Ask Question
1 April, 18:33

Two airstreams are mixed steadily and adiabatically. The first stream enters at 35°C and 30 percent relative humidity at a rate of 15 m3/min, while the second stream enters at 12°C and 90 percent relative humidity at a rate of 25 m3/min. Assuming that the mixing process occurs at a pressure of 90 kPa, determine the specific humidity, the relative humidity, the dry-bulb temperature, and the volume flow rate of the mixture. Use data from the tables.

a) The specific humidity is __kg H2O/kg dry air.

b) The relative humidity is __ %.

c) The dry-bulb temperature is __ °C.

d) The volume flow rate of the mixture is __m3/min.

+1
Answers (1)
  1. 1 April, 19:43
    0
    a) The specific humidity is _0.00881_kg H2O/kg dry air.

    b) The relative humidity is _59.8_ %.

    c) The dry-bulb temperature is _20.2_ °C.

    d) The volume flow rate of the mixture is _39.77_m3/min

    Explanation:

    First let determine the enthalpy h1 and absolute humidity W1 from the psychometric chart using temperature T1 = 35°C and relative humidity Q1 = 30%

    h1 = 62.2kJ/kg

    V1 = 0.89m³/kg

    W1 = 0.01054

    The volume flow rate V'1 = 15m³/min, the mass flow rate m'1 is,

    m'1 = V'1/v1 = 15/0.89 = 16.85kg/min

    Also, From temperature T2 = 12°C and relative humidity Q2 = 90% and volume flow rate V'2 = 25m³/min

    h2 = 31.9kJ/kg

    V2 = 0.82m³/kg

    W2 = 0.00785

    the mass flow rate m'2is,

    m'2 = V'2/v2 = 25/0.82 = 30.49kg/min

    To get the absolute humidity W3 and the enthalpy h3,

    m'1/m'2 = (W2-W3) / (W3-W1)

    16.85/30.49 = (0.00785-W3) / (W3-0.01054)

    0.5526 (W3-0.01054) = (0.00785-W3)

    0.5526W3 - 0.00582 = 0.00785-W3

    0.5526W3 + W3 = 0.00785 + 0.00582

    1.5526W3 = 0.01367

    W3 = 0.01367/1.5526 = 0.00881

    absolute humidity W3 = 0.00881

    m'1/m'2 = (h2-h3) / (h3-h1)

    16.85/30.49 = (31.9 - h3) / (h3 - 62.2)

    0.5526 (h3 - 62.2) = (31.9 - h3)

    0.5526h3 - 34.37 = 31.9 - h3

    0.5526h3 + h3 = 31.9 + 34.37

    1.5526h3 = 66.27

    h3 = 66.27/1.5526 = 42.68kJ/kg

    enthalpy h3 = 42.68kJ/kg

    Use W3 and h3 to determine temperature T3, relative humidity Q3 and and specific volume V3 all through psychometric chart

    T3 = 20.2°C

    Q3 = 59.8%

    V3 = 0.84m³/kg

    By last of mass conversations, we can determine m3,

    m3 = m1 + m2

    m3 = 16.85 + 30.49 = 47.34kg/min

    Then, the volume flow rate V'3 can be found by,

    V'3 = m3*V3

    V'3 = 47.34*0.84 = 39.77m³/min.

    volume flow rate V'3 = 39.77m³/min.
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “Two airstreams are mixed steadily and adiabatically. The first stream enters at 35°C and 30 percent relative humidity at a rate of 15 ...” in 📘 Engineering if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers