the side panels could interfere with access

generally, you want the hood as close as you can get it to the pots - again, given the access constraint - 42" seems high

not sure that Hugh & I are talking about the same thing - my attic fan will suck the doors in the house shut & I use it to remove smoke when the hood is inadequate

one way to think about the hood is that it is a pool - the smoke goes up and collects in the pool, the blower then sucks the smoke out of the pool

i.e. it is like a buffer

you might look at the fans on a big restaurant kitchen hood - get the info off of them & see what you need to be able to match

another approach would be to talk to an HVAC engineering Prof. somewhere (assuming you don't want to hire one)

since no one other Hugh & I has chimed in, I suspect that Pelican is not exactly crawling HVAC engineers...

I know there are a bunch of Architects on here -- don't you guys learn this stuff?

No HVAC engineer here. Isn't it over kill going through all this for a kitchen hood? Know the size of the fan and mate the duct size to it. there are only so many ways to mount a fan. right over the hood or exterior. I suppose you can have one in mid section, or like I suggested, have it on both ends ( I have never done it and think its crazy in a kitchen). One to push and one to pull. I am still not sold on how all this engineering on ducts are going to solve the problem. I am sure what John is trying to achieve? Not to draw outside air inward through various openings in the house or is he worry about having too many bends in the venting ducts? At the end, most of the air that is being drawn is going to be at the mouth of the hood opening. (that isn't most of the stinky air or grease from cooking, but hot air with grease will and does rise so a lot of it will be capture as it goes up toward the hood) He can isolate himself in a small room with a stove and a small vent to the outside with a huge hood.

Jeff

Don't forget about the pot filler back there. KWC makes a really nice residential one that folds in and out. That thing is massive. I know you like to cook, damn you don't mess around. if you are going to put in a monster like that, you better add the wok stove in there. forget the viking ones. Go small restaurant Wok burner (cost less then Viking but fit and finish isn't as nice, looks very restaurant). Will it pass inspection in a residential application? They need a lot of BTU. Around here, you can call the gas company to give you a bigger residential meter. I did at my house. It fires the BBQ, and an outdoor stove next to it, fire pit, two heating units, and a big tankless water heater.

When I build my outdoor kitchen, I really wanted that commercial wok burner. due to space and appearance, I decided to install a DCS burner instead. to fire up a wok in the spring outside will attract all the coyotes, in the canyon. Too bad.

Jeff

Jeff (171) "Isn't it over kill going through all this for a kitchen hood? Know the size of the fan and mate the duct size to it. there are only so many ways to mount a fan. right over the hood or exterior." "I am [not] sure what John is trying to achieve? "

Sorry, I know that in some ways this is overthinking a simple range hood. Basically here's my motivation:
- Learn about how these things are designed and calculated - quest for self-betterment through knowledge, blah blah, but I am genuinely interested in how guys like Hugh and you and our lurking HVAC engineers figure these things out.
- See how much I can lower CFM reguired via optimized design - cut down cost of fan (a kind person sent me links to fan models, the 1500 cfm stuff is kinda spendy and, just as bad, really bulky - a 1500 cfm side-blast fan will look like a trash can sprouting from the side of my house), size of ducts, noise, etc.
- See how much I can lower MUA required, and MUA's impact on temperature of my house, via lower exhaust CFM and optimized location of MUA. My house is cold in winter (typically 62F) and hot in summer (typ 80F) - 100 y/o, no insulation, too cheap to run central heat/ac all the time. I cook a lot. I worry that if the house is already 62F and I start a 1 hour sear-wok-fry-broil session, with fan pulling in ~1500 cu ft of 30F air every minute . . . my cooking privileges may get cut off.

Randy, my attic fan will do the same thing, suck ashes from the fire place if I don't close the damper. Its different when you have a column of air, that has weight and keeping it moving.

The following comes from "Industrial Hygiene Engineering" by John Talty:

In discussing energy requirements for local exhaust ventilation (LEV) a logical place to begin is with the still air out in front of the hood. This air, into which the contaminant is being generated, must be induced to flow into the LEV system in accordance with the calculated air flow requirements. To fulfill the objective of LEV, this air must be "energized"; in other words, enough energy must be supplied to the system by the fan to overcome the inertia of the still air so that the specified volumetric flow rate will move into the hood, hopefully entraining and transporting the contaminant along with it. Thus, the first component of energy overcomes the inertia of the still air and accelerates it to the desired system velocity. This energy represents the useful work accomplished by the system.

