParallelogramArea = Base X Height
A = b h∙
TriangleArea = 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 = 2rc = pd = 2 pr
A = pr2
Rectangular SolidVolume = Length X Width X Height
V = l w h∙ ∙Surface = 2 l w + 2 l h + 2 w h∙ ∙ ∙ ∙ ∙ ∙
Table 1.3 Geometric formulas
PrismsVolume = Base X Height
v=b h ∙Surface = 2b + Ph (b is the area of the base P is the perimeter of the
base)
CylinderVolume = pr2 x height
V = pr2 hSurface = 2p radius x height
S = 2prh + 2pr2
PyramidV = 1/3 b h∙
b is the area of the baseSurface 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
ConesVolume = 1/3 pr2 x height
V= 1/3 pr2h Surface = pr2 + prs
S = pr2 + prs =pr2 + pr
SphereVolume = 4/3 pr3
V = 4/3 pr3
Surface = 4pr2
S = 4pr2
Table 1.4 Unit conversions
_____________________________________________________________________________
Mass and Weight1 ounce = 437.5 grains = 28.35 grams1 pound = 16 ounces = 7,000 grains = 453.6 grams1 ton = 2,000 pounds1 kilogram = 1000 grams = 2.2046 pounds1 metric ton = 1000 kilograms
Length1 mile = 5,280 feet = 1,760 yards = 320 rods = 80 chains = 1.609 km1 chain = 66 feet = 22 yards = 4 rods = 100 links1 rod = 16.5 feet = 5.5 yards1 meter = 39.37 inch = 3.28 feet1 foot = 12 inches = 30.48 centimeters1 inch = 2.54 centimeters = 25.4 millimeters
Table 1.4 Unit conversions
Area1 hectare = 10,000 square meters = 2.47 acres1 acre = 160 square rods = 43,560 square feet = 0.405 hectares1 square mile = 640 acres
Volume1 cubic inch =16.39 cubic centimeters1 cubic foot = 1,728 cubic inches = 7.48 gallons1 cubic yard = 27 cubic feet1 gallon = 4 quarts = 8 pints = 231 cubic inches = 128 fluid ounces = 3.785
liters1 bushel = 1.244 cubic feet1 cubic meter = 1000 liters1 liter = 1000 cubic centimeters = 1000 milliliters
Time1 hour = 60 minutes = 3,600 seconds1 minute = 60 seconds
Table 1.4 Unit conversions
Water Volume/Mass Properties1 gallon = 8.34 pounds1 cubic inch = 0.03611 pounds1 cubic foot = 62.4 pounds 1 liter = 1 kilogram1 cubic centimeter = 1 gram1 cubic meter = 1,000 kilograms = 1 metric ton
Specific and Latent HeatSpecific Heat of Water – 1.0 Btu/pound/oF, 1.0 calorie/gram/oCSpecific Heat of Ice – 0.48 Btu/pound/oF, 0.48 calorie/gram/oCLatent Heat of Fusion for Ice – 144 Btu/pound, 80 calories/gramLatent Heat of Vaporization of Water – 970 Btu/pound, 540 calories/gram
Table 1.4 Unit conversions
Force, Work, Power and Energy1 calorie = 4.186 J1 Calorie = 1000 calories = 3.9683 Btu1 pound-force = 4.448 Newtons1 Newton = 1 kg-m/s2
1 Joule = 1 Newton-meter1 Btu = 1054 Joules = 252 calories = 780 feet-pounds-force1 kilowatt = 1000 Joules/second = 1 kilojoules/second1 kilowatt.hours = 3414 Btu 1 horsepower = 550 feet-pounds-force/second = 33,000 foot-pounds-force/minute1 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 ProcedureGiven: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 ProcedureSolution: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 swathwidth 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)?
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?