Problem Solving & Energy Balances 4: Those Pesky US Units ABET Course Outcomes: Important Resource: Class Plan distance Velocity time acceleration

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1 Problem Solving & Energy Balances 4: Those Pesky US Units ABET Course Outcomes: 1. solve and document the solution of problems involving elements or configurations not previously encountered (e) 2. solve problems using multiple approaches including varied analytic approaches, diagrams, formal solution steps or simple computer programs (e) By the end of this class you should be able to: Explain and use US units in problem solving for unit conversions identifying term type Solve energy balance problems using US units Important Resource: A brief paper on these unit systems with examples is posted on the website. 14 th Annual Opportunity Banquet Career Fair Banquet Speaker: P.E., Director of MDOT Games & Prizes!!! Ticket Price Goes Up Friday!!! Handout: US Unit Systems Class Plan Lecture Outline: 1. Introduction/review a. Review: Chain to Remember b. Closed System Energy Terms 2. English Unit systems (handout) a. Intro to English unit systems b. Problems: gravity vs. lateral movement c. units of groups energy 3. Intro to energy balances in US units a. potential energy b. kinetic energy c. combined distance time a. Energy b. Pressure c. Force d. Power e. Velocity Velocity Momentum A Chain to Remember time acceleration x mass x mass Force Area Pressure 1 x distance Energy 2 time Power 3

2 Next two slides: overview of English unit systems Note differences basis of two systems (US Engineering is over specified) different mass units need g c for US engineering system (Note for metric: g c = 1 kg m/(n s 2 ) both use of lb f as a basic unit (different from metric where F is a derived unit) Note revised 2 nd law formula changes F = ma to F = ma/g c Messier than metric (due to long history) Key complications: lb as both a force and a mass unit (g c ) Common units require knowing odd conversion factors + multiple units for derived units with a force) Primary Units SI System (mks) m, kg, s US Engineering ft, s, lbf & lbm (over specified) Mass kg lbm US Common/ Gravitational ft, s, & lb (as a force) slug lbf s 2 /ft Force N kg m/s 2 lbf lb 1 Unit Systems: Mass and Force US Unit Systems: Mechanical Quantities based on the lb force Quantity lb f based conversion to common force (F = ma) (a basic unit in the English/US systems Energy or Work (E = FL) power (P=E/t) pressure (p = F/L 2 ) or stress ( = F/L 2 ) lbf (lb force) ft lbf / 778 = BTU (ft lbf)/s / 550 = hp (lbf /ft 2 ) / 144 = psi Next slide two simple gravitation problem Problems use as you see fit eg. ask students for approach show example on board or copy solutions to slide F = mg = (30 kg)(9.81 m/s 2 ) = kg.m/s 2 = 294 N Put on board g c = slug = 1 lbf s 2 /ft F = mg/g c = (30 lbm) = 30 lbf 32.2 ft / s lbm ft/lbf s 2

3 Gravitational force A 30 kg block hangs from a cable. What is the tension force in the cable? (g = 9.81 m/s 2 ) Mass A 30 lbm block hangs from a cable. What is the tension force in the cable?(g = 32.2 ft/s 2 ) Next slide Problems using English units notice horizontal vs. vertical Have students try briefly and then discuss solution The idea here is for them to come to grips with the problem before seeing the solution OK if they do not answer at first. Note English easier for vertical lift (there is a reason for lb f ) Messier for horizontal movement Can keep track of what to do with careful analysis of units Answers follow slide (note if unhidden this slide has the blue answers animated) Example Using g c : A 100 lb m crate rests on the floor. How much work (F * d) is required to move it: a. 10 feet across the floor against a 2 lb f frictional force? b. 10 feet straight up? note:./. = 1 Example Using g c : A 100 lb m crate rests on the floor. How much work (F x d) is required to move it at a constant speed: a. 10 feet across the floor against a 2 lbf frictional force? (2 lb f )(10 ft) = 20 ft lbf b. 10 feet straight up? note: F = = 100 lbm = 100 lbf W = F x d = 100 lbf x 10 ft = 1000 ft lbf./. = 1 What force is required to accelerate that crate at 10 ft/s 2, on a frictionless surface? c. What force is required to accelerate that crate at 10 ft/s 2, on a frictionless surface? F = = / = 31 lbf.

