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olga_2 [115]
3 years ago
15

The reversible and adiabatic process of a substance in a compressor begins with enthalpy equal to 1,350 kJ/kg, and ends with ent

halpy equal to 3,412 kJ/kg. If the compressor efficiency is 0.85, find the actual specific work required by the compressor to operate, in kJ/kg.
Engineering
1 answer:
notsponge [240]3 years ago
7 0

Answer:

the actual specific work required by the compressor to operate is 2425.88 kJ/kg

Explanation:

Given that;

h₁ = 1350 kJ/kg

h₂₅ = 3412 kJ/kg

compressor efficiency П_ise = 0.85

we know that;

compressor efficiency П_ise = isentropic work / actual work

П_ise = (h₂₅ - h₁) / (h₂ - h₁ )

so

0.85 =  (h₂₅ - h₁) / (actual work )

Actual work = (h₂₅ - h₁) / 0.85

Actual work = (3412 - 1350) / 0.85

Actual work = 2062 / 0.85

Actual work = 2425.88 kJ/kg

Therefore the actual specific work required by the compressor to operate is 2425.88 kJ/kg

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A cylindrical 1045 steel bar is subjected to repeated compression-tension stress cycling along its axis. If the load amplitude i
Gre4nikov [31]

Answer:

13.4 mm

Explanation:

Given data :

Load amplitude ( F )  = 22,000 N

factor of safety ( N )= 2.0

Take ( Fatigue limit stress amplitude for this alloy ) б = 310 MPa

<u>calculate the minimum allowable bar diameter to ensure that fatigue failure will not occur</u>

minimum allowable bar diameter = 13.4 * 10^-3 m ≈ 13.4 mm

<em>attached below is a detailed solution</em>

3 0
2 years ago
simple Brayton cycle using air as the working fluid has a pressure ratio of 10. The minimum and maximum temperatures in the cycl
Irina18 [472]

Answer:

a) 764.45K

b) 210.48 kJ/kg

c) 30.14%

Explanation:

pressure ratio = 10

minimum temperature = 295 k

maximum temperature = 1240 k

isentropic efficiency for compressor = 83%

Isentropic efficiency for turbine = 87%

<u>a) Air temperature at turbine exit </u>

we can achieve this by interpolating for enthalpy

h4 = 783.05 kJ/kg ( calculated in the background ) at state 4 using Table A-17  for  Ideal gas properties of air

T4 ( temperature at Turbine exit ) = 760 + ( 780 - 760 ) (\frac{783.05-778.18}{800.13-778.18} ) = 764.45K

<u>b) The net work output </u>

first we determine the actual work input to compressor

Wc = h2 - h1  ( calculated values )

     = 626.57 - 295.17 =  331.4 kJ/kg

next determine the actual work done by Turbine

Wt = h3 - h4  ( calculated values )

     = 1324.93 - 783.05 = 541.88 kJ/kg

finally determine the network output of the cycle

Wnet = Wt - Wc

         = 541.88 - 331.4  = 210.48 kJ/kg

<u>c) determine thermal efficiency </u>

лth = Wnet / qin  ------ ( 1 )

where ; qin = h3 - h2

<em>equation 1 becomes </em>

лth = Wnet / ( h3 - h2 )

      = 210.48 / ( 1324.93 - 626.57 )

      = 0.3014  =  30.14%

6 0
3 years ago
A vacuum gage connected to a tank reads 30 kPa at a location where the barometric reading is 755 mmHg. Determine the absolute pr
navik [9.2K]

Answer:

Absolute pressure=70.72 KPa

Explanation:

Given that Vacuum gauge pressure= 30 KPa

Barometer reading =755 mm Hg

We know that barometer always reads atmospheric pressure at given situation.So  atmospheric pressure is equal to  755 mm Hg.

We know that P= ρ g h

Density of Hg=13600 \frac{kg}{m^3}

So P=13600 x 9.81 x 0.755

P=100.72 KPa

We know that

Absolute pressure=atmospheric pressure + gauge pressure

But here given that 30 KPa is a Vacuum pressure ,so we will take it as negative.

Absolute pressure=atmospheric pressure + gauge pressure

Absolute pressure=100.72 - 30   KPa

So

Absolute pressure=70.72 KPa

8 0
3 years ago
An array of electronic chips is mounted within a sealedrectangular enclosure, and colling is implemented by attaching analuminum
Licemer1 [7]

Answer:

Base temperature is 46.23 °C

Explanation:

I've attached explanations

6 0
3 years ago
A gas stream flowing at 1000 cfm with a particulate loading of 400 gr/ft3 discharges from a certain industrial plant through an
Makovka662 [10]

<u>Solution and Explanation:</u>

Volume of gas stream = 1000 cfm (Cubic Feet per Minute)

Particulate loading = 400 gr/ft3 (Grain/cubic feet)

1 gr/ft3 = 0.00220462 lb/ft3

Total weight of particulate matter = 1000 \mathrm{cfm} \times 400 \mathrm{gr} / \mathrm{tt} 3 \times .000142857 \mathrm{lb} / \mathrm{ft} 3 \times 60=3428.568 \mathrm{lb} / \mathrm{hr}

Cyclone is to 80 % efficient

So particulate remaining = 0.20 \times 3428.568 \mathrm{lb} / \mathrm{hr}=685.7136

emissions from this stack be limited to = 10.0 lb/hr

Particles to be remaining after wet scrubber = 10.0 lb/hr

So particles to be removed = 685.7136- 10 = 675.7136

Efficiency = output multiply with 100/input = 98.542 %

4 0
3 years ago
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