HOMEWORK 4 SOLUTIONS

1.  Using the psychrometric chart,
    Tdp = 19.5° C dew point temp
     Twb = 21.9° C wet bulb temp
     w = 0.0142 kg H2O/kg dry air humidity ratio
     h = 65 kJ/kg dry air enthalpy
     v = 0.872 m3/kg dry air volume

2.  P_atm = 101.32 kPa
     P = P_a + P_v
     w = P_v / 1.605 (P_atm – P_v)
     0.0142 kg H₂O/kg dry air = P_v / 1.605 (101.32 kPa – P_v)
     0.0142 kg H₂O/kg dry air (1.605) (101.32 kPa – P_v) = P_v
     2.31– 0.0227 P_v = P_v
      P_v = 2.26 kPa
      P = P_a + P_v
      101.32 = P_a + 2.26
       P_a = 99.06 kPa

3.  M_e = m/(1-m)
     M_e = 0.185/(1-0.185)
     M_e = 0.227
     1-rh = exp [-cTM_eⁿ]
     c = 1.98 * 10^-5
     n = 1.9
     1-rh = exp [-1.98*10^-5 (21+273)K (22.7)^1.9]
     rh = 88.2%

     Yes. There will be condensate because the dew point temperature of the
     surrounding air is greater than 4° C. It is approximately 19.2° C.

4.  N₂: 78.084 % * 28 g/mol = 21.8635 g/mol
     O₂: 20.946 % * 32 g/mol = 6.7027 g/mol
     Ar: 0.934 % * 40 g/mol = 0.3736 g/mol
     CO₂: 0.033 % * 44 g/mol = 0.0145 g/mol
     Ne : 0.0018 % * 20 g/mol = 0.00036 g/mol
     He: 0.0005 % (negligible)
     CH₄: 0.0002 % (negligible)
     MW_air = (21.8635 + 6.7027 + 0.3736 + 0.0145 + 0.00036) g/mol
     MW_air = = 28.9547 g/mol

5. 
From Psychrometric chart:
Hi = 0.0143 kg H₂O/kg dry air @ T = 28° C
Rh = 60%
Ho = 0.013 kg H₂O/kg dry air @ T =18° C
Mass Balance:


(100 kg dry air/hr) (0.0143 – 0.013) kg H₂O/kg dry air
0.13 kg H₂O/hr

6. 
Assume 1 kg of corn
Y = X + 1
Mass Balance:
0.12X +m0.185(1) = .14(X+1)
0.12X + 0.185 = .14X +.14
0.02X = 0.045
X = 2.25 kg per kg of 18.5% moisture corn

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