Question

A well insulated turbine operates at steady state. Steam enters the turbine at 4 MPa with a specific enthalpy of 3015.4 kJ/kg and a velocity of 10 m/s. The steam expands to the turbine exit where the pressure is 0.07 MPa, specific enthalpy is 2431.7 kJ/kg, and the velocity is 90 m/s. The mass flow rate is 11.95 kg/s. Neglecting potential energy effects, determine the power developed by the turbine, in kW.

Answer 1

Answer:

power developed by the turbine = 6927.415 kW

Explanation:

given data

pressure = 4 MPa

specific enthalpy h1 = 3015.4 kJ/kg

velocity v1 = 10 m/s

pressure = 0.07 MPa

specific enthalpy h2 = 2431.7 kJ/kg

velocity v2 = 90 m/s

mass flow rate = 11.95 kg/s

solution

we apply here  thermodynamic equation that

energy equation that is

$h1 + \frac{v1}{2}  + q = h2 + \frac{v2}{2}  + w$

put here value with

turbine is insulated so q = 0

so here

$3015.4 *1000 + \frac{10^2}{2}  =  2431.7 * 1000 + \frac{90^2}{2}  + w$

solve we get

w = 579700 J/kg = 579.7 kJ/kg

and

W = mass flow rate × w

W = 11.95 × 579.7

W = 6927.415 kW

power developed by the turbine = 6927.415 kW

Answer 2

The power developed by the turbine will be 6927,415 kW.

We can arrive at this answer through the following calculation:

• Use the thermodynamic equation:

$h1+\frac{v1}{2} +q=h2+\frac{v2}{2} +w$

• Considering that $q=0$ we can solve the equation as follows:

$3015.4*1000+\frac{10^2}{2}=2431.7*1000+\frac{90^2}{2}+w\\3015400+\frac{10^2}{2}=2431700+\frac{90^2}{2}+w\\w=579.7 kJ/kg$

• Then we can use this value to find the value of W, which is the power developed by the turbine. This will be done as follows:

$W=mass flow rate×w\\W = 11.95 × 579.7\\W = 6927.415 kW$

More information:

https://brainly.com/question/1874753?referrer=searchResults