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3 years ago

What is meant by thermodynamic property? How do you classify the property? Explain with specific examples.

Engineering

1 answer:

Sergeu [11.5K]3 years ago

4 0

Answer:

The condition of very system can be identified by some physical characteristics of the system known as Thermodynamic Property of the system. For example, pressure, temperature, volume, etc.

Explanation:

Thermodynamic properties are nothing but the physical characteristics of any body by which the physical condition of the body can be described. properties describes the different states of a system. They are macroscopic.

Properties are of two types ---

Intensive property

Extensive property

Intensive Property : Those properties of a system which does not depend upon mass of the system are known Intensive property.

For example, pressure, temperature, density,specific volume, etc

Extensive Property : Properties that depends on the mass oh the system is called Extensive properties.

For example, energy, volume, etc.

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A 7-cm diameter horizontal pipe connected to the side of a tank at 4 m below the water surface in the tank discharges water to t

Anna11 [10]

Answer:

47.10 m/s

Explanation:

Diameter of the horizontal pipe, d = 7 cm = 0.07 m

Area of the horizontal pipe, A = $\frac{\pi}{4}d^2$

A = $\frac{\pi}{4}0.07^2$ = 0.00384 m²

Head between the water surface and the pipe level, z = 4 m

Head added by the pump, hp = 8 - 200Q²

where, Q is the discharge

losses, hL = 5 m

now,

applying the Bernoulli's theorem between the water surface and the pipe outlet,

we have

$\frac{P_1}{\rho}+\frac{V_1^2}{2g}+z+h_P=\frac{P_2}{\rho}+\frac{V_2^2}{2g}+h_L$

where,

P₁ = pressure at the water surface = 0 (as atmospheric pressure only)

V₁ = Velocity at the free surface = 0

ρ is the density of the water

P₂ = pressure at the outlet = 0 (as atmospheric pressure only)

V₂ = Velocity at the outlet

on substituting the values, we get

$z+h_P=\frac{V_2^2}{2g}+h_L$

$4+(8-200Q^2)=\frac{V_2^2}{2\times9.81}+5$

$7-200Q^2=\frac{V_2^2}{2\times9.81}$

also,

Q = A × V₂

thus,

$7-200\times(0.00384\times\ V_2)^2=\frac{V_2^2}{2\times9.81}$

V₂ = 47.10 m/s

3 0

3 years ago

If the 1550-lb boom AB, the 190-lb cage BCD, and the 169-lb man have centers of gravity located at points G1, G2 and G3, respect

Natasha2012 [34]

Answer:

hello the required diagram is missing attached to the answer is the required diagram

7.9954 kip.ft

Explanation:

AB = 1550-Ib ( weight acting on AB )

BCD = 190 - Ib ( weight of cage )

169-Ib = weight of man inside cage

Attached is the free hand diagram of the question

calculate distance $x!$

= cos 75⁰ = $\frac{x^!}{10ft}$

$x! = 10 * cos 75^{o}$ = 2.59 ft

calculate distance x

= cos 75⁰ = $\frac{x}{30ft}$

x = 30 * cos 75⁰ = 7.765 ft

The resultant moment produced by all the weights about point A

∑ Ma = 0

Ma = 1550 * $x!$ + 190 ( x + 2.5 ) + 169 ( x + 2.5 + 1.75 )

Ma = 1550 * 2.59 + 190 ( 7.765 + 2.5 ) + 169 ( 7.765 + 2.5 + 1.75 )

= 4014.5 + 1950.35 + 2030.535

= 7995.385 ft. Ib ≈ 7.9954 kip.ft

6 0

4 years ago

_____ is a process for identifying and modeling business events, who initiated them, and how the system should respond to them.

Sladkaya [172]

Answer: The answer is A). Use case modelling

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3 years ago

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Andrews [41]

Answer:

Its data

Explanation:

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3 years ago

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krek1111 [17]

Answer:Videos

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3 years ago