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RoseWind [281]
3 years ago
7

An ideal vapor-compression refrigeration cycle that uses refrigerant-134a as its working fluid maintains a condenser at 800 kPa

and an evaporator at -20°C. Determine this system’s COP and the amount of power required to service a 150 kW cooling load. (Take the required values from saturated refrigerant-134a tables.)

Engineering
1 answer:
mote1985 [20]3 years ago
4 0

Answer:

COP = 3.828

W' = 39.18 Kw

Explanation:

From the table A-11 i attached, we can find the entropy for the state 1 at -20°C.

h1 = 238.43 KJ/Kg

s1 = 0.94575 KJ/Kg.K

From table A-12 attached we can do the same for states 3 and 4 but just enthalpy at 800 KPa.

h3 = h4 = hf = 95.47 KJ/Kg

For state 2, we can calculate the enthalpy from table A-13 attached using interpolation at 800 KPa and the condition s2 = s1. We have;

h2 = 275.75 KJ/Kg

The power would be determined from the energy balance in state 1-2 where the mass flow rate will be expressed through the energy balance in state 4-1.

W' = m'(h2 - h1)

W' = Q'_L((h2 - h1)/(h1 - h4))

Where Q'_L = 150 kW

Plugging in the relevant values, we have;

W' = 150((275.75 - 238.43)/(238.43 - 95.47))

W' = 39.18 Kw

Formula foe COP is;

COP = Q'_L/W'

COP = 150/39.18

COP = 3.828

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In a creep test, increasing the temperature will (choose the best answer) A. increase the instantaneous initial deformation B. i
Hitman42 [59]

Answer:

All of the above

Explanation:

firstly, a creep can be explained as the gradual deformation of a material over a time period. This occurs at a fixed load with the temperature the same or more than the recrystallization temperature.

Once the material gets loaded, the instantaneous creep would start off and it is close to electric strain. in the primary creep area, the rate of the strain falls as the material hardens. in the secondary area, a balance between the hardening and recrystallization occurs. The material would get to be fractured hen recrstallization happens.  As temperature is raised the recrystallization gets to be more.

8 0
2 years ago
A coal fired power plant geneartes 2.4 lbs. of CO2 per kWh. A lighting system consumes 300,000kWh per year. A corporation is con
Serjik [45]

Answer:

The perceived economic impact of CO2 generated per year by lighting sstem is $8164.67.

Explanation:

The CO2 requirement for the plant is:

Amount of CO2 per year = (2.4 lb / KWh)(300,000 KWh)

Amount of CO2 per year = (720000 lb)(1 ton/ 2204.62 lb)

Amount of CO2 per year = 326.59 ton

The perceived economic impact of CO2 generated per year will then be:

Economic Impact = ($25 / ton)(326.59 ton)

<u>Economic Impact = $8164.67</u>

7 0
3 years ago
An aluminium alloy tube has a length of 750 mm at a temperature of 223°C. What will be its length at 23°C if its coefficient of
uranmaximum [27]

Answer:

Final length= 746.175 mm

Explanation:

Given that Length of aluminium at 223 C is 750 mm.As we know that when temperature of material increases or decreases then dimensions of material also increases or decreases respectively with temperature.

Here temperature of aluminium decreases so the final length of aluminium decreases .

As we know that

\Delta L=L\alpha\Delta T

Now by putting the values

\Delta L=750\times \25.5\times 10^{-6}\times 200

ΔL=3.82 mm

So final length =750-3.82 mm

Final length= 746.175 mm

3 0
3 years ago
Water at atmospheric pressure boils on the surface of a large horizontal copper tube. The heat flux is 90% of the critical value
masya89 [10]

Answer:

The tube surface temperature immediately after installation is 120.4°C and after prolonged service is 110.8°C

Explanation:

The properties of water at 100°C and 1 atm are:

pL = 957.9 kg/m³

pV = 0.596 kg/m³

ΔHL = 2257 kJ/kg

CpL = 4.217 kJ/kg K

uL = 279x10⁻⁶Ns/m²

KL = 0.68 W/m K

σ = 58.9x10³N/m

When the water boils on the surface its heat flux is:

q=0.149h_{fg} \rho _{v} (\frac{\sigma (\rho _{L}-\rho _{v})}{\rho _{v}^{2} }  )^{1/4} =0.149*2257*0.596*(\frac{58.9x10^{-3}*(957.9-0.596) }{0.596^{2} } )^{1/4} =18703.42W/m^{2}

For copper-water, the properties are:

Cfg = 0.0128

The heat flux is:

qn = 0.9 * 18703.42 = 16833.078 W/m²

q_{n} =uK(\frac{g(\rho_{L}-\rho _{v})     }{\sigma })^{1/2} (\frac{c_{pL}*deltaT }{c_{fg}h_{fg}Pr  } \\16833.078=279x10^{-6} *2257x10^{3} (\frac{9.8*(957.9-0.596)}{0.596} )^{1/2} *(\frac{4.127x10^{3}*delta-T }{0.0128*2257x10^{3}*1.76 } )^{3} \\delta-T=20.4

The tube surface temperature immediately after installation is:

Tinst = 100 + 20.4 = 120.4°C

For rough surfaces, Cfg = 0.0068. Using the same equation:

ΔT = 10.8°C

The tube surface temperature after prolonged service is:

Tprolo = 100 + 10.8 = 110.8°C

8 0
3 years ago
Ti-6Al-4V has a fracture toughness of 74.6 MPa-m0.5. How much stress (in MPa) would it take to fail a plate loaded in tension th
Nikitich [7]

Answer:

critical stress  = 595 MPa

Explanation:

given data

fracture toughness =  74.6 MPa-\sqrt{m}

crack length = 10 mm

f = 1

solution

we know crack length = 10 mm  

and crack length = 2a as given in figure attach

so 2a = 10

a = 5 mm

and now we get here with the help of plane strain condition , critical stress is express as

critical stress  = \frac{k}{f\sqrt{\pi a}}    ......................1

put here value and we get

critical stress  = \frac{74.6}{1\sqrt{\pi 5\times 10^{-3}}}

critical stress  = 595 MPa

so here stress is change by plane strain condition because when plate become thinner than condition change by plane strain to plain stress.

plain stress condition occur in thin body where stress through thickness not vary by the thinner section.

6 0
3 years ago
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