The Choice of Supply Design Conditions
If there is a continuous source of heat having an output of Q in a hermetically sealed room the temperature within the room, fr, will rise until the flow of heat through the walls, of area A and thermal transmittance U, equals the output of the source:
Q = AU(t’-t0) (6.1)
In which t0 is the outside air temperature.
It then follows that
Tr = t0 + Q/AU (6.2)
And hence tt will always exceed t0.
EXAMPLE 6.1
Calculate the temperature maintained in a room of dimensions 3 m x 3 m x 3 m if the average f/-value of the walls, floor and ceiling is 1.1 W m-2 K-1, the outside temperature is 27°C and the heat gains within the room are 2 kW.
Answer
By equation (6.2):
Tr = 27 + 2000/(6 x 32 x 1.1) = 27 + 33.7 = 60.7°C
This is clearly too high for the sustained comfort of any of the occupants and the first possible remedy to consider is mechanical ventilation. If outside air with a density p and a specific heat capacity c is supplied to the room at a rate of n air changes per hour, it will absorb some of the heat gain as its temperature rises to the value of tT before being discharged to waste. If we write V for the volume of the room a new heat balance can be established:
Q = pcnV(tT — t0)/3600 + AU(tT — t0) (6.3)
And hence a new expression for room temperature is
Tr = t0 + 3600g/(pcnV + 3600AU) (6.4)
Although tr still exceeds t0 the excess will not be as great since the denominator of the second term in equation (6.4) is larger than that in equation (6.2).
For the conditions of example 6.1 calculate tx if the room is ventilated at a rate of 10 air changes per hour, the density of air being 1.15 kg m-3 and its specific heat 1.034 kJ kg“1 K"1.
Answer
By equation (6.4):
Fr = 27 + 3600 x 2000/(1.15 x 1034 x 10 x 33 + 3600 x 6 x 32 x 1.1)
= 27 + 13.5°C = 40.5°C
The answers to examples 6.1 and 6.2 are numerically correct but misleading. First, the supply air temperature rises as it passes through the fan used to deliver it to the room (see section 6.5) and, secondly, the thermal capacity of the room walls, floor and ceiling will act as a reservoir for the heat gain, storing some of it for a while before releasing it to the space and mitigating the temperature rise of the air (see sections 7.17 to 7.20). Equation (6.4) shows that increasing the air change rate of mechanical ventilation gives diminishing returns. Supplying more than about ten air changes per hour is seldom worthwhile, without the use of mechanical refrigeration.
EXAMPLE 6.3
For the conditions of example 6.2 calculate the value of the steady outside temperature which will result in an inside steady-state value of 22°C. Assume the air delivered rises in temperature by 0.5°C as it is handled by the supply fan but ignore the effect of the thermal capacity of the room.
Answer
By equation (6.4)
22° = tQ+ 13.5 + 0.5 t0 = 8.0°C
The heat loss through the room fabric plus the cooling effect of 10 air changes per hour delivered at 8.0°C exactly balance the internal heat gain of 2 kW. It is clear that for 22°C to be maintained within the room, in the presence of 2 kW of heat gain, when the outside temperature is greater than 8.1°C the air must be artificially cooled. This can be generalised as follows:
Sensible heat gain = mass flow rate of supply air
X specific heat capacity x temperature rise = m x c x (fr — fs) (6.5)
Where m is the supply mass flow rate and ts is the supply air temperature. The volumetric flow rate of air at temperature t, that corresponds to the mass flow rate m, is v, and equals the quotient mlp, where p, is the air density at temperature t. We can now apply a Charles’ law correction (see section 2.5) and write p( = p0(273 +10)/(273 +1) where p0 is a standard air density at a standard temperature t0. Hence,
Sensible heat gain = [v, xpcx (273 + t0)/(273 + ()] x c x (tT — ts)
If we choose p0 = 1.191 kg m 3 at 20°C dry-bulb and 50 per cent saturation, and put c =
1.026 kJ kg-1 then
. sensible heat gains (273 + t) _
—— XT58- <6-6)
If the sensible heat gain is expressed in kW then v, is in m3 s-1 but if the gain is in W then v, is in litres s“1. Equation (6.6) can equally be adopted for a heat loss when ts must then be greater than tr. In using equation (6.6) it is important to note that it is based on the fundamental principle shown by equation (6.5). The reason why it is necessary to establish equation (6.6) is because it is so very useful: air distribution fittings, duct systems, air handling plants, fans and other items of equipment are all sized and selected in terms of volumetric flow rate, not mass flow rate. It is also very important to understand that, since it is based on a mass flow rate absorbing the sensible heat gains, the volumetric flow rate must be associated with the corresponding air density. Thus the value of t in the expression (273 + t) is always the temperature at which the volumetric airflow rate, v;, is expressed. Very often t is also fs, but not always.
EXAMPLE 6.4
A room measuring 3mx3mx3m suffers sensible heat gains of 2 kW and is to be maintained at 22°C dry-bulb by a supply of cooled air. If the supply air temperature is 13°C dry-bulb, calculate
(a) the mass flow rate of air which must be supplied in kg s-1,
(,b) the volumetric flow rate which must be supplied in m3 s-1,
(c) the volumetric flow rate which must be extracted, assuming that no natural infiltration or exfiltration occurs.
Take the specific heat of air as 1.012 kJ kg-1 K-1.
Answer
(a) By equation (6.5)
M = 2/[ 1.012 x (22 — 13)] = 0.2196 kg s“1 (.b) By equation (6.6)
V 2 (273 + 13)
13 (22 — 13) 358
= 0.1775 m3 s’1 at 13°C
It is important to observe that, for a given mass flow rate, a corresponding volumetric flow rate has meaning only if its temperature is also quoted. This principle is not always adhered to since it is often quite clear from the context what the temperature is.
(c) That a variation of calculable magnitude exists is shown by evaluating the amount of air leaving the room. Since no natural exfiltration or infiltration occurs, the mass of air extracted mechanically must equal that supplied.
By equation (6.6)
. 2 w (273 + 22)
V22 (22 — 13) X 358
= 0.1831 m3 s’1 at 22°C
|
. _ . w (273 + 22) У22 — Vi3 X (273 + 13)
= 0.1831 m3 s-1 at 22°C |
Equation (6.6) can be used then to express the volumetric flow rate at any desired temperature for a given constant mass flow rate. The equation is sometimes simplified. A value of 13°C is chosen for the expression of the volumetric flow rate. We then have
|
Flow rate (m3 s *) = |
Sensible heat gain (kW) _ (273 + 13) Ц7Ц X 358
|
(6.7) |
Sensible heat gain (kW) „
———- ;——- ————— x 0.
This version suffers from the disadvantage that a Charles’ law correction for temperature cannot be made and so the answers it gives are correct only for air at 13°C. The inaccuracy that may result from this is usually unimportant in most comfort conditioning applications.
Posted in Air Conditioning Engineering