Answer:
speeds above 343 m/s
Explanation:
I have taken the test got 100%
(b) the exergy destroyed during this process.
A) The exergy of the refrigerant at the initial and final states are :
B) The exergy destroyed during this process is : - 1048.4397 kJ
Given data :
Mass ( M ) = 5 kg
P1 = 0.7 Mpa = P2
T1 = 60°C = 333 k
To = 24°C = 297 k
P2 = 100 kPa
A) Determine the exergy at initial and final states
At initial state :
U = 274.01 kJ/Kg , V = 0.034875 m³/kg , S = 1.0256 KJ/kg.k
exergy ( Ф ) at initial state = M ( U + P₂V - T₀S )
= 5 ( 274.01 + 100* 10³ * 0.034875 - 297 * 1.0256)
≈ - 135.5285 kJ
At final state :
U = 84.44 kJ / kg , V = 0.0008261 m³/kg, S = 0.31958 kJ/kg.k
exergy ( ( Ф ) at final state = M ( U + P₂V - T₀S )
= -51.96 kJ
B) Determine the exergy destroyed
exergy destroyed = To * M ( S2 - S1 )
= 297 * 5 ( 0.31958 - 1.0256 )
= - 1048.4397 KJ
Hence we can conclude that A) The exergy of the refrigerant at the initial and final states are : Initial state = - 135.5285 kJ, Final state = -51.96 kJ and The exergy destroyed during this process is : - 1048.4397 kJ
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Exergy of refrigerant-134a at initial and final states is obtained from property tables and by multiplying the mass of the refrigerant with its specific exergy at each state. The difference in exergy between the two states represents the exergy destroyed.
To solve the given question, we need the property values of
refrigerant-134a
at the initial and the final states.
At an initial state of 0.7 MPa and 60°C, the specific exergy for refrigerant-134a can be obtained from property tables which are standard in thermodynamics textbooks. Same for the final state at 0.7 MPa and 24°C, the specific exergy can be obtained from the same property tables.
The exergy of the refrigerant at the initial and the final states can be calculated by multiplying the mass of the refrigerant with its specific exergy at each state.
Exergy destruction during this process can be calculated using the relation between exergy change and exergy destruction. The exergy change of a system between initial and final states is equal to the difference of the exergy of the system at final and initial states.
Based on the second law of thermodynamics, the difference in exergy should be equal to the exergy destroyed during the process.
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Answer:
???
Explanation:
Answer:
λ = -47 nC / m
Explanation:
The missing question is as follows:
" The potential difference between the surface of a 2.2 cm -diameter power line and a point 1.9 m distant is 3.8 kV. What is the magnitude of the line charge density on the power line? Express your answer using two significant figures. "
Given:
- The Diameter of the power line D = 2.2 cm
- The distance between two ends of power line L = 1.9m
- The potential difference across two ends V = 3.8 KV
Find:
What is the magnitude of the line charge density on the power line?
Solution:
- The derivation of the line of charges for a length L oriented along any axis centered at origin and the potential difference between two ends is as follows:
V = 2*k*λ*Ln( D / L )
Where,
k : Coulomb's Constant = 8.99*10^9
λ : The line charge density
- Re-arrange and solve for λ:
λ = V / 2*k*Ln( D / L )
Plug in the values:
λ = 3800 / 2*8.99*10^9*Ln( 2.2 / 190 )
λ = -4.74022*10^-8 C / m
λ = -47 nC / m
Line charge density is the total charge distributed along the length of a wire, expressed in coulombs per meter. To calculate it, divide the total charge by the total length of the wire. Without specific numbers for charge and length, a numerical value can't be given.
To calculate the magnitude of the line charge density of a power line, you need to know the total charge (Q) distributed along the total length (L) of the wire. The line charge density (λ) is then defined as λ = Q/L. Unfortunately, without any specific numbers provided for these parameters, I can't provide a numerical answer.
Line charge density is a significant concept in electromagnetism and is measured in coulombs per meter (C/m).
Remember that the charge can be uniform or non-uniform along the length of the line.
For example, if a power line has a total charge of 0.02 C spread along its length of 50 m, it would have a line charge density of λ = Q/L = 0.02 C / 50 m = 0.0004 C/m
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Answer:
The height is
The kinetic energy during collision is not conserved
The Mechanical energy during the collision is not conserved
The mechanical energy after the collision is not conserved
Explanation:
From the question we are told that
The mass of the block is
The mass of the wad of putty is
The speed o the wad of putty is
The law of momentum conservation can be mathematically represented as
Where is the initial momentum which is mathematically represented as
While is the initial momentum which is mathematically represented as
Where s the final velocity
So
Making the subject
substituting values
According to the law of energy conservation
Where KE is the kinetic energy of the system which is mathematically represented as
And PE is the potential energy of the system which is mathematically represented as
So
Making h the subject of the formula
substituting values
Now the kinetic energy is conserved during collision because the system change it height during which implies some of the kinetic energy was converted to potential energy during collision
The the mechanical energy of the system during the collision is conserved because this energy consists of the kinetic and the potential energy.
Now after the collision the mechanical energy is not conserved because the external force like air resistance has reduced the mechanical energy of that system
Collecting data and analyzing results
Designing and implementing systems
Maintaining and using diagnostic equipment
Designing and using laboratory equipment
Mark this and return
Another thing that Ernie put in the common section is collecting data and analyzing results.
A Venn diagram is used to show a representation of data. The center of the Venn diagram is often used to indicate the data set that is the same.
Looking at the Venn diagram, another thing that Ernie put in the common section is collecting data and analyzing results.
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Answer:
P = 1.29625 W
Explanation:
Given
Q = 85.0 cm ³ / s, p₁ = 110 mmHg, u₁ = 25.0 cm / s, h = 5.0 cm
Also knowing the density of the blood is
ρₐ = 1.05 x 10 ³ kg / m³
Δp₁ = 110 mmHg * 133.322 Pa / 1 mmHg
Q = 85.0 cm³ / s = 85.0 x 10 ⁻⁶ m³ / s
To calculated the power
P = H * Q
H = Δp₁ + ¹/₂ * ρₐ * ( u₁² - v₂²) + ρₐ * g *Δh
H = 14.666 x 10 ³ Pa + 0.5 * 1.05 x 10 ³ kg / m³ * ( 25 x 10 ² m /s )² + 1.05 x 10 ³ kg / m³ * 9.8 m /s² * 0.05 m
H = 15.25 x 10 ³ Pa
P = 15.25 x 10 ³ Pa * 85.0 x 10 ⁻⁶ m³ / s
P = 1.29625 W