4 m [=] slugs ( = lbf s 2 /ft) g [=] ft/s 2 t [=] s h (height) [=] ft v [=] ft/s What are the resulting (English) basic units for: mv [=] mg [=] Next Slide: Results for units exercise review groups (units practice) lead into focusing on energy units Group Units Type mv [=] (lbf s 2 /ft) (ft/s) = lbf s momentum ½ mv 2 [=] (lbf s 2 /ft) (ft 2 /s 2 ) = ft lbf energy ½mv 2 [=] mgh [=] mv/t [=] (lbf s) / s = lbf force mv/t [=] g t 2 [=] What types of quantities are these (e.g. a force, an energy )? mg [=] (lbf s 2 /ft) (ft/s 2 ) = lbf force mgh [=] (lbf s 2 /ft) (ft/s 2 )ft = ft lbf energy g t 2 [=] (ft/s 2 ) (s 2 ) = ft distance U g = m (g) h Datum (h = 0) Gravitational Potential Energy (Ug) m Where g is the magnitude of acceleration of gravity (i.e ft/s 2 ) h Calculate Ug for: 5 lb m bag of flour on a high dive (10 ft above the water, m*g = 5 lb f ) 1 gallon of water (m*g = 8.34 lbf) falling 520 ft (the average drop to a generator on the Hover Dam) Calculate Ug for: 5 lb bag of flour on a high dive (i.e., 10 ft above the water, mg = 5 lbf) Ug = (mg)(h) = (5 lbf ) (10 ft) = 50 ft lbf 1 gallon of water (mg = 8.34 lbf) falling 520 ft (the average drop to a generator on the Hover Dam) Ug = (8.34 lbf) (520 ft) = 4337 ft lbf (= 5.57 BTU) Notice: a lb force is a weight, the force exerted by an object in a gravitational field. So one way to look at a force in lbf is that it is mass times gravitational acceleration, mg (because F=ma).

5 m K = ½ mv 2 v Kinetic Energy (K) Energy of a moving object Calculate the Energy of A 90 mph (132 ft/s) fast ball in ft lbf W=m*g = 5 oz = 0.31 lb f The first answer converted to BTUs A 3307 lb f car, traveling at 65 mph (95 ft/s) A 90 mph (132 ft/s) fast ball in ft lbf (mg = 5 oz = 0.31 lbf or ~ 0.01 slug (lbf s 2 /ft), Three variations: K = ½ ( 0.31 lbf )/(32.2 ft/s 2 ) (132 ft/s) 2 = 83.9 ft lbf or K = ½ (0.31 lbm)/(32.2 lbm ft/(lbf s 2 )(132 ft/s) 2 = 83.9 ft lbf or K = ½ ( lbf s 2 /ft) (132 ft/s) 2 = 83.6 ft lbf Notice in all cases the units work out as well as the number The above in BTUs A 3307 lbf car, traveling at 65 mph (95 ft/s) K = ½ (3307 lbf)/(32.2 ft/s 2 ) (95 ft/s) 2. = BTU = 4.6 x 10 5 ft lbf or BTU KE & PE: Equations and a Simple Energy Balance (Reminder) Kinetic Energy (K) Gravitational Potential Energy (U g ) K = ½ mv 2 U g = m (g) h m v In a closed system with no work or heat exchange the total energy change is zero: E = 0 K + U g = 0 ½ m (v 2 ) + m (g) h = 0 or ½ m v 2 + m (g) h = constant m h Previous Slide: KE/PE Review Balance Preview 1. Formula: Review formula for KE and PE 2. Units: note how they work out 3. Signs: need to make sure signs make sense notice in this case g is the magnitude of gravitation acceleration. you must keep track of its direction. 4. Balance Equations: Go through balance equations at bottom a) General concept sum of energy changes = 0 b) Summing types of equations c) Second equation form 5. System: this is a balance on an object (the system)

6 Energy in a Closed System System (of fixed mass) Energy = a property Balance Problem 2: Gravitational Potential and Kinetic Energy V xo = 0 h o = 70 ft If there are no losses (due to friction or drag), what is its velocity at the bottom of the ramp? (mechanical energy crossing a system boundary) Reference line or Datum line E o = E f V x =? Previous Slide Problem Setup (Ask students setup and review) Diagram: on previous slide Given: h 0 = 70 ft V x0 = 0 ft/s Find: Final velocity, V f =? Known: E 0 = E f for h 0 at bottom of ramp: (i.e., no KE at start no PE at final) E o = m ( g) h 0 E f = ½ m v 2 f Assumptions: no losses (friction and drag) Solution: Set two energy formula equal to each other (have students calc.) (i.e., combine known equations above) m (g) h o = ½ m v f 2 cancel m s and solve for v f 2 v f2 = 2 (g) h 0 = 2 (32.2 ft/s 2 ) 70 ft = 4508 ft 2 /s 2 v = 67 ft/s