Example of a Example of a Challenge Organized Challenge Organized Around the Legacy Around the Legacy

CycleCycle

Course: BiotransportCourse: Biotransport

Challenge: Post-mortem IntervalChallenge: Post-mortem Interval

The Challenge: The Challenge: Estimate the Time of DeathEstimate 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.

Generate IdeasGenerate 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

Research and Research and ReviseRevise

Examination of AssumptionsExamination of Assumptions

Model and Data used by forensic Model and Data used by forensic pathologist to estimate the time pathologist to estimate the time

of death:of death:

How did the coroner arrive at midnight as the time of death? K = ?

( )adT

K T Tdt

Body temperature at 6 AM (rectal) = 90.5Body temperature at 6 AM (rectal) = 90.5°F°F Ambient Temperature = Ambient Temperature = 6565°F°F Body removed to coroner’s office (Body removed to coroner’s office (6565°F)°F) Body temperature at 8 AM = 88.3°FBody temperature at 8 AM = 88.3°F Assumed pre-death body temperature = 98.6°FAssumed pre-death body temperature = 98.6°F

Thermal Energy Balance on Body:Thermal Energy Balance on Body:Macroscopic AnalysisMacroscopic Analysis

Rate ofAccumulation of

ThermalEnergy

=

ThermalEnergyentering

body

-

ThermalEnergyleavingbody

+

Rate ofProductionof Thermal

Energy

dt

dTmCp ( )ahS T T 0 +0

( )ahS T T

Newton’s Law of Cooling

neglect internal resistance to heat transfer: Tcore = Tsurface = T

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

case?case?

Your own InvestigationYour own Investigation

You visit the crime scene. What will you You visit the crime scene. What will you do there?do there?

You visit the coroner’s office. What You visit the coroner’s office. What information do you request?information do you request?

Any other information you might need?Any other information you might need?

Investigation determines:Investigation determines: When found, body was almost completely When found, body was almost completely

submergedsubmerged Body was pulled from the creek when discovered Body was pulled from the creek when discovered

at 5:30 AMat 5:30 AM Creek water temperature was 65Creek water temperature was 65°F°F No detectable footprints other than the victim’s No detectable footprints other than the victim’s

and the person that discovered the body.and the person that discovered the body. Water velocity was nearly zero.Water velocity was nearly zero. Victim’s body weight = 80 kgVictim’s body weight = 80 kg Victim’s body surface area = 1.7 mVictim’s body surface area = 1.7 m22

Cause of death: severe concussionCause of death: severe concussion Medical Records: victim in good health, normal Medical Records: victim in good health, normal

body temperature = 98.6body temperature = 98.6ºFºF

Your investigation also reveals Your investigation also reveals typical heat transfer coefficients: typical heat transfer coefficients:

Heat transfer from a body to a Heat transfer from a body to a stagnant fluid (W/(mstagnant fluid (W/(m22 °C)) °C))

h for air: h for air: 2 2 – 23 – 23 h for water: 100 - 700h 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

New estimate of time of deathNew estimate of time of death

Provide a procedure that can be used to find Provide a procedure that can be used to find the time of death assuming that:the time of death assuming that:

the body was in the creek (h = 100 the body was in the creek (h = 100 W/mW/m22ºC) from the time of death until ºC) from the time of death until discovered at 5:30 AM.discovered at 5:30 AM.

the body was removed from the creek at the body was removed from the creek at 5:30 AM and body temperature 5:30 AM and body temperature measurements made at 6 AM & 8 AM measurements made at 6 AM & 8 AM while the body cooled in air (h = 2.46 while the body cooled in air (h = 2.46 W/mW/m22ºC).ºC).

Summary: Macroscopic Approach Summary: Macroscopic Approach (Lumped Parameter Analysis)(Lumped Parameter Analysis)

Time of death estimated by coroner assuming Time of death estimated by coroner assuming cooling in air was about midnight cooling in air was about midnight (guilty!)(guilty!)

Time of death estimated by your staff Time of death estimated by your staff assuming initial cooling in water was about assuming initial cooling in water was about 5:22 AM 5:22 AM (innocent!)(innocent!)..

0

p

hSt

mCa

a

T Te

T T

2 3 4 5 6 7 8 AM

T=98.6°F

T=91.1°F

T=90.5°FT=88.3°F

0

ln a

a

T T

T T

112

T(5:30 AM)

An estimate of Post Mortem Interval An estimate of Post Mortem Interval (PMI) based on h(PMI) based on hwaterwater using this using this

method is probably:method is probably:

Acc

urate

Too lo

ng

Too s

hort

11

13

1

1.1. AccurateAccurate

2.2. Too longToo long

3.3. Too shortToo short

The prosecutor gets wise and The prosecutor gets wise and hires a biomedical engineer!hires a biomedical engineer!

Your model prediction is criticized because a Your model prediction is criticized because a lumped analysis (macroscopic) was used.lumped analysis (macroscopic) was used.

The witness states that:The witness states that: internal thermal resistance in the body cannot be internal thermal resistance in the body cannot be

neglected.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.

T vs t from different regionsT vs t from different regions

Single study

(leg)

How can we find the ratio of internal How can we find the ratio of internal to external thermal resistance for to external thermal resistance for

heat transfer from a cylinder?heat transfer from a cylinder?

int

c R R

ernal external

T T T TQ

R R

Tc

TS = TR

T∞

R

L

conduction to surface

c Rr R body body

r R

c Rr R body

internalbody

T TdTq k k

dr R

T TQ Sq Sk

R

RR

Sk

conduction & convection from surface

1

R

external

Q hS T T

RhS

Q

Biot Number (Bi)Biot Number (Bi)

If Bi<0.1, we can neglect internal resistance (5%)If Bi<0.1, we can neglect internal resistance (5%)

If Bi >0.1, we should account for radial variations If Bi >0.1, we should account for radial variations (low external resistance or high internal (low external resistance or high internal resistance)resistance)

body

hRBi

2kcylinder:

152(0.5)

)(100)(0.15

2k

hRBi

body

Cooling of Cylindrical Body: Cooling of Cylindrical Body: Assume Radial SymmetryAssume Radial Symmetry

2 2

p p r z 2 2 2

vT T T T 1 T 1 T Tc - c v v k r S

t r r z r r r r z

0vr 0

z

v z()

0v

0

()

0z

()0S

R T

h

TR

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)

1 p

T Tc k r

t r r r

Apply boundary & initial conditions:

00

( ( ) ), 0, (0)

R r

T Tk h T R T T Tr r

Cooling of a CylinderCooling of a CylinderCenterline TemperatureCenterline Temperature vs. Time vs. Time

x1 = R; = (k/ρCp)body; m = 1/2Bi = kbody/hR

Assuming Centerline Temperature = Rectal Temperature:Design a procedure to find the time of death from this chart

Using the Graphical Solution to Using the Graphical Solution to Estimate the Time of Death.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

Time of death = 12:30 AM +/-

Fo = 0.12t = FoR2/ = (.12)(.15m)2/(.54x10-3 m2/hr) = 5 hr

Guilty!

Should the Defense Should the Defense Rest?Rest?

Are there any other confounding Are there any other confounding factors?factors?

Different radiusDifferent hNot a cylinder

Module SummaryModule Summary

Models are valuable for predicting Models are valuable for predicting important biomedical phenomenaimportant biomedical phenomena

Models are only as accurate as the Models are only as accurate as the information provided and the validity of the information provided and the validity of the assumptions made.assumptions made.