Question

In a steam power plant, the steam 0.1 bar and 0.95 dry enters the condenser and leaves as saturated liquid at 0.1 bar and 45°C. Cooling water enters the condenser in separate steam at 20°C and leaves at 35° C without any loss of its pressure and no phase change. Neglecting the heat interaction between the condenser and surroundings and changes in kinetic energy and potential energy, determine the ratio of mass-flow rate of cooling water to condensing steam.

Answer 1

Inlet condition of steam : Pressure ‘P₁’ = 0.1 bar dryness fraction x = 0.95 

Exit condition of steam : saturated water at 0.1 bar and 45°C Inlet temperature of cooling water = 20°C Exit temperature of cooling water = 35°C Applying SFEE to control volume 

Q – W_s = m_f [(h₂ – h₁)+ 1/2(V₂² – V₁²) + g(z₂ – z₁)]

Neglecting the changes in kinetic and potential energies. 

And there is no shaft work i.e.; W_s = 0

Q = m_f (h₂ – h₁)

h₁ = h_f1 + xhf_g1

= 191.81 + 0.95 (2392.82) = 2464.99 kJ/kg

h₂ = h_f2 at 0. 1 bar (since the water is saturated liquid)

h₂ = 191.81 kJ/kg

 = m_f (191.81 – 2464.99)

= – 2273.18 m_f kJ/kg ...(i)

-ve sign shows the heat rejection.

By energy balance;Heat lost by steam = Heat gained by cooling water

Q = m_fw.C_w (T₄ – T₃) = m_fw x 4.1868(35 – 20) 

= 62.802m_fw ...(ii)

Equate equation (i) and (ii); 

We get Q = -2273.18m_f 

= 62.802m_fwm_fw/m_f = 36.19 

Ratio of mass flow rate of cooling water to condensing steam = m_fw/m_f = 36.19.