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Zina [86]
3 years ago
14

Show from the first principles that, for a perfect gas with constant specific heat capacities

Engineering
1 answer:
iVinArrow [24]3 years ago
5 0

The given question is incomplete as the equation given is not making sense.

From internet source i have taken the complete question and its been attached below.

The solution to the question is also attached in next image with complete detail.

<h3> I hope it will help you a lot!</h3>

You might be interested in
A well-insulated tank in a vapor power plant operates at steady state. Saturated liquid water enters at inlet 1 at a rate of 125
Gekata [30.6K]

Answer:

a) \dot m_{3} = 135\,\frac{lbm}{s}, b) h_{3}=168.965\,\frac{BTU}{lbm}, c) T = 200.829\,^{\textdegree}F

Explanation:

a) The tank can be modelled by the Principle of Mass Conservation:

\dot m_{1} + \dot m_{2} - \dot m_{3} = 0

The mass flow rate exiting the tank is:

\dot m_{3} = \dot m_{1} + \dot m_{2}

\dot m_{3} = 125\,\frac{lbm}{s} + 10\,\frac{lbm}{s}

\dot m_{3} = 135\,\frac{lbm}{s}

b) An expression for the specific enthalpy at outlet is derived from the First Law of Thermodynamics:

\dot m_{1}\cdot h_{1} + \dot m_{2} \cdot h_{2} - \dot m_{3}\cdot h_{3} = 0

h_{3} = \frac{\dot m_{1}\cdot h_{1}+\dot m_{2}\cdot h_{2}}{\dot m_{3}}

Properties of water are obtained from tables:

h_{1}=180.16\,\frac{BTU}{lbm}

h_{2}=28.08\,\frac{BTU}{lbm} + \left(0.01604\,\frac{ft^{3}}{lbm}\right)\cdot (14.7\,psia-0.25638\,psia)

h_{2}=29.032\,\frac{BTU}{lbm}

The specific enthalpy at outlet is:

h_{3}=\frac{(125\,\frac{lbm}{s} )\cdot (180.16\,\frac{BTU}{lbm} )+(10\,\frac{lbm}{s} )\cdot (29.032\,\frac{BTU}{lbm} )}{135\,\frac{lbm}{s} }

h_{3}=168.965\,\frac{BTU}{lbm}

c) After a quick interpolation from data availables on water tables, the final temperature is:

T = 200.829\,^{\textdegree}F

8 0
3 years ago
Read 2 more answers
If the contact surface between the 20-kg block and the ground is smooth, determine the power of force F when t = 4 s. Initially,
grigory [225]

Answer:

The power of force F is 115.2 W

Explanation:

Use following formula

Power  = F x V

F_{H} = F cos0

F_{H} = (30) x 4/5

F_{H} = 24N

Now Calculate V using following formula

V = V_{0} + at

V_{0} = 0

a = F_{H} / m

a = 24N / 20 kg

a = 1.2m / S^{2}

no place value in the formula of V

V = 0 + (1.2)(4)

V = 4.8 m/s

So,

Power = F_{H} x V

Power = 24 x 4.8

Power = 115.2 W

3 0
3 years ago
Prove the following languages are nonregular, once using the pumping lemma and once using the Myhill-Nerode theorem. When using
VashaNatasha [74]

Answer:

For any string, we use s = xyz

Explanation:

The pumping lemma says that for any string s in the language, with length greater than the pumping length p, we can write s = xyz with |xy| ≤ p, such that xyi z is also in the language for every i ≥ 0. For the given language, we can take p = 2.

