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3 years ago

Suppose that tank holds 1% liquid water by volume and 99% vapor water by volume at a temperature of 185oC. What is the quality?

Engineering

1 answer:

Vika [28.1K]3 years ago

4 0

Answer:

quantity = 0.645749

Explanation:

given data

holds liquid water = 1%

hold vapor water = 99%

temperature = 185°C

solution

we know that at temperature 185°C

Specific Volume of Saturated Liquid water (vf) = 0.0011343 m³/kg

and Specific Volume of Saturated Vapor (vg) = 0.17390 m³/kg

so quality will be

quality = $\frac{vapour*specific\ volume\ water}{water*specific\ volume\ vapour}$ ................1

put here value we get

quality = $\frac{99*0.0011343}{1*0.17390}$

quantity = 0.645749

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Explanation:

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3 years ago

(a) A non-cold-worked 1040 steel cylindrical rod has an initial length of 100 mm and initial diameter of 7.50 mm. is to be defor

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Answer:

A) 1040 steel is not a possible candidate for this application

B) 35.94%

Explanation:

Initial length = 100 mm = 0.1 m

Initial diameter ( d ) = 7.5 mm = 0.0075 m

Tensile load ( p ) = 18,000 N

Condition : The 1040 steel must not experience plastic deformation or a diameter reduction of more than 1.5 * 10^-5 m

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stress = p / A0 = ( 18000 ) / ( $\frac{\pi }{4}$ ) * 0.0075^2

= 18,000 / (4.418 * 10^-5 ) = 407.424 MPa

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stress = 407.424 MPa

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3 years ago

Select the best answer to the questo

Norma-Jean [14]

Answer:

Explanation:

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3 years ago

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A rigid tank whose volume is 2 m3, initially containing air at 1 bar, 295 K, is connected by a valve to a large vessel holding a

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Answer:

$Q_{cv}=-339.347kJ$

Explanation:

First we calculate the mass of the aire inside the rigid tank in the initial and end moments.

$P_iV_i=m_iRT_i$ (i could be 1 for initial and 2 for the end)

State1

$1bar*|\frac{100kPa}{1}|*2=m_1*0.287*295$

$m_1=232kg$

State2

$8bar*|\frac{100kPa}{1bar}|*2=m_2*0.287*350$

$m_2=11.946$

So, the total mass of the aire entered is

$m_v=m_2-m_1\\m_v=11.946-2.362\\m_v=9.584kg$

At this point we need to obtain the properties through the tables, so

For Specific Internal energy,

$u_1=210.49kJ/kg$

For Specific enthalpy

$h_1=295.17kJ/kg$

For the second state the Specific internal Energy (6bar, 350K)

$u_2=250.02kJ/kg$

At the end we make a Energy balance, so

$U_{cv}(t)-U_{cv}(t)=Q_{cv}-W{cv}+\sum_i m_ih_i - \sum_e m_eh_e$

No work done there is here, so clearing the equation for Q

$Q_{cv} = m_2u_2-m_1u_1-h_1(m_v)$

$Q_{cv} = (11.946*250.02)-(2.362*210.49)-(295.17*9.584)$

$Q_{cv}=-339.347kJ$

The sign indicates that the tank transferred heat to the surroundings.

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