Question

k) A stream of warm water is produced in a steady-flow mixing process by combining 1.0 kg s-1of cool water at 25°C with 0.8 kg s-1of hot water at 75°C. During mixing, heat is lost to the surroundings at the rate of 30 kJ s-1. What is the temperature of the warmwater stream? Assume the specific heat of water constant at 4.18 kJ kg-1K-1

Answer 1

Answer:

T_ww = 43,23°C

Explanation:

To solve this question, we use energy balance and we state that the energy that enters the systems equals the energy that leaves the system plus losses. Mathematically, we will have that:

E_in=E_out+E_loss

The energy associated to a current of fluid can be defined as:

E=m*C_p*T_f

So, applying the energy balance to the system described:

m_CW*C_p*T_CW+m_HW*C_p*T_HW=m_WW*C_p*T_WW+E_loss

Replacing the values given on the statement, we have:

1.0 kg/s*4,18 kJ/(kg°C)*25°C+0.8  kg/s*4,18 kJ/(kg°C)*75°C=1.8 kg/s*4,18 kJ/(kg°C)*T_WW+30  kJ/s

Solving for the temperature Tww, we have:

(1.0 kg/s*4,18 kJ/(kg°C)*25°C+0.8 kg/s*4,18 kJ/(kg°C)*75°C-30 kJ/s)/(1.8 kg/s*4,18 kJ/(kg°C))=T_WW

T_WW=43,23 °C

Have a nice day! :D