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.
[Illustration: Representing the air column lengths]
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?_
[Illustration: Outline of rectangular space with triangular roof]
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.