Answer:
Entropy generation==0.12 KW/K
Explanation:



Mass flow rate= 


mass flow rate=
So by putting the values
Mass flow rate=2.97 kg/s
So entropy generation=(2.97)(0.0412)
=0.12 KW/K
If we have 20-ampere circuit breakers. The number of circuits to the larger whole number is: 13.
<h3>Number of circuits</h3>
Receptacles on a single strap= 180 VA each.
Hence,
VA of the circuit=(Volts x Amperes)/One receptacle
Let plug in the formula
VA of the circuit=(120 volts x 20 amperes)/180 VA
VA of the circuit= 2,400 VA (circuit)/180 VA
VA of the circuit = 13 circuits
Therefore the number of circuits to the larger whole number is: 13.
Learn more about number of circuits here:brainly.com/question/2969220
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Answer:
= 1.47kJ/K
Explanation: <u>Entropy</u> is the measure of a system's molecular disorder, i.e, the unuseful work a system does.
The nitrogen gas in the insulated tank can be described as an ideal gas, so it can be used the related formulas.
For the entropy, the ratio of initial and final temperatures is needed and as volume is constant, we use:




<u>Specific</u> <u>Heat</u> is the quantity of heat required to increase the temperature 1 degree of a unit mass of a substance. Specific heat of nitrogen at constant volume is
0.743kJ/kg.K
The change in entropy is calculated by
![\Delta S= m[c_{v}ln(\frac{T_{2}}{T_{1}})-Rln(\frac{V_{2}}{V_{1}} )]](https://tex.z-dn.net/?f=%5CDelta%20S%3D%20m%5Bc_%7Bv%7Dln%28%5Cfrac%7BT_%7B2%7D%7D%7BT_%7B1%7D%7D%29-Rln%28%5Cfrac%7BV_%7B2%7D%7D%7BV_%7B1%7D%7D%20%29%5D)
For the nitrogen insulated in a rigid tank:
![\Delta S= m[c_{v}ln(\frac{T_{2}}{T_{1}})]](https://tex.z-dn.net/?f=%5CDelta%20S%3D%20m%5Bc_%7Bv%7Dln%28%5Cfrac%7BT_%7B2%7D%7D%7BT_%7B1%7D%7D%29%5D)
Substituing:
![\Delta S= 3[0.743ln(1.94)]](https://tex.z-dn.net/?f=%5CDelta%20S%3D%203%5B0.743ln%281.94%29%5D)
1.47
The entropy change of nitrogen in an insulated rigid tank is 1.47kJ/K
Answer:
0.31 μm
Explanation:
this question wants us to Determine the depletion region width, xn, in the n-side in unit of μm. using the information below.
density in the p-side = 5.68x10^16
density in the n-side = 1.42x10^16

= √(1.42x10⁵)(1.76056335x10⁻¹⁷ + 7.042253521x10⁻¹⁷)(1.2)
= √150.74x10⁻¹¹
= 3.882x10⁻⁵
approximately 0.39μm
xn = 0.39 x 0.8
= 0.31μm
0.31 um is the depletion region width. thank you!
Answer:
a. 2.08, b. 1110 kJ/min
Explanation:
The power consumption and the cooling rate of an air conditioner are given. The COP or Coefficient of Performance and the rate of heat rejection are to be determined. <u>Assume that the air conditioner operates steadily.</u>
a. The coefficient of performance of the air conditioner (refrigerator) is determined from its definition, which is
COP(r) = Q(L)/W(net in), where Q(L) is the rate of heat removed and W(net in) is the work done to remove said heat
COP(r) = (750 kJ/min/6 kW) x (1 kW/60kJ/min) = 2.08
The COP of this air conditioner is 2.08.
b. The rate of heat discharged to the outside air is determined from the energy balance.
Q(H) = Q(L) + W(net in)
Q(H) = 750 kJ/min + 6 x 60 kJ/min = 1110 kJ/min
The rate of heat transfer to the outside air is 1110 kJ for every minute.