Question 6.CSGP.90: Steam at 500 kPa, 300°C is used to heat cold water at 15°C t......

C.V. Each line separately. No work but there is heat transfer out of the steam flow and into the liquid water flow.

Water line energy Eq.:  \dot{ m }_{ liq } h _{ i }+\dot{ Q }=\dot{ m }_{ liq } h _{ e } \Rightarrow \dot{ Q }=\dot{ m }_{ liq }\left( h _{ e }- h _{ i }\right)

For the liquid water look in Table B.1.1

Steam line energy has the same heat transfer but it goes out

Steam Energy Eq.:    \dot{ m }_{\text {steam }} h _{ i }=\dot{ Q }+\dot{ m }_{\text {steam }} h _{ e } \Rightarrow \dot{ Q }=\dot{ m }_{\text {steam }}\left( h _{ i }- h _{ e }\right)

For the steam look in Table B.1.3 at 500 kPa

\Delta h _{\text {steam }}= h _{ i }- h _{ e }=3064.2-2748.67=315.53 \,kJ / kg

Now the heat transfer for the steam is substituted into the energy equation for the water to give

\dot{ m }_{\text {steam }} / \dot{ m }_{\text {liq }}=\Delta h _{\text {liq }} / \Delta h _{\text {steam }}=\frac{250.93}{315.53}= 0 . 7 9 5