CHAPTER XXI.

PROBLEMS AS TO VENTILATION.

The following formula enables many problems relating to ventilation to be solved. Let p = the amount of poison (carbonic acid) in every cubic foot of fresh air, viz. ·0004 cubic foot. Let A = the number of cubic feet of fresh air delivered or available, P = the amount of carbonic acid exhaled, and x = the amount of carbonic acid per cubic foot in the room at the end of a given time. Then—

x = p + P/A, whence A = P/(x - p).

_If the carbonic acid in the air of a room is ·75 per 1,000 volumes (that in the outer air being ·4 per 1,000 volumes), and there are five persons in the room, how much air is entering the room per hour?_

Here x = ·00075. p = ·0004. P = ·6 (_i.e._ number of cubic feet of carbonic acid expired by each person per hour). Now x = p + P/A.

·00075 = ·0004 + ·6/_A_.

Therefore _A_ = about 1,700.

Thus 1,700 cubic feet are required for each individual to keep the air within the given limit, and five times this amount will be required for five persons = +8,500+ cubic feet.

_A room has been occupied for one hour, at the end of which the total carbonic acid present was found to be 1·1 per 1,000 parts. The carbonic acid in the open air amounting to ·0004 per cubic foot, find the quantity of air supplied per hour._

Here x = ·0011. p = ·0004 and P = ·6.

Hence ·0011 = ·0004 + ·6/A.

Therefore A = +857+ cubic feet.

_If six persons are in a room containing 3,000 cubic feet, and there is a supply of 2,000 cubic feet of air per head per hour; how much carbonic acid is there in the air of the room at the end of 4 hours?_

Here p = ·0004. P = ·6 × 6 × 4 = 14·4. A = 2,000 × 6 × 4 + 3,000 = 51,000. x = ·004 + 14·4∕51,000 = ·000682 = 6·82 parts CO₂ in 10,000 of air.

_The air of a room occupied by 6 persons and containing 5,000 cubic feet of space, yields 7·5 parts of CO₂ per 10,000 parts of air. How much air is being supplied per hour?_

A = P/(x - p) = ·6 x 6/(·00075 - ·0004) = +10,280+ cubic feet.

_In the same room what would be the condition of the air at the end of 4 hours?_

x = ·0001 + ·6 × 6 × 4/(10280 × 4 + 5,000)

= ·0004 + 14·4∕46,120 = ·000712 = +7·12 of CO₂+ in 10,000 of air.

_Given two sleeping rooms, Y 10 ft. by 15 ft. and 10 ft. high, Z 15 ft. by 20 ft. and 12 ft. high, with three adults in each; how much fresh air would you supply in each? What would be the condition of the air of each of the rooms after, 1∕4, 1∕2, 1, and 2 hours respectively?_

Amount of fresh air to be supplied in _Y_—

A = P/(x - p) = ·6 × 3/(·0005 - ·0004) = _9,000_ cubic feet per hour.

Condition of air in Y after 1∕4 hour— Here p = ·0004. P = ·6 × 3∕4 = ·45. A = 9,000∕4 + 1,500 = 3,750. x = ·0004 + ·45∕3,750 = +·00052+.

At the end of 2 hours— x = ·0004 + 3·6/(18,000 + 1,500) = +·000584+.

And similarly for Z.

_Suppose two rooms, one 10 feet cube, the other 50 feet by 20 feet and 15 feet high, have continuously admitted into each of them a volume of fresh air containing ·04 parts carbonic acid per 100 parts, amounting to 2,000 cubic feet per hour, so as to replace to that extent the air of the room; suppose also that an average adult be placed in each room: show by detailed calculation what would be the condition of impurity of air in each room, as measured by carbonic acid, at the end of 4 hours and 12 hours respectively._

In the case of the first room— P = ·6 × 4 = 2·4. A = 2,000 × 4 + 1,000 = 9,000. p = ·0004. x = ·0004 + 2·4∕9,000 = +·000667+.

The amount of impurity at the end of 12 hours, and in the second room may be similarly ascertained.

=Ventilation in relation to Temperature.=—The temperature of a given atmosphere is a most important factor in determining the ease with which it is replenished from the external air. Speaking generally, the greater the difference between the temperature of two masses of air the more rapidly an interchange occurs.

Air has weight. A column of it one inch square and extending to the uppermost limit of the atmosphere weighs about 14·6 lbs., and exerts this pressure on all substances at the surface of the earth. This pressure is exerted uniformly in all directions; but for this fact our chests would be crushed in by the external pressure on them, which amounts to over four tons. If the atmospheric pressure is diminished at any point, it is evident that the surrounding air will tend to press in this direction. Now, when air is heated it expands, and consequently the heavier fresh air flows in from all sides and pushes the lighter air upwards.

