Answer:
a) 4.1 kw
b) 4.68 tons
c) 4.02
Explanation:
Saturated vapor enters compressor at ( p1 ) = 2.6 bar
Saturated liquid exits the condenser at ( p2 ) = 12 bar
Isentropic compressor efficiency = 80%
Mass flow rate = 7 kg/min
A) Determine compressor power in KW
compressor power = m ( h2 - h1 )
= 7 / 60 ( 283.71 - 248.545 )
= 4.1 kw
B) Determine refrigeration capacity in tons = m ( h1 - h4 )
= 7/60 ( 248.545 - 107.34 )
= 16.47 kw = 4.68 tons
C) coefficient of performance ( COP )
= Refrigeration capacity / compressor power
= 16.47 / 4.1 = 4.02
Attached below is the beginning part of the solution
In regards to high-wind installations, if an old composition asphalt shingle roof is going to be replaced, the roofing contractor should remove the existing shingles and underlayment.
<h3>What are High Wind
installations?</h3>
High Wind is known to be a newly created installation techniques that is often used in the installation of offshore wind turbines and this is often done at higher wind speeds.
Note that, In regards to high-wind installations, if an old composition asphalt shingle roof is going to be replaced, the roofing contractor should remove the existing shingles and underlayment.
Learn more about installations from
brainly.com/question/13093522
#SPJ1
Answer:
Accumulated depreciation at December 32, 2023 = 29679
Explanation:
machine = 160000
useful life = 40 years
residual value = 3000
straight-line depreciation rate = (160000 - 3000) ÷ 40
= 3925 per year
Annual depreciation rate = 100% ÷ 40 = 2.5%
double-declining rate = 2 × 2.5% = 5%
Depreciation for 2020 = 5% × 160000 = 8000
Depreciation for 2021 = 5% × (160000 - 8000) = 7600
Depreciation for 2022 = 5% × (160000 - 8000 - 7600) = 7220
Depreciation for 2023 = 5% × (160000 - 8000 - 7600 - 7220) = 6859
Accumulated depreciation at December 32, 2023
= 8000 + 7600 +7220 + 6859
= 29679
You will go 495 miles in 9 hours
Explanation & answer:
Assuming a smooth transition so that there is no abrupt change in slopes to avoid frictional loss nor toppling, we can use energy considerations.
Initially, the cube has a kinetic energy of
KE = mv^2/2 = 10 lbm * 20^2 ft^2/s^2 / 2 = 2000 lbm-ft^2 / s^2
At the highest point when the block stops, the gain in potential energy is
PE = mgh = 10 lbm * 32.2 ft/s^2 * h ft = 322 lbm ft^2/s^2
By assumption, there was no loss in energies, we equate PE = KE
322h lbm ft^2/s^2 = 2000 lbm ft^2/s^2
=>
h = 2000 /322 = 6.211 (ft)
distance up incline = h / sin(30) = 12.4 ft
