# An application of the decay equation to changes of enthalpy

Throughout this book it has been usual to consider the steady-state case. Air has been supplied to the room being conditioned at a nominally constant temperature and since this temperature is lower than the design temperature in the room, the sensible heat gains to the room are offset. Under these conditions the cooling capacity of the airstream exactly matches the sensible heat gain to the room, and if the sensible gain stays unaltered, the temperature of the air in the room also remains unaltered.

The picture is different, however, when the air conditioning plant is first started up. Suppose that with the plant not running the conditions within the room are equal to those outside and that no heat gains are occurring. When the air conditioning system is started it delivers air to the room at a temperature very much below the value of the temperature prevailing there. Thus, the initial cooling capacity of the airstream is very large and the temperature of the air in the room is rapidly reduced. As this reduction is effected, the difference between supply air temperature and room air temperature decreases, and so the cooling capacity of the supply airstream diminishes.

This is a somewhat simplified picture of what is occurring. For example, heat gains are not solely due to transmission; solar radiation, electric lighting and occupants provide additional sensible gains and there are, of course, latent gains as well. There are, thus, complicated changes of load occurring.

By a process similar to that used to derive equation (16.13) it is possible to formulate, according to Jones (1963), a differential equation which represents the physical situation and to obtain a solution to it in terms of the relevant heat exchanges, enthalpies and masses.

The solution is

H = M{[ha + tf(0](l — e-n)} + H0c~n In this equation the following notation has been used:

16.2 An application of the decay equation to changes of enthalpy 483 M = mass of air contained in the room, in kg

H0 = initial enthalpy of the air in the room, in kJ. (Hence H0 = Mh0, where h0 is the specific enthalpy in kJ kg-1) h. d = specific enthalpy of the air supplied to the room in kJ kg-1 H(t) = the enthalpy gain to the room, expressed in kJ s-1 per kg s_1 of air supplied, at any time t. (Hence the units of H(t) are kJ kg-1. In general, H(t) is a function of time and the ease with which equation (16.15) may be made to yield a useful answer depends on how complicated this function is.)

Ga = rate of mass flow of the air supplied to the room, in kg s’1

The use of this equation is not limited to problems relating to changes of temperature or moisture content in an air-conditioned room. It may be used to solve a variety of problems involving unsteady state operation. In using the equation it must be remembered that the influence of the thermal inertia of the building itself (as expressed by the concept of admittance in the CIBSE Guide) is ignored. The errors arising from this are usually quite small in the short term, particularly if the floor is carpeted and the walls are lightweight.

EXAMPLE 16.8

A room measures 3mx6mx3m high and is air conditioned. Making use of the information given below, determine the dry-bulb temperature and relative humidity in the room three minutes after the air conditioning plant has been started, assuming that the initial state in the room is the same as the state outside. Ignore the effect of the thermal inertia of the building.

 2 kW 0.2 kW 28°C dry-bulb, 20°C wet-bulb (screen) 22°C dry-bulb, 50 per cent saturation 13°C dry-bulb, 8.055 g kg“1 0.217 kg s“1 1.025 kJ kg“1 K’1 Sensible heat gain Latent heat gain Outside state Design inside state Constant supply air state Constant supply air quantity Specific heat capacity of humid air

The humid volume at a state midway between the inside and the outside design states is about 0.8572 m3 kg-1 of dry air and this establishes the fact that the mass of air contained in the room is about 63 kg.

Consider first the change of temperature that occurs when the air conditioning plant is started. Since temperature alone is the concern, the terms in equation (16.15) involving enthalpy may be conveniently modified so as to express only the sensible components of the enthalpy.

The information available may now be summarised—

Ga = 0.217 kg s“1 M = 63 kg

H0 = 63 x 1.025 x (28 — 0) = 1807 kJ = 9.22 kJ kg-1

/ia = 1 x 1.025 x (13 — 0) = 13.32 kJ kg-1 n = Gtt/M = 0.217f/63 = 0.00344?

Then, from equation (16.15),

H = 63(13.32 + 9.22)(1 — e-0 003 44′) + 1807 e’0’003 44t

As t approaches a value of infinity, H approaches a value of 63 x 22.54 kJ, and so the specific sensible enthalpy of the air in the room tends towards a value of 22.54 kJ kg-1. This means that the dry-bulb temperature of the air in the room eventually reaches a value of 22°C, as designed for. This is its ‘potential value’ (see section 13.10).

This may not seem very satisfactory. It must be remembered though, that when the system is started up it does not usually face its full design load. There is, thus, an opportunity for the system to pull down the room air temperature to 22°C under conditions of partial load. After three minutes, for the case under consideration—

T = 3 minutes = 180 seconds

E-0.003 44 x 180 = e-0.62 = 0-538

Hence,

H = (63 x 22.54)(1 — 0.538) + 1807 x 0.538 = 1627 kJ

Thus

Dry-bulb temperature = 53 ^fll)25 =

A similar approach, with appropriate modifications to the values of enthalpy so that latent heat is taken into account, but not the sensible component, yields a figure for the change in moisture content. From tables the specific enthalpy of dry air at 28°C is 28.17 kJ kg“1 and at the outside design state it is 55.36 kJ kg-1, hence

Ga =0.217 kg s"1 M = 63 kg

= 63 x (55.36 — 28.17) = 1712 kJ H(t) = 0.2/0.217 = 0.9215 kJ kg"1 hA = 33.41 — 13.08 = 20.33 kJ kg-1 where 33.41 kJ kg-1 is the enthalpy of the

Supply air and 13.08 kJ kg-1 is the enthalpy of dry air at 13°C

Then, from equation (16.15), as before,

H = 63(20.33 + 0.9215)(1 — e0 003 44t) + 1712 e“0 003 44t

As t approaches infinity, H tends to 63 x 21.25, and so the specific enthalpy in the room

Tends to 21.25 + 22.13 = 43.38 kJ kg-1, the design value, where 22.13 kJ kg-1 is the

Enthalpy of dry air at 22°C, after an infinitely long period of time. After three minutes, as before, we get a solution as follows—

H = 63 x 21.25(1 — e“0 62) + 1712 e“062 = 1541 kJ or 1541/63 = 24.5 kJ kg“1 At 25.2°C the sensible component of enthalpy, from tables, is 25.25 kJ kg-1 and so the total enthalpy is 49.75 kJ kg“1.

At a dry-bulb temperature of 25.2°C and an enthalpy of 49.75 kJ kg the relative humidity is about 48 per cent.

Exercises

1. A classroom having a cube of 283 m3 undergoes 1 y air changes per hour from natural ventilation sources. The concentration of C02 in the outside air is 0.03 per cent and the production of C02 per person is 4.72 x 10“6 m3 s-1. Ignore the influence of the ventilation effectiveness coefficient, ev.

(a) What is the maximum occupancy if the C02 concentration is to be less than 0.1 per cent at the end of the first hour, assuming that the initial concentration is 0.03 per cent?

(b) What is the maximum occupancy if the classroom is continuously occupied and the concentration must never exceed 0.1 per cent?