Question

0.25Kg/sec of water is heated from 30°C to 60°C by hot gases that enter at 180°C and leaves at 80°C. Calculate the mass flow rate of gases when its C_P = 1.08KJ/kg-K. Find the entropy change of water and of hot gases. Take the specific heat of water as 4.186KJ/kg-k.

Answer 1

Mass of water m_w = 0.25Kg/sec 

Initial temperature of water T_W1 = 30°C 

Final temperature of water T_W2 = 60°C 

Entry Temperature of hot gas = T_g1 = 180°C 

Exit Temperature of hot gas = T_g2 = 80°C 

Mass flow rate m_f =? 

Specific heat of gas C_Pg = 1.08KJ/kg-K 

Specific heat of water C_W = 4.186KJ/kg-K 

Heat gives by the gas = Heat taken by water

m_s C_Ps.dT_s = m_WC_PW.dT_W

m_s × 1.08 × (180 – 100) = 0.25 × 4.186 × (60 – 30)

Mass flow rate of gases = ms = 0.291 Kg/sec

Change of Entropy of water = ds_W = m_W.C_PW. ln (T₂/T₁)

= 4.186 × 0.25 × ln [(60 + 273)/ (30 + 273)]

Change of Entropy of water = 0.099 KJ/°K

Change of Entropy of Hot gases = ds_g = m _g.C_Pg. ln (T₂/T₁)

= 0.291 × 1.08 × ln [(80 + 273)/ (180 + 273)]

Change of Entropy of Hot gas = -0.0783 KJ/°K