Once the proper flow rate of air in in motion, it must remain in motion until such time as it is discharged from the workplace through the properly designed stack. However, as the air makes its way through the system, additional energy is consumed or lost due to factors such as turbulence and friction. Although this energy is lost in the sense that it contributes nothing toward the useful work of accelerating air, its magnitude must be taken tallied into the overall energy requirements of the system.

Generally speaking, LEV system energy loses can be broken down into the following categories:

* Turbulence losses as the air enters the hood,
*Friction losses as the air moves through the ductwork,
*Losses in fittings, (elbows, contractions, expansions, orifices, exhaust caps, etc.,
*System loses at the fan, and
*Loses across air cleaning devices (filters, grease screens).

If these losses are not accounted for its very easy to select a fan and motor that are insufficient to 1) accelerate the air in the system to the required velocity and 2, to keep the air at that velocity sufficient to keep the contaminants in suspension until ejected at the exhaust stack.

An LEV system has the downstream and the upstream side of the fan; or suction (-) and pressure (+) sides. The fan "scoops up" "buckets" of air with each fan blade creating a slightly lower pressure immediately in front of the fan blade. Air pressure rushes in towards that lower pressure area to fill the partial vacuum. There is at the same time an increase in air pressure on the pressure side of the fan. This fan-generated differential pressure between the inside of the LEV system and the outside ambient air is called Static Pressure (SP).

The Static Pressure represents the energy that continually delivered to the system by the fan. It is analogous in concept to potential energy in that it indicates the capability of the system to do work (i.e., overcome the inertia of still air, accelerate it to the desired system velocity, and convey it through the system until it is finally discharged into the ambient environment).

LEV system pressures are not measured in PSI, because of the small pressure differentials. Rather, they are measured in height of a liquid water column. Air pressure at sea level is nominally 14.7 psi, which equates to 406 inches of water column. If the SP existing in a LEV system at a given point would support a 2-inch water column, then the system has 2-inches of water gage (wg).

The fan must be capable of generating a sufficient magnitude of Static Pressure (in inches wg) at the hood opening so that the desired quantity of still air will be accelerated into the hood and up to the proper system velocity. The important question is how much Static Pressure the fan must deliver at the hood to accomplish this. To answer this question, another important pressure in an LEV system must be introduced Velocity Pressure (VP). As the fan blades turn and continually generate static pressure (potential energy) the pressure differential created begins to induce the still air to move as discussed above. As the fan blades begin spinning, a portion of the static pressure generated is immediately converted into pressure exerted by moving the air. In other words, some of the potential energy created by the fan is converted into kinetic energy of air in motion. The pressure exerted by this moving air is referred to as velocity pressure. In an operating LEV system, negative static pressure at the hood face directs room air into the system and accelerates it to the desired system velocity. Thus, a portion of the fan-generated static pressure available at the hood face is continuously converted into velocity pressure as the air is accelerated o the desired system velocity.

Essentially, velocity pressure is the kinetic energy generated in a LEV system as the result of air movement. The wind you feel on a windy day (or when you hold your hand out of a window of a moving car (Porsche obviously) is due to velocity pressure. Unlike static pressure, which acts in all directions, velocity pressure acts only in the direction of flow. Unlike static pressure, velocity pressure is always positive in a LEV system, regardless of the side of the fan on which it is measured. It is the pressure exerted by moving air and cannot be negative. If no air is moving, the velocity pressure is 0. (more)

I know this is pathetic, but I find this stuff you are posting to be fascinating.

Oh, Randy, the reason the (proposed) hood is so high is to clear a 6' 6" head. I could bring it down, but with less overhang vs the front of the range. I'm not sure which is better.

Up to this point, I have used the term "desired system velocity" without explanation. Stating that the static pressure must be sufficient to accelerate still air to some desired system velocity implies that the LEV designer has in mind some target velocity at which the designer desires the air to be moving as it makes its way through the system. This is in fact the case.

Since it is the total fan static pressure requirement that the LEV designer is calculating, there must be some relationship between the static pressure at the hood opening and the desired velocity of air in the duct leading to the hood. To understand the relationship between static pressure and velocity, one must understand the relationship between velocity and velocity pressure. VP is defined in the Industrial Ventilation Manual as "that pressure required to accelerate air from zero velocity to some velocity (V) and is proportional to the kinetic energy of the air stream."

