Answer:
a. 33120MJ
b. 5.6
c. 48
Explanation:
∆U= 120,000 KJ/h
Since
1 day = 24 hrs
14 days =24 x 14 hrs
14 days = 336 hrs
∆U = 120000 x 336 KJ
=40320000KJ
K = 1000
M = 1000000
∆U = 403200 MJ
Work done
W= 2000 KW.h ( 1 h = 3600 s)
W= 7200 MJ
According to the first law of thermodynamics
∆U = Q+W
Q= 40320 - 7200 MJ
1)Qa=33120 MJ
Coefficient of performance
COP = ∆U/W
COP= 40320 / 7200
2)COP = 5.6
3)COP of ideal heat pump
Th/(Th - Tl)
Th = 15°C
Tl = 9°C
Convert Celsius to Kelvin
273 + 15 = 288
288/(15-9)
288/6
48
COP= 48
Answer:
61.5 °C
Explanation:
The resistance of the relay coil at 15 °C is R = V/I = 6/0.12 = 50 ohms. In order for the coil current to remain above 0.10 A, the resistance must remain below R = 6/0.10 = 60 ohms.
At some temperature difference ΔT from 15 °C, the resistance of the coil will be ...
R = R0(1 +α·ΔT)
where R0 is the resistance at 15 °C, α is the temperature coefficient of resistance, and ΔT is the temperature change. We want to solve this for ΔT:
R/R0 = 1 +α·ΔT
(R/R0 -1)/α = ΔT = (60/50 -1)/0.0043 ≈ 46.5 . . . . °C
The relay may fail to operate at temperatures above (15 +46.5) °C = 61.5 °C.
Answer:
Explanation:
The bolt is under double shear because it connects 3 plates.
Now the shear force resisted by each shear joint(V):
V=P/(No. of shear area)=185/2=92.5N.
Answer:
Instrument Landing System
Explanation:
The ILS works by sending radio waves from the runway to the aircraft. Which is then intercepted and is used to guide the aircraft onto the runway.