Initial velocity = zero (i.e., free fall)
Final velocity = 30m/s
Acceleration due to gravity = 10m/s²
For a body falling freely under the action of gravity, g is taken positive
◈ <u>First equation of kinenatics</u> :
⇒ v = u + gt
⇒ 30 = 0 + 10t
⇒ t = 30/10
⇒ <u>t = 3s</u>
Hence, object will attain a speed of 30m/s after 3s.
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:
We can answer the questions with this data:
A)
B)
C)
D)
