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ioda
3 years ago
15

A piston-cylinder assembly has initially a volume of 0.3 m3 of air at 25 °C. Mass of the air is 1 kg. Weights are put on the pis

ton until the air reaches to 0.1 m3 and 1,000 °C, in which the air undergoes a polytropic process (PV" const). Assume that heat loss from the cylinder, friction of piston, kinetic and potential effects are negligible. 1) Determine the polytropic constant n. 2) Determine the work transfer in ki for this process, and diseuss its direction. 3) sketch the process in T-V (temperature-volume) diagram.

Engineering
1 answer:
kogti [31]3 years ago
5 0

Answer:

n=2.32

w= -213.9 KW

Explanation:

V_1=0.3m^3,T_1=298 K

V_2=0.1m^3,T_1=1273 K

Mass of air=1 kg

For polytropic process  pv^n=C ,n is the polytropic constant.

  Tv^{n-1}=C

  T_1v^{n-1}_1=T_2v^{n-1}_2

298\times .3^{n-1}_1=1273\times .1^{n-1}_2

n=2.32

Work in polytropic process given as

       w=\dfrac{P_1V_1-P_2V_2}{n-1}

      w=mR\dfrac{T_1-T_2}{n-1}

Now by putting the values

w=1\times 0.287\dfrac{289-1273}{2.32-1}

w= -213.9 KW

Negative sign indicates that work is given to the system or work is done on the system.

For T_V diagram

  We can easily observe that when piston cylinder reach on new position then volume reduces and temperature increases,so we can say that this is compression process.

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Please help I need it by today!!!
AfilCa [17]

Answer:

if engineering disappeared for a day i would be at a loss. i wouldnt know what to do with myself considering engineering is my life. one way that engineers improve my life is they help me to understand enything end everything

Explanation:

7 0
2 years ago
For some metal alloy, a true stress of 345 MPa (50040 psi) produces a plastic true strain of 0.02. How much will a specimen of t
saveliy_v [14]

Complete Question

For some metal alloy, a true stress of 345 MPa (50040 psi) produces a plastic true strain of 0.02. How much will a specimen of this material elongate when a true stress of 411 MPa (59610 psi) is applied if the original length is 470 mm (18.50 in.)?Assume a value of 0.22 for the strain-hardening exponent, n.

Answer:

The elongation is =21.29mm

Explanation:

In order to gain a good understanding of this solution let define some terms

True Stress

       A true stress can be defined as the quotient obtained when instantaneous applied load is divided by instantaneous cross-sectional area of a material it can be denoted as \sigma_T.

True Strain

     A true strain can be defined as the value obtained when the natural logarithm quotient of instantaneous gauge length divided by original gauge length of a material is being bend out of shape by a uni-axial force. it can be denoted as \epsilon_T.

The mathematical relation between stress to strain on the plastic region of deformation is

              \sigma _T =K\epsilon^n_T

Where K is a constant

          n is known as the strain hardening exponent

           This constant K can be obtained as follows

                        K = \frac{\sigma_T}{(\epsilon_T)^n}

No substituting  345MPa \ for  \ \sigma_T, \ 0.02 \ for \ \epsilon_T , \ and  \ 0.22 \ for  \ n from the question we have

                     K = \frac{345}{(0.02)^{0.22}}

                          = 815.82MPa

Making \epsilon_T the subject from the equation above

              \epsilon_T = (\frac{\sigma_T}{K} )^{\frac{1}{n} }

Substituting \ 411MPa \ for \ \sigma_T \ 815.82MPa \ for \ K  \ and  \  0.22 \ for \ n

       \epsilon_T = (\frac{411MPa}{815.82MPa} )^{\frac{1}{0.22} }

            =0.0443

       

