Question

) Water flows through a horizontal coil heated from the outside by high-temperature flue gases. As it passes through the coil the water changes state from liquid at 250 kPa and 70 to vapor at 150 kPa and 120 . Its entering velocity is 5 and its exit velocity is 220 . Determine the heat transferred through the coil per unit mass of water.

Answer 1

Step-by-step explanation:

Formula for steady flow energy equation for the flow of fluid is as follows.

`m[h_(1) + (V^(2)_(1))/(2)] + z_(1)g] + q = m[h_(1) + (V^(2)_(1))/(2) + z_(1)g] + w`

Now, we will substitute 0 for both
`z_(1)` and
`z_(2)`, 0 for w, 334.9 kJ/kg for
`h_(1)`, 2726.5 kJ/kg for
`h_(2)`, 5 m/s for
`V_(1)` and 220 m/s for
`V_(2)`.

Putting the given values into the above formula as follows.

`m[h_(1) + (V^(2)_(1))/(2)] + z_(1)g] + q = m[h_(1) + (V^(2)_(1))/(2) + z_(1)g] + w`

`1 * [334.9 * 10^(3) J/kg + ((5 m/s)^(2))/(2) + 0] + q = 1 * [2726.5 * 10^(3) + ((220 m/s)^(2))/(2) + 0] + 0`

q = 6597.711 kJ

Thus, we can conclude that heat transferred through the coil per unit mass of water is 6597.711 kJ.