Homework 1 2015 Fall solutions.pdf

# Homework 1 2015 Fall solutions.pdf - Engineering 100...

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Unformatted text preview: Engineering 100 Homework Homework 1 Date Issued: Thursday, September 10 Due Date: Thursday, September 17 (11:40 am in class) Notes on homework: • Homework does not need to be in a formal report format • Put your name in the upper right hand corner • Title the assignment “Homework X” • Homework should be neat and legible • May be handwritten, typed, or a combination of the two. • Show all work for calculations, including any supporting data, programming code, etc, from Excel, Matlab, or other software used for computation. • If your homework covers more than 1-­‐page, it should be stapled. It is your responsibility to staple before class, we do not carry staplers (do not ask us for one)! • Failure to comply with the above will result in point deductions 1) Perform the following units conversions and indicate the type of unit. (e.g. length, volume, etc). You likely will not know the conversion factors from memory; look them up and indicate conversion factors used. a. 5 inches = ______ mm 5 inches x (2.54 cm/inch) x (10 mm / 1 cm) = 127 mm Unit of length b. 2 gallons = ______ liters 2 gallons x (3.785 gallons/liter) = 7.57 liters Unit of volume c. 60 miles per hour = ______ meters/second 60 miles/hr x (1609 m / mile) x (1 hr / 3600 s) = 26.82 m/s Unit of speed d. 100 kilowatt hour (kWh) = ______ joules 100 kWh x (1000 W/kW) x (3600 s/h) = 3.6x108 J Unit of energy e. 300 horsepower = ______ kilowatts 300 hp x (0.7457 kW/hp) = 223.7 kW, Unit of power 1 Engineering 100 Homework 2) An RTD (temperature sensor) has a temperature coefficient of 0.004/oC and resistance of 106 ohms at 20 oC. a. What is its resistance at 25oC? For each oC of temperature increase, the resistance will change 0.004 of its nominal value. At 25 oC, the resistance will be 106 + 5(0.004)(106) = 108.12 ohms. b. If the RTD is placed in the circuit below, what would be the voltage Vout at 25oC? Assume that R1, R2, and R3 are all 106 ohms. The current on the left side will be 10V/(R1+R3) = 10V/(212 ohms) = 0.0472 A The current on the right side will be 10V/(R2+RTD) = 10V/(214.12 ohms) = 0.0467 A Vout will be the difference between the voltage across RTD and voltage across R3. Vout = (I_RTD)(RTD) -­‐ (I3)(R3) = (0.0467)(108.12) -­‐ (0.0472)(106) = 0.0460 V 3) Calculate the unknown current I in the circuit below, and calculate the power absorbed or supplied by each element in the circuit. 2 Engineering 100 Homework From KCL, I = 4A – 1A = 3A P (4 A source) = 4 x 9 = 36 W supplied P (9 V element) = 1 x 9 = 9 W absorbed P (3 V element) = 3 x 3 = 9 W absorbed P (6 V source) = 3 x 6 = 18 W absorbed Note power is conserved. The 6 V source (battery) is absorbing power, this is bad! 4) Calculate Vo for the circuit below This is a series circuit, the current is the same everywhere. We can find the current I using KVL. -­‐9 + 6I + 2I + 3 = 0 I = 6/8 A Now that we know I, we can determine Vo (9V battery minus drop across 6 ohm resistor, or drop across 2 ohm resistor and 3V battery) Vo = 2 (6/8) + 3 = 4.5 V 5) Define the following terms related to data collection and analysis: a. Accuracy How close the measurement data is to the actual value b. Precision The predictability in the measurement. Precision determines the number of significant figures used in the measurement. c. How can you represent the uncertainty in accuracy and precision in tables and graphs? Use error bars to represent uncertainty in accuracy, choose significant figures according to the precision. 3 ...
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