From the definition we mentioned instantaneous length and this can be  obtained mathematically as follows

           l_i = l_o e^{\epsilon_T}

Where

       l_i is the instantaneous length

      l_o is the original length

Substituting  \ 470mm \ for \ l_o \ and \ 0.0443 \ for  \ \epsilon_T

             l_i = 470 * e^{0.0443}

                =491.28mm

We can also obtain the elongated length mathematically as follows

            Elongated \ Length =l_i - l_o

Substituting \ 470mm \ for l_o and \ 491.28 \ for \ l_i

          Elongated \ Length = 491.28 - 470

                                       =21.29mm

4 0
3 years ago
A cylindrical brass rod has a length of 5.00cm extending from a holder and a diameter of 4.50mm. Its Young's modulus is 98.0GPa.
Galina-37 [17]

Answer:

elongation of the brass rod is 0.01956 mm

Explanation:

given data

length = 5 cm = 50 mm

diameter = 4.50 mm

Young's modulus = 98.0 GPa

load = 610 N

to find out

what will be the elongation of the brass rod in mm

solution

we know here change in length formula that is express as

δ = \frac{PL}{AE}    ................1

here δ is change in length and P is applied load  and A id cross section area and E is Young's modulus and L is length

so all value in equation 1

δ = \frac{PL}{AE}  

δ = \frac{610*50}{\frac{\pi}{4} * 4.50^2 * 98*10^3}  

δ = 0.01956 mm

so elongation of the brass rod is 0.01956 mm

7 0
3 years ago
A coil having resistance of 7 ohms and inductance of 31.8 mh is connected to 230v,50hz supply.calculate 1. The circuit current 2
lora16 [44]

(1) The current in the circuit is 18.87 A,

(2) The phase angle is 54.97°

(3) The power factor is 0.574

(4) The power consumed is 2491.2 W

(1) To calculate the current in the circuit, first, we need to find the overall impedance of the circuit.

We can calculate the overall impendence of the circuit using the formula below.

  • Z = √[R²+(2πfL)²]........................ Equation 1

Where:

  • R = resistance of the coil
  • f = Frequency
  • L = Inductance of the coil
  • Z = Overall impedance of the circuit

From the question,

Given:

  • R = 7 ohms
  • L = 31.8 mH = 0.0318 H
  • f = 50 Hz
  • π = 3.14

Substitute these values into equation 1

  • Z = √[7²+(2×3.14×50×0.0318)²]
  • Z = √(49+99.7)
  • Z = √(148.7)
  • Z = 12.19 ohms.

Therefore we use the formula below to calculate the current in the circuit.

  • I = V/Z.................. Equation 2

Where:

  • V = Voltage
  • I = current in the circuit.

Given:

  • V = 230 V.

Substitute into equation 2

  • I = 230/12.19
  • I = 18.87 A

(2) To calculate the phase angle, we use the formula below.

  • ∅ = tan⁻¹(2πfL/R)............... Equation 3

Where:

  • ∅ = Phase angle.


Substitute into equation 3

  • ∅ = tan⁻¹(2×3.14×50×0.0318/7)
  • ∅ = tan⁻¹(9.9852/7)
  • ∅ = tan⁻¹(1.426)
  • ∅ = 54.97°

(3) To calculate the power factor, we use the formula below.

  • pf = cos∅............ Equation 4

Where:

  • pf = power factor.

Substitute the value of ∅ into equation 4

  • pf = cos(54.97°)
  • pf = 0.574.

(4) And Finally to calculate the power consumed we use the formula below.

  • P = V×I×pf................ Equation 5

Where:

  • P = The power consumed

Substitute the values into equation 5

  • P = 230(18.87)(0.574)
  • P = 2491.22 W


Hence, (1) The current in the circuit is 18.87 A, (2) The phase angle is 54.97° (3) The power factor is 0.574 (4) The power consumed is 2491.2 W

Learn more about Impedance here: brainly.com/question/13134405

5 0
2 years ago
Show the ERD with relational notation with crowfoot. Your ERD must show PK, FKs, min and max cardinality, and correct line types
zhenek [66]

Answer

The answer and procedures of the exercise are attached in the following archives.

Step-by-step explanation:

You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.  

3 0
3 years ago
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