and some common sense….....heat transfer evap.(water spray) 50 55 60 65 70 75 80 egg weight (g)...

Poultry Performance Plus'Value added' advice for poultry production

Ron Meijerhof

Physics of incubation

Purpose of this presentation:

Not to scare you!

Not to turn you into engineers!

Not to impress you with how difficult it is!

But to show you that a lot can be calculated

A big part of incubation is basic physics

And some common sense…..

What does the embryo need?

Turning

Gas exchange

uptake of oxygen

loss of carbondioxide

Moisture loss

about 12-15%

Heat transfer

heat balance

Process controlled by temperature

but which temperature: external (air) or internal (embryo)?

C6H12O6

+

6O2

6CO2

+

6H20

+

e

Carbohydr/yolk fat

+

oxygen

carbon dioxide

+

water

+

heat

Air temperature and internal temperature are not equal

They are often only slightly related….

embryo temperature is result of a balance between

heat production

heat transfer temperature, air velocity, water spray, (heat capacity)

dT R.H

.

Air velo

Heat transfer

Evap.(water spray)

50 55 60 65 70 75 80

Egg weight (g)

0.0

0.5

1.0

1.5

2.0

2.5

Tegg -

Tair (

ºC)

0.5 m/s 2.0 m/s

Meijerhof and van Beek, 1993

Effect of air velocity on internal egg temperature (18 days)

Air velocity lowers temperature difference egg-shell (dT)

- because air velocity increases heat transfer

- Egg temperature depends on air velocity

Air velocity should be uniform!

50 55 60 65 70 75 80

Egg weight (g)

0.0

0.5

1.0

1.5

2.0

2.5

Tegg -

Tair (

ºC)

0.5 m/s 2.0 m/s

Meijerhof and van Beek, 1993

Effect of air velocity on egg shell temperature uniformity (18 days)

Air velocity

Tegg - tair

High heat production

Low heat production

Air velocity equalizes temperature!

Air velocity lowers temperature difference egg-shell (dT)

- because air velocity increases heat transfer

- Egg temperature depends on air velocity

Air velocity should be uniform!

Air velocity equalizes temperature difference egg-shell (dT)

- because warm eggs lose relatively more heat than cold eggs

- egg shell temperature uniformity depends on air velocity

Air velocity should be high

Especially when egg uniformity is low

- Different age flocks, different lines

Air velocity: not always recognized

transfer

E. O. Oviedo-Rondón et al, 2011

+ 80 g BW

Air velocity: not always recognized

transfer

E. O. Oviedo-Rondón et al, 2011

But is this an experiment about removing infertiles/deads?

or about increasing air velocity by creating space…

+ 80 g BW

But also: turning

Turning every hour or more: better results

• because air velocity equalizes hot/cold spots

Not turning (after 13-14 days): better results

• because of increase in air flow

• Depending on machine type

Smaller turning angle (38 vs 45): better results

• more space between trays, more air velocity

Evaporation: coolingEach gram of water costs 2.26 kJ energy to evaporate

Eggs lose 12% until 18 days

100,000 eggs x 60 g = 6000 kg

6000 kg x 12% = 720 kg (liter)

720 kg / 18 days / 24 hours = 1.7 l/hr

100,000 eggs => 1.7 litre moisture loss/hr

constant cooling of –0.3oF

Evaporation: cooling

Each gram of water costs 2.26 kJ energy to evaporate

100.000 eggs => 1.7 litre moisture loss/hr

constant cooling of –0.3oF

Ventilating 500 m3/hr (23oC, 50% R.H.)

6 litre of spraying water/hr

Constant cooling…

-1oF if all eggs equally contribute

but do all eggs contribute???

Egg size, EST and position in incubator

Position relative to fan

Egg size Near Far

Large (68.8 g) 38.7b 39.4a

Average (65.3 g) 38.8 38.8

Small (62.4 g) 38.6 38.8

Elibol and Brake, 2008

Egg size and position in incubator

Position Egg weight

Large Average Small

Early dead, % Near 3.93 4.24 4.52

Far 5.73 5.10 4.35

Mid dead, % Near 0.22 0.14 0.49

Far 0.57 0.43 0.28

Late dead, % Near 3.00b 3.96 2.25

Far 6.87a 3.79 3.57

Second grade, % Near 1.15b 1.91 1.55

Far 4.58a 2.13 2.51

Elibol and Brake, 2008

The main effect of spraying is cooling…

Not only the eggs, the thermostat as well…

Let’s do some calculations….

How much ventilation do we need?

