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:
a)
, b)
, c) Irreversible.
Explanation:
a) The power input required by the refrigerator is:



b) The Coefficient of Performance of the refrigerator is:



c) The maximum ideal Coefficient of Performance of the refrigeration is given by the inverse Carnot's Cycle:



The refrigeration cycle is irreversible, as
.
Answer:
1. Slope = 53.3 x 10⁻⁶
2. Deflection = -0.00016m
Explanation:
given:
let L = 4 m (span of cantilever beam)
let w = 300 N/m (distributed load)
let EI =60 MNm² (flexural stiffness)
dy w * L³ 300 x 4³
1. slope = ------- = --------- = ------------------- = 53.3 x 10⁻⁶
dx 6EI 6 x 60x10⁶
wL⁴ 300 x 4⁴
2. Deflection = y = - ----------- = - ------------------ = -0.00016m
8EI 8 x 60x10⁶
therefore the deflection is 0.16mm downwards.
Answer:
(A) Adding Nitrogen
Explanation:
R11 or R123 are flammable substance under certain conditions when mixed with Oxygen, adding Nitrogen is a safety procedure used in displacing any R11 or R123 which are considered as hazardous when mixed with Oxygen.