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## Q. 2.15

A steam turbine operates adiabatically with a power output of 4000 kW. Steam enters the turbine at 2100 kPa and 475°C. The exhaust is saturated steam at 10 kPa that enters a condenser, where it is condensed and cooled to 30°C. What is the mass flow rate of the steam, and at what rate must cooling water be supplied to the condenser, if the water enters at 15°C and is heated to 25°C?

## Verified Solution

The enthalpies of entering and exiting steam from the turbine are found from the steam tables:

$H_{1} = 3411.3 kJ⋅ kg^{−1} and H_{2} = 2584.8 kJ⋅ kg^{−1}$

For a properly designed turbine, kinetic- and potential-energy changes are negligible, and for adiabatic operation Q = 0. Equation (2.32) becomes simply $W_{s} = ΔH$.

$\Delta H = Q + W_{s}$     (2.32)

Then $\dot{W_{s} } = \dot{m}(\Delta H)$ , and

$\dot{m}_{steam} = \frac{\dot{W_{s} } }{\Delta H} = \frac{− 4000 kJ⋅ s^{−1}}{( 2584.8 − 3411.3 ) kJ⋅ kg^{−1}} = 4.840 kg⋅ s^{−1}$

For the condenser, the steam condensate leaving is subcooled water at 30°C, for which (from the steam tables) $H_{3}= 125.7 kJ·kg^{−1}$. For the cooling water entering at 15°C and leaving at 25°C, the enthalpies are

$H_{in} = 62.9 kJ⋅ kg^{−1} and H_{out} = 104.8 kJ⋅ kg^{−1}$

Equation (2.29) here reduces to

$\dot{m}_{steam}(H_{3} – H_{2} ) +\dot{m}_{water}(H_{out} – H_{in} )= 0$

$4 .840 ( 125.7 − 2584.8 ) + ​​\dot{m}_{water} ( 104.8 − 62.9 ) = 0$

Solution gives,

​​$\dot{m}_{water} = 284.1 kg⋅ s^{−1}$