Example of a Challenge Organized Around the Legacy Cycle

1. Example of a Challenge Organized Around the Legacy Cycle Course: Biotransport Challenge: Post-mortem Interval

2. The Challenge: Estimate the Time of Death As a biomedical engineer, you are called to testify as an expert witness on behalf of the defendant, who is accused of murder. The body of her boyfriend was found at 5:30 AM in a creek behind her house. The prosecutor’s expert witness places the time of death at about midnight. The defendant has witnesses that account for her whereabouts before 11 PM and after 2 AM, but she cannot provide an alibi for the period between 11 PM and 2 AM.

3. Generate Ideas How did the prosecutor’s expert witness arrive at the time of death? What information will you need to challenge the time of death estimate? Discussion Results: How? Rate of Body Cooling. Info? Temperature measurements

4. Research and Revise Examination of Assumptions

5. Model and Data used by forensic pathologist to estimate the time of death: • Body temperature at 6 AM (rectal) = 90.5°F • Ambient Temperature = 65°F • Body removed to coroner’s office (65°F) • Body temperature at 8 AM = 88.3°F • Assumed pre-death body temperature = 98.6°F How did the coroner arrive at midnight as the time of death? K = ?

6. Thermal Energy Balance on Body:Macroscopic Analysis Newton’s Law of Cooling neglect internal resistance to heat transfer: Tcore = Tsurface = T Rate of Accumulation of Thermal Energy Thermal Energy entering body Thermal Energy leaving body Rate of Production of Thermal Energy - + = 0 +0

7. Are there any assumptions made in deriving the equation used by the pathologist that may be inappropriate for this case?

8. Your own Investigation • You visit the crime scene. What will you do there? • You visit the coroner’s office. What information do you request? • Any other information you might need?

9. Investigation determines: • When found, body was almost completely submerged • Body was pulled from the creek when discovered at 5:30 AM • Creek water temperature was 65°F • No detectable footprints other than the victim’s and the person that discovered the body. • Water velocity was nearly zero. • Victim’s body weight = 80 kg • Victim’s body surface area = 1.7 m2 • Cause of death: severe concussion • Medical Records: victim in good health, normal body temperature = 98.6ºF

10. Your investigation also reveals typical heat transfer coefficients: Heat transfer from a body to a stagnant fluid (W/(m2°C)) • h for air: 2 – 23 • h for water: 100 - 700 • Based on these coefficients, you might expect temperature of a body in stagnant water at 65°F to fall at: • About the same rate as in air at 65°F • At a faster rate than in air at 65°F • At a slower rate than in air at 65°F

11. New estimate of time of death Provide a procedure that can be used to find the time of death assuming that: • the body was in the creek (h = 100 W/m2ºC) from the time of death until discovered at 5:30 AM. • the body was removed from the creek at 5:30 AM and body temperature measurements made at 6 AM & 8 AM while the body cooled in air (h = 2.46 W/m2ºC).

12. Summary: Macroscopic Approach (Lumped Parameter Analysis) • Time of death estimated by coroner assuming cooling in air was about midnight (guilty!) • Time of death estimated by your staff assuming initial cooling in water was about 5:22 AM (innocent!). T=98.6°F T(5:30 AM) T=91.1°F T=90.5°F T=88.3°F 12 1 2 3 4 5 6 7 8 AM

13. 0 25 of 31 An estimate of Post Mortem Interval (PMI) based on hwater using this method is probably: • Accurate • Too long • Too short

14. The prosecutor gets wise and hires a biomedical engineer! • Your model prediction is criticized because a lumped analysis (macroscopic) was used. • The witness states that: • internal thermal resistance in the body cannot be neglected. • the body takes longer to cool than you predicted . • body temperature varies with position and time.

15. (leg) T vs t from different regions Single study

16. How can we find the ratio of internal to external thermal resistance for heat transfer from a cylinder? T∞ TS = TR R conduction to surface Tc L conduction & convection from surface

17. Biot Number (Bi) If Bi<0.1, we can neglect internal resistance (5%) If Bi >0.1, we should account for radial variations (low external resistance or high internal resistance) cylinder:

18. Cooling of Cylindrical Body: Assume Radial Symmetry We wish to find how temperature varies in the solid body as a function of radial position and time. Evaluate equation term by term h R T TR T(r,t) Apply assumptions: Apply boundary & initial conditions:

19. Cooling of a CylinderCenterline Temperature vs. Time Assuming Centerline Temperature = Rectal Temperature: Design a procedure to find the time of death from this chart x1 = R;  = (k/ρCp)body; m = 1/2Bi = kbody/hR

20. Using the Graphical Solution to Estimate the Time of Death. Core Temperature at 5:30 AM = 91.1°F (Tc-T∞)/(T0-T∞)=(91.1-65)/(98.6-65)=0.777 m=k/hR=0.5/(100 x .15) = 0.033 Fo = 0.12 t = FoR2/ = (.12)(.15m)2/(.54x10-3 m2/hr) = 5 hr Time of death = 12:30 AM +/- Guilty!

21. Should the Defense Rest? Are there any other confounding factors? Different radius Different h Not a cylinder

22. Module Summary • Models are valuable for predicting important biomedical phenomena • Models are only as accurate as the information provided and the validity of the assumptions made.