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Gelneren [198K]
3 years ago
6

Water at 70 kPa and 1008C is compressed isentropically in a closed system to 4 MPa. Determine the final temperature of the water

and the work required, in kJ/kg, for this compression.
Engineering
1 answer:
Nikitich [7]3 years ago
3 0

Answer:

The answer is "909.3928  KJ".

Explanation:

70 \ kPa  \ \ and \ \ 100^{\circ}C \\\\s_i= 7.56162\ \frac{kJ}{kgK}\\\\u_i= 2509.39 \ \frac{kJ}{kg}\\\\

The method is isentropic since the cylinders are shielded.

Calculating the work:

w= u_2-u_i \\\\

   = 3418.7728-2509.38 \\\\=909.3928 \ KJ

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What is flow energy? Do fluids at rest possess any flow energy?
anzhelika [568]

Answer:

Flow energy is defined as, flow energy is the energy needed to push fluids into control volume and it is the amount of work done required to push the entire fluid. It is also known as flow work. Flow energy is not the fundamental quantities like potential and kinetic energy.

Fluid at state of rest do not possess any flow energy. It is mostly converted into internal energy as, rising in the fluid temperature.

8 0
3 years ago
The fracture strength of glass may be increased by etching away a thin surface layer. It is believed that the etching may alter
Korvikt [17]

Answer:

the ratio of the etched to the original crack tip radius is 30.24

Explanation:

Given the data in the question;

we determine the initial fracture stress using the following expression;

(σf)₁ = 2(σ₀)₁ [ α₁/(p_t)₁ ]^{1/2 ----- let this be equation 1

where; (σ₀)₁ is the initial fracture strength

(p_t)₁ is the original crack tip radius

α₁ is the original crack length.

first, we determine the final crack length;

α₂ = α₁ - 16% of α₁

α₂ = α₁ - ( 0.16 × α₁)

α₂ = α₁ - 0.16α₁

α₂ = 0.84α₁

next, we calculate the final fracture stress;

the fracture strength is increased by a factor of 6;

(σ₀)₂ = 6( σ₀ )₁

Now, expression for the final fracture stress

(σf)₂ = 2(σ₀)₂ [ α₂/(p_t)₂ ]^{1/2 ------- let this be equation 2

where (p_t)₂ is the etched crack tip radius

value of fracture stress of glass is constant

Now, we substitute 2(σ₀)₁ [ α₁/(p_t)₁ ]^{1/2 from equation for (σf)₂  in equation 2.

0.84α₁ for α₂.

6( σ₀ )₁ for (σ₀)₂.

∴

2(σ₀)₁ [ α₁/(p_t)₁ ]^{1/2  = 2(6( σ₀ )₁) [ 0.84α₁/(p_t)₂ ]^{1/2  

divide both sides by 2(σ₀)₁

[ α₁/(p_t)₁ ]^{1/2  =  6 [ 0.84α₁/(p_t)₂ ]^{1/2

[ 1/(p_t)₁ ]^{1/2  =  6 [ 0.84/(p_t)₂ ]^{1/2

[ 1/(p_t)₁ ]  =  36 [ 0.84/(p_t)₂ ]

1 / (p_t)₁ = 30.24 / (p_t)₂

(p_t)₂ = 30.24(p_t)₁

(p_t)₂/(p_t)₁ = 30.24

Therefore, the ratio of the etched to the original crack tip radius is 30.24

6 0
3 years ago
________ are written to “maximize” or “minimize” a specific value associated with the product needs in order to define the goal
Anna35 [415]

Answer:

Objective statements.

Explanation:

An objective statement can be defined as a short statement that explicitly states or describes what a person wants exactly or is looking out for in a particular item.

Objective statements are written to “maximize” or “minimize” a specific value associated with the product needs in order to define the goal or aim of the design process.

This ultimately implies that, objective statements are used by various manufacturing industries or companies to explicitly define the minimum or maximum requirements for the production of its goods.

4 0
3 years ago
Read 2 more answers
ILL GIVE YOU BRAINLIEST PLEASE HELP PLEASE
Lesechka [4]

Answer:

try to pop it back in good luck im scared for you

4 0
3 years ago
Heats of Reaction and Hess's Law experiment tells you to look up three values. Write down these values. Be sure to clearly indic
Anna [14]

Answer:

Enthalpy of reaction (kJoules/mole)

Heat of formation of products (kJoules/mole)

Heat of reaction of reactants (kJoules/mole)

Explanation:

The general expression for calculating the overall enthalpy of reaction is given as following:

ΔH = ∑ΔH[producst] - ∑Δ[reactants]

Thus, the heat of reaction is given as the difference between the formation of the products and the formation of the reactants. The units are expressed as kJ/mol of reactants or products.

Thus, the three values are fundamental in the determination of the overall energy of the reaction from Hess' Law.

8 0
3 years ago
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