Answer:
Explanation:
Cop of reversible refrigerator = TL / ( TH - TL)
TL = low temperature of freezer = 20 °F
TH = temperature of air around = 75 °F
Heat removal rate QL = 75 Btu/min
W actual, power input = 0.7 hp
conversion on F to kelvin = (T (°F) + 460 ) × 5 / 9
COP ( coefficient of performance) reversible = (20 + 460) × 5/9 / (5/9 ( ( 75 +460) - (20 + 460) ))
COP reversible = 480 / 55 = 8.73
irreversibility expression, I = W actual - W rev
COP r = QL / Wrev
W rev = QL / COP r where 75 Btu/min = 1.76856651 hp where W actual = 0.70 hp
a) W rev = 1.76856651 hp / 8.73 = 0.20258 hp is reversible power
I = W actual - W rev
b) I = 0.7 hp - 0.20258 hp = 0.4974 hp
c) the second-law efficiency of this freezer = W rev / W actual = 0.20258 hp / 0.7 hp = 0.2894 × 100 = 28.94 %
Answer:
1) a. Customers requiring AC electric power transmission for powering remote devices which may include a subsea transmission system where power is distributed to subsea devices
b. Customers that utilize commutator-type motors
2) The primary voltage is 150 volts
Explanation:
The parameters given are;
Number of primary winding,
= 50 turns
Number of secondary winding,
= 150 turns
Voltage in secondary winding,
= 450 volts
Voltage in primary winding = 
The relation between
,
,
and
is as follows;

Which gives;

From which we have;

The primary voltage = 150 volts.
1) PRIVATE
2) PRIVATE FIRST CLASS
3) CORPAL
4) SPECIALIST
5) SERGEANT
6) STAFF SEGEANT
7) SEREANT 1ST CLASS
8) MASTER SERGENT
9) FIRST SEGEANT
10) SEGEANT MAJOR
11) COMMAND SERGEANT
12) SEGREANT MAJOR OF THE ARMY