Answer:
heat transfer for the process is - 643.3 kJ
Explanation:
given data
mass m = 2 kg
pressure p1 = 500 kPa
temperature t1 = 400°C = 673.15 K
temperature t2 = 40°C = 313.15 K
pressure p2 = 300 kPa
to find out
heat transfer for the process
solution
we know here mass is constant so
m1 = m2
so by energy equation
m ( u2 - u1 ) = Q - W
Q is heat transfer
and in process P = A+ N that is linear spring
so
W = ∫PdV
= 0.5 ( P1+P2) ( V1 - V2)
so for case 1
P1V1 = mRT
put here value
500 V1 = 2 (0.18892) (673.15)
V1 = 0.5087 m³
and
for case 2
P2V2 = nRT
300 V2 = 2 (0.18892) (313.15)
V2 = 0.3944 m³
and
here W will be
W = 0.5 ( 500 + 300 ) ( 0.3944 - 0.5087 )
W = -45.72 kJ
and
Q is here for Cv = 0.83 from ideal gas table
Q = mCv ( T2-T1 ) + W
Q = 2 × 0.83 ( 40 - 400 ) - 45.72
Q = - 643.3 kJ
heat transfer for the process is - 643.3 kJ
Answer:
pneumatic power system
Explanation:
pneumatic power can be used to quietly operate power windows, door locks, power mirrors, and much much more, also negative pressure pneumatics (vacuum) is used to control many engine and fuel systems
Answer:
1. Buy Quiet – select and purchase low-noise tools and machinery
2. Maintain tools and equipment routinely (such 3. as lubricate gears)
3. Reduce vibration where possible
4. Isolate the noise source in an insulated room or enclosure
5. Place a barrier between the noise source and the employee
6. Isolate the employee from the source in a room or booth (such as sound wall or window
Explanation:
Hope my answer will help u.
Answer:
The correct answer is option (A) 0.060 uF
Note: Kindly find an attached image of the complete question below
Sources: The complete question was well researched from Quizlet.
Explanation:
Solution
Given that:
C₁ = 0.1 μF
C₂ =0.22 μF
C₃ = 0.47 μF
In this case, C₁, C₂ and C₃ are in series
Thus,
Their equivalent becomes:
1/Ceq = (1/C₁ + 1/C₂ +1/C₃
1/Ceq =[ (1/0.1 + 1/0.22 +1/0.47)]
1/Ceq =[(0.22 * 0.47) + (0.1 * 0.47) + (0.1 * 0.22)/(0.1 * 0.22 *0.47)]
1/Ceq =[(0.1034 + 0.047 + 0.022)/(0.01034)
1/Ceq =[(0.1724)/(0.01034)]
1/Ceq = [(16.67)]
1/Ceq =(1/16.67) = 0.059μf
Ceq = 0.059μf ≈ 0.060μf
Therefore the equivalent capacitance of the three series capacitors is 0.060μf