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Room Pressure
2

Room Pressure

Room Pressure

(OP)
Hello,
Please can you help me, I have a room 2,2m x 2,3m x 2,55 and I need it to be pressurized to 15 Pascals, the volume of the room would be 12.9 or say 13 Cubic m

RE: Room Pressure

and your question is??

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

RE: Room Pressure

(OP)
What air volume do I need to achieve the 15 Pa

RE: Room Pressure

In order to pressurize a room you have to calculate the inflow and the outflow, if the outflow exceeds the inflow you will never achieve pressurization.
B.E.

You are judged not by what you know, but by what you can do.

RE: Room Pressure

it will hardly go that way, you would have to make testing of room cracks. the easier way is to install adjustable relief dampers.

RE: Room Pressure

In theory its simple, (P1 x V1)/T1 = (P2 x V2)/T2. Remember these are all in absolute units (hence p1 is 101,325 Pa a at sea level). You know the change in pressure, temp stays the same so you can work out how much volume it needs to change the pressure. To get to 15 Pa, it won't be a lot. Are you sure you don't mean 15kPa?

15pa is equivalent to 1 point 53 mm water guage (!)

Of course this only applies if not a single molecule of air leaks out of your room.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

RE: Room Pressure

Is this a homework problem?

RE: Room Pressure

(OP)
No I have an application where we need to pressurize an area to 15 Pa, I have the volume of the room and I know how much air to supply to achieve the 3-5 air changes required in the room but I don't know how much air to supply to achieve the 15 Pa room pressure

RE: Room Pressure

Is the room constructed already? If it is, test it with a door blower and a Dwyer magnahelic.

RE: Room Pressure

When we pressurize rooms for pharma, we use:

Airflow Through Large Intentional Openings
The relationship describing the airflow through a large intentional opening is based on the Bernoulli equation with steady, incompressible flow. The general form that includes stack, wind,
and mechanical ventilation pressures across the opening is
Q = 776*C*A * SQRT(2* delta-p / density
(This is equation 36 in Ch 16 of 2013 Ashrae Fundamentals in English units)
where:
Q = airflow rate, cfm
C = discharge coefficient for opening, dimensionless (usually 1.0 for a door)
A = cross-sectional area of opening, ft2 (Single door with 1/4 crack, = 0.23 ft^2)
p = air density, lbm/ft3 (0.075)
dp = pressure difference across opening, in. of water (convert 15 pascals to English)
776 = unit conversion factor

knowledge is power

RE: Room Pressure

cdxx, i am wondering whether you proved that approach in practice?

your maths works well for discharging into open space, which i doubt you use as a method of pressurizing pharma areas.

normally, there you have rooms with excess supply and rooms with excess exhaust where pressurizing is coming out of flow differences and pressure loss rates related to flow.

you can desire any flow through transition opening, but real flow will depend on real pressure loss, which makes process iterative, and i doubt much you can achieve exact overpressure without pressure dampers, differential pressure-controlled supply and exhaust fans, or both. you cannot use dp control for central fans if you have rooms with differing requirements so damper remains the only option imho.

RE: Room Pressure

for real world practice you can't really calcualte. wha tyou need to do it:
- build the rooms needing positive or negative pressure as tight as possible
- use exhaust and inflow flow meters, like Accutrol air valves (can't use pitot-tube devices for dirt and accuracy reasons)
- supply or exhaust a few % of flow more.
- see if the pressure relationship is as you want it.
- in some cases you also may use dP sensors between those rooms.

You also need to set up control so that if a flow station or a dP sensor fails, the room fails in the extreme of the pressure. i mean if you space is supposed to be 0.01" negative, failure of control should not make it positive. i once had to fix a shooting range pressure control (designed by someone else) and that would fail in being positive to adjacent rooms, not good for a room that is supposed to be negative. Murphy's law dictates failures will happen, prepare for them.

RE: Room Pressure

I cannot imagine anybody using inlet and out let flow meters to control room pressurization when a pressure sensor is far less expensive, simpler, and directly measures what is to be controlled.

RE: Room Pressure

Drazen, in theory this is the best way I know how to calculate. You need some starting point. In practice there are many variables, as you imply, which affects the accuracy of your calculations. Room construction was usually the culprit. I would size the return conservatively, and balance the return to maintain the pressurization required.

Energy explains the issues that you need to think of to make it work in practice.

I have only done "static" systems with a hard balance. There are "dynamic" systems that are constantly hunting. You need top quality air valves, and even then it is difficult, because the accuracy is so small that the sensor technology is not there yet, when put in practice (from what I was told).

Berk, that switch does sot have the accuracy required. You usually have rooms at 0.03 in wc difference required.

knowledge is power

RE: Room Pressure

I've done this many times, for positive and negative relative differential pressure. The easiest way is to set your required minimum room air changes, then use a manual damper and a Dwyer to field set to desired relative differential pressure.

RE: Room Pressure

I agree with urgross, drazen, and composite pro.

So here is an interesting question with an answer based in experience:

If you are pressurizing a room positive (lets say a typical amount is 0.03-0.07" w.c.), how much "less" return air do you return in rooms for balancing?

I have been taught 100 CFM per door, or 10% less than SA, with judgement being used for both.

Anyone else?

RE: Room Pressure

(OP)
Hello all,
I have my answer to my question, can someone tell me how to close this topic

RE: Room Pressure

I TAKE 10 CFM FOR 1 m2 OF FLOOR + 150 CFM FOR EACH DOOR IN THE ROOM.
THIS IS THE OUTDOOR AIR.
YOU MUST ALSO SEAL THE CEILING AND CRACKS WITH SIKA OR SILICONE.
IF YOU HAVE SUCTION AIR FROM THE ROOM ADD IT TO THE FRESH AIR.
ALSO NOTTICE THAT THE DOORS ARE OPENED TO THE ROOM AND NOT OPENED TO THE OUTSIDE OF THE ROOM BECAUSE , SO THE HIGH PRESSURE WILL "CLOSE" THE DOOR AND NOT "OPEN" IT (IN THIS CASE AIR WILL RUN AWAY FROM THE DOOR

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