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:
f=1.59 Hz
Explanation:
Given that
Simple undamped system means ,system does not consists any damper.If system consists damper then it is damped spring mass system.
Velocity = 100 mm/s
Maximum amplitude = 10 mm
We know that for a simple undamped system spring mass system
now by putting the values
100 = ω x 10
ω = 10 rad/s
We also know that
ω=2π f
10 = 2 x π x f
f=1.59 Hz
So the natural frequency will be f=1.59 Hz.
Answer:
A counter which counts from 0 to 255 with seven segment display
Timer Mode Control (TMOD)
Explanation:
Answer:
The minimum particle diameter that is removed at 85% is 1.474 * 10 ^⁻4 meters.
Solution
Given:
Length = 48 m
Width = 12 m
Depth = 3m
Flow rate = 4 m 3 /s
Water density = 10 3 kg/m 3
Dynamic viscosity = 1.30710 -3 N.sec/m
Now,
At the minimum particular diameter it is stated as follows:
The Reynolds number= 0.1
Thus,
0.1 =ρVTD/μ
VT = Dp² ( ρp- ρ) g/ 10μ²
Where
gn = The case/issue of sedimentation
VT = Terminal velocity
So,
0.1 = Dp³ ( ρp- ρ) g/ 10μ²
This becomes,
0.1 = 1000 * dp³ (1100-1000) g 0.1/ 10 *(1.307 * 10 ^⁻3)²
= 3.074 * 10 ^⁻6 = dp³ (.g01 * 10^6)
dp³=3.1343 * 10 ^⁻12
Dp minimum= 1.474 * 10 ^⁻4 meters.