100,000 eggs of 60g in machine

Single stage/multi stage

Let’s do some calculations….

How much ventilation do we need?

100,000 eggs of 60g in machine

Single stage/multi stage

Minimum ventilation for day old chicks:

0.36 m3/kg/hr

So 100,000 x 0.060 kg x 0.36 m3/kg/hr?

2160 m3/hr? No, not really….

First of all:

Eggs are not embryos

What is the weight of an embryo?

In a machine? MS (average age) vs SS (18 days)

In a hatchery….



And did it change over the years?



Romanoff (1938) Nangsuay et al (2015)

Day 1: 0.002 g

Day 6: 0.5 g

Day 9: 1.5 g

Day 11: 4 g 5 g

Day 14: 10 g 14 g

Day 18: 22 g 30 g



Romanoff (1938) Nangsuay et al (2015)

Day 1: 0,002 g

Day 6: 0,5 g

Day 9: 1,5 g

Day 11: 4 g 5 g

Day 14: 10 g 14 g

Day 18: 22 g 30 g

Yes, over the years eggs have become heavier

But also: the same egg size has a heavier embryo earlier

Maximum HP is still 0,15 W, but total HP is higher

So we better calculate with today’s embryos.



Nangsuay et al (2015)

Day 1:

Day 6:

Day 9:

Day 11: 5 g

Day 14: 14 g

Day 18: 30 g

So what are the dimensions for heat production, ventilation?




Day 1:

Day 6:

Day 9:

Day 11: 5 g

Day 14: 14 g

Day 18: 30 g

In SS machines:

Max capacity for

30 g, 18 days

MS machines and hatcheries:

Average capacity for

5-6 g, 10-12 days


So embryo weight instead of egg weight

We said: 0.36 m3/kg/hr ventilation needed

So 100,000 x 0.030 kg x

0.36 m3/kg/hr max?

1080 m3/hr?

and 100,000 x 0.006 kg x

0.36 m3/kg/hr average?

216 m3/hr?

MS machines and hatcheries:

Average capacity for

6 g, 10-12 days

In SS machines:

Max capacity for

30 g, 18 days

Solution:

Eggs are not embryos (not egg weight but embryo weight)

But are embryos equal to chicks?

they are not chicks, they are embryos!

Don’t walk, don’t play, don’t scream

Just sit in the egg and grow….

We know what eggs do

Embryos transfer nutrients into:

tissue, CO2, H2O and heat (fixed ratio)

If you know one, you know the others

Heat output (mJ) = 16.20 x O2 (l) + 5 x CO2 (l)

If we know heat output we know CO2 output

And we know max CO2 conc we want (4000 ppm)

We can calculate ventilation requirement

Eggs (+ air) give CO2, “diluted” with ventilation

500 ppm CO2 4000 ppm CO2

4000 ppm CO2

CO2

Solution:

CO2 production:

HP egg = 0.15 W

0.15 W is 0.15 J/s

mJ = 16.2 x O2 + 5 x CO2

0.15 J/s = is 1850 l CO2 / 100,000 eggs / hr

Ventilation:

Max CO2 concentration 4000 ppm

Incoming air 500 ppm

Theor. ventilation: 529 m3 / hr / 100,000 eggs (18 days)

(1,000,000 / (4000-500) x (1850/1000)

ppm/m3 diff in-out CO2 prod

So 500-550 m3/hr for 100,000 eggs at 18 days…

How does that compare to broilers?

100,000 broilers of 30 g

Minimum ventilation (O2 and CO2) required: 0.36 m3 / kg / hr

100,000 x 0.03 x 0.36 = 1080 m3/hr

So theoretically:

2 times more for a kg of bird than a kg of embryo…

Is this realistic?

Is this realistic?

Eggs 0.15 W max heat output

Day old chicks produce (literature)

0.3-0.35 W (without feed and water)

So yes, in theory doc produce 2 times more heat as embryos of same weight

and heat means CO2 means ventilation

Single stage:

100,000 eggs with max 30 g embryo

100,000 x 0.030 x 0.36 x 1/2 = 500 m3/hr (max)

But at day 1? And day 11?

Single stage:


100,000 x 0.030 x 0.36 x 1/2 = 500 m3/hr (max)



Day 1:

Day 6:

Day 9:

Day 11: 5 g

Day 14: 14 g

Day 18: 30 g

If day 18 = 30g = 500 m3/hr

day 11 = 5g = 500 x (5/30) = 83 m3/hr

Single stage:


100,000 x 0.030 x 0.36 x 1/2 = 500 m3/hr (max)


Multi stage:

100,000 eggs with average 5 g embryo

100,000 x 0.005 x 0.36 x 1/2 = 83 m3/hr (constant) ?

