Answer to Question #213401 in Molecular Physics | Thermodynamics for Yomna

Question

An insulated water heater operating at steady state has two inlets and one 

outlet. At inlet 1, liquid water enters at 7 bar, 42oC, and a mass flow rate of 70 

kg/s. At inlet 2, wet vapor at 7 bar with a quality of 98% is injected. At outlet, 

saturated liquid at 7 bar exits the water heater. Ignoring heat transfer with the 

surroundings and neglecting kinetic and potential energy effects, determine 

the mass flow rate, in kg/s, at inlet 2

Answer 1

Answer:-

At inlet 1 ,

$P_1= 7 \ bar \\ T_1 =42^o C \\ m_1=70 kg/s$

At inlet 2

$P_2=7 \ bar\\ x_2=0.98\\$

at exit 3 , $p_3=7 \ bar$

at T₁ = 42^o C from saturated stream tables ,

h_f1=167.57 + (188.45-167.57) $[\frac{42-40}{45-40}]=175.9kJ/kg$

$V_{f1}=1.0078\times 10^{-3}+(1.0099\times10^{-3}-1.0078\times10^{-3})\times [\frac{42-40}{45-40}]\\=1.0086\times10^{-3}m^3/kg$

$P_{Sat1}=0.07384+(0.09593-0.07384)\times [\frac{42-40}{45-40}] \\=0.08268 bar$

At $P_2=7$ bar from saturated steam tables,

$h_{f2}=697.22 kJ/kg\\ h_{fs2}=2066.3 KJ/Kg\\ h_2=h_{f2}+X_2h_{fs2}$

$h_2=697.22+0.98\times2066.3\\ h_22722.2 KJ/Kg$

at $p_3 =7 \ bar$ from saturated steam tables ,

$h_3=h_{f3}$

$h_3=697.22KJ/kg$ ($\therefore$ saturated liquid state)

By mass rate balance under steady state

$m_1+m_2+m_3=0.....(1)$

By energy rate balance equation

$0=Q_{cv}-W_{cv}+m_1(h_1+\frac{v_1^2}{2}+gz_1)+m_2(h_2+\frac{v_2^2}{2}+gz_2)-m_3(h_3+\frac{v_3^2}{2}+gz_3)$

$\therefore$ work done , heat transfer , kinetic and

potential energies are negligible

$m_1h_1+m_2h_2+m_3h_3=0...........(2)$

using (1) in (2)

$m_1h_1+m_2h_2-(m_1+m_2)h_3=0\\$

on putting the values

$m_2=70\times\frac{(697.22-176.6)}{(2722.2-697.22)}$

$m_2=18 \ kg/s$

Mass flow rate 2 is $m_2=18 \ kg/s$