(32) Determine tlwb by drawing a straight line between the adp and the entering conditions to the apparatus (GSHF line). Where tldb intersects this line, read the
tlwb (Fig. 52).
tlwb = 58 F wb
The straight line connecting the leaving conditions at the apparatus with the theoretical condition of the air entering the evaporative humidifier represents the theoretical process line of the air. This theoretical condition of the air entering the humidifier represents what the room conditions are if the humidifier is not operating. The slope of this theoretical process line is the same as RSHF (.94).
The heavy lines on Fig. 52 illustrate the theoretical air cycle as air passes through the conditioning apparatus to the evaporative humidifier, then to the room, and finally back to the apparatus where the return air is mixed with the ventilation air. Actually, if a straight line were drawn from the leaving conditions of the apparatus (58.4 F db, 58 F wb) to the room design conditions, this line would be the RSHF line and would be the process line for the supply air as it picks up the sensible and latent loads in the space (including the latent heat added by the sprays).
The following two methods of laying out the system are recommended when the humidifier is to be used for both partial load control and reducing the air quantity.
1. Use two humidifiers; one to operate continuously, adding the moisture to reduce the air quantity, and the other to operate intermittently to control the humidity. The humidifier used for partial load is sized for the effective room latent load, not including that produced by the other humidifier. If the winter requirements for moisture addition are larger than summer requirements, then the humidifier is selected for these conditions. This method of using two humidifiers gives the best control.
2. Use one humidifier of sufficient capacity to handle the effective room latent heat plus the calculated amount of latent heat from the added moisture required to reduce the air quantity. In Part B, Step 5, the humidifier would be sized for a latent load of 48,600 Btu/hr.
(4000 x 95) + (6000 x 70)
tedb = = 80 F db10,000
tewb = 69.8 F wb
Part 1. Load Estimating | Chapter 8. Applied Psychrometrics Sensible Cooling
A sensible cooling process is one that removes heat from the air at a constant moisture content, line (1-2, Fig. 48. Sensible cooling occurs when either of the following conditions exist:
1. The GSHF as calculated or plotted on the psychrometric chart is 1.0.
2. The ESHF calculated on the air conditioning load estimate form is equal to 1.0.
In a sensible cooling application, the GSHF equals 1.0. The ESHF and the RSHF may equal 1.0. When only the RSHF equals. 1.0, however, it does not necessarily indicate a sensible cooling process because latent load, introduced by outdoor air can give a GSHF less than 1.0. The apparatus dewpoint is referred to as the effective surface temperature (tes) in sensible cooling applications. The effective surface temperature must be equal to, or higher than, the dewpoint temperature of the entering air. In most instances, the tes does not lie on the saturated line and, therefore, will not be the dewpoint of the apparatus. However, the calculations for ESHF, tadp and cfmda may
still be performed on the term tes for tadp. The use of the
term cfm da in a sensible cooling application should not be construed to indicate that dehumidification is occurring. It is used in the “Air Conditioning Load Estimate” form and in Example 5 to determine the air quantity required thru the apparatus to offset the conditioning loads.
The leaving air conditions from the coil are dictated by the room design conditions, the load and the required air quantity. The effective surface temperature may be found by using equation 36.
Example 5 illustrates the method of determining the apparatus dewpoint or the effective surface temperature for a sensible cooling application.
Example 5- Sensible Cooling
Given:
Location – Bakersfield, California Summer design – 105 F db, 70 F wb Inside design – 75 F db, 50% maximum rh RSH – 200,000 Btu/hr
RLH – 50,000 Btu/hr Ventilation – 13,000 cfmoa Find:
1. Outdoor air load (OATH) 2. Grand total heat (GTH)
3. Grand sensible heat factor (GSHF) 4. Effective sensible heat factor (ESHF)
5. Apparatus dewpoint (tadp) or the effective surface temp. (tes)
6. Dehumidified air quantity (cfmda)
7. Entering and leaving conditions at the apparatus (tedb, tewb, tldb, tlwb)
Solution:
1. OASH = 1.08×(105-75)×(13,000 = 420,000 Btu/hr (14) OALH = .68×(54-64)×13,000 = -88,500 Btu/hr (15) The latent load is negative and a greater absolute value than the room latent load. Therefore, the inside design conditions must be adjusted unless there is a means to humidify the air.
Room latent heat = 50,000 Btu/hr
Room moisture content = 54+ = 59.65 grains Adjusted inside design –75 F db, 59.65 grains
OALH = .68× (54-59.65) ×13,000 = -50,000 Btu/hr (15) OATH = 420,000+(-50,000)= 370,000 Btu/hr (17) 2. TSH = 200,000+420,000 = 620,000 Btu/hr (7) TLH = 50,000+(-50,000) = 0 (8) GTH = 620,000+0 = 620,000 Btu/hr (9) 3. GSHF = = 1 (27)
This is a sensible cooling application since GSHF=1
4. Assume a bypass factor of 0.05 from tables 61 and 62.
ESHF=
(26) 5. Plot the ESHF to the saturation line on the
psychrometric chart. The apparatus dewpoint is read as tadp = 48.8 F, fig. 53.
(36) 6. cfmda = =
(36) Since the dehumidified air quantity is less than the outdoor ventilation requirements, substitute the cfmoa for cfmda. This results in a new effective surface temperature which does not lie on the saturated line. tes = 75 - = 58.4 F (36) This temperature, tes, falls on the GSHF line.
7. This is an all outdoor air application since the cfmda is less than the ventilation requirements therefore:
tedb = toa = 105F
tewb = 70F
Calculate the tsa which equals the tldb by subsituting tes for tadb in equation (28).
tldb = tsa = 105-(1-.05) (105-58.4) = 60.7 F (28) Determine the tlwb by drawing a straight line between the tes and the entering conditions at the apparatus. (This is the GSHF line.) Where tldb intersects this line, read tlwb tlwb=54.6 F
In Example 5, the assumed .05 bypass factor is used to determine tes and dehumidified air quantity. Since the dehumidified air quantity is less than the
NOTE: Numbers in parentheses at right edge of column refer to equations beginning on page 150.
.68 x 13,00050,000 620,000 620,000 200,000 + (.05) 420,000 + 50,000 + (.05) (-50,000) = .823200,000 + (.05) 420,000 200,000 + (.05) 420,000 221,000 1.08 x (75 – 48.8) (1 - .05) 26.9= 8,230 CFM 200,000 + (.05) 420,000 1.08 x (1 - .05) x 13,000
ventilation air requirement, the .05 bypass factor is used again to determine a new tes, substituting the ventilation air requirement for the dehumidified air quantity. The new
tes is 58.4 F.