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Ket [755]
3 years ago
13

Refrigerant-134a is throttled from the saturated liquid state at 800 kPa to a pressure of 140 kPa. Determine the temperature dro

p during this process and the final specific volume of the refrigerant.
Engineering
1 answer:
zavuch27 [327]3 years ago
5 0

Answer:

The temperature drop is 61.1 °C

The final specific volume of the refrigerant is 1.236 m^3/kg

Explanation:

Initial pressure of refrigerant = 800 kPa = 800/100 = 8 bar

Final pressure of refrigerant = 140 kPa = 140/100 = 1.4 bar

From steam table

At 8 bar, initial saturated temperature is 170.4 °C

At 1.4 bar, final saturated temperature is 109.3 °C

Temperature drop = initial saturated temperature - final saturated temperature = 170.4 - 109.3 = 61.1 °C

Also, from steam table

At 1.4 bar, specific volume is 1.236 m^3/kg

Final specific volume of the refrigerant is 1.236 m^3/kg

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Determine the specific volume of superheated water vapor at 15 MPa and 350°C, using a. The ideal-gas equation Answer: 0.01917 m3
ollegr [7]

Answer:

specific volume by ideal gas equation = 0.01917 m³/kg

specific volume by compressibility chart = 0.01246 m³/kg

specif volume by super heated stream table is 0.0114810 m³/kg

Explanation:

given data

temperature  T = 350°C = 623 K

pressure P = 15 MPa = 15000 kPa

to find out

specific volume by  ideal-gas equation ,generalized compressibility chart and steam tables

solution

we will apply here ideal gas equation that is

specific volume = \frac{R*T}{P}   ..............1

here P is pressure and T is temperature and R is gas constant i.e 0.4615 kJ/kg-K

specific volume =  \frac{0.4615*623}{15000}

specific volume = 0.01917 m³/kg

and

by the compressibility chart

critical pressure of water Pcr = 22.06 Mpa

and critical temperature of water Tcr = 647.1 K

so

reduced pressure will be = \frac{P}{Pcr}

reduced pressure = \frac{15}{22.06} = 0.68 Mpa

and

reduced temperature will be = \frac{T}{Tcr}

reduced pressure = \frac{623}{647.1} = 0.963 K

so by compressibility chart pressure 0.68 Mpa and temperature 0.963 K

compressibility factor Z is 0.65

so

specific volume = compressibility factor Z × ideal specific volume

specific volume = 0.65 × 0.01917

specific volume = 0.01246 m³/kg

and

by the steam table

use here super heated stream table for

pressure = 15 Mpa

ans temperature = 350°C

so

specif volume by super heated stream table is 0.0114810 m³/kg

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A liquid stream containing 52.0 mole% benzene and the balance toluene at 20.0°C is fed to a continuous single-stage evaporator a
OLga [1]

Answer:

Operating Pressure P = 793.716 mmHg

Y_Benzene y1 = 0.541

Explanation:

Given that;

liquid phase leaving the evaporator = 32.5 mole%

Equi Temp T = 99.0°C = 99 + 273.15 = 372.15 K

Now let 1 and 2 represent Benzene and Toluene respectively.

Antoine's Constant for these components are;

COMPONENETS        A                B                    C

Benzene 1             4.72583     1660.652        -1.461

Toluene  2            4.07827     1343.943         -53.773

Antoine's equation is expressed as;

Ps = 10^(A - (B/(T+C)))

Ps is in Bar and T is in Kelvin

so

P1s = 10^( 4.72583 - (1660.652/(372.15 - (-1.461)))) = 1.7617 Bar

P2s = 10^( 4.07827 - (1343.943/(372.15 - (-53.773)))) = 0.7195 Bar

now here, liquid leaving and vapor are both in equilibrium

composition of liquid leaving are;

X1 = 32.5%    = 0.325

X2 = 1 - X1 = 1 - 0.325 = 0.675

Now

Raoult's Law is expressed as;

p × y1=x1 × pis     for all components

So for Benzene ; p × y1=x1 × p1s   ------let this be equation 1

for Toluene ; p × y2=x2 × p2s   ------let this be equation 2

lets add equ 1 and 2

p × y1=x1 × p1s + p × y2=x2 × p2s

p(y1 + y2) = x1 × p1s + x2 × p2s

buy y1 + y2 = 1

therefore we substitute

p(1) = 0.325 × 1.7617 + 0.675 × 0.7195 = 1.0582 Bar

we know that 1 Bar = 750.062 mmHg

so p = 1.0582 × 750.062

p = 793.716 mmHg

Also from equation 1

p × y1=x1 × p1s

y1 = (x1 × p1s) / p

y1 = (0.325 × 1.7617) / 1.0582

y1 = 0.541

Therefore;

Operating Pressure P = 793.716 mmHg

Y_Benzene y1 = 0.541

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