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

Explanation:

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

    $m[h_{1} + \frac{V^{2}_{1}}{2}] + z_{1}g] + q = m[h_{1} + \frac{V^{2}_{1}}{2} + z_{1}g] + w$

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.

     $m[h_{1} + \frac{V^{2}_{1}}{2}] + z_{1}g] + q = m[h_{1} + \frac{V^{2}_{1}}{2} + z_{1}g] + w$  

     $1 \times [334.9 \times 10^{3} J/kg + \frac{(5 m/s)^{2}}{2} + 0] + q = 1 \times [2726.5 \times 10^{3} + \frac{(220 m/s)^{2}}{2} + 0] + 0$

                q = 6597.711 kJ

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