Water flows through a horizontal coil heated from outside. During its passage, it changes state from liquid at 200 kPa and 80^{o}C to vapor at 100 kPa and 125^{o}C. The entering and exit velocities are 3 m/s and 200 m/s respectively. Determine the heat transferred through the coil per unit mass of water. H_{inlet} = 334.9 kJ/kg; H_{outlet} = 2726 kJ/kg.

Calculations:

For the steady flow process, the first law is written as

DH + Du^{2}/2 + gDz = Q + W_{s}

since there is no shaft work, W_{s} = 0

and flow is horizontal, Dz = 0

Therefore,

DH + Du^{2}/2 = Q

substituting for the quantities,

(2726 - 334.9) x 1000 + (200^{2} - 3^{2})/2 = Q (in terms of J/kg)

Q = 2411.1 kJ/kg

Heat transferred through the coil per unit mass of water = 2411.1 kJ