Mass balance:

m1 + m2 = m3 = m(say)

1.237

Let,

1.238

By first law of thermodynamics,

m1h1 + m2h2 = (m1 + m2)h3

1.239

or,

xh1 + (1 - x)h2 = h3

1.240

Since,

h = h(T)

1.241

From Eq. (1.240)

xT1 + (1-x)T2 = T3

1.242

1.243

where

1.244

By second law of thermodynamics,

1.245

1.246

1.247

1.248

or,

1.249

Let, N_s Entropy generation number =

Now, substituting from Eq. (1.243) in Eq. (1.249), we get

1.250

1.251

For, x = 1, N_s = 0 ⇒ m₂ = 0, the system is with single stream.

For, , N_s = 0, temperature of both the streams are same.

Rate of exergy loss due to mixing is,

Wlost = I = T0Sgen

1.252

1.253