Answer:
I=9.6×e^{-8} A
Explanation:
The magnetic field inside the solenoid.
B=I*500*muy0/0.3=2.1×e ^-3×I.
so the total flux go through the square loop.
B×π×r^2=I×2.1×e^-3π×0.025^2
=4.11×e^-6×I
we have that
(flux)'= -U
so differentiating flux we get
so the inducted emf in the loop.
U=4.11×e^{-6}×dI/dt=4.11×e^-6×0.7=2.9×e^-6 (V)
so, I=2.9×e^{-6}÷30
I=9.6×e^{-8} A
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.
Answer:
1. Equatorial Evergreen or Rainforest
2. Tropical forest
3. Mediterranean forest
4. Temperate broad-leaved forest
5. Warm temperate forest
Explanation:
Answer:
a) Please see attached copy below
b) 0.39KJ
c) 20.9‰
Explanation:
The three process of an air-standard cycle are described.
Assumptions
1. The air-standard assumptions are applicable.
2. Kinetic and potential energy negligible.
3. Air in an ideal gas with a constant specific heats.
Properties:
The properties of air are gotten from the steam table.
b) T₁=290K ⇒ u₁=206.91 kj/kg, h₁=290.16 kj/kg.
P₂V₂/T₂=P₁V₁/T₁⇒ T₂=P₂T₁/P₁ = 380/95(290K)= 1160K
T₃=T₂(P₃/P₂)⁽k₋1⁾/k =(1160K)(95/380)⁽⁰°⁴/₁.₄⁾ =780.6K
Qin=m(u₂₋u₁)=mCv(T₂-T₁)
=0.003kg×(0.718kj/kg.k)(1160-290)K= 1.87KJ
Qout=m(h₃₋h₁)=mCp(T₃₋T₁)
=0.003KG×(1.005kj/kg.k(780.6-290)K= 1.48KJ
Wnet, out= Qin-Qout = (1.87-1.48)KJ =0.39KJ
c)ηth= Wnet/W₍in₎ =0.39KJ/1.87KJ = 20.9‰
Answer:
final pressure = 200KPa or 29.138psia
Explanation:
The detailed step by step calculations with appropriate conversion factors applied are as shown in the attachment.