Calculating heat Transfer native a heavy steam Turbine

Steam enters a generator operating at stable state with a mass flow rate that 4600 kg/h. The turbine creates a strength output of 1000 kW. In ~ the inlet, the pressure is 60 bar, the temperature is 400°C, and the velocity is 10 m/s. In ~ the exit, the pressure is 0.1 bar, the high quality is 0.9 (90%), and the velocity is 30 m/s. Calculate the rate of warmth transfer in between the turbine and surroundings, in kW.

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**Known **A vapor turbine operates at steady state. The mass circulation rate, strength output, and also states that the heavy steam at the inlet and also exit are known.

**Find **Calculate the price of warm transfer.

**Schematic and also Given Data:**

**Engineering Model**

1. The manage volume shown on the accompanying figure is at steady state.

2. The adjust in potential energy from inlet to departure can it is in neglected.

**Analysis **To calculate the warm transfer rate, begin with the one-inlet, one-exit form of the energy rate balance because that a manage volume at stable state, Eq. 4.20a. The is,

0=\dotQ_ cv -\dotW_ cv +\dotm\left<\left(h_1-h_2\right)+\frac\left( V _1^2- V _2^2\right)2+g\left(z_1-z_2\right)\right> (4.20a)

0=\dotQ_ cv -\dotW_ cv +\dotm\left<\left(h_1-h_2\right)+\frac\left( V _1^2- V _2^2\right)2+g\left(z_1-z_2\right)\right>where \dotm is the mass flow rate. Fixing for \dotQ_ cv and also dropping the potential energy adjust from inlet to exit

\dotQ_ cv =\dotW_ cv +\dotm\left<\left(h_2-h_1\right)+\left(\frac V _2^2- V _1^22\right)\right> (a)

To to compare the magnitudes the the enthalpy and also kinetic power terms, and also stress the unit counter needed, every of these terms is evaluated separately.

First, the specific enthalpy difference h_2-h_1 is found. Using Table A-4, h_1=3177.2 kJ/kg. State 2 is a two-phase liquid–vapor mixture, so v data native Table A-3 and also the offered quality

h_2=h_ f 2+x_2\left(h_ g 2-h_ f 2\right) =191.83+(0.9)(2392.8)=2345.4 kJ / kgHence,

h_2-h_1=2345.4-3177.2=-831.8 kJ / kgConsider next the details kinetic power difference. Making use of the given values because that the velocities,

1 \left(\frac V _2^2- V _1^22\right)=\left<\frac(30)^2-(10)^22\right>\left(\frac m ^2 s ^2\right)\left|\frac1 N 1 kg \cdot m / s ^2 \right| \left|\frac1 kJ 10^3 N \cdot m \right|

= 0.4 kJ/kg

Calculating \dotQ_ cv indigenous Eq. (a),

2 \dotQ_ cv =(1000 kW )+\left(4600 \frac kg h \right)(-831.8+0.4)\left(\frac kJ kg \right)\left|\frac1 h 3600 s \right| \left|\frac1 kW 1 kJ / s \right|

= -62.3 kW

1 The size of the readjust in specific kinetic power from inlet to exit is lot smaller 보다 the details enthalpy change. Keep in mind the use of unit conversion factors here and also in the calculate of \dotQ_ cv to follow.

2 The an adverse value the \dotQ_ cv way that there is heat transfer from the wind turbine to that is surroundings, as would certainly be expected. The size of \dotQ_ cv is small relative to the power developed.

**Skills Developed**

**Ability to…**

• apply the steady-state energy rate balance to a manage volume.

• develop an engineering model.

• retrieve building data for water.

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**Quick Quiz**

If the readjust in kinetic power from inlet to departure were neglected, evaluate the heat transfer rate, in kW, keeping all other data unchanged. Comment. Ans. −62.9 kW.