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asambeis [7]
3 years ago
14

Water vapor at 6 MPa, 600 degrees C enters a turbine operating at steady state and expands to 10kPa. The mass flow rate is 2 kg/

s, and the power developed is 2626 kW. Stray heat transfer and kinetic and potential energy effects are negigible.
Determine :

(a) the isentropic turbine efficiency

(b) the rate of entropy production within the turbine, in kW/K.
Engineering
1 answer:
kirill115 [55]3 years ago
6 0

Answer:

Explanation:

Obtain the following properties at 6MPa and 600°C from the table "Superheated water".

h_1=3658.8KL/Kg\\s_1=7.1693kJ/kg.k

Obtain the following properties at 10kPa from the table "saturated water"

h_{f2}=191.81KJ/Kg.K\\h_{fg2}=2392.1KJ/Kg\\s_{f2}=0.6492KJ/Kg.K\\s_{fg2}=7.4996KJ/Kg.K

Calculate the enthalpy at exit of the turbine using the energy balance equation.

\frac{dE}{dt}=Q-W+m(h_1-h_2)

Since, the process is isentropic process Q=0

0=0-W+m(h_1-h_2)\\h_2=h_1-\frac{W}{m}\\\\h_2=3658.8-\frac{2626}{2}\\\\=2345.8kJ/kg

Use the isentropic relations:

s_1=s_{2s}\\s_1=s_{f2}+x_{2s}s_{fg2}\\7.1693=6492+x_{2s}(7.4996)\\x_{2s}=87

Calculate the enthalpy at isentropic state 2s.

h_{2s}=h_{f2}+x_{2s}.h_{fg2}\\=191.81+0.87(2392.1)\\=2272.937kJ/kg

a.)

Calculate the isentropic turbine efficiency.

\eta_{turbine}=\frac{h_1-h_2}{h_1-h_{2s}}\\\\=\frac{3658.8-2345.8}{3658.8-2272.937}=0.947=94.7%

b.)

Find the quality of the water at state 2

since h_f at 10KPa <h_2<h_g at 10KPa

Therefore, state 2 is in two-phase region.

h_2=h_{f2}+x_2(h_{fg2})\\2345.8=191.81+x_2(2392.1)\\x_2=0.9

Calculate the entropy at state 2.

s_2=s_{f2}+x_2.s_{fg2}\\=0.6492+0.9(7.4996)\\=7.398kJ/Kg.K

Calculate the rate of entropy production.

S=\frac{Q}{T}+m(s_2-s_1)

since, Q = 0

S=m(s_2-s_1)\\=2\frac{kg}{s}(7.398-7.1693)kJ/kg\\=0.4574kW/k

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Answer:

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Steam enters a turbine steadily at 7 MPa and 600°C with a velocity of 60 m/s and leaves at 25 kPa with a quality of 95 percent.
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a) The mass flow rate of the steam is:

\dot m = \frac{v_{in}\cdot A_{in}}{\nu_{in}}

\dot m = \frac{\left(60\,\frac{m}{s} \right)\cdot (0.015\,m^{2})}{0.055665\,\frac{m^{3}}{kg} }

\dot m = 16.168\,\frac{kg}{s}

b) The exit velocity of steam is:

\dot m = \frac{v_{out}\cdot A_{out}}{\nu_{out}}

v_{out} = \frac{\dot m \cdot \nu_{out}}{A_{out}}

v_{out} = \frac{\left(16.168\,\frac{kg}{s} \right)\cdot \left(5.89328\,\frac{m^{3}}{kg} \right)}{0.14\,m^{2}}

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\dot W_{out} = \dot m \cdot (-q_{loss} + h_{in}-h_{out})

\dot W_{out} = \left(16.168\,\frac{kg}{s} \right)\cdot \left(-20\,\frac{kJ}{kg} + 3650.6\,\frac{kJ}{kg} - 2500.2\,\frac{kJ}{kg}\right)

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