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1 Physics 7B - AB Lecture 2 April 10 Recap + examples Steady-State Energy Density Model Applied to Fluid Circuit & Electrical Circuit Lecture slides available.

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Presentation on theme: "1 Physics 7B - AB Lecture 2 April 10 Recap + examples Steady-State Energy Density Model Applied to Fluid Circuit & Electrical Circuit Lecture slides available."— Presentation transcript:

1 1 Physics 7B - AB Lecture 2 April 10 Recap + examples Steady-State Energy Density Model Applied to Fluid Circuit & Electrical Circuit Lecture slides available at

2 2 Course Website Click on Physics 7B-A/B Next week Lecturer 3 Dr. Kevin Klapstein April 16 My office hr is cancelled. Today Quiz 1!

3 Current I VS Fluid velocity v Remember Energy Density Equation ? a.k.a. Fluid Transport Equation/Extended Bernoulli eq. ∆P + (1/2)  ∆(v 2 ) +  g∆h + IR = E pump /V Current I is “ Volumetric flow rate” How much water (in terms of volume) flows through a section of a pipe per unit time. So it has the unit of volume per second [m 3 /s]. Fluid velocity is how fast fluid is moving. So it has the unit of distance per second. [velocity] = [m/s] I vs v

4 4 Recap on Fluids Basic rules for looking at fluids: 1. Energy (density) conservation a.k.a. Energy Density Model (P 2 – P 1 ) + (1/2)  (v 2 2 – v 1 2 ) +  g( h 2 – h 1 ) + IR 1  2 =  1  2 2. Current Conservation Current entering = current leaving 3. Pressures where two fluids systems touch are equal 4. Resistance is proportional to length 5. ALWAYS, pick two points within the SAME fluid system to apply 1 & 2!

5 5 Current is shown flowing down the pipe shown. Choose the correct statement. 1 2 (1) I 1 = I 2 (2) I 1 > I 2 (3) I 1 < I 2 (4) Cannot tell

6 6 Current is shown flowing down the pipe shown. Choose the correct statement. 1 2 (1) I 1 = I 2 (2) I 1 > I 2 (3) I 1 < I 2 (4) Cannot tell

7 7 Current is shown flowing down the (R=0) pipe shown. Fluid is flowing from 1 to 2. Choose the correct statement: 1 2 (1) I 1 = I 2 (2) I 1 > I 2 (3) I 1 < I 2 (4) Cannot tell Pumping this way

8 8 Current is shown flowing down the (R=0) pipe shown. Fluid is flowing from 1 to 2. Choose the correct statement: 1 2 (1) I 1 = I 2 (2) I 1 > I 2 (3) I 1 < I 2 (4) Cannot tell Pumping this way

9 9 Current is shown flowing down the pipe shown. Choose the correct statement. 1 2 (1) v 1 = v 2 (2) v 1 > v 2 (3) v 1 < v 2 (4) Cannot tell

10 10 Current is shown flowing down the pipe shown. Choose the correct statement. 1 2 (1) v 1 = v 2 (2) v 1 > v 2 (3) v 1 < v 2 (4) Cannot tell

11 11 Current is shown flowing down the (R=0) pipe shown. Fluid is flowing from 1 to 2. Choose the correct statement: 1 2 Pumping this way (1) v 1 = v 2 (2) v 1 > v 2 (3) v 1 < v 2 (4) Cannot tell

12 12 Current is shown flowing down the (R=0) pipe shown. Fluid is flowing from 1 to 2. Choose the correct statement: 1 2 Pumping this way (1) v 1 = v 2 (2) v 1 > v 2 (3) v 1 < v 2 (4) Cannot tell What energy system changed ?

