# Engine Room ventilation - Air out issue

Discussion in 'Boat Design' started by Lucya, May 20, 2018.

1. Joined: May 2018
Posts: 28
Likes: 0, Points: 1
Location: Croatia

### LucyaJunior Member

Hi all,

ventilation ducting design.

Engine Room - about 150 m3 volume.
On the starboard there are 2 Inlet ducts; supported with fans - duct A=0,5 m2
On the portside there are 2 Outlet ducts; without fans - duct A=0,3 m2 (smaller than Air-in duct)

Required E.R. airflow 40 m3/h EDIT: 40000 (11 m3/s); resulting 5,5 m3/s for duct (two ducts)
Air in per fan, flow rate: 8 m/s
Grilles: 6 m/s
Each air OUT duct must exhaust around 6 m3/s - how can I easily check this?
E.R. temp: 40 C
Out temp: 30 C
Temp difference: 10 C

My concerns are:

How can I calculate whether this volume/cross section "A" of exhaust duct air will be able to escape the engine room fast enough and hence result in excessive back pressure in ER through the duct of above mentioned "A"?
How can I calculate ER pressure; and maintain positive/negative pressure in ER? Does it need to be positive pressure in ER?
How can I calculate that the estimated back pressure of the air out system is high enough to prevent the engine room ventilation system from delivering the flow rate required to maintain the Delta-T value at <=10⁰C, when vessel machinery is operating at 100%?
P.S. Preferring not to use fans in Air-out ducting, maybe re-designing duct area...

Last edited: May 20, 2018
2. Joined: Aug 2002
Posts: 15,246
Likes: 949, Points: 123, Legacy Rep: 2031
Location: Milwaukee, WI

### gonzoSenior Member

To start with, the air will expand when it heats in the engineroom, so the volume exiting will be larger than the volume entering. From that volume, you may subtract the volume used by engines and generators (don't for a more conservative calculation). In this system, the exhaust duct is the limiting factor. However, area alone is not the only restriction, by the geometry of the ducting and grill is too. Turbulence and friction will also reduce flow. I think that the exhaust should always be larger than the intake. A pressurized engineroom will blow fumes and hot air into the rest of the boat , which is not a good thing. However, if a pressurized engineroom is acceptable, what is the maximum pressure you can allow?

3. Joined: May 2018
Posts: 28
Likes: 0, Points: 1
Location: Croatia

### LucyaJunior Member

I must highlight that my field of work is hull/outfitting, mach and hvac are new to me so I am currently figuring out all of this.

If I have understood it right, how the volume of air is expanding since the engines (2) are using air for combustion and have separate exhaust system...?

What about the pressure in ER, what are accepted parameters regarding it? How can I even (theoreticaly) calculate it?
Does the ER need negative or positive pressure regarding out pressure to suck the air out of ER through ducts without fans?

So sorry for my pure understanding about the field, that's why I am asking questions.

Thank you in advance for any feedback.

4. Joined: Apr 2015
Posts: 393
Likes: 69, Points: 28, Legacy Rep: 37
Location: Berlin, Germany

### HeimfriedSenior Member

You should check your numbers. The air flows given above are not correctly converted.
Is it
40000 m³/h = 11 m³/s
or is it
40 m³/h = 11 dm³/s ?

Regarding the volume of the ER, the first mentioned is more probable.

5. Joined: May 2018
Posts: 28
Likes: 0, Points: 1
Location: Croatia

### LucyaJunior Member

Since I have just converted hours into seconds, 40000 (m3/h) /3600=11 (m3/s), as I wrote on the 1st post; what is wrong with this?

6. Joined: Apr 2015
Posts: 393
Likes: 69, Points: 28, Legacy Rep: 37
Location: Berlin, Germany

### HeimfriedSenior Member

The dimension m/s indicates a velocity (of the air stream). If a flow rate of 6 m³/s is to pass a a duct with a cross section area of 0.3 m², the required average air velocity is (6 m³/s) / 0.3 m² = 20 m/s .

Lucya likes this.
7. Joined: Apr 2015
Posts: 393
Likes: 69, Points: 28, Legacy Rep: 37
Location: Berlin, Germany

### HeimfriedSenior Member

You wrote 40 m³/h in the 1st post, not 40000 m³/h. That's quite a difference.

8. Joined: May 2018
Posts: 28
Likes: 0, Points: 1
Location: Croatia

### LucyaJunior Member

OMG sorry, yes, you are right.
I meant 40000. Terrible typo mistake :/ But OK thank you for the mark-up, I'll edit.

9. Joined: May 2018
Posts: 28
Likes: 0, Points: 1
Location: Croatia

### LucyaJunior Member

Mr. Heimfried, thank you.
How can I check what will be the air velocity in the duct of 0,3 m2 cross section without the support of the fan - natural exhaust....
Is there any way to calculate it (theoretically) without measurement instruments, and regarding the pressure in the ER since the info of Air in and ER volume is known?

10. Joined: Apr 2015
Posts: 393
Likes: 69, Points: 28, Legacy Rep: 37
Location: Berlin, Germany

### HeimfriedSenior Member

The expansion of the air, when heated from 30 °C to 40 °C is T2 / T1 = (273 K + 40 K) / (273 K + 30 K) = 1.033.

