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4 years ago

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Air is compressed in a reversible, isothermal, steady- flow process from 15 psia, 100°F to 100 psia. Calculate the work of compr

ession per pound, the change of entropy, and the heat transfer per pound of air compressed.

1 answer:

mixas84 [53]4 years ago

3 0

Answer:

|W|=169.28 KJ/kg

ΔS = -0.544 KJ/Kg.K

Explanation:

Given that

T= 100°F

We know that

1 °F = 255.92 K

100°F = 310 .92 K

$P _1= 15 psia$

$P _1= 100 psia$

We know that work for isothermal process

$W=mRT\ln \dfrac{P_1}{P_2}$

Lets take mass is 1 kg.

So work per unit mass

$W=RT\ln \dfrac{P_1}{P_2}$

We know that for air R=0.287KJ/kg.K

$W=RT\ln \dfrac{P_1}{P_2}$

$W=0.287\times 310.92\ln \dfrac{15}{100}$

W= - 169.28 KJ/kg

Negative sign indicates compression

|W|=169.28 KJ/kg

We know that change in entropy at constant volume

$\Delta S=-R\ln \dfrac{P_2}{P_1}$

$\Delta S=-0.287\ln \dfrac{100}{15}$

ΔS = -0.544 KJ/Kg.K

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One kilogram of water fills a 150-L rigid container at an initial pressure of 2 MPa. The container is then cooled to 40∘C. Deter

lukranit [14]

The pressure of water is 7.3851 kPa

<u>Explanation:</u>

Given data,

V = 150× $10^{-3}$ $m^{3}$

m = 1 Kg

$P_{1}$ = 2 MPa

$T_{2}$ = 40°C

The waters specific volume is calculated:

$v_{1}$ = V/m

Here, the waters specific volume at initial condition is $v_{1}$ , the containers volume is V, waters mass is m.

$v_{1}$ = 150× $10^{-3}$ $m^{3}$ /1

$v_{1}$ = 0.15 $m^{3}$ / Kg

The temperature from super heated water tables used in interpolation method between the lower and upper limit for the specific volume corresponds 0.15 $m^{3}$ / Kg and 0.13 $m^{3}$ / Kg.

$T_{1}$ = 350+(400-350) $\frac{0.15-0.13}{0.1522-0.1386}$

$T_{1}$ = 395.17°C

Hence, the initial temperature is 395.17°C.

The volume is constant in the rigid container.

$v_{2}$ = $v_{1}$ = 0.15 $m^{3}$ / Kg

In saturated water labels for $T_{2}$ = 40°C.

$v_{f}$ = 0.001008 $m^{3}$ / Kg

$v_{g}$ = 19.515 $m^{3}$ / Kg

The final state is two phase region $v_{f}$ < $v_{2}$ < $v_{g}$ .

In saturated water labels for $T_{2}$ = 40°C.

$P_{2}$ = $P_{Sat}$ = 7.3851 kPa

$P_{2}$ = 7.3851 kPa

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3 years ago

The heat transfer rate due to free convection from a vertical surface, 1 m high and 0.6 m wide, to quiescent air that is 20 K co

andreyandreev [35.5K]

Answer:

The ratio of heat transfer rate is 0.88

Explanation:

Given;

Case1 :

height of vertical surface, L = 1 m

width of vertical surface, w = 0.6 m

Case 2:

height of vertical surface, L = 0.6 m

width of vertical surface, w = 1 m

At an assumed film temperature of air = 300 K

then, read off from heat transfer table, temperature inverse β, surface area flow rate v, and Pr, to determine Rayleigh number for the two cases.

β = 1/300 = 0.00333 K⁻¹

v = 15.89 x 10⁻⁶ m²/s

Pr = 0.69

Case 1, L = 1 m

$R_a = \frac{g\beta TL^3P_r}{v^2}$

$R_a = \frac{9.8*0.00333* 20*1^3*0.69}{(15.89x10^{-6})2} \\\\R_a = 1.784 *10^9$

Case 2, L = 0.6 m

$R_a = \frac{g\beta TL^3P_r}{v^2} \\\\R_a = \frac{9.8*0.00333* 20*0.6^3*0.69}{(15.89*10^{-6})^2}\\\\ R_a = 3.853 *10^8$

From the values of Rayleigh numbers above, case 1 is Turbulent flow while case 2 is laminar flow

Thus: C₁ = 0.1, n₁ = ¹/₃

C₂ = 0.59, n₂ = 1/4

Ratio of heat transfer rate is given as:

$\frac{q_1}{q_2} = \frac{h_1 \delta T}{h_2 \delta T} \\\\\frac{q_1}{q_2} = \frac{h_1}{h_2} \\\\But, \frac{hL}{k} = CR_a^n L, \ \ h=\frac{k}{L}(CR_a^n L)\\\\\frac{q_1}{q_2} = \frac{C_1R_a_1^n L_2}{C_2R_a_2^n L_1} = \frac{0.1(1.784*10^9)^{\frac{1}{3}} *0.6}{0.59(3.853*10^8)^{\frac{1}{4}} *1} \\\\\frac{q_1}{q_2} = \frac{72.76}{82.66} = 0.88$

Therefore, the ratio of heat transfer rate is 0.88

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