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3 years ago

What is the value of the work interaction in this process?

Engineering

1 answer:

Cloud [144]3 years ago

7 0

Answer:

The answer is " $-121\ \frac{KJ}{Kg}$ ".

Explanation:

Please find the correct question in the attachment file.

using formula:

$\to W=-P_1V_1+P_2V_2 \\\\When \\\\\to W= \frac{P_1V_1-P_2V_2}{n-1}\ \ or \ \ \frac{RT_1 -RT_2}{n-1}\\\\$

$W =\frac{R(T_1 -T_2)}{n-1}\\\\$

$=\frac{0.287(25 -237)}{1.5-1}\\\\=\frac{0.287(-212)}{0.5}\\\\=\frac{-60.844}{0.5}\\\\=-121.688 \frac{KJ}{Kg}\\\\=-121 \frac{KJ}{Kg}\\\\$

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The time factor for a doubly drained clay layer

Margarita [4]

Answer with Explanation:

Assuming that the degree of consolidation is less than 60% the relation between time factor and the degree of consolidation is

$T_v=\frac{\pi }{4}(\frac{U}{100})^2$

Solving for 'U' we get

$\frac{\pi }{4}(\frac{U}{100})^2=0.2\\\\(\frac{U}{100})^2=\frac{4\times 0.2}{\pi }\\\\\therefore U=100\times \sqrt{\frac{4\times 0.2}{\pi }}=50.46%$

Since our assumption is correct thus we conclude that degree of consolidation is 50.46%

The consolidation at different level's is obtained from the attached graph corresponding to Tv = 0.2

i) $\frac{z}{H}=0.25=U=0.71$ = 71% consolidation

ii) $\frac{z}{H}=0.5=U=0.45$ = 45% consolidation

iii) $\frac{z}{H}=0.75U=0.3$ = 30% consolidation

Part b)

The degree of consolidation is given by

$\frac{\Delta H}{H_f}=U\\\\\frac{\Delta H}{1.0}=0.5046\\\\\therefore \Delta H=50.46cm$

Thus a settlement of 50.46 centimeters has occurred

For time factor 0.7, U is given by

$T_v=1.781-0.933log(100-U)\\\\0.7=1.781-0.933log(100-U)\\\\log(100-U)=\frac{1.780-.7}{0.933}=1.1586\\\\\therefore U=100-10^{1.1586}=85.59$

thus consolidation of 85.59 % has occured if time factor is 0.7

The degree of consolidation is given by

$\frac{\Delta H}{H_f}=U\\\\\frac{\Delta H}{1.0}=0.8559\\\\\therefore \Delta H=85.59cm$

5 0

3 years ago

Air is saturated with water vapor at 35.0 oC and a total pressure of 1.50 atmospheres. If the molar flow rate of the dry air in

Marizza181 [45]

Answer:

11.541 mol/min

Explanation:

temperature = 35°C

Total pressure = 1.5 * 1.013 * 10^5 = 151.95 kPa

note : partial pressure of water in mixture = saturation pressure of water at T = 35°c )

from steam table it is = 5.6291 Kpa

calculate the mole fraction of H $_{2}o$ ( YH $_{2}o$ )

= 5.6291 / 151.95

= 0.03704

calculate the mole fraction of air ( Yair )

= 1 - mole fraction of water

= 1 - 0.03704 = 0.9629

Now to determine the molar flow rate of water vapor in the stream

lets assume N = Total molar flow rate

NH $_{2}o$ = molar flow rate of water

Nair = molar flow rate of air = 300 moles /min

note : Yair * n = Nair

therefore n = 300 / 0.9629 = 311.541 moles /min

Molar flowrate of water

= n - Nair

= 311.541 - 300 = 11.541 mol/min

4 0

3 years ago

Isn't this website cheating?

Orlov [11]

Answer: What website?

Explanation:

8 0

3 years ago

Read 2 more answers

6msection of 150lossless line is driven by a source with vg(t) = 5 cos(8π × 107t − 30◦ ) (V) and Zg = 150 . If the line, which h

Rainbow [258]

Answer:

a. 5m

b. r = 0.16 e^-80.5◦

c. Zpn = (115.7 + j27.4) ohms

d. Vi = 2.2e^-j22.56◦ volts

e. Vi(t) = 2.2 cos (8π × 107t − 22.56◦ ) Volts

Explanation:

In this question, we are tasked with calculating a series of terms.

Please check attachment for complete solution and step by step explanation

7 0

3 years ago

Please help me, Engineering.

antiseptic1488 [7]

Answer:

15. R = R₁ + R₂ + R₃

16. $R = \dfrac{1}{R_1} +\dfrac{1}{R_2} = \dfrac{R_2 + R_1}{R_1 \cdot R_2}$

Explanation:

15. The equation for the sum of resistances arranged in series is given as follows;

$R_{Network}$ (Series) = R₁ + R₂ + R₃ + ... + $R_N$

Therefore, the equation for the total resistance, 'R' of 3 series resistors is given as follows;

R = R₁ + R₂ + R₃

Where;

R₁, R₂, and R₃ are the three resistances arranged in series

16. The equation for the sum of resistances arranged in parallel is given as follows;

$R_n = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3} + ... + \dfrac{1}{R_n}$

Therefore, the total resistance, R, of two resistance arranged in parallel is given as follows;

$R = \dfrac{1}{R_1} +\dfrac{1}{R_2} = \dfrac{R_2 + R_1}{R_1 \cdot R_2}$

8 0

2 years ago