So for an intake of 11 m³/s there will be an exaust of 11.36 m³/h (without consideration of the air take in of the combustion engines). Not that much.

The air flow rate of a ventilator (fan) is given as the (theoretical) air flow rate without any air current resistance. In real circumstances the fan will be not able to deliver this flow rate, but in most cases only a smaller part of it.

The manufacturer will provide a table and/or diagram showing the relation of pressure and flow rate. You have to calculate the flow resistance of the whole system (intake to outlet). For the components, e.g. duct, bow, grill this resistance is mostly given in terms of pressure loss, related to flow rate.

Last edited: May 20, 2018
11. Joined: Apr 2015
Posts: 393
Likes: 69, Points: 28, Legacy Rep: 37
Location: Berlin, Germany

### HeimfriedSenior Member

It is possible to calculate this. But you have to consider a real lot of things. Look at the link, if you like (pages 10, 12, 13, 508):

http://www.heliosfans.co.uk/Files/HeliosUK/pdf/helios_catalogue_4_0_GB.pdf

The booklet is not about ships ventilation but building ventilation. Because the physics of air is not different, it may give you a first idea of what to calculate.

Regarding the Delta-T: if engine runs at 100 % there should be given an max. value of waste heat to be discharged by air. Round about 1 kg of air will need a thermal energy of 1 kJ to rise its temperature for 1 Kelvin.
So 11 m³/s from 30 °C to 40 °C means about 12.7 kg/s * 10 K * 1 kWs/(kg * K) = 127 kW (if I didn't mistake).

Lucya likes this.
12. Joined: May 2018
Posts: 28
Likes: 0, Points: 1
Location: Croatia

### LucyaJunior Member

Thank you, the attached PDF seems worth to study....

Fan Performance in my case:
Air Flow : 8.14 m³/s
Static Pressure : 250 Pa
Air Density: 1.204 kg/m³

Can you help me on how can I use this data?

Regarding the resistances; according to the Fig.9, sheet 12 from Helios PDF - I do not have duct bended parts, batteries, attenuators etc.
There is "only"an ER of 150 m3 volume; air intake of 8 m3/s - per fan (2 pcs) (assuming fans WILL deliver this flow rate).
The engines and other equipment...
AIRFLOW FOR COMBUSTION: 1,5 m3/s
AIRFLOW FOR EVACUATION OF HEAT EMISSION: 8,5 m3/s
TOTAL AIRFLOW REQUIRED: 11 m3/s - data already given on 1st post.

Current ducts Air OUT data:
TWO PCS of 0,3 m2 (0,7 m x 0,45 m), 2m height above ER (from the ER ceiling to the Grille point L=2m), straight, steel/structural ducts.

The grille-out size is adaptable. Only the duct size concerns me.

13. Joined: Oct 2008
Posts: 6,973
Likes: 916, Points: 113, Legacy Rep: 2488
Location: Japan

Ok...apart from the excellent advice from Heimfreid...let's take a step back.

What is the purpose of an air-in and an air-out....well, it is because you have engines in a confined space, that require cooling and to breathe.

So...in order to ascertain if the numbers you have posted are correct - which could be the source of your dilemma in trying to calculate required air flow etc - what size and type of engines are in your ER..and what other, if any, sources of combustion too..such as gensets etc.

14. Joined: Apr 2015
Posts: 393
Likes: 69, Points: 28, Legacy Rep: 37
Location: Berlin, Germany

### HeimfriedSenior Member

Following my technical book a rectangular duct 0.7 m * 0.45 m has the same pressure loss as an round pipe of a diameter of 0.55 m (cross section areas are not the same, but the friction is). If the walls of the duct (leghth 2 m) are fair, there will be a pressure loss of 16 Pa = 16 N/m² . (A coarse surface would signifantly increase the pressure loss.) The grill resistance is to add. Is the grill horizontally oriented (rain in the ER)? If it is vertically oriented, the change of the direction of the airstream causes an additional resistance.

15. Joined: May 2018
Posts: 28
Likes: 0, Points: 1
Location: Croatia

### LucyaJunior Member

Hi Ad Hoc, thanks for joining the discussion, I am desperate

As I have written above:
AIRFLOW REQUIRED FOR ENGINES COMBUSTION: 1,5 m3/s
AIRFLOW REQUIRED FOR EVACUATION OF HEAT EMISSION (engines, pipes, cabinets....): 8,5 m3/s

Those are the results for needed air to work properly.
And if I already have required airflow for ER, do I need to spec the engines?
(assuming the results are correct, given from the Equipment supplier, exposed pipes heat emission calculation, exhaust pipes etc).

My dilemma is not the required air for ER, my dilemma is whether the Outlet duct 0,3m2 x 2 PSC (0,6m2 SUM) are enough to exhaust around 6 m3/s of air?
Saying that the velocity in each duct must be 10 m/s to satisfy this requirement (there are two ducts of 0,3m2).
How can I check what air velocity will be in the duct without fan in Outlet duct (natural non-forced air out)..?

Maybe I do not understand you all right....

Forum posts represent the experience, opinion, and view of individual users. Boat Design Net does not necessarily endorse nor share the view of each individual post.
When making potentially dangerous or financial decisions, always employ and consult appropriate professionals. Your circumstances or experience may be different.