Here are the cases:

  • Consider any string a i b j c k in the language. If i = 1 or i > 2, we take x = \epsilon   and y = a. If i = 1, we must have j = k and adding any number of a’s still preserves the membership in the language. For i > 2, all strings obtained by pumping y as defined above, have two or more a’s and hence are always in the language.
  • For i = 2, we can take    and y = aa. Since the strings obtained by pumping in this case always have an even number of a’s, they are all in the language.
  • Finally, for the case i = 0, we take x = \epsilon  , and y = b if j > 0 and y = c otherwise. Since strings of the form b j c k are always in the language, we satisfy the conditions of the pumping lemma in this case as well.
8 0
2 years ago
Consider a single crystal of nickel oriented such that a tensile stress is applied along a [001] direction. If slip occurs on a
Elena L [17]

Answer:

\mathbf{\tau_c =5.675 \ MPa}

Explanation:

Given that:

The direction of the applied tensile stress =[001]

direction of the slip plane = [\bar 101]

normal to the slip plane = [111]

Now, the first thing to do is to calculate the angle between the tensile stress and the slip by using the formula:

cos \lambda = \Big [\dfrac{d_1d_2+e_1e_2+f_1f_2}{\sqrt{(d_1^2+e_1^2+f_1^2)+(d_2^2+e_2^2+f_2^2) }} \Big]

where;

[d_1\ e_1 \ f_1] = directional indices for tensile stress

[d_2 \ e_2 \ f_2] = slip direction

replacing their values;

i.e d_1 = 0 ,e_1 = 0 f_1 =  1 & d_2 = -1 , e_2 = 0 , f_2 = 1

cos \lambda = \Big [\dfrac{(0\times -1)+(0\times 0) + (1\times 1) }{\sqrt{(0^2+0^2+1^2)+((-1)^2+0^2+1^2) }} \Big]

cos \ \lambda = \dfrac{1}{\sqrt{2}}

Also, to find the angle \phi between the stress [001] & normal slip plane [111]

Then;

cos \  \phi = \Big [\dfrac{d_1d_3+e_1e_3+f_1f_3}{\sqrt{(d_1^2+e_1^2+f_1^2)+(d_3^2+e_3^2+f_3^2) }} \Big]

replacing their values;

i.e d_1 = 0 ,e_1 = 0 f_1 =  1 & d_3 = 1 , e_3 = 1 , f_3 = 1

cos \  \phi= \Big [ \dfrac{ (0 \times 1)+(0 \times 1)+(1 \times 1)} {\sqrt {(0^2+0^2+1^2)+(1^2+1^2 +1^2)} } \Big]

cos \phi= \dfrac{1} {\sqrt{3} }

However, the critical resolved SS(shear stress) \mathbf{\tau_c} can be computed using the formula:

\tau_c = (\sigma )(cos  \phi )(cos \lambda)

where;

applied tensile stress \sigma = 13.9 MPa

∴

\tau_c =13.9\times (  \dfrac{1}{\sqrt{2}} )( \dfrac{1}{\sqrt{3}})

\mathbf{\tau_c =5.675 \ MPa}

3 0
3 years ago
An inductor (L = 400 mH), a capacitor (C = 4.43 µF), and a resistor (R = 500 Ω) are connected in series. A 44.0-Hz AC generator
MakcuM [25]

Answer:

(A) Maximum voltage will be equal to 333.194 volt

(B) Current will be leading by an angle 54.70

Explanation:

We have given maximum current in the circuit i_m=385mA=385\times 10^{-3}A=0.385A

Inductance of the inductor L=400mH=400\times 10^{-3}h=0.4H

Capacitance C=4.43\mu F=4.43\times 10^{-3}F

Frequency is given f = 44 Hz

Resistance R = 500 ohm

Inductive reactance will be x_l=\omega L=2\times 3.14\times 44\times 0.4=110.528ohm

Capacitive reactance will be equal to X_C=\frac{1}{\omega C}=\frac{1}{2\times 3.14\times 44\times 4.43\times 10^{-6}}=816.82ohm

Impedance of the circuit will be Z=\sqrt{R^2+(X_C-X_L)^2}=\sqrt{500^2+(816.92-110.52)^2}=865.44ohm

So maximum voltage will be \Delta V_{max}=0.385\times 865.44=333.194volt

(B) Phase difference will be given as \Phi =tan^{-1}\frac{X_C-X_L}{R}=\frac{816.92-110.52}{500}=54.70

So current will be leading by an angle 54.70

5 0
3 years ago
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