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Chapter 2

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?

Step-by-Step

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}