The expansion of air for every increase of 1° Cent. is ·003665 (1∕273), for every increase of 1° Fahr. is ·00203 (1∕492). Thus if the air in a room is 20° F. warmer than that outside, it will be expanded to 1∕25 additional bulk.

Thus if M = volume of a given air at 32°, with the barometer at 30 inches, and M₁ = volume at temperature t° above 32°, while _a_ = co-efficient of expansion for each degree of elevation of temperature, then the dilatation effected by heat will be expressed by the formula—

M₁ = M (1 + at).

When the temperature is decreasing

M₁ = M (1 -at).

If the air in a chimney flue is cooler than the air of the room with which it communicates, it will flow down into the room. It is the object of an economical fire-place to cause the chimney to act as an outlet for the products of combustion and for the impurities of the air of the room with the smallest possible waste of heat. Short of producing a down draught of cold air and smoke, the smaller the difference between the temperature of the air of a room and of the air escaping near the top of the chimney, the greater the economy of fuel.

The =movement of air in flues and other outlets= is governed by general laws, like those governing the general movements of fluids, but allowances require to be made for friction in the channels of entrance and outlet.

The theoretical velocity, when friction is not taken into account, may be calculated by a formula based on what is known as the _law of Montgolfier_, or the law of spouting fluids. According to this law, fluids pass through an opening in a partition with the same velocity as a body would attain in falling through a height equal to the difference in depth of the fluid on the two sides of the partition, _i.e._ to the difference of pressure on the two sides. Thus, if AB equals the height of a column of air at, say, 50° F., and AC is the height of the same quantity of air heated to 60°, then the velocity with which the warmer air ascends will be that which a body would acquire in falling from C to B.

Now the velocity in feet per second of falling bodies is about eight times the square root of the height from which they have fallen; and the formula for determining this is—

v = c √(2gh) = 8·2c √h.

In this formula v = required velocity in feet per second;

g = 32·17 feet per second;

h = distance fallen through by the body;

c = a constant determined by experiment, and expressing the proportion of the actual to the theoretical velocity.

Adapting this formula to the special circumstances under which Montgolfier’s formula holds, we find that the force which drives the warm air up the flue is the force of gravity, _i.e._ of the excess of the weight of a column of cold air over the weight of a column of warm air of exactly the same size (represented by BC in the preceding diagram). The difference of the two weights or pressures is found by multiplying the distance from the point of escape of heated air out of the room (fire-place or elsewhere) to the point of escape into the outer air (top of chimney or other point of exit), by the difference in temperature inside and outside, and again multiplying this product by 1∕492 for degrees of Fahrenheit temperature, or 1∕273 for degrees Centigrade.

Thus omitting c for the present, we have—

v = √(2g_h_(t - t^1)/492) = 8·2√(h(t - t^1)/492)

Where t = temperature in the chimney,

t^1 = temperature of the external air, and

h = height of chimney.

_Example.—The chief means of ventilating a given room is by its open fire-place. The temperature in the chimney is 100° F., that of the external air 40°, and the height of the chimney 50 feet; what is the velocity with which air is leaving the room?_

v = 8·2 √((100 - 40) × 50∕492)

= +20+.

This gives the theoretical velocity, but the real velocity will differ from the theoretical by an amount varying from 20 to 50 per cent.

It will be evident, from what has been said, that the movements of the air in a confined space are dependent upon (1) the difference between the internal and external temperatures; (2) the area and friction at the apertures through which air enters and leaves the room; and (3) the height of the column of ascending warm air. The higher the chimney (assuming it to contain warm air), the greater the draught and the more efficient the ventilation of the room communicating with it. Hence ventilation is more difficult in upper rooms of large houses and in single-storeyed houses than in the lower storeys of large houses.

=Allowance for Friction.=—Practically the friction varies greatly according to the size, form, and material of outlet for air. A rough or sooty or angular chimney greatly impedes the outgoing current of air.

It is usual to reduce the theoretical velocity by 20 to 50 per cent. Apart from the friction which is governed by roughness and length of channels, that due to bends in the channel may be calculated by the formula 1/(1-_sin_^2 θ), θ being the angle at any bend in this channel.

(It may be convenient to note that—

_sin_^2 90° = 1, _sin_^2 60° = 3∕4, _sin_^2 45° = 1∕2, _sin_^2 30° = 1∕4.)

Thus every right angle in a bent shaft reduces the velocity in it by one-half.

