Efficiency is the minimum use of energy to accomplish the task. The wasted energy will be 375 J when 750 J of energy is given.
<h3>What is wasted energy?</h3>
Wasted energy is energy that is not useful when the transformation in the system occurs.
Total energy = 750 J
The efficiency of the system = 50 %
Output work (OW) is calculated as:
Efficiency = output work ÷ input work × 100%
750 × 50 = 100 OW
OW = 375 J
Wasted energy = Total energy - output work
= 750 - 375
= 375 J
Therefore, the machine is 50 % inefficient and has wasted energy of 375 J.
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Answer:
i dont know but i will take the points tho hahah
Explanation:
Answer:
-25.63°C.
Explanation:
We know that throttling is a constant enthalpy process
From steal table
We know that if we know only one property in side the dome then we will find the other property by using steam property table.
Temperature at saturation pressure 1 bar is 99.63°C and Temperature at saturation pressure 0.35 bar is about 74°C .
So from above we can say that change in temperature is -25.63°C.
But there is no any option for that .
Answer:
M2 = 0.06404
P2 = 2.273
T2 = 5806.45°R
Explanation:
Given that p1 = 10atm, T1 = 1000R, M1 = 0.2.
Therefore from Steam Table, Po1 = (1.028)*(10) = 10.28 atm,
To1 = (1.008)*(1000) = 1008 ºR
R = 1716 ft-lb/slug-ºR cp= 6006 ft-lb/slug-ºR fuel-air ratio (by mass)
F/A =???? = FA slugf/slugaq = 4.5 x 108ft-lb/slugfx FA slugf/sluga = (4.5 x 108)FA ft-lb/sluga
For the air q = cp(To2– To1)
(Exit flow – inlet flow) – choked flow is assumed For M1= 0.2
Table A.3 of steam table gives P/P* = 2.273,
T/T* = 0.2066,
To/To* = 0.1736 To* = To2= To/0.1736 = 1008/0.1736 = 5806.45 ºR Gives q = cp(To* - To) = (6006 ft-lb/sluga-ºR)*(5806.45 – 1008)ºR = 28819500 ft-lb/slugaSetting equal to equation 1 above gives 28819500 ft-lb/sluga= FA*(4.5 x 108) ft-lb/slugaFA =
F/A = 0.06404 slugf/slugaor less to prevent choked flow at the exit
Answer:
I forget the word for it, but probably the guys who set up the power lines in the city.
Explanation:
