|Carleton Engineering SCE Faculty A. Adler Courses MCG4150 Final 2006|
MCG4150: Final Exam 2006
Directions: You have 180 minutes (3 hours) to complete this exam. The exam has six questions; you are required to answer any five of them. Each question is worth equal marks. This is a closed book exam; however, you are permitted to bring an 8.5"×14" 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 as they don't change the calculations by more than 10%. You may (or may not) use the following conditions and equations for your calculations:
1. A total-body plethysmograph is used to measure the breathing of a mouse for a pharmaceutical company testing asthma medications. The box is designed to fit tightly around the mouse with just the nose breathing external air. (This works quite well, since mice feel comfortable in small holes) Mice breathe with their noses; assume no air goes through the mouth.
1A. In order to measure tidal volume (VT) we measure box temperature (Tb) and pressure (Pb). The box volume is Vb and mouse volume is Vm. Develop an equation for VT based on these variables, given end-inspiration pressure (Pb,i) and end-expiration pressure (Pb,e).
1B. One problem with this apparatus is that the mouse warms the air in the box over time. Briefly describe one way in which warming will impact measurement of absolute lung volume (ie. total air volume in the lung)
1C. In class we reviewed temperature measurement using thermistors and thermocouples. Briefly describe each and compare them in terms of linearity and speed of response. Which has the largest operating range?
2. One important technique for measurement of lung volume is the nitrogen-washout technique. A patient breathes a maximal breath of pure O2 and exhales slowly to residual volume into a fast gas analyser.
2A. Sketch a graph of the fE,N2 versus time for the breath for a normal patient. Label the dead space and the closing volume.
2B. Consider an ideal case where a patient exhales at a constant rate of 250 mL/s. fE,N2 is initially zero. After 2 s, fE,N2 then abrubtly changes from 0.0 to 0.15 for 18 s, at which point the breath stops. What is the dead space volume?, What is the vital capacity?
answer: Dead space is 2s*250ml/s = 500ml. Vital capacity is (2s+18s)*250ml/s = 5L
2C. Using the data from the previous question What is the residual volume?
answer: RC + 4.5L = V RC*.79 + 4.5L*0 = V*.15 --- RC*5.27 = V = RC + 4.5L RC = 4.5L / 4.27 = 1.05
2D. Absolute lung volumes can be measured by gas dilution techniques (like the one analysed here) or by total-body (box) plethysmograph techniques. If there are non-communicating gas volumes in the lung, then the values measured by these techniques will differ. Briefly, explain the difference in measurements. Which will be larger?
3. One interesting physiological question is the measurement of the mechanical efficiency of activities. Instrumentation for this measurement is called a calorimeter. In an experiment, we ask a person to ride a bicycle at a constant pedal rotation rate against a constant load.
3A. Indirect calorimetry measures the rate of consumption of oxygen by the subject. The subject breathes room air at 15 breaths per minute at a tidal volume of 2 L with an fe,O2 of 0.15. What is the rate of oxygen consumption (VO2)? Assume each litre of oxygen consumed releases 21.1 kJ of energy. What is the rate of energy consumption?
answer: VO2= (fio2 - feo2)xVTxfB VO2 = (0.21 - 0.15)*15*2 = 1.8L/min Energy = 21.1kJ/L*1.8L/min = 38.0 kJ/min 38.0kJ/min * (min/60s) = 0.63 kW
3B. The subject is cycling at 120 rpm against a rotational load of 10.0 N-m. What is the mechanical efficiency of the work?
answer: Work= 120rpm x 2π x (min/60s) x 10.0 Nm = 0.126 kW Efficiency = 0.126/0.63 = 20%
3C. In order to do direct calorimetry, we wish to measure the total heat generated. To do this, we blow air through the experimental chamber at a constant rate and measure the change in air temperature. (The exercise bike is in the chamber, so the heat from its work contributes to the total) Air flows through the chamber at 200 L/s, and temperature changes from 20°C to 22°C. The heat capacity of dry air is 1.005 kJ/kg/K. (Note that we don't account for evaporation of sweat) What is the rate of energy consumption? Would this result increase or decrease if our model included the evaporation of sweat?
answer: Work = 200L/s*(1.21g/L)*(1.005J/g/K)*2K= 486W
3D. Describe direct calorimetry in terms of whether it is a zero-, first- or second-order system. Is there a time delay between the activity and measurement? Discuss two reasons why or why not.
