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Cloud [144]
3 years ago
7

The compressor in a refrigerator compresses saturated R-134a vapor at 0°F to 200 psia. Calculate the work required by this compr

essor, in Btu/lbm, when the compression process is isentropic. Use the tables for R-134a.
Engineering
1 answer:
kvasek [131]3 years ago
6 0

Answer:

The work required is W =  20.2 BTU per lbm

Explanation:

The value of entropy & enthalpy at initial conditions are

h_{1} = 103.1

S = 0.225

Final enthalpy

h_{2} = 123.3

Therefore work done

W = h_{1} - h_{2}

W = 103.1 - 123.3

W = - 20.2 BTU per lbm

Therefore the work required is W =  20.2 BTU per lbm

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Technician A states that air tools generally produce more noise than electric tools, so wear ear protection when using air tools
dusya [7]

Answer:

Both Technician A and Technician B are correct

Explanation:

Air tools and electric tools are both power tools as they are used to make work easier. Air tools generally use an air compressor that powers the motor of the tool making it possible to use it while electric tools as the name implies are powered by an electric source which in this case is batteries. An example of an air tool is the nail gun which can be used by furniture makers to drive nails and they are often louder than electric tools because of vibrations caused by the compressor making it necessary to use ear protection when using the tool for ear safety.

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3 years ago
use engineering judgement to estimate the size of cooling and heating equipment that is needed for a three story, 31,000 square
Blizzard [7]

Answer: operating cost = 22820.736 $

Energy = 32.291 KBtu/sf.

Explanation:

Total heating load is given as = 31000 * 31.25  

 = 968.75* 103 Btu

From the cooling capacity application;

If 1000ftsquare = 2.8 TR

Therefore 30000ftsquare = x

Where x is the total cooling load,

Therefore x = (30000 * 2.8) / 1000

x = 84TR.

Therefore, the total cooling load, x = 84TR

Using conversion factor;

i.e. converting Btu/hr to MBH

1 MBH = 1000 Btu per hour

we have 968.75* 103 Btu/hr = 968.75 MBH

let us proceed.

Estimating the “Annual operating cost” we first calculate the maximum operating cost (Avg/Annum)

For cooling load:

Max. operatn. Cost = 14 * 84 = 1176$

For heating load:

The average heating loading hours = 1000 hrs.

Conversion to Btu/yr gives = 968.75* 103 * 1000  

 = 968.75* 106 Btu/yr.

Conversion of 968.75* 103Btu to Kwh gives  

1Btu = 0.000293

968.75* 103 Btu = 283.84 Kwh  

Therefore, average cost gives;

(8.04/100) $ = 1Kwh

X = 283.84*103

X = 22820.736 $

Operating cost = 22820.736 $

Energy use in KBtu/sf is given as  

E = 968.75* 103 / 30000  

E = 32.291 KBtu/sf.

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