I have a "proof" of the following equation which is a little tedious to reproduce here, if I get bored I may labor through it later, but it involves the equation for a freely falling body the density of air and the density of water to arrive at the following equation

V=4005 vp^1/2

Where Velocity =V, and vp=velocity pressure.

Bernoulli's Theorum which is related to fluid energy states that the total energy in a system remains constant if friction is ignored. This can be applied to LEV systems. The static pressure represents the potential energy within the LEV system and the velocity pressure represents the kinetic energy. From elementary physics, we know that the sum of the potential and kinetic energy represents the total energy in the system. Thus, total pressure (TP equals static pressure (SP) plus velocity pressure (VP), or

TP = SP + VP

The concept of positive and negative pressure is important when designing a LEV system. You have to be careful to use the appropriate sign (for the pressure which exists). The normal convention of plus and minus for the signs is used. Thus, a static pressure in an LEV system would be represented as a negative value, which a velocity pressure in either an exhaust or supply system will be represented by a positive, or plus value. For example, if a exhaust system has a total pressure measured as -2 inches of water and the velocity pressure is calculated as 1 inch of water then the static pressure would be

TP=VP +SP
-2= 1 + SP
SP=-3 inches wg.

When the fan is turned on, energy is required to accelerate the air outside of the system to a given velocity t be maintained within the system. This energy is stated in inches of water static pressure. Once the air reaches the design velocity, this velocity will remain constant for a given duct size throughout the system. Thus, the velocity pressure will also remain constant because of the direct relationship:

v=4005 vp^1/2 (for standard temperature and pressure only)

The rate at which air is captured and flows through an LEV system is stated in cubic feet of air per minute (CFM). If Q denotes the rate of flow within the system then:

Q=vA

where v is the velocity of air in feet per minute, and
A is the cross-sectional area of the duct in sq. ft.

(more)

Its a good refresher for me. I haven't really delved into this in about 20 years since I sat for an passed (first time) the two-day exam to become a Certified Industrial Hygienist. Anyway, I'm just killing time before I have to start drinking my gallon of pre-colonoscopy tonic.

bring the cook top UP - the distance between them is a significant factor in "air" capture

I'll assert again that the cheap way to do this is to use a large attic fan and neck down the inlet section (which would be the exhaust section for you)

we've built a lot of wind tunnels that way and they always include diffusers to make the flow laminar

you may not care re cheap, or may not care to experiment, but if you cannot be certain of your paper design, then just build a cheap one and then modify if needed

personally, I'd use a commercial restaurant hood with an external commercial restaurant blower

As an example, I get into how you design a system later, but its easier to show how the math works using a hypothetical existing system.

Let's try an example.

Find the rate of flow in CFM of air traveling at 4,000 feet per minute in a circular duct 4 inches in diameter.

Using Q=vA, the area of the circular duct is A=pi*r^2, or A-pi*r^2/144

A=3.14 * 2^2/144 or 0.087 ft^2

Substituting into the equation

Q=4000 fpm x 0.087ft^2

Q=348 cfm

I mention measuring TP, SP and VP in inches of water, you do that by drilling a small hole in the pressure side of the duct work. A manometer is used to measure the static pressure in the system. I assume you know what a manometer is. One end of the tube is inserted into a flush fitting in the duct work and the other end of the manometer is open to the air outside the LEV system. It is zeroed with the system off and then the LEV system started. The differential change in water elevation across the manometer is the static pressure in inches wg. To obtain the TP, an impact tube. The impact tube is oriented inside the duct such as to have the opening point upstream in the air current. It is attached to the manometer to obtain TP in inches of wg.

Use of these two tubes allows you to indirectly calculated velocity pressure since TP= SP + VP.

PS - re education - Steven Vogel's "Physical Biology of Flow" is very fun & very readable - it's avail. in pb; I have 2 others when you are done w/ that book

As air flows through the LEV system frictional loses (or drag) act to retard the flow as a result of turbulence within the duct. That turbulence is related to the Reynolds number (Re), which was named after a British guy who studied drag on particulates.

Re=f(D.vavg .p u

Re=Dvavgp/u

Where:
D=Duct diameter in feet,
vavg =average linear velocity (ft/sec)
p=fluid density (lbs/ft^3)
u = fluid viscosity (lbs/ft-sec)


The Re is dimensionless, for streamline flow (laminar) the RE is <2100. The transition region between streamline and turbulent flow exists where the Re number has values 2100< Re > 4000. Turbulent flow exists for Re> 4000.

