Engine Room ventilation - Air out issue

[Lucya]

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.

How can I have no ambiguity?

Regardless, thank you very much for your elaboration.
Will study your input.

Can you help me as well with understanding "the estimated the back pressure for the ER air out system" - and securing the max of 10 degree Celsius temp difference (Delta T - ER-Outside)...?

 

[Heimfried]

There are two ducts, so 2 x 0,3 m2 - resulting 10 m/s each? (Sum of Air-out cross section is 0,6 m2 - 6/0,6=10)... or I am doing it wrong, again?

Do you mix the calculation between 1 and 2 ducts? You said there is a required air flow of 11 m³/s (intake?, so 11,4 m³/h at 40 °C). If you take the half of it, because of 2 ducts, you should calculate only with cross section of 1 duct.

(12 m³/h) / (0,6 m²) = 20 m/s

(6 m³/h) / (0.3 m²) = 20 m/s

 

[Lucya]

Do you mix the calculation between 1 and 2 ducts? You said there is a required air flow of 11 m³/s (intake?, so 11,4 m³/h at 40 °C). If you take the half of it, because of 2 ducts, you should calculate only with cross section of 1 duct.

(12 m³/h) / (0,6 m²) = 20 m/s

(6 m³/h) / (0.3 m²) = 20 m/s

Heimfried you are right, ty.
It was way past midnight in my country, my brain was overloaded, stayed till 5 AM studing this issue lol.

Can you please tell me whether the 20 m/s velocity in the duct can be obtain without the fan support? Is this the "chimney effect" due the temperature difference between in and out?
Will this effect of hotter air getting out up obtain the 20 m/s velocity given the current duct cross section....
I am trying to figure out will I need fan support in outlet duct to push out the air to obtain 20 m/s.

TY.

 

[Heimfried]

[...]
Can you please tell me whether the 20 m/s velocity in the duct can be obtain without the fan support? Is this the "chimney effect" due the temperature difference between in and out?
[...]

No, the chimney effect is by far not enough to cause this airflow rate. Both, Delta-T of 10 K and chimney height of 2 m is not much.
Either you use the pressure caused by the inlet fans or you will need additional exhaust fans.
The former requires a tight ER, so the air intake fans are able to create a sufficent pressure inside the ER.

The way to check I would go: sketch the complete path of the air flow. Complete means from outside free air at intake side to outside free air after outlet. Sum up all the frictions, resistances, obstacles in form of pressure losses. The volume of the ER dosn't matter much in this regard, if there is no bottleneck. But e.g. the airflow transit from ER into exhaust duct alone causes according to my tables a Delta p = 0.7 * rho/2 *w² = 0.7 * 0.55 kg/m³ * (20 m/s)² = 154 Pa (assumed a simple rectangular opening in the ceiling of the ER identical with the start of the duct). If I understand it right, your fan will only be able to build up a pressure of 250 Pa, hopefully connected with the required airflow. This fan pressure has to overcome all pressure losses: air inlet grill, transit to outlet duct, duct, outlet grill. I suspect it will be enough.

Pleas note, that I'm not a kind of expert regarding ships, engines or engine rooms. I just know some about the physics of air and technical items regarding building ventilation.

 

[Lucya]

Heimfried thank you very much.
Your reply is the exact thing I was looking for at the first place, while opening this discussion.

I will make the vent sketch and calculate all suggested parameters. I am expecting "negative" results - by negative I mean resizing the duct and arrange additional fan for exhaust.

And for the conclusion, if anybody knowns something about the calculation of additionally cooling the ER through shell plate (bottom plate, in touch with the sea), suppose this will also give some heat loss, but I do not know how to correctly calculate this, given the fact that parameters as shell plate area, shell material and sea temperature are well known.

Heimfriend thanks again.
Best Regards.

 

[Lucya]

Edit:

 

[Heimfried]

What kind of heat exchanger is intended? Thickness 8 m? Cross sectional area 30 m²?

Sketch or describe the intended system in more detail.

 

[Lucya]

Input:

- Thermal conductivity of material (W/mK) for shipbuilding steel: 41,84
- Cross Sectional Area (m2) (suppose area of steel from ship plate in bottom where E.R. is located; in touch with sea): 50
- Thickness (m): 8,0 mm plate resulting 0,008 m plate
- Hot Side Temperature (°C) - Engine room: 40
- Cold Side Temperature (°C) - Average summer sea temp: 20

Confused by the "W" heat results, suppose I read this wrong.

 

[Heimfried]

The calculation you made is not really helpful and - as it seems to me - wrong. My calcs result in 5,230,000 W, the tenfold value.

But the pure heat transfer through a large steel plate is not a critical point at all.

If you think about a system like this:
1. hot air in ER is the source
2. heat transfer to a heat carrier (water or so)
3. circular flow of heat carrier from ER to "bottom plate"
4. heat transfer to surrounding body of water

... the bottleneck will be the transfer of heat from air ER to heat carrier. The Delta-T is low, the specific heat capacity of air is low. You would need a large heat excanging device with fans.

 

[Lucya]

Hi Heimfried,
thank you very much for your reply.
Also have to mentioned that you helped me big time regarding the previous discussion we had on ER pressure, ducting etc....

Regarding the results of conduction heat transfer - we got the same result but the "calculator result rectangle" do not show all digits
(see, I moved the cursor to the end so the first number 5 now doesn't appear anymore, it shows only 6 digits max, that's why one zero was missing)

But the pure heat transfer through a large steel plate is not a critical point at all.

Could you please explain this to me more?
Heat exchanging device with fans?

I was only thinking about the natural heat loss from ER 40C while the sea water is max 20C thought the shell plate - in subject of cooling the ER (additionally!).
I mean, does it need to be included into the ventilation calculation.
So... the thing is not working this way suppose?

 

[BlueBell]

"W" is most likely watts, no?

 

[Lucya]

Yes, like power unit, as rate of energy transferred per unit of time (sec).
The energy Joules => one Joule per second => one Watt
Suppose, or this is wrong.

 

[Heimfried]

The calculated heat transfer through the bottom panel is not considering both the transition resistances from air to plate and from plate to water.
These transition resistances are far greater than the resistance of the pure transfer through the plate. Additional you have to consider the coats of paint and so on which are covering the plate inside and outside.
Additionally the heat stream is directed downwards, which is adverse because no convective flow of heat. The air ER near the bottom plate will cool down to the temperatur of the plate and build up a kind of layer ("cold air lake"). Likewise the bottom plate hinders the heated layer of the surrounding water to well up.

 

[Dragonpoint]

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

There is "only"an ER of 150 m3 volume; air intake : 8 m3/s - per fan

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.
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.

 

[Dragonpoint]

Parameter : 1.5 Cubic Metres tells you nothing.
8.5 Metres ( Cubic Metres ) per Second is a high airflow , very high ,what you want evacuating exhaust out of an Engine Room .
Two Fans , what you want to find out is Radius of Each Fan .