Answer: you can watch a video on how to solve this question on you tube
Answer:
96.1%
Explanation:
We know that lift force
------------(1)
Where is the lift force coefficient .
ρ is the density of fluid.
A is the area.
V is the velocity.
Now when speed is increased by 2 % and all other parameter remains constant except .
Let;s take new value of lift force coefficient is .
-----------(2)
Now from equation 1 and 2
⇒
So we can say that revised value of lift force coefficient is 96.1% of original value.
Answer:
COP = 3.828
W' = 39.18 Kw
Explanation:
From the table A-11 i attached, we can find the entropy for the state 1 at -20°C.
h1 = 238.43 KJ/Kg
s1 = 0.94575 KJ/Kg.K
From table A-12 attached we can do the same for states 3 and 4 but just enthalpy at 800 KPa.
h3 = h4 = hf = 95.47 KJ/Kg
For state 2, we can calculate the enthalpy from table A-13 attached using interpolation at 800 KPa and the condition s2 = s1. We have;
h2 = 275.75 KJ/Kg
The power would be determined from the energy balance in state 1-2 where the mass flow rate will be expressed through the energy balance in state 4-1.
W' = m'(h2 - h1)
W' = Q'_L((h2 - h1)/(h1 - h4))
Where Q'_L = 150 kW
Plugging in the relevant values, we have;
W' = 150((275.75 - 238.43)/(238.43 - 95.47))
W' = 39.18 Kw
Formula foe COP is;
COP = Q'_L/W'
COP = 150/39.18
COP = 3.828