Wasteful reheat

If an arbitrary choice of supply air temperature is made, it is almost certain that the resultant design will not be an economical one, either in capital or in running cost. The reason for this is that wasteful reheat may have to be offset by the refrigeration plant. The point is best illustrated by means of an example.

EXAMPLE6.il

The sensible and latent heat gains to a room are 10 kW and 1 kW, respectively. Assuming that the plant illustrated in Figure 6.5(a) is used and that the cooler coil has a contact factor of 0.85, calculate the cooling load and analyse its make-up if the outside condition is 28°C dry-bulb, 20.9°C wet-bulb (sling) and the condition maintained in the room is 22°C dry — bulb and 50 per cent saturation. The supply air temperature is arbitrarily fixed at 16°C. Assume that the temperature rise due to fan power is IK.

Answer

From equation (6.6)

0

O

O

W

0

0

(i)

<D

(3

Q>

Sz

2

CL

©

O

O

O

<0

<D

.c

0)

Њ

Q.

Q.

3

CO

<D

подпись: <d

Supply

State

S

подпись: supply
state
s

Conditioned

Room

R

подпись: conditioned
room
r
Wasteful reheat

Plant arrangement (a)

подпись: plant arrangement (a)

12.59 g kg-1

подпись: 12.59 g kg-1

8.366 g kg’1 8.115 g kg"1 7.33 g kg-1

подпись: 8.366 g kg'1 8.115 g kg"1 7.33 g kg-1

CM

CM

O>

CM

Wasteful reheat

Psychrometry

(b)

Fig, 6.5 Plant arrangement and psychrometry for Example 6.11.

Supply air quantity = (22° 16°) X ^358^ = m3 S * at

From equation (6.8):

, • • 1 (273 + 16)

Supply air moisture content = gT — y-^r x————————— = gs

From CIBSE tables

Gr = 8.366 g kg“1 gs = 8.366-0.251 = 8.115 g kg“1

The supply air state is thus 16°C dry-bulb and 8.115 g kg“1. This is denoted by the letter

S in Figure 6.5(b). The line joining the state point of the air entering a cooler coil to the state point of the air leaving the coil must always cut the 100 per cent relative humidity curve when suitably produced (see section 3.4). Since the line joining O and S does not cross the saturation curve when produced, the cooler coil will not give state S directly. The coil must therefore be chosen to give the correct moisture content, gs, and reheat must be used to get the correct temperature, ts.

Four things are now established for the behaviour of the cooler coil:

(i) The state of the air entering the coil is known.

(ii) It has a contact factor of 0.85.

(iii) The moisture content of the air leaving the coil must be 8.115 g kg-1.

(iv) The straight line joining the state points of the air entering and leaving the coil cuts the saturation curve at the apparatus dew point.

Making use of the above information and employing the definitions of contact factor expressed by equations (3.1) and (3.3), the state of air leaving the cooler coil W can be worked out.

Denote the apparatus dew point by A. Then, on a moisture content basis

Go — g

подпись: go - g

W

подпись: w0.85 =

So 8 a

12.59 — 8.115

12.59 — ga

Whence

Јa = 7.33 g kg’1

Thus, A is located at 100 per cent relative humidity and 7.33 g kg-1.

W can now be marked in on the line joining O to A at a moisture content of 8.115 g kg-1. The temperature of A, from tables (or from a chart), is 9.4°C and its enthalpy is 27.97 kJ kg“1.

Using the definition of contact factor once more, but in terms of differences of enthalpy: q hn hw 60.3 hw

H0 — ha 60.3 — 27.87 where h0 has been determined from tables or a chart as 60.3 kJ kg-1. Hence K = 32.82 kJ kg-1 Using the approximate definition of contact factor we have:

98° — T °’85 = 28^

Whence

Tw = 12.2°C

Since the supply airflow of 1.345 m3 s_1 is at 16°C and 8.115 g kg-1 its specific volume is 0.8294 m3 kg-1, from tables, and because we know the enthalpies at states O (on the cooler coil) and W (off the cooler coil) we can calculate the cooling load as:

Cooling load = x (60.30 — 32.82) = 44.56 kW

It is instructive to analyse the load:

Wasteful reheat

= 27.42 kW

Sensible heat gain latent heat gain

 

= 10.00 kW = 1.00 kW

 

1.345

Fan power =

подпись: fan power =

= 1.66 kW

подпись: = 1.66 kwX 1.026 x (13.2° — 12.2°) 0.8294

1.345

Reheat wasted =

подпись: reheat wasted =

= 4.66 kW

подпись: = 4.66 kwX 1.026 x (16° — 13.2°)

0.8294

Total = 44.74 kW

(This is to be compared with 44.56 kW, computed by using the difference of enthalpy across the cooler coil.)

The analysis is illustrated in Figure 6.5(b). It can be seen that about 4.66 kW plays no useful thermodynamic part. The fact that reheat was needed arose from the arbitrary decision to make the supply air temperature 16°C.

A more sensible approach would be to dispense with reheat entirely and to use the lowest practical temperature for the supply air consistent with the contact factor of the cooler coil. The temperature difference between tr and ts would then be increased and less supply air would be needed. The consequence of this would be to reduce the cooling load because, although the sensible and latent heat gains are unaffected, the fresh-air load and the cooling dissipated in offsetting fan power are reduced, since less air is handled. This is clarified in the next section.

Posted in Air Conditioning Engineering


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