Question

An isentropic steam turbine processes 2 kg/s of steam at 3 MPa, which is exhausted at50 kPa and 100C. Five percent of this flow is delivered for feedwater heating at 500 kPa. Determine the power produced by this turbine, in kW.

Answer 1

Answer:

2285kw

Explanation:

since it is an isentropic process, we can conclude that it is a reversible adiabatic process. Hence the energy must be conserve i.e the total inflow of energy must be equal to the total outflow of energy.

Mathematically,

$\\ E_{inflow} = E_{outflow}$

Note: from the question we have only one source of inflow and two source of outflow (the exhaust at a pressure of 50kpa and the feedwater at a pressure of 5ookpa). Also the power produce is another source of outgoing energy    $\\ E_{inflow} = m_{1} h_{1}$ .

$E_{outflow} = m_{2} h_{2} + m_{3} h_{3} + W_{out}$

Where $m_{1} h_{1}$ are the mass flow rate and the enthalpies at the inlet  at a pressure of 3Mpa ,

$m_{2} h_{2}$ are the mass flow rate and the enthalpies  at the outlet 2 where we have a pressure of 500kpa respectively.,

and  $m_{3} h_{3}$   are the mass flow rate and the enthalpies  at the outlet 3 where we have a pressure of 50kpa respectively.,

We can now express write out the required equation by substituting the new expression for the energies

$m_{1} h_{1} = m_{2} h_{2} + m_{3} h_{3} + W_{out}$  

from the above equation, the unknown are the enthalpy values and  the mass flow rate.

first let us determine the enthalpy values at the inlet and the out let using the Superheated water table.  

It is more convenient to start from outlet 3 were we have a temperature $100^{0}C$ and pressure value of (50kpa or 0.05Mpa ). using double interpolation method  on the superheated water table to determine the enthalpy value with careful calculation we have  

$h_{3}$  = 2682.4 KJ/KG , at this point also from the table the entropy value ,$s_{3}$ value is 7.6953 KJ/Kg.K.

Next we determine the enthalphy value at outlet 2. But in this case, we don't have a temperature value, hence we use the entrophy value since the entropy  is constant at all inlet and outlet.

So, from the superheated water table again, at a pressure of 500kpa (0.5Mpa) and entropy value of  7.6953 KJ/Kg.K with careful  interpolation we arrive at a enthalpy value of 3206.5KJ/Kg.

Finally for inlet one at a pressure of 3Mpa, interpolting with an entropy value of 7.6953KJ/Kg.K  we arrive at enthalpy value of 3851.2KJ/Kg.

Now we determine the mass flow rate at each inlet and outlet. since  mass must also be balance, i.e  $m_{1} = m_{2} + m_{3}$

From the question the, the mass flow rate at the inlet $m_{1}}$  is 2Kg/s

Since 5% flow is delivered into the feedwater heating,  

$m_{2} = 0.05m_{1} = 0.05 *2kg/s = 0.1kg/s$

Also for the outlet 3 the remaining 95% will flow out. Hence

$m_{3} = 0.95m_{1} = 0.95 *2kg/s = 1.9kg/s$

Now, from $m_{1} h_{1} = m_{2} h_{2} + m_{3} h_{3} + W_{out}$   we substitute values

$W_{out} = m_{1} h_{1}-m_{2} h_{2}-m_{3} h_{3}$

$W_{out} = (2kg/s)(3851.2KJ/Kg) - (0.1kg/s)(3206.5kJ/kg)- (1.9)(2682.4kJ/kg)$

$W_{out} = 2285.19 kW$.

Hence the power produced is 2285kW