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 prosecutors 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 prosecutors 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: How did the coroner arrive at midnight as the time of death? K = ? Body temperature at 6 AM (rectal) = 90.5°F Body temperature at 6 AM (rectal) = 90.5°F Ambient Temperature = 65°F Ambient Temperature = 65°F Body removed to coroners office (65°F) Body removed to coroners office (65°F) Body temperature at 8 AM = 88.3°F Body temperature at 8 AM = 88.3°F Assumed pre-death body temperature = 98.6°F Assumed pre-death body temperature = 98.6°F

6. Thermal Energy Balance on Body: Macroscopic Analysis Rate of Accumulation of Thermal Energy = Thermal Energy entering body - Thermal Energy leaving body + Rate of Production of Thermal Energy 0+0 Newtons Law of Cooling neglect internal resistance to heat transfer: T core = T surface = T

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 crime scene. What will you do there? You visit the coroners office. What information do you request? You visit the coroners office. What information do you request? Any other information you might need? Any other information you might need?

9. Investigation determines: When found, body was almost completely submerged When found, body was almost completely submerged Body was pulled from the creek when discovered at 5:30 AM Body was pulled from the creek when discovered at 5:30 AM Creek water temperature was 65°F Creek water temperature was 65°F No detectable footprints other than the victims and the person that discovered the body. No detectable footprints other than the victims and the person that discovered the body. Water velocity was nearly zero. Water velocity was nearly zero. Victims body weight = 80 kg Victims body weight = 80 kg Victims body surface area = 1.7 m 2 Victims body surface area = 1.7 m 2 Cause of death: severe concussion Cause of death: severe concussion Medical Records: victim in good health, normal body temperature = 98.6ºF 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/(m 2 °C)) h for air: 2 – 23 h for air: 2 – 23 h for water: 100 - 700 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: 1.About the same rate as in air at 65°F 2.At a faster rate than in air at 65°F 3.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/m 2 ºC) from the time of death until discovered at 5:30 AM. the body was in the creek (h = 100 W/m 2 ºC) 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/m 2 ºC). 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/m 2 ºC).

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 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!). Time of death estimated by your staff assuming initial cooling in water was about 5:22 AM (innocent!). 23 45678 AM T=98.6°F T=91.1°F T=90.5°F T=88.3°F 112 T(5:30 AM)

13. An estimate of Post Mortem Interval (PMI) based on h water using this method is probably: 1. Accurate 2. Too long 3. Too short

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

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

16. How can we find the ratio of internal to external thermal resistance for heat transfer from a cylinder? TcTc T S = T R T R L conduction to surface 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 R T h TRTR Apply assumptions: We wish to find how temperature varies in the solid body as a function of radial position and time. Evaluate equation term by term T(r,t) Apply boundary & initial conditions:

19. Cooling of a Cylinder Centerline Temperature vs. Time x 1 = R; = (k/ρC p ) body ; m = 1/2Bi = k body /hR Assuming Centerline Temperature = Rectal Temperature: Design a procedure to find the time of death from this chart

20. Using the Graphical Solution to Estimate the Time of Death. Core Temperature at 5:30 AM = 91.1°F (T c -T )/(T 0 -T )=(91.1-65)/(98.6-65)=0.777 m=k/hR=0.5/(100 x.15) = 0.033 Time of death = 12:30 AM +/- Fo = 0.12 t = FoR 2 / = (.12)(.15m) 2 /(.54x10 -3 m 2 /hr) = 5 hr 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 valuable for predicting important biomedical phenomena Models are only as accurate as the information provided and the validity of the assumptions made. Models are only as accurate as the information provided and the validity of the assumptions made.