A feedwater heater operates at steady state with liquid water entering at inlet 1 at 7 bar, 42C, and a mass flow rate of 70 kg/s. A separate

Question

A feedwater heater operates at steady state with liquid water entering at inlet 1 at 7 bar, 42C, and a mass flow rate of 70 kg/s. A separate stream of water enters at inlet 2 as a two-phase liquid–vapor mixture at 7 bar with a quality of 98%. Saturated liquid at 7 bar exits the feedwater heater at 3. Ignoring heat transfer with the surroundings and neglecting kinetic and potential energy effects, determine the mass flow rate, in kg/s, at inlet 2.

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Khánh Gia 5 years 2021-08-29T17:15:07+00:00 1 Answers 15 views 0

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  1. thugiang
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    2021-08-29T17:16:08+00:00
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    Answer:

    18.76kg/s

    Explanation:

    From our data, we have:

    m_1=70kg/s, \ p_1=7bar,\ T_1=40\textdegree C\\p_2=7bar\\p_3=3bar

    The steady-state mass rate rate equation gives

    \sum_i \dot m_i=\sum_e \dot m_e\\\dot m_1 +\dot m_2=\dot m_3

    The steady-state energy balance gives;

    0=\dot Q_c_v-\dot W_c_v+\sum_i \dot m_i(h_i+0.5V_i^2+gz_i)-\sum_e \dot m_e(h_e+0.5V_e^2+gz_e)

    #Neglect heat transfer and KE and PE effects

    0=\sum_i \dot m_ih_i-\sum_e \dot m_eh_e\\0=\dot m_1h_1+\dot m_2h_2-\dot m_3h_3

    #Substitute for \dot m_3 from the mass rate balance:

    0=\sum_i \dot m_ih_i-\sum_e \dot m_eh_e\\0=\dot m_1h_1+\dot m_2h_2-\dot m_3h_3\\\\\dot m_2=\frac{\dot m_1(h_3-h_1)}{h_2-h_3}

    Inlet 1: water is a compressed liquid. From Table A-2,h_1\approx h_f_1=209.33kJ/kg

    inlet 2,steam is super-heated. From Table A-4, h_2=2827.9kJ/kg

    inlet 3: water is a saturated liquid. From A-3 table,h_3=h_f_3=762.81kJ/kg

    Substitute values in \dot m_2=\frac{\dot m_1(h_3-h_1)}{h_2-h_3} to find the mass flow rate at inlet 2:

    \dot m_2=\frac{(70kg/s)(762.81kJ/kg-209.33kJ/kg)}{2827.9kJ/kg -762.81kJ/kg}\\=18.76kg/s

    Hence, the mass flow rate at inlet 2 is 18.76kg/s

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