Answer:
-0.1006Kw/K
Explanation:
The rate of entropy change in the air can be reduced from the heat transfer and the air temperature. Hence,
ΔS = Q/T
Where T is the constant absolute temperature of the system and Q is the heat transfer for the internally reversible process.
S(air) = - Q/T(air) .......1
Where S.air =
Q = 30-kW
T.air = 298k
Substitute the values into equation 1
S(air) = - 30/298
= -0.1006Kw/K
Answer:
Answer: ±0.02 units or 20±0.02 units or 19.98-20.02 units depending on how they prefer its written (typically the first or second one)
Explanation:
says on the sheet. Unless otherwise stated 0.XX = ±0.02 tolerance
(based on image sent in other post)
Answer:
Steps:
1. Create a text file that contains blade diameter (in feet), wind velocity (in mph) and the approximate electricity generated for the year
2. load the data file for example, in matlab, use ('fileame.txt') to load the file
3. create variables from each column of your data
for example, in matlab,
x=t{1}
y=t{2}
4. plot the wind velocity and electricity generated.
plot(x, y)
5. Label the individual axis and name the graph title.
title('Graph of wind velocity vs approximate electricity generated for the year')
xlabel('wind velocity')
ylabel('approximate electricity generated for the year')
Answer:
Context
Explanation:
It is of great value for an engineer to keep the context of his/her experiment in mind.