13 13 The following shows a pipe without dissipation (R=0) with a pump attached. On which side is the pressure greater? 12 (1) Side 1 (2) Side 2 (3) Not enough information Pumping this way

14 14 The following shows a pipe without dissipation (R=0) with a pump attached. On which side is the pressure greater? 12 (1) Side 1 (2) Side 2 (3) Not enough information Pumping this way

15 15 The following shows a resistanceless pipe (R=0) with a pump attached. On which side is the pressure greater? 1 2 (1) Side 1 (2) Side 2 (3) Need to know (4) Even knowing would not be enough info Pumping this way * (The change in height is 2 m from 1 to 2)

16 16 The following shows a resistanceless pipe (R=0) with a pump attached. On which side is the pressure greater? 1 2 (1) Side 1 (2) Side 2 (3) Need to know (4) Even knowing would not be enough info Pumping this way * (The change in height is 2 m from 1 to 2)

17 How does P 1, P 2, P 3 compare? Assume no dissipation, i.e., R = 0.

18 18 Same height (approximately) 1,3 have the same speed (same area, current conserved) 2 is faster (smaller area, current conserved) 13 2 No pumps between 1 and 3 Assume no resistance between 1 and 3. This implies P 1 = P 3 > P 2.

19 Small amount of resistance modifies this. Pressures are P 1 > P 3 > P 2

20 Large amount of resistance modifies this. Pressures are P 1 > P 2 > P 3

21 21 Electrical circuits Recap and Examples on

22 22 Analogies between Fluid and Electrical circuits FluidsElectricity Deals with Fluid flowCharge flow Current Water flowing through a cross- section of pipe per unit time Electric charge flowing through a wire per unit time v Indicator of kinetic energy (neglected) h Indicator of gravitational potential energy (neglected) R Relates current and thermal energy loss Energy stored /unit volume or charge PressureVoltage (P 2 – P 1 ) + (1/2)  (v 2 2 – v 1 2 ) +  g( h 2 – h 1 ) + IR 1  2 =  1  2

23 23 Analogies between Fluid and Electrical circuits FluidsElectricity Deals with Fluid flowCharge flow Current Water flowing through a cross- section of pipe per unit time Electric charge flowing through a wire per unit time v Indicator of kinetic energy (neglected) h Indicator of gravitational potential energy (neglected) R Relates current and thermal energy loss Energy stored /unit volume or charge PressureVoltage (V 2 – V 1 ) + (1/2)  (v 2 2 – v 1 2 ) +  g( h 2 – h 1 ) + IR 1  2 =  1  2

24 24 Analogies between Fluid and Electrical circuits FluidsElectricity Deals with Fluid flowCharge flow Current Water flowing through a cross- section of pipe per unit time Electric charge flowing through a wire per unit time v Indicator of kinetic energy (neglected) h Indicator of gravitational potential energy (neglected) R Relates current and thermal energy loss Energy stored /unit volume or charge PressureVoltage Energy Density Equation a.k.a. transport equation (V 2 – V 1 ) + IR 1  2 =  1  2 Or ∆V =  – IR

25 25 Energy Density Equation applied to Electrical Circuits a.k.a. transport equation, loop rule (V 2 – V 1 ) + IR 1  2 =  1  2 Or ∆V =  – IR The change in the electrical potential energy per charge, (what we call voltage, or voltage drop), as we move from one point to another point will increase due to energy added by a battery or generator and will decrease due to the transfer of electric potential energy per charge to thermal energy system.

26 26 Current in = current out From rest of circuit To rest of circuit AB Current into A = I 1 Current out of A = I 2 + I 3 Current into B = I 2 + I 3 Current out of B = I 4 I 1 = I 2 + I 3 I 2 + I 3 = I 4 = I 1 Current (charge) conservation a.k.a. junction rule (everything that flows into a junction must be equal to everything that flows out)

27 27 Electric Power Elecric energy is useful to us because it can be easily transformed into other forms of energy, e.g. hair dryer, lightbulb… Power: rate at which energy is transformed by an electric device = Energy transformed/unit time Ex. 120Watt = 120 Joule/sec 1200 Watts Watts Watts

