III. During January, at a location in Alaska winds at −27°C can be observed. However, several meters below ground the temperatur
1 answer:
Answer:
Not possible.
Explanation:
According to second law of thermodynamics, the maximum efficiency any heat engine could achieve is Carnot Efficiency η defined by:
Where
and
are temperature (in Kelvin) of heat source and heatsink respectively
In our case (I will be using K = 273+°C) :
In percentage, this is 14.28% efficiency, which is the <em>maximum</em> theoretical efficiency <em>any</em> heat engine could have while working between -27 and 14 °C temperature. Any claim of more efficient heat engine between these 2 temperature are violates the second law of thermodynamics. Therefore, the claim must be false.
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Answer:
33.56 ft^3/sec.in
Explanation:
Duration = 6 hours
drainage area = 185 mi^2
constant baseflow = 550 cfs
<u>Derive the unit hydrograph using the inverse procedure </u>
first step : calculate for the volume of direct runoff hydrograph using the details in table 2 attached below
Vdrh = sum of drh * duration
= 29700 * 6 hours ( 216000 secs )
= 641,520,000 ft^3.
next step : Calculate the volume of runoff in equivalent depth
Vdrh / Area = 641,520,000 / 185 mi^2
= 1.49 in
Finally derive the unit hydrograph
Unit of hydrograph = drh / volume of runoff in equivalent depth
= 50 ft^3 / 1.49 in = 33.56 ft^3/sec.in
Answer:
a)
b) attached below
c) type zero system
d) k >
e) The gain K increases above % error as the steady state speed increases
Explanation:
Given data:
Motor voltage = 12 v
steady state speed = 200 rad/s
time taken to reach 63.2% = 1.2 seconds
<u>a) The transfer function of the motor from voltage to speed</u>
let ; be the transfer function of a motor
when i/p = 12v then steady state speed ( k1 ) = 200 rad/s , St ( time constant ) = 1.2 sec
hence the transfer function of the motor from voltage to speed
=
<u>b) draw the block diagram of the system with plant controller and the feedback path </u>
attached below is the remaining part of the detailed solution
c) The system is a type-zero system because the pole at the origin is zero
d) ) k >