Example

Assume a ventilation system is operational with a 4-inch diameter duct. At what rate of flow (cfm) will the upper limit of streamline flow be reached.

Re=Dvavgp/u

Re=2100
D=4 inches or 0.333 ft
p =0.075lbs/cu ft. for air
u= 0.0178 centipose at 70 degrees F

Since u=0.0178 cP (6.719 x 10^-4 lb/ft-sec/cP
u-1.2x 10^-5lb/ft-sec

Substituting in the formula for Re gives

2100=0.33ft x v ft/sec x 0.075 lbs/ft^3/1.2 x 10^-5 lbs/ft-sec

v=2100 x (1.2 x10^-5) ft/sec/0.33 x 0.075

v=1.01 feet per second or 60.6 feet per minute

Then

Q=vA
Q=60.6 ft/min x 0.087 ft^2
Q=5.27 cfm

Thus, for this ventilation system the flow will always be turbulent.

Ok, lets look further to determine the effect of this turbulent flow and the resulting frictional drag upon the operation of the LEV system. Since the rate of flow (Q) remains constant for a system, if the cross-sectional are of the duct remains constant, then the velocity will remain constant throughout the system.

This can be stated as Q=constant=vA

So, for a given cross-sectional area

v=constant

Since v=4005 VP^1/2
VP=constant

When frictional loses occur in the system, the velocity pressure remains constant as show above. The total pressure must change to overcome the losses within the system from the following:

TP=VP + SP + losses

The losses will then be reflected in the static pressure of the system, since the velocity pressure is constant. Thus, static pressure can be thought of as being composed of two components. These components are:

1. the pressure to accelerate the air from rest to a given velocity, and
2. the pressure necessary to keep the air moving at this velocity by overcoming frictional losses.

Example

Assume a LEV system is designed to exhaust air at a rate of 200 cfm through a 4-inch diameter duct. If the dust is 40 feet long and no losses occur at the inlet or hood, what are the two components of static pressure (the static pressure necessary to accelerate the air to a given velocity and the static pressure necessary to overcome losses)/

Solution

To obtain v, substitute in the formula Q=vA

obtaining

200 cfm=vx 0.87ft^2

v=2,300 fpm

I'm looking at a "Friction loss in inches of water per 100 ft" table which shows friction loss in inches of water per 100 ft on the "Y" scale, Cubic feet of Air per minute on the "X" scale and scale lines for various duct diameters and face velocities. Kind of like this one Air Ducts Friction Loss Diagram

With a 4-inch duct, 2,300 fpm and 200 cfm the chart shows about 2.5 inches of friction loss in inches of water per 100 feet of duct.

You can get the velocity pressure from v=4005 vp^1/2

VP=(2300/4005)^2

VP=0.33 inches of water

In order to accelerate air to the velocity of 2300 fpm from rest and assuming no inlet losses,

TP=SP + VP

however, air at rest has TP=0, so

0=SP + VP
SP=-VP
SP=-0.33 inches of water.

Again referring to the imbedded link above, for the intersection of Q and D at 200 cfm and 4 inches of wg we see loses of about 2.5 inches per 100 feet of duct.

so the loss will be
SP=-2.5 x 40/100 (round duct has about 40 joints per 100 feet of duct).

SP=-1 inch of water to overcome losses.

I'm getting to the end of my expertise on this and depending on your filtration, fan blade, and duct work you select, you'll need to also figure out what motor and rain cap you're going to slap on the end of this thing. Make any curves (elbows) as arcing as possible with as large a radius as you can put in.

CFM varies directly with fan speed,

Q2=RPM2
Q1 RPM2

SP varies as the square of the fan speed

SP2=(RPM2/RPM1)^2
SP1

Horsepower varies as the cube of the fan speed

HP2= (RPM2/RPM1)^2
HP1

SmileWavy

How's that "gin and tonic" tasting, minus the gin? Are you nice and clean yet?

Getting there, man I could use the gin right about now.

This is similar to what I have on mine. I have a 6" vent from outside my basement. goes under the kitchen floor to behind the cabinet my cooktop sits on. it pulls air from the space behind the cooktop. I have a 48" 800cfm hood.