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    Tutorial PTT 108 - 1

    Uploaded by

    Husna Hafiza Bt. R.Azami
    1. The rate of kinetic energy transported by the methanol is 0.0105 J/s. 2. At 30°C and 20% relative humidity: the absolute humidity is 0.0053 kg H2O/kg dry air, the wet-temperature is 15.6°C, the humid volume is 0.8665 m3/kg dry air, the dew point is 4.4°C, and the specific enthalpy is 43.6 kJ/kg dry air. 3. For the steam condenser problem, the required heat input is 8.9695 kW.

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    Tutorial PTT 108 - 1

    Uploaded by

    Husna Hafiza Bt. R.Azami
    21 views6 pages
    1. The rate of kinetic energy transported by the methanol is 0.0105 J/s. 2. At 30°C and 20% relative humidity: the absolute humidity is 0.0053 kg H2O/kg dry air, the wet-temperature is 15.6°C, the humid volume is 0.8665 m3/kg dry air, the dew point is 4.4°C, and the specific enthalpy is 43.6 kJ/kg dry air. 3. For the steam condenser problem, the required heat input is 8.9695 kW.

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    Tutorial PTT 108_1

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    1. The rate of kinetic energy transported by the methanol is 0.0105 J/s. 2. At 30°C and 20% relative humidity: the absolute humidity is 0.0053 kg H2O/kg dry air, the wet-temperature is 15.6°C, the humid volume is 0.8665 m3/kg dry air, the dew point is 4.4°C, and the specific enthalpy is 43.6 kJ/kg dry air. 3. For the steam condenser problem, the required heat input is 8.9695 kW.

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    21 views6 pages

    Tutorial PTT 108 - 1

    Uploaded by

    Husna Hafiza Bt. R.Azami
    1. The rate of kinetic energy transported by the methanol is 0.0105 J/s. 2. At 30°C and 20% relative humidity: the absolute humidity is 0.0053 kg H2O/kg dry air, the wet-temperature is 15.6°C, the humid volume is 0.8665 m3/kg dry air, the dew point is 4.4°C, and the specific enthalpy is 43.6 kJ/kg dry air. 3. For the steam condenser problem, the required heat input is 8.9695 kW.

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    of 6

    Tutorial PTT 108

    Material and Energy Balance

    1. Liquid methanol is pumped from a storage through a 1-in.ID pipe at a rate of 3.00
    gal/min. At what rate is the kinetic energy being transported by the methanol in the
    pipe. (Answer should be in SI unit).

    Answer:
    Check conversion unit table (all units should be converted to SI unit):
    1 m = 39.37 in.
    1 m3 = 264.17 gal
    1 min = 60 s
    ID = Inner diameter
    = 1-in.

    Volume flowrate
    = 3.00 gal/min +
    A =πr2

    Given that, velocity (m/s) = volume flowrate (m3/s) / cross-sectional area (m2)

    [ ]

    [ ]

    Given that, mass flowrate (kg/s) = [volume flowrate (m3/s)] x [density (kg/m3)]

    First determine the density of methanol: (Confirm the substance that you deal with in
    the problem, don’t just simply take the density of water!)
    Given that from Table B.1, the specific gravity (SG) for methanol is 0.792, thus the
    density of methanol is 0.792 g/cm3.

    ̇ [ ] [ ]

    Given that, rate of kinetic energy (J/s) = ½ x [mass flowrate (kg/s)] x [velocity (m/s)]2
    ̇ [ ] [ ]

    ̇ = 0.0105 J/s

    2. Use the psychrometric chart (given in the Appendix) to estimate (at temperature 30°C
    and 20% relative humidity (refer Example 8.4-5 as a guideline):

    i) The absolute humidity


    ii) The wet-temperature
    iii) Humid volume
    iv) The dew point
    v) Specific enthalpy

