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:
1170 km/hr
Explanation:
Let's first state the formula to be used
c = √(KRT)
The temperature at an altitude of 15km is -56.5° C
Let's not convert this to °K, we have
-56.5° + 273.15 = 216.65° K
Also, the temperature at 8km is -36.9° C, on converting to °K we have
-36.9° + 273.15 = 236.25° K
Then again, we look for the speed at both 15 km and 8 km both of which are 295 m/s and 308 km
Finally, we use the mach similarity formula
(V/c) of 15km = (V/c) of 8km
V of 8km = c of 8km * (V/c) of 15km
V of 8km = 1170 km/hr
Answer:
a) the maximum tensile stress due to the weight of the wire is 1361.23 psi
b) the maximum load P that could be supported at the lower end of the wire is 624.83 lb
Explanation:
Given the data in the question;
Length of wire L = 400 ft = ( 400 × 12 )in = 4800 in
Diameter d = 3/16 in
Unit weight w = 490 pcf
First we determine the area of the wire;
A = π/4 × d²
we substitute
A = π/4 × (3/16)²
A = 0.0276 in²
Next we get the Volume
V = Area × Length of wire
we substitute
V = 0.0276 × 4800
V = 132.48 in³
Weight of the steel wire will be;
W = Unit weight × Volume
we substitute
W = 490 × ( 132.48 / 12³ )
W = 490 × 0.076666
W = 37.57 lb
a) the maximum tensile stress due to the weight of the wire;
σ
= W / A
we substitute
σ
= 37.57 / 0.0276
= 1361.23 psi
Therefore, the maximum tensile stress due to the weight of the wire is 1361.23 psi
b) the maximum load P that could be supported at the lower end of the wire. Allowable tensile stress is 24,000 psi
Maximum load P that the wire can safely support its lower end will be;
P = ( σ
- σ
)A
we substitute
P = ( 24000 - 1361.23 )0.0276
P = 22638.77 × 0.0276
P = 624.83 lb
Therefore, the maximum load P that could be supported at the lower end of the wire is 624.83 lb