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Marina86 [1]
3 years ago
7

What are you calculating when forces are added together

Physics
1 answer:
ra1l [238]3 years ago
8 0

Answer:A

Explanation:

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Water enters the turbine of an ideal Rankine cycle as a superheated vapor at 18 MPa and 550°C. If the condenser pressure is 9 kP
ivolga24 [154]

Answer:

For this ideal Rankine cycle:

A) The rate of heat addition into the cycle Qh is 3214.59KJ/Kg.

B) The thermal efficiency of the system εt=0.4354.

C) The rate of heat rejection from the cycle Qc is 1833.32KJ/Kg.

D) The back work ratio is 0.013

Explanation:

To solve the ideal Rankine cycle we have to determinate the thermodynamic information of each point of the cycle. We will use a water thermodynamic properties table. In an ideal Rankine cycle, the process in the turbine and the pump must be isentropic. Therefore S₃=S₄ and S₁=S₂.

We will start with the point 3:

P₃=18000KPa  T₃=550ºc ⇒ h₃=3416.12 KJ/Kg  S₃=6.40690 KJ/Kg

Point 4:

P₄=9KPa S₄=S₃ ⇒ h₄=2016.60 (x₄=0.7645 is wet steam)

Point 1:

P₁=P₄ in the endpoint of the steam curve. ⇒ h₁= 193.28KJ/Kg  S₁=0.62235KJ/Kg  x₁=0

Point 2:

P₂=P₃ and S₂=S₁ ⇒ h₂=201.53KJ/Kg

With this information we can obtain the heat rates, the turbine, and the pump work:

Q_h=Q_{2-3}=h_3-h_2=3214.59KJ/Kg

Q_c=Q_{4-1}=h_4-h_1=1833.32KJ/Kg

W_{turb}=W_{3-4}=h_3-h_4=1399.52KJ/Kg

W_{pump}=W_{1-2}=h_2-h_1=18.25KJ/Kg

We can answer the questions with this data:

A) Q_h=Q_{2-3}=h_3-h_2=3214.59KJ/Kg

B) \epsilon_{ther}=\displaystyle\frac{W_{turb}}{Q_h}=0.4354

C)Q_c=Q_{4-1}=h_4-h_1=1833.32KJ/Kg

D) r_{bW}=\displaystyle\frac{W_{pump}}{W_{turb}}=0.013

3 0
3 years ago
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