Answer:
18.76kg/s
Explanation:
From our data, we have:
[tex]m_1=70kg/s, \ p_1=7bar,\ T_1=40\textdegree C\\p_2=7bar\\p_3=3bar[/tex]
The steady-state mass rate rate equation gives
[tex]\sum_i \dot m_i=\sum_e \dot m_e\\\dot m_1 +\dot m_2=\dot m_3[/tex]
The steady-state energy balance gives;
[tex]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)[/tex]
#Neglect heat transfer and KE and PE effects
[tex]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[/tex]
#Substitute for [tex]\dot m_3[/tex] from the mass rate balance:
[tex]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}[/tex]
Inlet 1: water is a compressed liquid. From Table A-2,[tex]h_1\approx h_f_1=209.33kJ/kg[/tex]
inlet 2,steam is super-heated. From Table A-4, [tex]h_2=2827.9kJ/kg[/tex]
inlet 3: water is a saturated liquid. From A-3 table,[tex]h_3=h_f_3=762.81kJ/kg[/tex]
Substitute values in [tex]\dot m_2=\frac{\dot m_1(h_3-h_1)}{h_2-h_3}[/tex] to find the mass flow rate at inlet 2:
[tex]\dot m_2=\frac{(70kg/s)(762.81kJ/kg-209.33kJ/kg)}{2827.9kJ/kg -762.81kJ/kg}\\=18.76kg/s[/tex]
Hence, the mass flow rate at inlet 2 is 18.76kg/s