4. Many different instruments have been invented to measure arterial oxygen. Consider:
5. Electrical measurement of muscle activity.
5A. In order to obtain improved measurements of the ECG, one can put an electrode into an esophageal sensor. The advantage is that 1) the esophagus passes very close to the heart, and that 2) esophagus does not sweat, or have a layer of dead cells like the skin. Briefly discuss each effect, and why it allows improved measurements
5B. The electrical signal from heart muscle is the ECG; the electrical signal from skeletcal muscle is the EMG. Interestingly, the ECG and EMG are very different. The ECG is a structured, low frequency signal, while the EMG is a stochastic, high frequency signal. Briefly explain this difference, based on the different physiology of the muscle types.
5C. The pressure pulse from the systolic contraction propagates at a finite speed in the body, known as the pulse transit time (PTT). Using an ECG sensor and a stethoscope, design a system to measure the PTT to the wrist. Sketch a block diagram of your system, sketch the waveforms you expect in each instrument, and briefly discuss how it functions.
6. As a person ages, the systolic blood pressure increases, largely due to stiffening of the aorta. In order to measure this stiffening, we calculate the aortic compliance.
6A. One study (Watkins et al., Arch Int Pharmacodyn Ther., 1988) measured aortic compliance in units of ACU, where they defined the aortic compliance unit (ACU) to be an expansion in diameter of 0.001 mm/mmHg. If we consider the aorta to be cylindrical tube of 20 cm length, with a 1 cm inner diameter, calculate the conversion factor between compliance in ACU and in units of m³/Pa
answer: Compliance = ΔV/ΔP Change in area = 2πR dR ΔA/mmHg = 2π(.005m)(1e-6m) = 3.14e-8 m²/mmHg C= (3.14e-8 m²/mmHg)(760mmHg/101300Pa)(.2m) 1 ACU = 4.71e-11 m³/Pa
6B. Using this cylindrical model for the aorta, we can analyse it as a second order system and calculate the natural frequency and damping ratio. If the compliance of the aorta is 5.0 ACU at age 20 and 4.0 ACU at age 40, what is change in natural frequency between these ages?
answer: Age 20 => Compliance = 5 ACU = 2.34e-10 m³/Pa Age 40 => Compliance = 4 ACU = 1.89e-10 m³/Pa Inertance = ρ*L/π/r² = (1060 kg/m³)*(.2m)/π/(.005m)² = 2.70e6 kg/m^4 Natural freq = 1/2/π/√(LC) = 1/2/π/√( 2.34e-10 m³/Pa × 2.70e6 kg/m^4 ) = 6.33 Hz (Age 20) = 1/2/π/√( 1.89e-10 m³/Pa × 2.70e6 kg/m^4 ) = 7.05 Hz (Age 40)
6C. Cardiac output may be measured by thermodilution. Consider a system to measure left ventricular output by injecting cold saline into the pulmonary vein, and measuring the temperature at the end of the aorta. (it would be difficult to position the catheter in this position, but ignore that difficulty for this question). Briefly sketch the temperature profile measured at the temperature sensor, and describe how it would be possible to measure left ventricular output this way.
6D. Describe what (if any) differences changes in aortic compliance would make to the temperature profile from thermodilution.
Last Updated: $Date: 2008-10-20 18:46:08 -0400 (Mon, 20 Oct 2008) $