Question

Consider the mixing of 0.8 kg/s of hot water at 348 K and 1 kg/s of cool water at 298 K that is generating warm water. Assume no work is being done and the system is in steady state, but heat is lost at the rate of 30 kJ/s during this mixing. Find the temperature of the warm water flow stream? Assume Cp =4.18 kJkg

Answer 1

Answer:

T_warm = 47.22 C

Explanation:

Using energy balance for the system:

m_1*h_1 + m_2*h_2 = m_3*h3   ... Eq1

h_i = c_p. T_i   ... Eq 2

m_1 + m_2 = m_3   ... steady flow system (Eq 3)

Substitute Eq 2 and Eq3 in Eq1

m_3 = 0.8 + 1 = 1.8 kg/s

(0.8)*(4.18)*( 348-273) + (1)*(4.18)*( 298-273) = 1.8 * 4.18 *T_3

T_3 = 355.3 / (1.8*4.18) = 47.22 C