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Nataly_w [17]
3 years ago
6

5) Initially, the pressure and temperature of steam inside a solid capsule is at 100-pound force per square inch absolute (psia)

, and 600 degrees Fahrenheit (°F), respectively. Because heat is gradually removed from this container, the pressure inside the capsule drops by the amount of 10-pound force per square inch absolute (psia). Answer the following questions, A. (10 points) The change of entropy per unit mass between the initial and final states B. (5 points) The amount of heat transfer per unit mass for the process C. (5 points) Sketch the T-s diagram for the process, showing the associated values of the thermodynamic properties for states and 2 on your sketch.

Engineering
1 answer:
kvv77 [185]3 years ago
8 0

Answer:

Check the explanation

Explanation:

Kindly check the attached image below for the full step by step explanation to your question.

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Supercharged engine what it does to the car
geniusboy [140]

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A supercharger is an air compressor that increases the pressure or density of air supplied to an internal combustion engine. This gives each intake cycle of the engine more oxygen, letting it burn more fuel and do more work, thus increasing power.

Explanation:

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RoseWind [281]

Answer:

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Explanation:

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3 years ago
2. A
Charra [1.4K]
Answer: Projectile

Explanation:
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8 0
3 years ago
Which is/are not a mechanism commonly associated with tool wear (mark all that apply)?a. Adhesion b. Attrition c. Abrasion d. Co
butalik [34]

Answer: d)Coercion

Explanation:Tool wear is defined as the situation when the cutting tool is subjected to the regular process of cutting metal then they tend to wear because of the continuous action of cutting and facing stresses and pressure . The mechanism that does not happen during this process are coercion that means the process of exerting forces on any material forcefully against the will or need. Therefore, adhesion,attrition and abrasion are the process of tool wear .So the correct option is (d)

5 0
3 years ago
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 = − −
sergij07 [2.7K]

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,

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

 

     

7 0
3 years ago
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