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Rufina [12.5K]
3 years ago
6

Q1. Basic calculation of the First law (2’) (a) Suppose that 150 kJ of work are used to compress a spring, and that 25 kJ of hea

t are given off by the spring during this compression. What is the change in internal energy of the spring during the process? (b) Suppose that 100 kJ of work is done by a motor, but it also gives off 10 kJ of heat while carrying out this work. What is the change in internal energy of the motor during the process?
Engineering
1 answer:
Bogdan [553]3 years ago
4 0

Answer:

(a) ΔU = 125 kJ

(b) ΔU = -110 kJ

Explanation:

<em>(a) Suppose that 150 kJ of work are used to compress a spring, and that 25 kJ of heat are given off by the spring during this compression. What is the change in internal energy of the spring during the process?</em>

<em />

The work is done to the system so w = 150 kJ.

The heat is released by the system so q = -25 kJ.

The change in internal energy (ΔU) is:

ΔU = q + w

ΔU = -25 kJ + 150 kJ = 125 kJ

<em>(b) Suppose that 100 kJ of work is done by a motor, but it also gives off 10 kJ of heat while carrying out this work. What is the change in internal energy of the motor during the process?</em>

<em />

The work is done by the system so w = -100 kJ.

The heat is released by the system so q = -10 kJ.

The change in internal energy (ΔU) is:

ΔU = q + w

ΔU = -10 kJ - 100 kJ = -110 kJ

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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 = − −
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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 closed, rigid, 0.50 m^3 tank is filled with 12 kg of water. The initial pressure is p1 = 20 bar. The water is cooled until the
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Answer:

X1= 41%

heat transfer = -3450.676 KJ

Explanation:

To get the properties for pure substance in a system we need to know at least to properties. These are usually pressure and temperature because they’re easy to measure. In this case we know the initial pressure (20 bar) which is not enough to get all the properties, but they ask to determine quality, this a property that just have meaning in the two-phase region (equilibrium) so with this information we can get the temperature of the system and all its properties.

There is another property that we can calculate from the data. This is the specific volume. This is defined as \frac{volume}{mass}. We know the mass (12 Kg) and we can assume the volume is the volume of the tank  (0.5 m^{3}) because they say that the tank was filled.  

With this we get a specific volume of  

Specific volume = \frac{0,5 m^{3}}{ 12 kg}= 0.04166667 \frac{m^{3}}{Kg}

From the thermodynamic tables we can get the data for the saturated region with a pressure of 20 bar.

Temperature of saturation = 212.385 °C  

Specific volume for the saturated steam (vg) = 0.0995805 \frac{m^{3}}/{Kg}[\tex]Specific volume for the saturated liquid (vf)= 0.00117675 [tex]\frac{m^{3}}/{Kg}[\tex] The specific volume that we calculate before 0.04166667 m^3/Kg is between 0.00117675 m^3/Kg and 0.0995805 m^3/Kg so  we can be sure that we are in two-phase region (equilibrium).The quality (X) is defined as the percentage in mass of saturated steam in a mix (Two-phase region) The relation between specific volume and quality is  [tex]v = (1-x)*v_{f} + x*v_{g}[\tex]  where  v in the specific volume in the condition (0.04166667 m^3/Kg)  vf = Specific volume for the saturated liquid (0.00117675 m^3/Kg)vg = Specific volume for the saturated steam (0.0995805 m^3/Kg)x = qualityclearing the equation we get:[tex]X = \frac{(v-v_{f})}{(v_{g}-v_{f})} 

X =\frac{(0.04166667- 0.00117675)}{ (0.0995805 – 0.00117675)} = 0.411

The quality is 41%

To calculate the heat transfer we use the next equation.  

Q = m * Cp * delta T  

Where  

Q = heat transfer (Joules, J)

m= mass of the substance (g)

Cp = specific heat (J/g*K) from tables  

Delta T = change in temperature in K for this equation.  

The mass of the substance is 12 kg or 12000 g for this equation  

Cp from tables is 4,1813 J/g*K. You can find this value for water in different states. Here we are using the value for liquid water.  

For delta T, we know the initial temperature 212.385 °C.

We also know that the system was cooled. Since we don’t have more information, we can assume that the system was cooled until a condition where all the steam condensates so now we have a saturated liquid. Since we know the pressure (4 bar), we can get the temperature of saturation for this condition from the thermodynamics tables. This is 143.613 °C, so this is the final temperature for the system.  

T(K) = T°C +273  

T1(K) = 212.385 + 273.15 = 485.535 K

T2 (K) = 143.613 +273.15= 416.763 K

Delta T (K) = (T2-T1) =416.763 K - 485.535 K = -68.772 K

Now we can calculate Q

Q = 12000g * 4,1813 J/g*K* (-68.772 K) = -3450676.36 J or -3450.676 KJ

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