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Dahasolnce [82]
3 years ago
7

A two-phase mixture of water and steam with a quality of 0.63 and T = 300F expands isothermally until only saturated vapor rema

ins. The initial volume is 0.114 ft3 . During the process, 16.2 Btu of heat are added. Find the work done. Also draw this process in the P-v space indicating the initial and final states with respect to the liquid-vapor dome.

Engineering
1 answer:
VMariaS [17]3 years ago
5 0

Answer:

Explanation:

Hello!

To solve this problem you must follow the following steps, which are fully registered in the attached image.

1. Draw the complete outline of the problem.

2. Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)

through prior knowledge of two other properties.

3. Use temodynamic tables to find the density of water in state 1, by means of temperature and quality, with this value and volume we can find the mass.

3. Use thermodynamic tables to find the internal energy in state 1 and two using temperature and quality.

4. uses the first law of thermodynamics that states that the energy in a system is always conserved, replaces the previously found values ​​and finds the work done.

5. draw the pV diagram using the 300F isothermal line

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8. Two 40 ft long wires made of differing materials are supported from the ceiling of a testing laboratory. Wire (1) is made of
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Answer:

Material K has a modulus of elasticity E=3.389× 10¹¹ Pa

Material H has a modulus of elasticity E=1.009 × 10⁹ Pa

Material K has higher value of modulus of elasticity than material H

Material K is stiffer.

Explanation:

Wire 1 material H

Length=L = 40 ft =12.192 m

Diameter= 3/8 in = 0.009525 m

Area= A= πr²,where r=0.009525/2 =0.004763

A=3.142*0.004763² =0.00007126 m²

Force, F= 225 lb=  225*4.45 =1001.25 N

Change in length =Δ L= 0.10 in = 0.00254

To find modulus of elasticity apply'

E=F*L/A*ΔL

E=1001.25*12.192/(0.004763*0.00254)

E= 1009027923.58 Pa

E=1.009 × 10⁹ Pa

For Wire 2 material K

Length=L= 40 ft =12.192 m

Diameter = 3/16 in = 0.1875 in = 0.004763 m

Area= πr² = 3.142 * (0.004763/2)² = 0.00000567154 m²

Force, F= 225 lb=  225*4.45 =1001.25 N

Change in length =Δ L= 0.25 in =0.00635 m

To find modulus of elasticity apply'

E=F*L/A*ΔL

E= (1001.25*12.192)/(0.00000567154 * 0.00635 )

E=338955422575 Pa

E=3.389× 10¹¹ Pa

Material  K has a greater modulus of elasticity

The material with higher value of E is stiffer than that with low value of E.The stiffer material is K.

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A particle is emitted from a smoke stack with diameter of 0.05 mm. In order to determine how far downstream it travels it is imp
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Answer: downward velocity = 6.9×10^-4 cm/s

Explanation: Given that the

Diameter of the smoke = 0.05 mm = 0.05/1000 m = 5 × 10^-5 m

Where radius r = 2.5 × 10^-5 m

Density = 1200 kg/m^3

Area of a sphere = 4πr^2

A = 4 × π× (2.5 × 10^-5)^2

A = 7.8 × 10^-9 m^2

Volume V = 4/3πr^3

V = 4/3 × π × (2.5 × 10^-5)^3

V = 6.5 × 10^-14 m^3

Since density = mass/ volume

Make mass the subject of formula

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Using the formula

V = sqrt( 2Mg/ pCA)

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g = 9.81 m/s^2

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p = density = 1200 kg/m3

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Substitute all the parameters into the formula

V = sqrt[( 2 × 7.9×10^-11 × 9.8)/(1200 × 24 × 7.8×10^-9)]

V = sqrt[ 1.54 × 10^-9/2.25×10-4]

V = 6.9×10^-6 m/s

V = 6.9 × 10^-4 cm/s

6 0
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