A moving-coil instrument, which gives full-scale deflection with 0.015 A has a copper coil having resistance of 1.5 Ohm at 15°C
1 answer:
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Answer: The exit temperature of the gas in deg C is .
Explanation:
The given data is as follows.
= 1000 J/kg K, R = 500 J/kg K = 0.5 kJ/kg K (as 1 kJ = 1000 J)
= 100 kPa,
We know that for an ideal gas the mass flow rate will be calculated as follows.
or, m =
=
= 10 kg/s
Now, according to the steady flow energy equation:
= 5 K
= 5 K + 300 K
= 305 K
= (305 K - 273 K)
=
Therefore, we can conclude that the exit temperature of the gas in deg C is .
Answer:
16 minutes
Explanation:
This is an example of a class of problems in which two quantities start with different initial values and change at different rates. In such problems, the rates of change are generally ones that cause the values to converge.
The question usually asks when the values will be the same. The generic answer is, "when the difference in rates makes up the difference in initial values."
Here the tanks differ in initial fill height by 12 -8 = 4 ft. The rates of change differ by 0.5 -0.25 = 0.25 ft/min. The more filled tank is draining faster (important), so the fill heights will converge after ...
(4 ft)/(0.25 ft/min) = 16 min
The level in the two tanks will be the same after 16 minutes.
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<em>Additional comment</em>
The oil levels at that time will be 4 ft.
You can write two equations for height:
y = 12 -0.5x . . . . . . . height in feet after x minutes (tank A)
y = 8 -0.25x . . . . . . height in feet after x minutes (tank B)
These will be equal when ...
y = y
12 -0.5x = 8 -0.25x
4 = 0.25x . . . . . . . . . . add 0.5x -8
16 = x . . . . . . . . . . . . multiply by 4 . . . . time to equal height
The graph shows when the tanks will have equal heights and when they will be drained.