Single stage:


100,000 x 0,030 x 0.36 x 1/2 = 500 m3/hr (max)


Multi stage:

100,000 eggs with average 5 g embryo

100,000 x 0,005 x 0,36 x 1/2 = 83 m3/hr (constant) ?

Reality is 200 m3/hr, as growth rate is not lineair

So requirement for fresh air is relatively low…

Average 200 m3/hr for a machine of 100,000 eggs

But we do not ventilate only for oxygen/carbondioxide..

We also ventilate for humidity

We want to create 37.5oC and 55% R.H. in the machine

To let eggs lose 12% weight in 18 days

Outside air is 23oC and 50% R.H.

Lets assume we ventilate 200 m3/hr for 100.000 eggs

Question:

Do we have to add water (spray)?

How much?

Solution:

Water content of air: Mollier diagram

Solution:


Or: use your smartphone/Ipad/tablet….For instance Klimacalc (www.sercom.nl)

Solution:


- 23oC and 50% RH contains:

10 (10.3) g H20 per m3 (kg) of air

- 37.5oC and 55% RH contains:

25 (24.7) g H2O per m3 (kg) of air

- So we need to add 25-10=15 (14.4) g H20 per m3

Coming from the eggs and/or the sprayer

23oC , 50% RH

= 10 g H2O/m3

23oC , 50% RH

= 10 g H2O/m3

37.5oC , 55% RH

= 25 g H2O/m3

Deficit: 25-10

= 15 g/m3

23oC , 50% RH

= 10 g H2O/m3

37.5oC , 55% RH

= 25 g H2O/m3

Solution (cont)

How much do the eggs add?- 100.000 eggs x 60 gram = 6000 kg

- Lose 12% in 18 days

12% of 6000 kg is 720 kg in 18 days

- How much per hour? 720 kg in 18 days

720 kg / 18 days / 24 hrs = 1.7 kg/hr

So every hour 1700 g from the eggs

Deficit: 25-10

= 15 g/m3

1700 g/hr

100.000 egg machine, 60 g

23oC , 50% RH

= 10 g H2O/m3

37.5oC , 55% RH

= 25 g H2O/m3

Solution (cont)

How much do the eggs add per m3?

- 100.000 eggs x 60 gram = 6000 kg

- Lose 12% (720 kg) in 18 days

- Lose 720 / 18 / 24 = 1700 g/hr

- Ventilation: 200 m3/hr (average)

So each m3 gets 1700 / 200 = 8.5 gram from eggs

Deficit: 25-10

= 15 g/m3


200 m3/hr 200 m3/hr

1700 g / 200 m3/hr

= 8.5 g/m3

1700 g/hr

23oC , 50% RH

= 10 g H2O/m3

37.5oC , 55% RH

= 25 g H2O/m3

Deficit: 25-10

= 15 g/m3 – 8.5 g/m3

= 6.5 g/m3 shortage


200 m3/hr 200 m3/hr

1700 g / 200 m3/hr

= 8.5 g/m3

1700 g/hr

23oC , 50% RH

= 10 g H2O/m3

37.5oC , 55% RH

= 25 g H2O/m3

Deficit: 25-10

= 15 g/m3 – 8.5 g/m3

= 6.5 g/m3 shortage

1700 g/hr


200 m3/hr 200 m3/hr

1700 g / 200 m3/hr

= 8.5 g/m3

6.5g * 200 m3/hr

= 1.3 l/hr

23oC , 50% RH

= 10 g H2O/m3

37.5oC , 55% RH

= 25 g H2O/m3

Solution (cont)

How much do the eggs add?- 100.000 eggs x 60 gram = 6000 kg

- Lose 12% (720 kg) in 18 days

- Lose 720 / 18 / 24 = 1.7 kg/hr

- Ventilation: 200 m3/hr

- So each m3 gets 1700 / 200 = 8.5 gram from eggs

- Sprayer needs to add 15 – 8.5 = 6.5 g/m3

Sprayer: 6.5 g/m3 x 200 m3/hr= 1300 g (1.3 l) / hr

Many machines: not enough cooling

To maintain temperature:

- More ventilation

- Often machines ventilate 2-5 times more than needed

especially older multi stage machines

What happens if: 1000 m3/hr?

