Dry-bulb temperature(T) =24°C
Wet-bulb temperature(Tw) = 17°C
Pressure ,P = 1 atm
As we know that psychrometric chart are drawn at consistent pressure.
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From the diagram
ω= particular humidity
Lets take these two lines Dry-bulb temperature(T) line and also Wet-bulb temperature(Tw) reduced at allude P
From chart at allude P
Specific humidity,ω = 0.00922 kg/kg
The enthalpy ( h)
The loved one humidity, RH
RH= 49.58 %
Specific volume ,
v= 0.853 m³/kg
After it"s lengthy use, the bridge consistently undergoes via alternating strains. in outcome, the elastic stamina of the bridge gets diminished. while the process goes on the bridge renders elastic tiredness and a permanent adjust happens which makes it collapse which is the reason why bridges are claimed unsafe after a lengthy use :)
Indicate the result of each of the adhering to unit vector cross assets (unit vector hat-signs not shown): . kxi = . jxi= -jxk
The cross product of two vectors is given by
Wright here, θ be the angle between the two vectors and also widehatn be the unit vector along the direction of cross product of two vectors.
Here, K x i = - j
As K is the unit vector alengthy Z axis, i is the unit vector alengthy X axis and j be the unit vector alengthy axis.
The direction of cross product of 2 vectors is given by the best hand palm rule.
So, k x i = j
j x i = - k
- j x k = - i
i x i = 0
A platinum wire is provided to determine the melting allude of indium. The resistance of the platinum wire is 2.000 Ω at 20°C and also inc
The melting suggest of indium is 157.436 levels Celsius.
The resistance of the platinum wire, R1 = 2
The temperature at R1 is, T1 = 20 degrees Celsius.
The enhanced resistance, R2 = 3.072
Let the temperature at 3.072 = T2
Now discover the temperature at which the indium starts melting.
We recognize that α = ( R2 - R1 ) / < R1 × ( T2 - T1 ) >
Given, α = 3.9 x 10^-3/ levels Celsius.
T2- T1 = ( R2 - R1 ) / R1 α
T2 – T1 = (3.072 – 2) / (2 × 3.9 x 10^-3)
T2 – T1 = 137.436
T2 = T1 + 137.436
T2 = 20 + 137.436
T2 = 157.436 level Celsius
11 months ago
The air in this room consists of numerous tiny, independent molecules. You have the right to be sure that these air molecules will not all shift
(B) be incredibly unmost likely and also therefore violate the 2nd law of thermodynamics.
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This process is highly unlikely and if happens would violet the second regulation of thermodynamics.
The regulations of motion do not ascendancy out a relocate to one side of the room of all the air molecules. But it would be exceptionally unlikely to occur that. The second law of thermodynamics states that the universe responds to scenarios that end up being more and more most likely, not improbable.
11 months ago
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