    Answer: (Refer to the attached chart)

    i) The absolute humidity =0.0053 kg H2O/kg dry air


    ii) The wet-temperature = 15.6°C
    iii) Humid volume = 0.8665 m3/kg dry air
    iv) The dew point = 4.4°C
    v) Specific enthalpy = (44 – 0.39) = 43.6 kJ/ kg dry air
    3. Superheated steam at 10 bar absolute and 400°C is feed to a condenser at a flowrate
    of 100 mole/h. The steam is cooled to 150 °C at 1 bar. Calculate the heat (Q) that
    must be transferred to or from the condenser in kW

    Inlet stream = 100 mol/h Outlet stream = 100 mol/h


    Condenser
    Water (400°C, 10 bar, v) Water (150°C, 1 bar, v)

    Using Steam Tables (Table B.7 in Appendix)


    Answer: Be careful with the unit,
    it is in kJ/kg not kJ/mol!
    At P = 10 bar, T = 400°C, Ĥ = 3264 kJ/kg
    At P = 1 bar, T =150°C, Ĥ = 2776 kJ/kg

    Substance ṅin (mol/h) Ĥin(kJ/kg) ṅout (mol/h) Ĥout (kJ/kg)


    Water (v) 100 3264 100 2776

    Given that ̇ ̇
    Thus,

    The rate change of the enthalpy:

    ̇ ∑ ̇ ̂ ∑ ̇ ̂

    Solve the energy balance equation (open system): Neglecting kinetic and potential
    energy, no shaft work, thus the required heat input:
    0 0 0
    ∆Ḣ + ∆Ek +∆Ep = Q -Ws

    ̇ ̇
    Using heat capacity data (Table B.2 in Appendix)
    Answer:
    ∆ĤH2O
    Water (400°C, 10 bar, v) Water (150°C, 1 bar, v)

    ∆Ĥ1 ∆Ĥ2
    Water (400°C, 1 bar, v)

    ∆ĤH2O = ∆Ĥ1 + ∆Ĥ2 Refer to your hypothetical process paths for water.

    ∆Ĥ1 : Change in P from 10 bar to 1 bar at constant T. Assuming ideal gas, thus ∆Ĥ1
    =0 kJ/mol
    ∆Ĥ2: Change in T from 400°C to 150°C at constant P.

    Thus,
    ∆Ĥ2 ∫
    From Table B.2 use equation Form 1 to determine the Cp:
    Cp = a + bT + cT2 + dT3
    H2O (v): Cp =33.46 x 10-3 + 0.6880 x 10-5T + 0.7604 x 10 -8T2 + (-3.593 x 10-12)T3
    So do integrate,
    =∫ ) dT

    = [ ]
    = -8.9695 kJ/mol

    Thus,

    ̇ ̇̂

    Solve the energy balance equation (open system): Neglecting kinetic and potential
    energy, no shaft work, thus the required heat input:
    0 0 0
    ∆Ḣ + ∆Ek +∆Ep = Q -Ws

    ̇ ̇

    Using specific enthalpy data (Table B.8 in Appendix)


    Answer:
    Assuming ideal gas (H2O):
    At T = 400°C, Ĥ = 13.23 kJ/mol
    At T =150°C, Ĥ = ?
    Do interpolation, since the T = 150°C falls between 100°C and 200°C:
    At 100°C, Ĥ = 2.54 kJ/mol
    At 200°C, Ĥ = 6.01 kJ/mol
    Thus,
    ̂

    Substance ṅin (mol/h) Ĥin(kJ/mol) ṅout (mol/h) Ĥout (kJ/mol)


    Water (v) 100 13.23 100 4.275

    The rate change of the enthalpy:

    ̇ ∑ ̇ ̂ ∑ ̇ ̂

    Solve the energy balance equation (open system): Neglecting kinetic and potential
    energy, no shaft work, thus the required heat input:
    0 0 0
    ∆Ḣ + ∆Ek +∆Ep = Q -Ws

    ̇ ̇

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