Carleton Engineering SCE Faculty A. Adler Courses MCG4150 Midterm 2006 |

## MCG4150: Midterm 2006
Directions:
You have 80 minutes to complete this exam.
The exam has four questions; you are required to answer any three of them.
Each question is worth equal marks.
This is a closed book exam; however,
you are permitted to bring a 8.5"×11"
sheet of notes into the exam.
You are permitted to use a calculator.
You may not communicate with anyone during the exam except the instructor.
You may make assumptions to simplify the problems as long a they don't change the calculations by more than 10%. You may assume the following conditions and constants: - Atmospheric pressure, P
_{atm}: 101.3 kPa - Density of air at P
_{atm}: 1.21 kg/m^{3} - 1 cmH
_{2}O = 98.5 Pa - 760 mmHg = 101.3 kPa
- 1 Amp= 6.24×10
^{18}e^{−}/s
### 1. Flow Transducer DesignsConsider the following types of air flow transducers: −Pneumotachometer (which measures pressure differences accross a flow resistance) −Hot-wire anemometer (which measures the current required to maintain a thermistor at constant temperature) −Rotating-vane flowmeter (which measures rotation of a turbine in the air stream)**1A.****Sketch each transducer type and describe how it works**.**1B.****Compare each type in terms of: 1) accuracy, 2) frequency response, 3) linearity of the underlying sensor eleement, 4) directional response (response to flow in different directions)**### 2. Flow Transducer Frequency LimitsIn this question, we want to compare the frequency response of the hot-wire anemometer and pneumotachometer.**2A.**The pneumotachometer can be modelled as a 15 cm long tube of inner diameter 3 cm. Atmospheric pressure is 101.3 kPa.**Calculate the compliance and inertance of the gas in pneumotachometer.****answer:**Compliance = V/P_{atm}= πr²L/P_{atm}= π*(0.015m)²*(0.15m)/101300Pa = 1.0467e-009 m^5/N Inertance = ρ*L/π/r² = (1.21 kg/m³)*0.15m/(π*(0.015m)²) = 256 kg/m^4**2B.**The hot-wire anemometer acts as a first order system with time constant τ=0.5s. The pneumotachometer acts as a second order system with natural frequency*f*_{n}.**Compare the frequency response of the hot-wire anemometer and pneumotachometer. Which is the highest?****answer:**ω_{n}= (L_{c}C_{c})^{−½}=( 1.0467e-009m^5/N * 256 kg/m^4)^(-.5) =( 2.6876e-007 m-kg/N)^(-.5) = (2.6876e-007 m-kg-s²/kg-m)^(-.5) =1929 1/s*f*_{n}= ω_{n}/2/π= 307 Hz. pneumotachometer is much higher freq than anemometer### 3. Electrode TypesConsider the following two electrode types: −Ag/AgCl electrode −Stainless steel (SS) electrode.**3A.****What is a polarizable electrode? Compare these two electrodes in terms of polarizablility.****3B.**An ECG amplifier is connected to an Ag/AgCl electrode and has an input resistance of 1 MΩ. The QRS peak of the ECG lasts 50 ms and has an average signal level of +20 mV.**How many CL**^{−}ions enter or leave the body from the electrode during the QRS peak?**answer:**V=IR I=V/R= 20mV/1MΩ= 20nA Electrons = (20e-9 A)*(6.24e18 e/s/A)*(50e-3 s) = 6.24e9 electrons Valence of CL^{−}is one. Thus 6.24e9 ions leave body**3C.**Consider a body which touches an exposed live wire carrying 120V at 60Hz..**Rank the following in terms of danger.** −Touching an SS electrode lead to the wire. −Touching an Ag/AgCl electrode lead to the wire. −Touching dry skin to the wire. −Touching wet skin to the wire.### 4. Nerve Conduction VelocityConsider the classic "knee-jerk" reflex test.**4A.****Sketch a diagram and describe the nerve signal conduction pathway during this test. What accounts for the time delay between the hammer tap and muscle contraction?****4B.**We wish to measure the time between hammer tap and muscle contraction.**Design a system to measure this time delay. Sketch your system and describe how it works?**
Last Updated: $Date: 2006-03-10 09:30:52 -0500 (Fri, 10 Mar 2006) $ |