ASHRAE HVAC APPLICATIONS SI ADD AND CORR-2015 Additions and Corrections.pdf

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1、A.1ASHRAE HANDBOOKAdditions and CorrectionsThis report includes additional information, and technicalerrors found between June 15, 2012, and April 1, 2015, in the SIeditions of the 2012, 2013, and 2014 ASHRAE Handbook vol-umes. Occasional typographical errors and nonstandard symbollabels will be cor

2、rected in future volumes. The most current list ofHandbook additions and corrections is on the ASHRAE web site(www.ashrae.org).The authors and editor encourage you to notify them if youfind other technical errors. Please send corrections to: HandbookEditor, ASHRAE, 1791 Tullie Circle NE, Atlanta, GA

3、 30329, ore-mail mowenashrae.org. 2012 HVAC Systems and Equipmentp. 6.6, Table 1. Units for rsshould be (mK)/W.pp. 7.26-27, Figs. 36 and 38. The SI versions of these figures areprovided below and at right.p. 19.2, Table 1. For the rightmost four columns, the headingsshould be +1500, 1500, +2500, and

4、 2500 Pa.p. 25.1, Fig. 1. Point E was omitted. The correct figure appearsbelow right.p. 26.2, Eq. (2a). In the rightmost fraction, change msto me.p. 26.4, 1st col., top. Units for the results for w2and w4should bekg/kg of dry air.p. 40.19, Table 1. In fifth column, change value in row 6 to 0.0345,an

5、d in row 7 to 0.0334.p. 41.3, Fig. 3. Replace parts B and C of the graphic with the hori-zontal polymer tube shown on p. A.2.p. 44.8, Table 1. In both equations for flow and pressure, change theminuses to equals signs.p. 51.8. In the first column, change “44 153 kWh” to “44 513 kWh”in two places: th

6、e first line, and in the paragraph before Eq. (4). In thesecond column, second paragraph, equation for TES tank volume,change “44 153” to “44 513” and add “ 0.9” after “998 kg/m3.” Theresult remains 4738 m3.p. 51.13, Table 4. For hour 6, Storage should be 1400 kW.p. 51.24, Fig. 25. The bottom right

7、line should have the dischargearrow pointing to the left, and the charge arrow pointing to the right.The corrected figure is shown on p. A.2.Fig. 36 Effect of Inlet Pressure andSuperheat on Condensing Turbine(2012 HVAC Systems and Equipment, Ch. 7, p. 26)Fig. 38 Single-Stage Noncondensing Turbine Ef

8、ficiency(2012 HVAC Systems and Equipment, Ch. 7, p. 27)Fig. 1 Dehumidification Process Points(2012 HVAC Systems and Equipment, Ch. 25, p. 1)A.2 20122014 ASHRAE Handbook Additions and Corrections (SI)2013 Fundamentalsp. 1.9, 2nd col. Above Eq. (38), change “td( p, w)” to “td( p, W ).”p. 1.10, 1st col

9、. In the table for Situation 2, 3rd line, Comments col-umn, change “td” to “t.”p. 4.21, Solution for Example 10, starting in 3rd para. Compute hiusing Eq. (T8.5a). For cpw, units should be J/(kgK). Delete fsand itsvalue. Nudshould be 248.3, and hishould be 7087 W/(m2K). Com-pute housing Eq. (T9.10a)

10、. Change Ra to 74 574, Nu to 7.223, hoto3.65 W/(m2K), and the equation for hotto 3.65 + 4.3 =7.95 W/(m2K). Change the last paragraph in the example as follows:Solving for domakes the left side of Equation (42) equal to the rightside, and gives do= 0.040 03 m. Now, using the new value of 0.040 03 mfo

11、r the outer diameter, the new values of hoand hotare 3.86 W/(m2K)and 8.20 W/(m2K), respectively. The updated value of dois 0.044 03 m.Repeating the process several times results in a final value of do=0.047 17 m. Thus, an outer diameter of 0.045 m (corresponding to aninsulation radial thickness of 1

12、2.5 mm) keeps the outer surface tempera-ture at 24.1C, higher than the dew point. Another method is to useEquation (42) to solve for tofor values of docorresponding to availableinsulation thicknesses and using the insulation thickness that keeps toabove the dew-point temperature.p. 9.19, 1st col. Fo

13、r Eq. (71), remove the minus sign before “(tsk34).” In the following definitions, add that tbsetshould be 36.49C. InEq. (72), change the minus in the denominator to a plus. In the para-graph below Eq. (74), blshould be 1 kg/L.p. 9.23, after Eq. (83). WCI should be multiplied by 1.163 to getW/m2.Ch.

14、14, climate data table for Esbjerg, Denmark. Please use the2009 data for this location.p. 14.3, Table 1A. Change “Hours 8/4 for inner radius, it should be r1.p. 27.2, 2nd col. Data for elements 2 and 5 and the total should be asfollows: p. 27.3, 2nd col. For element 1, wind speed should be 6.7 m/s.p

15、. 27.4, 1st col., Example 3. In the elements table, wind speed forelement 1 should be 6.7 m/s. After the equation for UAssembly, change“300 mm” to “600 mm.”p. 27.4, 1st col., Example 4. On fourth line, face shell thicknessshould be 32 mm. In the Solution, wind speed for Roshould be6.7 m/s.p. 27.4, 1

16、st col., 2nd full para. The steel member should be 90 mmdeep.2014 Refrigerationp. 12.14, Fig. 13. Change “polyoxypropylene” to “polyalkylene.”p. 24.1, 1st paragraph under Eq. (2). Change 1.6 to 9.1, and 6 to34.p. 24.6, definitions for Eq. (16). In the definition for qs/A, the ref-erence should be to Figure 9.Element R, (m2K)/W2. 100 mm concrete, 1920 kg/m3, k = 1.1 0.095. 10 mm built-up roof membrane 0.06Total 5.83