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bonufazy [111]
3 years ago
10

Air enters a well-insulated turbine operating at steady, state with negligible velocity at 4 MPa, 300°C. The air expands to an e

xit pressure of 100kPa. The exit velocity and temperature are 90 m/s and 100°C, respectively. The diameter of the exit is 0.6 m. Determine the power developed by the turbine (in kW). You may assume air behaves like an ideal gas throughout the process
Engineering
1 answer:
svetoff [14.1K]3 years ago
5 0

Answer:

-3744.45 kW

Explanation:

The energy balance for a turbine is the following:

U_{e}+P_{e}*v_{e}+gz_{e}+\frac{C_{e}^{2}}{2}+q=W+U_{s}+P_{s}*v_{s}+gz_{s}+\frac{C_{s}^{2}}{2}

Where:

U = internal energy, P= pressure, v=volume, g= gravity, z= height, C= velocity, q= heat, W= power.  

Since it is a well-insulated turbine q= 0.

In steady state m (mass flow) is constant.

m=\frac{1}{v_{s}}*A*C

Where v_{s} is the specific volume and A = area.

There is no information about a change in height during the process so we can say that the term (gz) in the equation both in the inlet (e) and outlet (s) is zero.

In the inlet the velocity is negligible so \frac{C_{e}^{2}}{2} is zero.

Also, enthalpy (h) is:

h=U+P*v

Reorganizing the equation with this information we have

h_{e} =W + h_{s} +\frac{C_{s}^{2}}{2}

W= m*(h_{e}- h_{s}-\frac{C_{s}^{2}}{2}) in kW

We can get the enthalpy from thermodynamic tables for the air with the conditions in the inlet and in outlet. These are:

h_{e} (300 C) = 578.81 \frac{kJ}{kg}

h_{s} (100 C) = 273.26 \frac{kJ}{kg}

Mass flow  

The area is the area of the circle A=pi()*r^{2}

A=pi()*(0.3m)^{2}=0.282 m^{2}

We can get the specific volume from thermodynamics tables for air at 100°C. We use this condition because we know the diameter and the velocity in the outlet. This value is  

1.057 \frac{m^{3}}{kg}  

Mass flow is  

m=\frac{1}{1.057\frac{m^{3}}{kg}} * 0.282 m^{2}* 90\frac{m}{s}=24.072 \frac{kg}{s}

So power is:

W= 24.072 \frac{kg}{s}*(578.81 \frac{kJ}{kg}-273.26 \frac{kJ}{kg}+\frac{(90\frac{m}{s})^{2}}{2}=-3744.45 Kw

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Answer:

1) The exergy of destruction is approximately 456.93 kW

2) The reversible power output is approximately 5456.93 kW

Explanation:

1) The given parameters are;

P₁ = 8 MPa

T₁ = 500°C

From which we have;

s₁ = 6.727 kJ/(kg·K)

h₁ = 3399 kJ/kg

P₂ = 2 MPa

T₂ = 350°C

From which we have;

s₂ = 6.958 kJ/(kg·K)

h₂ = 3138 kJ/kg

P₃ = 2 MPa

T₃ = 500°C

From which we have;

s₃ = 7.434 kJ/(kg·K)

h₃ = 3468 kJ/kg

P₄ = 30 KPa

T₄ = 69.09 C (saturation temperature)

From which we have;

h₄ = h_{f4} + x₄×h_{fg} = 289.229 + 0.97*2335.32 = 2554.49 kJ/kg

s₄ =  s_{f4} + x₄×s_{fg} = 0.94394 + 0.97*6.8235 ≈ 7.563 kJ/(kg·K)

The exergy of destruction, \dot X_{dest}, is given as follows;

\dot X_{dest} = T₀ × \dot S_{gen} = T₀ × \dot m × (s₄ + s₂ - s₁ - s₃)

\dot X_{dest} = T₀ × \dot W×(s₄ + s₂ - s₁ - s₃)/(h₁ + h₃ - h₂ - h₄)

∴ \dot X_{dest} = 298.15 × 5000 × (7.563 + 6.958 - 6.727 - 7.434)/(3399 + 3468 - 3138  - 2554.49) ≈ 456.93 kW

The exergy of destruction ≈ 456.93 kW

2) The reversible power output, \dot W_{rev} = \dot W_{} + \dot X_{dest} ≈ 5000 + 456.93 kW = 5456.93 kW

The reversible power output ≈ 5456.93 kW.

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A geothermal heat pump absorbs 15 KJ/s of heat from the Earth 15 m below a house. This heat pump uses a 7.45 kJ/s compressor.
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Answer:

COP of the heat pump is 3.013

OP of the cycle is  1.124

Explanation:

W = Q₂ - Q₁

Given

a)

Q₂ = Q₁ + W

     = 15 + 7.45

     = 22.45 kw

COP = Q₂ / W = 22.45 / 7.45 = 3.013

b)

Q₂ = 15 x 1.055 = 15.825 kw

therefore,

Q₁ = Q₂ - W

Q₁ = 15.825 - 7.45 = 8.375

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Answer:

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Explanation:

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wall thickness 0.30 inc

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