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Andrej [43]
3 years ago
12

Argon contained in a closed, rigid tank, initially at 33.7°C, 2.1 bar, and a volume of 4.2 m3, is heated to a final pressure of

7.2 bar. Assuming the ideal gas model with k = 1.65 for the argon, determine the heat transfer, in kJ.
Engineering
1 answer:
Gelneren [198K]3 years ago
4 0

Answer:

total heat required to raise pressure from 2.1 bar to 7.2 bar is 3295.336 kJ

Explanation:

given data

temperature t1 =  33.7°C = 306.7 K

pressure p1 = 2.1 bar

volume v1 = 4.2 m³

pressure p2 = 7.2 bar

ideal gas model k = 1.65

solution

we know heat transfer in constant volume is express as

Q = m × Cv × (t2-t1)   .............1

and here gas constant is 208.13 J/Kg.k

specific heat constant volume Cv = \frac{R}{K-1}  

put here value

Cv = \frac{208.13}{1.65-1}  

Cv = 320.2 J/Kg.k  

and

we get mass of argon gas m is

P1 × V1 = m × R × T1

m = \frac{2.1\times 10^5 \times 4.2}{208.13\times 306.7}  

m = 13.817 kg

and

t2 = \frac{7.2\times 306.7}{2.1}  

t2 = 1051.54  K

and

put equation 1 we get

Q = 13.81 × 320.2 × (1051.54-306.7)

Q = 3295.336 kJ

so total heat required to raise pressure from 2.1 bar to 7.2 bar is 3295.336 kJ

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In the first-order process,a blue dye reacts to form a purple dye. The amount of blue dye at the end of 1 hr is 480 g and the en
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Answer:

The answer is 960 kg

Explanation:

Solution

Given that:

Assume the initial dye concentration as A₀

We write the expression for the dye concentration for one hour as follows:

ln (C₁) = ln (A₀) -kt

Here

C₁ = is the concentration at 1 hour

t =time

Now

Substitute 480 g for C₁ and 1 hour for t

ln (480) = ln (A₀) -k(1) ------- (1)

6.173786 =  ln (A₀) -k

Now

We write the expression for the dye concentration for three hours as follows:

ln (C₃) = ln (A₀) -k

Here

C₃ = is the concentration at 3 hour

t =time

Thus

Substitute 480 g for C₃ and 3 hour for t

ln (120) = ln (A₀) -k(3) ------- (2)

4.787492 = ln (A₀) -3k

Solve for the equation 1 and 2

k =0.693

Now

Calculate the amount of blue present initially using the expression:

Substitute 0.693 for k in equation (2)

4.787492 = ln (A₀) -3 (0.693)

ln (A₀) =6.866492

A₀ =e^6.866492

= 960 kg

Therefore, the amount of the blue dye present from the beginning is  960 kg

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An air-standard Carnot cycle is executed in a closed system between the temperature limits of 350 and 1200 K. The pressures befo
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This question is incomplete, the complete question is;

An air-standard Carnot cycle is executed in a closed system between the temperature limits of 350 and 1200 K. The pressures before and after the isothermal compression are 150 and 300 kPa, respectively. If the net work output per cycle is 0.5 kJ, determine (a) the maximum pressure in the cycle, (b) the heat transfer to air, and (c) the mass of air. Assume variable specific heats for air.

Answer:

a) the maximum pressure in the cycle is 30.01 Mpa

b) the heat transfer to air is 0.7058 KJ

c) mass of Air is 0.002957 kg

Explanation:

Given the data in the question;

We find the relative pressure of air at 1200 K (T1) and 350 K ( T4)

so from the "ideal gas properties of air table"

Pr1 = 238

Pr4 = 2.379

we know that Pressure P1 is only maximum at the beginning of the expansion process,

so

now we express the relative pressure and pressure relation for the process 4-1

P1 = (Pr2/Pr4)P4

so we substitute

P1 = (238/2.379)300 kPa

P1 = 30012.6 kPa = 30.01 Mpa

Therefore the maximum pressure in the cycle is 30.01 Mpa

b)

the Thermal heat efficiency of the Carnot cycle is expressed as;

ηth = 1 - (TL/TH)

we substitute

ηth = 1 - (350K/1200K)

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ηth = 0.7084

now we find the heat transferred

Qin = W_net.out / ηth

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Qin = 0.5 / 0.7084

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c)

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= -0.1989 Kj/Kg.K = S1 - S2

so that; S2 - S1 = 0.1989 Kj/Kg.K

Next we find the net work output per unit mass for the Carnot cycle

W"_netout = (S2 - S1)(TH - TL)

we substitute

W"_netout = ( 0.1989 Kj/Kg.K )( 1200 - 350)K

= 169.065 kJ/kg

Finally we find the mass

mass m = W_ net.out /  W"_netout

we substitute

m = 0.5 / 169.065

m = 0.002957 kg

Therefore, mass of Air is 0.002957 kg

5 0
3 years ago
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