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:
2062 lbm/h
Explanation:
The air will lose heat and the oil will gain heat.
These heats will be equal in magnitude.
qo = -qa
They will be of different signs because one is entering iits system and the other is exiting.
The heat exchanged by oil is:
qo = Gp * Cpo * (tof - toi)
The heat exchanged by air is:
qa = Ga * Cpa * (taf - tai)
The specific heat capacity of air at constant pressure is:
Cpa = 0.24 BTU/(lbm*F)
Therefore:
Gp * Cpo * (tof - toi) = Ga * Cpa * (taf - tai)
Ga = (Gp * Cpo * (tof - toi)) / (Cpa * (taf - tai))
Ga = (2200 * 0.45 * (150 - 100)) / (0.24 * (300 - 200)) = 2062 lbm/h
Answer:
See explaination
Explanation:
2. 0-1 km shear value: taking winds at 1000mb and 850 mb
15 kts south easterly and 50 kts southerly
Vector difference 135/15 and 180/50 will be 170/61 or southerly 61 kts
3. 0-6 km shear value: taking winds at 1000 mb and 500 mb
15 kts south easterly and 40 kts westerly
Vector difference 135/15 and 270/40 will be 281/51 kts
please see attachment
Explanation:
The unit refrigeration is generally is given in terms of tons.In refrigeration compressor consume some amount of work to produce the cooling effect with the help of evaporator and condenser.
In the simple words ton is the cooling load of refrigeration system.
So
1 ton = 3.5 KW
1 ton = 12,000 BTU/hr
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