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The question begins this way: “The performance of your unit is not correct.” Why? “Because when I calculate the coil load from the stated conditions I do not get the capacity shown” Answer: When calculating the Total Capacity do not use Qt = 4.5 * cfm * (h1 – h2) Because 4.5 is derived for standard air as follows: At 100 DB and 78 WB, the W = .015601 lbw/lba h = ha + Whg = cpa*T + W*(1061 + .444*T) Btu/lba h1 =.24*100 + .015601*(1061 + .444*100) = 41.25 Btu/lba At 57.50 DB and 57.30 WB, the W = .0099622 lbw/lba h2 = .24*57.50 + .0099622*(1061 + .444*57.50) = 24.62 Btu/lba If you use this equation then you will get the following: Qt = 4.5 * 15000 * (41.25 – 24.62) = 1,122,525 Btu/hr, or a perceived error of 8.3 At 100 DB and 78 WB, the W = .015601 lbw/lba, Density = .06914 ft³/lba h1 = 41.25 Btu/lba from above At 57.50 DB and 57.30 WB, the W = .0099622 lbw/lba h2= 24.62 Btu/lba above ma = CFM*Density*60 = 15000 * .06914 * 60 = 62,226 lba/hr Qt = ma * (h[1] – h[2]) = 62226*(41.25 – 24.62) = 1,034,818 Btu/hr
Because the 1.1 is derived from standard air as follows: ma * Cpm = SCFM*DensityStd*(Cpa + W*Cpw) ma * Cpm = SCFM *.075 * 60 * (.24 + .0093*.444) = 1.1*SCFM If you use this equation then you will get the following: Qs = 1.1 * 15000 * (100-57.50) = 701,250 Btu/hr Coil programs use actual conditions: At 100 DB and 78 WB, the W = .015601 lbw/lba, Density = .06914 ft³/lba At 57.50 DB and 57.30 WB, the W = .0099622 lbw/lba ma = CFM*Density*60 = 15000 * .06914 * 60 = 62226 lba/hr Cpm = (Cpa + W*Cpw) Btu/(lba – F) Qs = ma * Cpm * (T[1] – T[2]) Qs = 622226*(.24+.015601*.444)*(100-57.5) = 652,988 Btu/hr The Total Latent Load is computed from: QL = Qt – Qs Using standard Air QL = 1,034,818 - 701,250 = 333,568 Btu/hr Using actual air: QL = 1,034,818 - 652,988 = 381,830 Btu/hr The above shows a sea level calculation. The same equations apply for altitude however the true density must include DB, WB, and PB = altitude PB. PB altitude is calculated from the following equation: PB = 14.696 * (1 - ALTITUDE*6.8753E-6)^5.2559 |
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