Air still goes from 23oC-50% RH to 37.5oC-55% RH

So air still needs to get 15 gram/m3

Eggs still lose 12% moisture in 18 days, 1700 g/hr

Deficit: 25-10

= 15 g/m3

1700 g/hr


1000 m3/hr 1000 m3/hr

23oC , 50% RH

= 10 g H2O/m3

37.5oC , 55% RH

= 25 g H2O/m3




Eggs still lose 12% moisture in 18 days, 720 kg

But now divided over 1000 m3

1700 / 1000 m3 (instead of 1700 / 200)= 1.7g

Deficit: 25-10

= 15 g/m3

1700 g/hr


1000 m3/hr 1000 m3/hr

1700 g / 1000 m3/hr

= 1.7 g/m3

23oC , 50% RH

= 10 g H2O/m3

37.5oC , 55% RH

= 25 g H2O/m3

Deficit: 25-10

= 15 g/m3 – 1,7 g/m3

= 13.3 g/m3 shortage

1700 g/hr


1000 m3/hr 1000 m3/hr

1700 g / 1000 m3/hr

= 1.7 g/m3

23oC , 50% RH

= 10 g H2O/m3

37.5oC , 55% RH

= 25 g H2O/m3

Deficit: 25-10

= 15 g/m3 – 1,7 g/m3


1.7 l/hr


1000 m3/hr 1000 m3/hr

1700 g / 1000 m3/hr

= 1.7 g/m3

13.3g * 1000 m3/hr

= 13.3 l/hr

23oC , 50% RH

= 10 g H2O/m3

37.5oC , 55% RH

= 25 g H2O/m3




Eggs still lose 12% moisture in 18 days, 720 kg

But now divided over 1000 m3

1700 / 1000 m3 (instead of 1700 / 200)=1.7g

So sprayer adds 15 – 1.7 = 13.3 g/m3,

13.3 g x 1000 m3 / hr = 13.3 l / hr (instead of 1.3 l / hr)

Problem?

high air velocity and spraying water: very cold spot

and as reaction, machine start to heat, overheating the other eggs…

Modern single stage machines:

- Enough cooling capacity

- Ventilation determined by CO2 production

At start: no ventilation required

At 18 days: 500 m3/100.000 eggs

What happens if:

Incoming air: 23oC, 50% RH

Machine: 37.5oC, 55% RH

Eggs are 18 days in incubation

37.5oC , 55% RH

= 25 g H2O/m3Deficit: 25-10

= 15 g/m3

1700 g/hr


500 m3/hr 500 m3/hr

1700 g / 500 m3/hr

= 3.4 g/m3

23oC , 50% RH

= 10 g H2O/m3

37.5oC , 55% RH

= 25 g H2O/m3Deficit: 25-10

= 15 g/m3 – 3.4 g/m3


1700 g/hr


500 m3/hr 500 m3/hr

1700 g / 500 m3/hr

= 3.4 g/m3

Sprayer: 11.6 g/m3 x 500 m3 = 5.8 l/hr

23oC , 50% RH

= 10 g H2O/m3

Modern single stage machines:

- Enough cooling capacity

- Ventilation not for cooling but for CO2

At start: no ventilation required

At 18 days: 500 m3/100.000 eggs

What happens if:

Incoming air: 23oC, 50% RH

Machine: 37.5oC, 55% RH

Eggs 18 days in incubation

Sprayer: 5.8 l water/hr. Significant local cooling…

and we don’t need it, as we have cooling capacity

37.5oC , 33% RH

= 13.4 g H2O/m310 g/m3 + 3.4 g/m3


1700 g/hr


500 m3/hr 500 m3/hr

1700 g / 500 m3/hr

= 3.4 g/m3

23oC , 50% RH

= 10 g H2O/m3

Sprayer off, RH drops

Solution: non-lineair moisture loss

- Start: dampers closed, high RH

- End: sprayers off, low RH

- Low moisture loss at start, high at end

Same amount of total moisture loss

But: - No ventilation at start: warming eggs at start

- No sprayer at end: no local cooling, stable machines

But only possible if cooling capacity is sufficient

- main effect of spraying is cooling….

To conclude:

Operating machines is a lot about physics

Control ventilation

Control relative humidity

Control carbon dioxide

To control heat exchange, control temperature

with 65 million years of experience….

birds only control temperature and turning…

To conclude:

Incubation:

A bit of embryology

A bit of physics

And a lot of common sense!

enjoy the miracle of turning an egg into a chicken…

Poultry Performance Plus'Value added' advice for poultry production

Thank you for your attention!

and some common sense….....heat transfer evap.(water spray) 50 55 60 65 70 75 80 egg weight (g)...

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