These parts are commonly called carburetor emulsion tubes. These tubes maintain the air-fuel ratio at different speeds.
The carburetor is a device of the combustion engine power supply system that mixes fuel and air in order to facilitate internal combustion.
The carburetor emulsion tubes are tubes that maintain the air-fuel ratio at different velocities.
These tubes (carburetor emulsion tubes) are small brass cylinders where the metering needle slides into them.
Learn more about carburetors here:
brainly.com/question/4237015
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
Answer: $17,206.13
Explanation:
Hi, to answer this question we have to apply the next formula:
Annual electricity cost = (P x 0.746 x Ckwh x h) /η
P = compressor power = 78 hp
0.746 kw/hp= constant (conversion to kw)
Ckwh = Cost per kilowatt hour = $0.11/kWh
h = operating hours per year = 2500 h
η = efficiency = 93% = 0.93 (decimal form)
Replacing with the values given :
C = ( 78 hp x 0.746 kw/hp x 0.11 $/kwh x 2500 h ) / 0.93 = $17,206.13
