PLSSSSS Help !!!!!!!!!!!. It's due today. <br> I will give brainliest to correct answer

Refrigerant 22 flows in a theoretical single-stage compression refrigeration cycle with a mass flow rate of 0.05 kg/s. The conde

Debora [2.8K]

Answer:

a. The work done by the compressor is 447.81 Kj/kg

b. The heat rejected from the condenser in kJ/kg is 187.3 kJ/kg

c. The heat absorbed by the evaporator in kJ/kg is 397.81 Kj/Kg

d. The coefficient of performance is 2.746

e. The refrigerating efficiency is 71.14%

Explanation:

According to the given data we would need first the conversion of temperaturte from C to K as follows:

Temperature at evaporator inlet= Te=-16+273=257 K

Temperatue at condenser exit=Te=48+273=321 K

Enthalpy at evaporator inlet of Te -16=i3=397.81 Kj/Kg

Enthalpy at evaporator exit of Te 48=i1=260.51 Kj/Kg

b. To calculate the the heat rejected from the condenser in kJ/kg we would need to calculate the Enthalpy at the compressor exit by using the compressor equation as follows:

w=i4-i3

W/M=i4-i3

i4=W/M + i3

i4=2.5/0.05 + 397.81

i4=447.81 Kj/kg

a. Enthalpy at the compressor exit=447.81 Kj/kg

Therefore, the heat rejected from the condenser in kJ/kg=i4-i1

the heat rejected from the condenser in kJ/kg=447.81-260.51

the heat rejected from the condenser in kJ/kg=187.3 kJ/kg

c. Temperature at evaporator inlet= Te=-16+273=257 K

The heat absorbed by the evaporator in kJ/kg is Enthalpy at evaporator inlet of Te -16=i3=397.81 Kj/Kg

d. To calculate the coefficient of performance we use the following formula:

coefficient of performance=Refrigerating effect/Energy input

coefficient of performance=137.3/50

coefficient of performance=2.746

the coefficient of performance is 2.746

e. The refrigerating efficiency = COP/COPc

COPc=Te/(Tc-Te)

COPc=255/(321-255)

COPc=3.86

refrigerating efficiency=2.746/3.86

refrigerating efficiency=0.7114=71.14%

8 0

4 years ago

What input is required from ADC to allow the INS to calculate W/V?

Kazeer [188]

TAS is an important input which is required from ADC to allow the INS calculate W/V.

INS is abbreviation for Inertial Navigation System and it can be defined as a navigation device that makes use of motion sensors, a computer, and rotation sensors, so as to continuously calculate by dead reckoning the velocity, position, and orientation of a moving object.

In the Aviation and Engineering filed, True Airspeed (TAS) is an important input which is required from ADC to allow the Inertial Navigation System (INS) calculate W/V.

Read more on Inertial Navigation System here: brainly.com/question/26052911

#SPJ12

4 0

2 years ago

What steel type and ASTM designation is preferred for W-shapes?

Ulleksa [173]

Answer:

The preferred steel type for W-shapes is structural steel and the its preferred ASTM designation is ASTM A992.

Explanation:

The ASTM A992 is a structural steel and it's the most available for w-shapes; besides its availabilty, its ductility improvements makes it the preferred choice; other common designations for this shapes are ASTM A572 Grade 50,0r ASTM A36, but this designations aren't as available as ASTM A992, and it has to be confirmed prior to their specification.

3 0

4 years ago

A wing component on an aircraft is fabricated from an aluminum alloy that has a plane strain fracture toughness of 27 . It has b

Mamont248 [21]

Answer:

The stress level at which fracture will occur for a critical internal crack length of 6.2mm is 135.78MPa

Explanation:

Given data;

Let,

critical stress required for initiating crack propagation Cc = 112MPa

plain strain fracture toughness = 27.0MPa

surface length of the crack = a

dimensionless parameter = Y.

Half length of the internal crack, a = length of surface crack/2 = 8.8/2 = 4.4mm = 4.4*10-³m

Also for 6.2mm length of surface crack;

Half length of the internal crack = length of surface crack/2 = 6.2/2 = 3.1mm = 3.1*10-³m

The dimensionless parameter

Cc = Kic/(Y*√pia*a)

Y = Kic/(Cc*√pia*a)

Y = 27/(112*√pia*4.4*10-³)

Y = 2.05

Now,

Cc = Kic/(Y*√pia*a)

Cc = 27/(2.05*√pia*3.1*10-³)

Cc = 135.78MPa

The stress level at which fracture will occur for a critical internal crack length of 6.2mm is 135.78MPa

For more understanding, I have provided an attachment to the solution.

4 0

4 years ago

Create a file named students containing the following data in your current directory. Each line in this file represents a studen

Ostrovityanka [42]

Answer:

#!/bin/bash

# Simple line count example, using bash

# Usage: ./line_count.sh file

# -----------------------------------------------------------------------------

# Link filedescriptor 10 with stdin

exec 10<&0

# stdin replaced with a file supplied as a first argument

exec < $1

# remember the name of the input file

in=$1

# init

file="current_line.txt"

let count=0

# this while loop iterates over all lines of the file

while read LINE

do

# increase line counter

((count++))

# write current line to a tmp file with name $file (not needed for counting)

echo $LINE > $file

# this checks the return code of echo (not needed for writing; just for demo)

if [ $? -ne 0 ]

then echo "Error in writing to file ${file}; check its permissions!"

fi

done

echo "Number of lines: $count"

echo "The last line of the file is: `cat ${file}`"

# Note: You can achieve the same by just using the tool wc like this

echo "Expected number of lines: `wc -l $in`"

# restore stdin from filedescriptor 10

# and close filedescriptor 10

exec 0<&10 10<&-

Explanation:

4 0

4 years ago

PLSSSSS Help !!!!!!!!!!!. It's due today. I will give brainliest to correct answer