17. Assignments thermoregulation (15) Flashcards
ME scheme: what does it look like? For thermoregulation assignments
ME
-> maintenance (heat 100%)
-> Production (product X%, Heat 100-X%)
-> Work (Power 25%, heat 75%)
Pig = 80 kg
Calculate total heat load of this pig, given..
- met weight = 80^0.75 = 26.7
- feed intake = 2.5 ME
- Production efficiency = 70%
Q = What is the total heat load?
BMR = brody formula = 400*80^0,75 = 10,680 kj/day = MEm
Food intake = 2.5ME = 2.510,680 = 26,700 kj/day
Work is not mentioned. Maintenance is 1*ME
Production is therefore 2.5-1 = 1.5x ME
Product efficiency = 70%, heat load is therefore 30%.
1.5 * 10,680 = 16020 kj. 30% = 4806 kj
+ 1*10,680 = 15486 kJ
Pig = 80 kg
Calculate total heat load of this pig, given..
- met weight = 80^0.75 = 26.7
- feed intake = 2.5 ME
- Production efficiency = 70%
Q = What is the composition of the heat, product and feed load?
Maintenance heat = 1ME
Production heat = 0.3 1.5 = 0,45ME
(total = 1,45)
Product = 0.7 1.5 = 1.05ME
Feed = 2.5ME
Pig = 80 kg
Calculate total heat load of this pig, given..
- met weight = 80^0.75 = 26.7
- feed intake = 2.5 ME
- Production efficiency = 70%
Q = How much O2 is consumed by the pig?
Thumb rule: 1 L O2 = 20 kJ/L
Oxygen is not needed to store nutrients. Product is not used for oxygen consumption.
You use the value of heat to calculate the total amount of oxygen consumed.
Energy of heat = 15,486 (kJ/day)/20(kJ/L) = 744 liter O2/day
1 kcal = … joules
4200
What is a formula to calculate the time (min) required for a body to increase its temperature with 1*C?
time (min) = 15*W^0.25 (weight in kg)
E.g. One-day chicken of 50 gr:
0.05 kg = 15*0.05^0.25 = 7.1 min
The larger you are, the less/more time you relatively need to warm up because …
you have way more heat production from the volume, compared to loss which is related to the surface area.
Available feed = 2ME.
What is the maximal energetic efficiency for production? When is it maximal? Use heat diagram p. 164
1ME for maintenance
Therefore 1ME for product
0.6ME = 60% for product,
04ME = for product heat + 1ME = 1.4ME for total heat.
Use heat diagram p. 164: what is the LCT for maintenance?
20*C
Is Thermoneutral zone fixed for a certain level of heat production (no species differences!)? True/false
False
Thermoneutral zone does not rely on environmental factors: true/false
False
Heat production is constant throughout the thermoneutral zone. True/false
True
Heat loss is constant throughout thermoneutral zone: true/false
False (once you pass your body temperature, besides your normal standard heat production, there is more heat load from the environment. Total heat load is more than you produce by yourself. Heat loss also needs to increase, by evaporation)
Is Heat production the same at all LCT’s?
No
Is Heat loss the same at all UCT’s?
YES
All UCP’s are on this line (B0 – Bx): on every UCP there is the same heat loss.
Heat loss at UCP is always at maximum evaporation capacity
Cyclist: maintenance 7kJ/min
UCT = 44 *C
Max evaporation = 22.4 ml/min
External work (power) = 14 kJ/min
Max skin temp = 37 *C
Heat of evaporation = 2.5 kJ/gr H2O
- Calculate heat production (times MEm)
Power = 14 kj/min = 25% of 2ME (of 14)
100% = 144 = 56 kj/min
Maintenance = 7 kj/min = 1*ME
Nothing is said about production, therefore we do not take this into account.
7 + 75% * 56 = 49 kj/min
Cyclist: maintenance 7kJ/min
UCT = 44 *C
Max evaporation = 22.4 ml/min
External work (power) = 14 kJ/min
Max skin temp = 37 *C
Heat of evaporation = 2.5 kJ/gr H2O
- Calculate energy expenditure
total EE = 56 + 7 = 63 kJ/min
Cyclist: maintenance 7kJ/min
UCT = 44 *C
Max evaporation = 22.4 ml/min
External work (power) = 14 kJ/min
Max skin temp = 37 *C
Heat of evaporation = 2.5 kJ/gr H2O
- Calculate max capacity for evaporation
heat of evap = 2.5 kj/gr H20
Max evap = 22.4 ml/min
2.5 * 22.4 = 56 kj/min
Cyclist: maintenance 7kJ/min
UCT = 44 *C
Max evaporation = 22.4 ml/min
External work (power) = 14 kJ/min
Max skin temp = 37 *C
Heat of evaporation = 2.5 kJ/gr H2O
- Determine (graphically) max Ta that allows long lasting activity
Look at all max values, look at the MEm the cycler is at and determine temperature when looking at the UCT line.
Cyclist: maintenance 7kJ/min
UCT = 44 *C
Max evaporation = 22.4 ml/min
External work (power) = 14 kJ/min
Max skin temp = 37 *C
Heat of evaporation = 2.5 kJ/gr H2O
- Determine external heat load
At cycling, look at total heat produced by cycler (7MEm) and max evaporative capacity (8ME)
8-1 = 1x MEm is external heat load (7kJ/min)
What can happen to the chemical (gross) energy of nutrients after ingestion?
Loss in faeces, ME anabolic + catabolic processes -> ATP, excretion of N via urea in urine
- What can happen to the energy trapped in ATP?
ATP: anabolic: storage, synthesis, heat, exercise
Which metabolic processes may contribute to extrta heat production at low environmental temperatures?
- Shivering thermogenesis: muscle contraction
- Non-shivering thermogenesis:
Metabolic basis (UCP: uncoupling proteins) and
Substrate cycling and oxidative processes (change of substrate to have higher release of heat)
What is the heat capacity of the human body?
4.2 J/g/*C
RQ (CO2/O2) of normal diet = ?
0.8 (protein is 0.82)
