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2 years ago

What is the relationship between mechanical stress inside of a cylindrical part with a given force applied to it

1 answer:

7 0

The relationship between mechanical stress inside of a cylindrical part with a given force applied to it is stress = force/area.

<h3>How is pressure associated with strain?</h3>

The utility of pressure offers upward thrust to strain, however now no longer the opposite manner around. A pressure is a push or pull upon an item attributable to the item's interplay with some other item while Stress is an inner resistance to supplied through the frame itself each time it's far beneath a few type of deformation.

Stress is denoted through σ. It is represented as N/m2. Stress formulation is made use of to locate carried out on any given frame if pressure and region on which pressure is exerted is given withinside the problem.

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Oil with a density of 800 kg/m3 is pumped from a pressure of 0.6 bar to a pressure of 1.4 bar, and the outlet is 3 m above the i

Naddik [55]

Answer:

23.3808 kW

20.7088 kW

Explanation:

ρ = Density of oil = 800 kg/m³

P₁ = Initial Pressure = 0.6 bar

P₂ = Final Pressure = 1.4 bar

Q = Volumetric flow rate = 0.2 m³/s

A₁ = Area of inlet = 0.06 m²

A₂ = Area of outlet = 0.03 m²

Velocity through inlet = V₁ = Q/A₁ = 0.2/0.06 = 3.33 m/s

Velocity through outlet = V₂ = Q/A₂ = 0.2/0.03 = 6.67 m/s

Height between inlet and outlet = z₂ - z₁ = 3m

Temperature to remains constant and neglecting any heat transfer we use Bernoulli's equation

$\frac {P_1}{\rho g}+\frac{V_1^2}{2g}+z_1+h=\frac {P_2}{\rho g}+\frac{V_2^2}{2g}+z_2\\\Rightarrow h=\frac{P_2-P_1}{\rho g}+\frac{V_2^2-V_1^2}{2g}+z_2-z_1\\\Rightarrow h=\frac{(1.4-0.6)\times 10^5}{800\times 9.81}+\frac{6.67_2^2-3.33^2}{2\times 9.81}+3\\\Rightarrow h=14.896\ m$

Work done by pump

$W_{p}=\rho gQh\\\Rightarrow W_{p}=800\times 9.81\times 0.2\times 14.896\\\Rightarrow W_{p}=23380.8\ W$

∴ Power input to the pump 23.3808 kW

Now neglecting kinetic energy

$h=\frac{P_2-P_1}{\rho g}+z_2-z_1\\\Righarrow h=\frac{(1.4-0.6)\times 10^5}{800\times 9.81}+3\\\Righarrow h=13.19\ m\\$

Work done by pump

$W_{p}=\rho gQh\\\Rightarrow W_{p}=800\times 9.81\times 0.2\times 13.193\\\Rightarrow W_{p}=20708.8\ W$

∴ Power input to the pump 20.7088 kW

6 0

3 years ago

Koch traded Machine 1 for Machine 2 when the fair market value of both machines was $60,000. Koch originally purchased Machine 1

Mariana [72]

Answer:

Koch's adjusted basis in machine 2 after the exchange is $60,000

Explanation:

given data

fair market value = $60,000

originally purchased Machine 1 = $76,900

Machine 1 adjusted basis = $40,950

Machine 2 seller purchase = $64,050

Machine 2 adjusted basis = $55,950

solution

As he exchanged machine for another at $60,000

and this exchanged in fair market

so adjusted basis = $50,000

Adjusted basis is the price of the item that affects the factors that are considered price. These factors usually include taxes, depreciation value, and other costs of acquiring and maintaining a given item. Adjusted basis is important so the right amount to sell

Adjusted basis increases when a person deducts expenses from factor taxes and operating statements

so Koch's adjusted basis in machine 2 after the exchange is $60,000

3 0

3 years ago

You are to design two CONCEPTUALLY different synchronous state machines (Mealy and Moore) that perform the task described below.

allochka39001 [22]

Answer:

Explanation:

I hope this helps!

3 0

3 years ago

A well insulated turbine operates at steady state. Steam enters the turbine at 4 MPa with a specific enthalpy of 3015.4 kJ/kg an

Anarel [89]

Answer:

power developed by the turbine = 6927.415 kW

Explanation:

given data

pressure = 4 MPa

specific enthalpy h1 = 3015.4 kJ/kg

velocity v1 = 10 m/s

pressure = 0.07 MPa

specific enthalpy h2 = 2431.7 kJ/kg

velocity v2 = 90 m/s

mass flow rate = 11.95 kg/s

solution

we apply here thermodynamic equation that

energy equation that is

$h1 + \frac{v1}{2} + q = h2 + \frac{v2}{2} + w$

put here value with

turbine is insulated so q = 0

so here

$3015.4 *1000 + \frac{10^2}{2} = 2431.7 * 1000 + \frac{90^2}{2} + w$

solve we get

w = 579700 J/kg = 579.7 kJ/kg

and

W = mass flow rate × w

W = 11.95 × 579.7

W = 6927.415 kW

power developed by the turbine = 6927.415 kW

7 0

3 years ago

The longest_word function is used to compare 3 words. It should return the word with the most number of characters (and the firs

NARA [144]

Answer:

len(word2) >= len(word1) and len(word2) >= len(word3):

Question with blank is below

def longest_word(word1,word2,word3):

if len(word1) >= len(word2) and len(word1) >= len(word3):

word = word1

elif _________________________________________

word = word2

else:

word = word3

return word

print(longest_word("chair","couch","table"))

print(longest_word("bed","bath","beyond"))

print(longest_word("laptop","notebook","desktop"))

print(longest_word("hi","cat","Cow"))

Explanation

In line 1 of the code word1, word2, and word3 are the parameters used to for the defining the longest_word function. They will be replaced by 3 words to be compared. The code that is filled in the blank is len(word2) >= len(word1) and len(word2) >= len(word3): It is a conditional statement that is true only if the number of characters in the string of word2 is greater than or equal to word1 and word2 is greater than that of word3 .

4 0

4 years ago