28 28 Electric Power Elecric energy is useful to us because it can be easily transformed into other forms of energy, e.g. hair dryer, lightbulb… Power = (∆V) x I = I 2 R 1200 Watts Watts Bright ness of a bulb Watts R hair dryer R clothes dryer R bulb 120V

29 R1R1 R2R2 Bulbs 1 and 2 are connected with a battery as shown to the right. The bulbs have different Resistances. Which statement must be true? (1)The two bulbs have the same voltage different, ∆V, across them and that ∆V is half the battery voltage. (2)The two bulbs are equally bright. (3)The two bulbs and the battery have all the same current through them. (4)All of the above must be true. (5)None of the above must be true. Note : For electrical circuits, wires connecting circuit elements are assumed to have zero resistance.

30 R1R1 R2R2 Bulbs 1 and 2 are connected with a battery as shown to the right. The bulbs have different Resistances. Which statement must be true? (1)The two bulbs have the same voltage different, ∆V, across them and that ∆V is half the battery voltage. (2)The two bulbs are equally bright. (3)The two bulbs and the battery have all the same current through them. (4)All of the above must be true. (5)None of the above must be true. Note : For electrical circuits, wires connecting circuit elements are assumed to have zero resistance.

31 31 If the four bulbs in the figure are identical and the batteries are identical, which circuit puts out more light? (Remember brightness depends on the power transformed) (1)1 emits more light. (2)The two emit the same amount of light. (1)2 emits more light. R R RR  

32 32 (Remember brightness depends on the power transformed) From today’s lecture, For circuit1: R 1equivalent = R/2, so I 1 =  /R 1 equi. = (2  )/R As bulbs are identical, half of I 1 flows through each bulb.  Power bulb in circuit 1 = ((I 1 /2) 2 R = {  /R} 2 R =  2 /R <= gives the measure of brightness of a bulb in circuit 1 For circuit2: R 2equivalent = 2R, so I 2 =  /R 2equi. =  /(2R) and this I 2 flows through both bulbs. Power bulb in circuit 2 = I 2 2 R = (  /2R) 2 R =  2 /4R <= gives the measure of brightness of a bulb in circuit 2 If the four bulbs in the figure are identical and the batteries are identical, which circuit puts out more light? R R RR   * For a review of circuit analysis, refer to P.21 of the course text (Method of Equivalent Reduction)

33 33 If the four bulbs in the figure are identical and the batteries are identical, which circuit puts out more light? (Remember brightness depends on the power transformed) (1)1 emits more light. (2)The two emit the same amount of light. (1)2 emits more light.

34 34 If the two bulbs in the circuit 1 are NOT identical, which bulb puts out more light? (1)Bulb A emits more light. (2)The two emit the same amount of light. (1)Bulb B emits more light. RARA RBRB (Remember brightness depends on the power transformed) R A > R B Example Solve this for  = 6V, R A = 3 Ohm, R B = 1 Ohm

35 35 If the two bulbs in the circuit 1 are NOT identical, which bulb puts out more light? (1)Bulb A emits more light. (2)The two emit the same amount of light. (1)Bulb B emits more light. RARA RBRB (Remember brightness depends on the power transformed) R A > R B Example Solve this for  = 6V, R A = 3 Ohm, R B = 1 Ohm You should find, I A = 2A, I B = 6A, and P A = 12Watts, P B = 36Watts

36 36 Be sure to write your name, ID number & DL section!!!!! 1MR 10:30-12:50 Dan Phillips 2TR 2:10-4:30Abby Shockley 3TR 4:40-7:00John Mahoney 4TR 7:10-9:30Ryan James 5TF 8:00-10:20Ryan James 6TF 10:30-12:50John Mahoney 7W 10:30-12:50Brandon Bozek 7F 2:10-4:30Brandon Bozek 8MW 8:00-10:20Brandon Bozek 9MW 2:10-4:30Chris Miller 10MW 4:40-7:00Marshall Van Zijll 11MW 7:10-9:30Marshall Van Zijll


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