Answer to Question #296289 in Mechanical Engineering for mahf

Question

Consider an ordinary shower where hot water at 60°C is mixed with cold water at 10°C. If it is desired that a steady stream of warm water at 45°C be supplied, determine the ratio of the mass flow rates of the hot to cold water. Assume the heat losses from the mixing chamber to be negligible and the mixing to take place at a pressure of 150 kPa. 

Answer 1

Solution;

Given;

$T_h=60°c$

$T_c=10°c$

$T_m=45°$

$P=150kPa$

Let $m_h,m_c,m_m$ be the respective mass flow rates.

For conservation of mass;

$m_m=m_h+m_c$

Applying the steady flow energy equation and assume;

Q=0

W=0

P.E=K.E=0

$m_ch_c+m_hh_h=m_mh_m$

$m_ch_c+m_hh_h=(m_c+m_h)h_m$

From the tables;

$h_h=h_{f_{60}}=251.18kJ/kg$

$h_c=h_{f_{10}}=42.022kJ/kg$

$h_m=h_{f_{45}}=188.44kJ/kg$

$42.022m_c+251.18m_h=188.44m_c+188.44m_h$

Hence;

$62.74m_h=146.418m_c$

$\frac{m_c}{m_h}=\frac{144.418}{62.74}=2.33$