The loss by friction in two similar tubes of equal sectional area varies (1) directly with the square of the velocity of the air currents; and (2) directly with the length of the outlet channel. In two similar tubes of unequal size the loss by friction is (3) inversely as the diameter of the cross-section in each.

When two tubes are of different shapes, the loss by friction is inversely as the square roots of the sectional areas.

Owing to the variable value of the co-efficient of friction (called _c_ in the first formula given), it is usually preferable to measure the actual rate of progress of air through a given flue by means of an anemometer (wind measure). Then the velocity of the current of air and the area of the cross section of the flue being given, the volume of air discharged in a given time is represented by the product of these two and the time which has elapsed.

Thus, q = a × v.

Where q = quantity of air discharged in a given time, a = area of cross section of flue, v = velocity of current.

By means of this formula, the area of chimney required to discharge a given volume of air at a given average velocity can be ascertained. Thus—

a = q/v.

The application of the preceding principles and formulæ will be rendered clearer by the following examples.

_How much inlet and outlet area per head will be required to give 10 persons in a room of 5,000 cubic feet capacity, 2,000 cubic feet of air per head per hour, supposing that the outside temperature is 40°, while the internal temperature is 60°, and the height of the heated column of air 20 feet?_

First ascertain the velocity of entrance and exit of air.

v = 8·2√(h(t - t^1)/492)

= 8·2√(20(60 - 40)/492) = 8·2 × ·902.

= 7·3964 = velocity in feet per second.

If we allow one-fourth for friction, then there remains a velocity of 5·5473 feet per second.

5·5473 feet per second = 19700·8 feet per hour.

Now, a = q/v

= 2,000∕19700·8 = ·1015 square feet.

= +14·6 square inches.+

Thus the size of the outlet required per head is 14·6 square inches. The size of the room and the number occupying it do not enter into the question, except for a short time at the beginning. (See page 135.)

The amount of inlet required will also be 14·6 square inches per head. Theoretically it ought to be slightly less than that required for outlet, as the outgoing air is more expanded than that entering the room; but practically no allowance need be made for this fact.

The total amount of inlet and outlet required per head = +29·2 square inches.+

_If the mean temperature of a room is 61°, the external temperature 45°, while the heated column of air is 50 feet, and the required delivery of air 2,000 cubic feet per hour, find the size of inlet and outlet._

v = 8·2√(h(t - t^1)/492)

= 8·2√(50(61 - 45)/492)

= _10·55 feet per second._

= _37,980 feet per hour._

If we make no allowance for friction, then

a = q/v

= 2,000∕37,980 square feet.

= 2,000 × 144∕37,980 = 7·58 square inches.

This gives the required size of outlet. The size of inlet and outlet together = _15·16 square inches._

_If 3,000 cubic feet of air are supplied in one hour through an aperture of 12 square inches to a room containing 1,000 cubic feet of space, at what rate does the air enter the room?_

12 square inches = 1∕12 square foot.

a = q/v

1∕12 = 3,000/v

Therefore v = 36,000 feet per hour.

= _10 feet per second._

_If a room is supplied with 3,000 cubic feet of air per hour, through a single opening, what must be its area, if the rate of movement of the air is 5 feet per second?_

5 feet per second = 18,000 feet per hour.

a = 3000∕18000 = 1∕6 square foot.

= _24 square inches._

As already stated, the difficulties connected with the estimation of amount of friction greatly detract from the practical value of the formulæ just given. Even the results given by anemometers are not always trustworthy, but by comparing the results given by them with those obtained by the use of Montgolfier’s formula an approximation to the truth can be obtained.

The ordinary _anemometer_ consists of four tiny vanes fixed to a spindle, so that revolutions are caused by the current of air the velocity of which is to be measured. The revolutions are counted by a mechanical arrangement. The value of the revolutions of the vanes has to be first determined by direct experiment; a known bulk of air being forced through a channel of known size at a uniform rate, and the instrument graduated accordingly. In Fletcher’s anemometer a modification of the manometer or pressure-gauge has been used for the same purpose.

=Inlets and Outlets.=—Having given the average velocity of the wind, the size of a room, and the number of persons occupying it, the size of inlet opening required can easily be calculated.

_Find the size of inlet for air in a room occupied by one person, the air moving at the average velocity of 5 feet per second, assuming that 3,000 cubic feet of air are to be supplied per hour._

Let x = size of inlet.

Then x × 60 × 60 × 5 = 3,000.

Therefore x = 3000∕18000 = 1∕6 square foot.

= _24 square inches._

_Given that the air moves at a velocity of 10 feet per second, and that the area of the inlet aperture into a room is 12 square inches, find how much air enters the room in an hour._

Let y = amount of air.

