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Allisa [31]
3 years ago
14

Derive an expression for the specific heat difference of a substance whose equation of state is 1 2 ( ) RT a P b b T ν ν ν = − −

+ where a and b are empirical constants.
Engineering
1 answer:
sergij07 [2.7K]3 years ago
7 0

Answer:

Given data:

Equation of the state p=\frac{RT}{v-b}-\frac{a}{v(v+b) T^{1/2} }

Where p = pressure of fluid, pα

T = Temperature of fluid, k

V = Specific volume of fluid m^{3} / k g

R = gas constant , j/k g k

a, b = Constants

Solution:

Specific heat difference, \begin{array}{c}c_{p}-c_{v}=-T\left(\frac{\partial v}{\partial T}\right)^{2} p \\\left(\frac{\partial P}{\partial v}\right)_{r}\end{array}

According to cyclic reaction

\left(\frac{\ dv}{\ dT}\right)_{p}=-\frac{\left(\frac{\ d P}{\ d T}\right)_{v}}{\left(\frac{\ d P}{\ d v}\right)_{v}}

Hence specific heat difference is

c_{p}-c_{v}=\frac{-T\left(\frac{\ d v}{\ d T}\right)_{p}^{2}}{\left(\frac{\ d p}{\ dv}\right)_{v}}

Equation of state, p=\frac{R T}{v-b}-\frac{a}{v(v+b)^{\ 1/2}}

Differentiating the equation of state with respect to temperature at constant volume,

\(\left(\frac{\ d P}{\ d T}\right)_{v}=\frac{R}{v-b}-\frac{1}{2}- \frac{a}{v(v+b)^} T^{\frac{-1}{2}}\)

\begin{aligned}&\left(\frac{\ dP}{\ dT}\right)_{V}=\frac{R}{v-b}+\frac{a}{2 v(v+b) T^{3 / 2}}\end{aligned}

Differentiating the equation of the state with respect to volume at constant temperature.

\(\left(\frac{\ dP}{\ dv}\right)_{\gamma}=+(-1) \times R T(v-b)^{-1-1}+\frac{a}{b T^{1 / 2}}\left(\frac{1}{v^{2}}-\frac{1}{(v+b)^{2}}\right)\)\\\(\left(\frac{\ dP}{\ dv}\right)_{r}=-\frac{R T}{(v-b)^{2}}+\frac{a}{T^{1 / 2}}\left(\frac{2 v+b}{v^{2}(v+b)^{2}}\right)\)

Substituting both eq (3) and eq (4) in eq (2)

We get,

       {cp{} - } c_{v}=\frac{T\left(\frac{R}{v-b}+\frac{a}{2 v(v+b) T^{3 / 2}}\right)^{2}}{\left(\frac{R T}{(v-b)^{2}}-\frac{a(2 v+b)}{T^{1 / 2} v^{2}(v+b)^{2}}\right)}

Specific heat difference equation,

\(c_{p} -c_{v}}=\frac{T\left(\frac{R}{v-b}+\frac{a}{2 v(v+b)^{T}^{3 / 2}}\right)^{2}}{\left(\frac{R T}{(v-b)^{2}}-\frac{a(2 v+b)}{T^{1 / 2} v^{2}(v+b)^{2}}\right)}\)

 

     

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A gas stream contains 18.0 mole% hexane and the remainder nitrogen. The stream flows to a condenser, where its temperature is re
Anna [14]

Answer:

A. 72.34mol/min

B. 76.0%

Explanation:

A.

We start by converting to molar flow rate. Using density and molecular weight of hexane

= 1.59L/min x 0.659g/cm³ x 1000cm³/L x 1/86.17

= 988.5/86.17

= 11.47mol/min

n1 = n2+n3

n1 = n2 + 11.47mol/min

We have a balance on hexane

n1y1C6H14 = n2y2C6H14 + n3y3C6H14

n1(0.18) = n2(0.05) + 11.47(1.00)

To get n2

(n2+11.47mol/min)0.18 = n2(0.05) + 11.47mol/min(1.00)

0.18n2 + 2.0646 = 0.05n2 + 11.47mol/min

0.18n2-0.05n2 = 11.47-2.0646

= 0.13n2 = 9.4054

n2 = 9.4054/0.13

n2 = 72.34 mol/min

This value is the flow rate of gas that is leaving the system.

B.

n1 = n2 + 11.47mol/min

72.34mol/min + 11.47mol/min

= 83.81 mol/min

Amount of hexane entering condenser

0.18(83.81)

= 15.1 mol/min

Then the percentage condensed =

11.47/15.1

= 7.59

~7.6

7.6x100

= 76.0%

Therefore the answers are a.) 72.34mol/min b.) 76.0%

Please refer to the attachment .

4 0
3 years ago
A differential amplifier is to have a voltage gain of 100. What will be the feedback resistance required if the input resistance
yuradex [85]

Answer:

required feedback resistance ( R2 ) = 100 k Ω

Explanation:

Given data :

Voltage gain = 100

input resistance ( R1 ) = 1 k ohms

calculate feedback resistance required

voltage gain of differential amplifier

\frac{Vout}{V2 - V1 }  = \frac{R2}{R1}

= Voltage gain =  R2/R1

= 100 = R2/1

hence required feedback resistance ( R2 ) = 100 k Ω

4 0
3 years ago
Calculate the LER for the rectangular wing from the previous question if the weight of the glider is 0.0500 Newton’s.
lapo4ka [179]

Answer:

0.2

Explanation:

Since the span and chord of the rectangular wing is missing, due to it being from the other question, permit me to improvise, or assume them. While you go ahead and substitute the ones from your question to it, as it's both basically the same method.

Let the span of the rectangular wing be 0.225 m

Let the chord of the rectangular wing be 0.045 m.

Then, the area of any rectangular chord is

A = chord * span

A = 0.045 * 0.225

A = 0.010 m²

And using the weight of the glider given to us from the question, we can find the LER for the wing.

LER = Area / weight.

LER = 0.010 / 0.05

LER = 0.2.

Therefore, using the values of the rectangular wing I adopted, and the weight of the glider given, we can see that the LER of the glider is 0.2

Please mark brainliest...

3 0
3 years ago
Read 2 more answers
Outline the process used to test the following hypothesis: Titanium cages are stronger than steel cages for hockey goalie masks.
son4ous [18]

Answer:

true

Explanation:

8 0
3 years ago
An adiabatic gas turbine expands air at 1300 kPa and 500◦C to 100 kPa and 127◦C. Air enters the turbine through a 0.2-m2 opening
Viktor [21]

Given:

Pressure, P_{1} = 1300 kPa

Temperature,  T_{1} = 500^{\circ}

P_{2} = 100 kPa

T_{2} = 127^{\circ}  

velocity, v = 40 m/s

A = 1m^{2}

Solution:

For air propertiess at

P_{1} = 1300 kPa

T_{1} = 500^{\circ}

h_{1} = 793kJ/K

v_{1} = 0.172\frac{m^{3}}{kg}

and also at

P_{2} = 100 kPa

T_{2} = 127^{\circ}  

h_{2} = 401 KJ/K

v_{2} =  1.15\frac{m^{3}}{kg}

a) Mass flow rate is given by:

m' = \frac{Av}{v_{1}}

Now,

m = \frac{0.2\times 40}{0.172} = 46.51 kg/s

b) for the power produced by turbine, P = m'(h_{1} - h_{2})

P = 46.51\times(793 - 401) = 18.231 MW

5 0
4 years ago
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