NP15 – Thermoregulation: how to use the heat
diagram and partitioning of ME
Heat diagram summary
• A and b defines metabolic border where
metabolism is at 1x maintenance
§ Line linking all As = line of passive heat
loss
§ Line linking all Bs = line of max
evaporation
§ TNZ can be beyond Tb (dotted line), e.g.
B0 is beyond Tb
§ After Bs, ME increases because active
sweating is implemented (uses ATP for more
heat loss)
A1 – Growing pigs (80 kg)
Calculate the total heat load of this pig, given:
metabolic weight = 800.75 = 26.7 Feed intake 2.5 MEm (eat a lot on a day)
Production efficiency = 70%
à take into account all the factors (start with the partitioning of energy )
§ We assumed no work from the pig
§ We have 2.5 ME à 1 ME for maintenance (at least once) (energy for maintenance is 100% as heat)
and 1.5 ME for production (of which 30% heat in production = 45% heat)
§ HEAT PROD: BMR = 400 * 26.7 = 10,680 kJ/day = MEm (maintenance or basal metabolic rate)
§ FEED INTAKE: 2.5 MEm = 2.5 * 10,680 = 26,700 kJ/day
à Total heat: 1 MEm + 0.45 MEm = 1.45 MEm = 1.45 * 10, 680 = 15, 486 kJ/day
What is the composition of the heat load?
Heat: 15,486 kJ/day = 1.45 * MEm
Product: 11,214 kJ/day = 1.05 * MEm
Feed: 26,700 kJ/day = 2.5 * MEm = heat + product
How much oxygen in consumed by the pig? L of O2?
You don’t use oxygen for the product à oxygen only related to the heat production
To have it in litre, use 20kJ/L O2 à 15 486 (kJ/day) / 20 (kJ/L) = 774 L O2/day
Brody formula (maintenance): BMR = H (kJ/day) = 400 * W0.75 = MEm
à applies to all animals to compare species
A2 – Brody formula
Use the Brody formula to develop a formula to calculate the time (min) required for a body to increase its
temperature with 1°C (assumed complete insulation)
Calculate the time required for a:
- One day chick (50g) 0.05
- Rat (500g) 0.5
- Baby (5kg) 5
- Adult human (50kg) 50
diagram and partitioning of ME
Heat diagram summary
• A and b defines metabolic border where
metabolism is at 1x maintenance
§ Line linking all As = line of passive heat
loss
§ Line linking all Bs = line of max
evaporation
§ TNZ can be beyond Tb (dotted line), e.g.
B0 is beyond Tb
§ After Bs, ME increases because active
sweating is implemented (uses ATP for more
heat loss)
A1 – Growing pigs (80 kg)
Calculate the total heat load of this pig, given:
metabolic weight = 800.75 = 26.7 Feed intake 2.5 MEm (eat a lot on a day)
Production efficiency = 70%
à take into account all the factors (start with the partitioning of energy )
§ We assumed no work from the pig
§ We have 2.5 ME à 1 ME for maintenance (at least once) (energy for maintenance is 100% as heat)
and 1.5 ME for production (of which 30% heat in production = 45% heat)
§ HEAT PROD: BMR = 400 * 26.7 = 10,680 kJ/day = MEm (maintenance or basal metabolic rate)
§ FEED INTAKE: 2.5 MEm = 2.5 * 10,680 = 26,700 kJ/day
à Total heat: 1 MEm + 0.45 MEm = 1.45 MEm = 1.45 * 10, 680 = 15, 486 kJ/day
What is the composition of the heat load?
Heat: 15,486 kJ/day = 1.45 * MEm
Product: 11,214 kJ/day = 1.05 * MEm
Feed: 26,700 kJ/day = 2.5 * MEm = heat + product
How much oxygen in consumed by the pig? L of O2?
You don’t use oxygen for the product à oxygen only related to the heat production
To have it in litre, use 20kJ/L O2 à 15 486 (kJ/day) / 20 (kJ/L) = 774 L O2/day
Brody formula (maintenance): BMR = H (kJ/day) = 400 * W0.75 = MEm
à applies to all animals to compare species
A2 – Brody formula
Use the Brody formula to develop a formula to calculate the time (min) required for a body to increase its
temperature with 1°C (assumed complete insulation)
Calculate the time required for a:
- One day chick (50g) 0.05
- Rat (500g) 0.5
- Baby (5kg) 5
- Adult human (50kg) 50