Then 10 × 60 × 60 × 12∕144 = y.

Therefore y = 3,000 cubic feet of air.

Calculations as to supply of air in a room founded on the _average_ velocity of air-currents are, however, much less trustworthy than when the velocity is determined, as previously explained, by means of Montgolfier’s formula, or, better still, by an anemometer.

The Commissioners on Improving the Sanitary Condition of Barracks and Hospitals, in their report (1861) recommended for _inlets_, one square inch for every 60 cubic feet in the contents of the room; or one square inch for every 120 cubic feet in the contents, if warm air is admitted round the fire-grate. For _outlet_ shafts on lower floors, one square inch to every 60 cubic feet, slightly increasing for the higher storeys.

=Amount of Air-space required.=—We may take 3,000 cubic feet of air as the average amount of air required hourly by each individual, and inasmuch as the air of a room cannot be changed oftener than three times an hour without producing an unpleasant draught, it follows that at least 1,000 cubic feet of space must be allowed per person.

This may be compared with the amount actually supplied under various circumstances.

In the British Army for each soldier— In permanent barracks 600 cubic ft. In wooden huts 400 „ In hospital wards at home 1,200 „ In hospital wards in the tropics 1,500 „

In general hospitals 1,000-1,500 „ In fever hospitals 2,000-3,000 „ In workhouse hospitals 850-1,200 „ In common lodging houses 300 or 350 „ Do., if occupied night and day 350 or 400 „ In workhouses 300 „

In schools— London School Board requires per scholar 130 „ English Educational Code per scholar (minimum), in old schools 80 „ Do., in new schools 120 „

_Floor-space_ has an important bearing on ventilation. In calculating the available cubic space of a room, the height over 12 feet should be disregarded. Thus, if 500 cubic feet is allowed for each individual, the floor-space should be 42 square feet. In barracks, soldiers are allowed 50 square feet of floor-space.

In the Government regulations for workhouses it is stated that there must not be more than two rows of beds, and that the height of rooms above 12 feet must not be reckoned. This gives a minimum floor-space of 25 square feet per occupant, or with dormitories 17 feet wide, a bed-space of about 3 feet.

In hospitals, the question of floor-space is extremely important, as it regulates the distance between the sick inmates and the convenience of nursing. Assuming each bed to be 3 feet wide and 6½ feet long, the distance between any two beds should be at least 5 feet. This makes the wall-space for each bed 8 feet long, and allows from 80 to 96 square feet of floor-space per bed. At St. Thomas’s Hospital, London, the floor-space is 112 square feet, and in fever hospitals it is from 150 to 300 square feet per bed. In regard to the ventilation of hospitals, it has been well said that nothing less than too much is enough.

=Means of ascertaining Cubic Space.=—Circumference of a circle = Diameter (D) × 3.1416.

Area of circle = D^2 × .7854.

Area of square = square of one of its sides.

Area of rectangle = product of two adjacent sides.

Area of triangle = base × 1∕2 height, or height × 1∕2 base.

Area of ellipse = product of the two diameters × ·7854.

Circumference of ellipse = half the sum of the two diameters × 3·1416.

Area of any polygon found by dividing into triangles, and taking the sum of their areas.

Cubic capacity of a cube found by multiplying the three dimensions together.

Cubic capacity of a cylinder = area of base × height.

Cubic capacity of a cone or pyramid = area of base × 1∕3 height.

Cubic capacity of a dome = area of base (circle) x 2∕3 height.

Cubic capacity of a sphere = D^3 x ·5236.

Area of segment of a circle found by adding to 2∕3 of product of chord and height, the cube of the height divided by twice the chord.

(Ch x H x 2∕3) + H^3∕2 Ch

_Give the dimensions of a circular ward for 12 patients, each to have 1,750 cubic feet of available air-space._

Capacity of ward = 1,750 X 12 = 21,000 cubic feet.

If we allow 120 square feet floor-space for each patient, then the total floor-space will be 1,440 square feet. Consequently the height of the ward = 21000∕1440 = 14·75 feet.

Area of circle = D^2 x ·7854.

1,440/·7854 = D^2.

Therefore D = +43·2 feet.+

Circumference of circle = D x 3·1416.

= 43·2 x 3·1416.

= +135·7 feet.+

The dimensions of the circular ward required are therefore a height of 14·75 feet, diameter of 43·2 feet, and circumference of 135·7 feet.

_Find the cubic capacity of a circular hospital ward 28 feet in diameter, 10 feet high, and with a dome-shaped roof 5 feet high._

Area of floor-space = D^2 x ·7854.

= 614·8 square feet.

Cubic capacity of the cylinder below the dome is 614·8 x 10 = 6,148 cubic feet.

Cubic capacity of dome = 614·8 x 2∕3 x 5 = 2049·3 cubic feet.

Total cubic capacity of the ward = 8197·3 cubic feet.

In practical measurements of rooms, deductions must be made from the cubic space for the furniture contained in it and for its inmates. About 10 cubic feet ought to be allowed for each bed and bedding, and 2½ to 4 cubic feet for each individual. Projecting surfaces must be allowed for by subtraction, and recesses by addition.

_A circular ward with a diameter of 36 feet has a dome-shaped roof, the height of whose centre is 18 feet. The height to the dome is 12 feet. Find the floor-space and total cubic contents. How many patients ought the ward to accommodate?_

Area of floor-space = (36)^2 x ·7854.

= 1017·8784 square feet.

Cubic capacity of cylinder below dome = 1017·87 x 12 = 12214·5 cubic feet.

Cubic capacity of dome = 1017·87 x 2∕3 x 6 = 4071·48 cubic feet.

Total cubic capacity of the ward = 16285·98 cubic feet.

Assuming that 1,500 cubic feet are required for each patient, then the ward is large enough for 10 _patients_. It is well to test this conclusion by calculating whether sufficient floor-space has been allowed for each patient. The floor-space has been found to be about 1,018 square feet, which would give 100 feet for each of 10 patients and more than the minimum standard previously stated.

_What number of people should be allowed to sleep in a dormitory 40 feet long, of which the accompanying sketch is a section?_

The cubic capacity of the quadrilateral space below the roof = 16 x 9 x 40 = 5,760 cubic feet.

Area of floor = area of base of roof = 40 x 10 = 640 square feet.

Cubic capacity of roof = 640 x 1∕3(13 - 9). = 853·3 cubic feet.

Total cubic capacity of dormitory = +6613·3 cubic feet.+

If we take the low standard of common lodging houses and allow 350 cubic feet of space for each inmate, then 18 _persons_ may be allowed to sleep in the dormitory.

_How much space would a man occupy supposing him to weigh 175 lbs.? How much is usually allowed for a man with his clothes, bed and bedding?_

The space occupied by a man is stated by Parkes to be from 2½ to 4 cubic feet (say 3 for the average). He gives the following rule: The weight of a man in stones divided by 4 gives the cubic feet he occupies. Thus, a man weighing 175 lbs. would occupy 3-1∕8 cubic feet of space.

About 10 additional cubic feet must be allowed for clothing, bedding, and bed for each person.

_What size of inlet and outlet aperture should be allowed per head? How large should each individual inlet be made? If an inlet aperture 100 square inches in area is divided into four, with apertures of 25 square inches each, what is the loss by friction?_

A size of 24 square inches per head for inlet and the same for outlet meets common conditions.

It is desirable to make each individual inlet not larger than 48 to 60 square inches in area, _i.e._ large enough for two or three men; and each outlet not larger than one square foot, or enough for six men (Parkes). This ensures more uniform diffusion of the air throughout a room. On the other hand, the loss by friction is greatly increased by having a number of small openings instead of one large opening. This loss is inversely to the square roots of the respective areas. Thus the square root of 100 is 10; the sum of the square roots of the four apertures of 25 square inches each is 20. The loss by friction is double in the second case what it was in the undivided opening. It is evident, therefore, that in order to get as much air through the four openings as through the original large opening, each must be equal in size to half the original opening.

_Why is ventilation more difficult in upper rooms of large houses and in single-storied houses than in the lower storeys of large houses?_

Cold external air being heavier than the internal warm air presses downwards to the lowest point, and pushes up the warmer air. If there were a vacuum in the room, air would rush into it with a velocity which, as seen before, is represented by the formula—

v = √(2_gs_).

Where g = 32, s = height of column of air, which we may take as roughly 5 miles.

From this formula we obtain v = 1,306 feet per second.

It is evident that in such a case the velocity of entry of air into a vacuum on the ground floor would be greater than into a vacuum on any of the higher storeys, owing to the greater velocity acquired through the increased action of gravity.

And the same increased facility of entry of air into lower rooms must hold good under ordinary circumstances, inasmuch as by Montgolfier’s formula (which is founded on the fundamental formula v = √(2_gs_)

v = √(2_gh_(t-t^1)/492)

h = distance between top of chimney and floor of room in question, and thus the velocity with which air enters is governed by the difference between the internal and external temperature, and the height from which the cold air descends in order to take the place of the air which has escaped.