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3 years ago

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Suppose a contract states that the designer should bear the responsibility if substantial differences were found between the des

ign drawings and the reality. What type of disagreement may happen due to this statement? Select one:
a. Disagreement about the definition of concepts and terminology
b. Disagreement about irrelevant facts
c. Disagreement about unknown facts
d. Disagreement about the definition of facts

1 answer:

Zolol [24]3 years ago

7 0

Answer:

C.) Disagreement about unknown facts

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Steam enters an adiabatic turbine at 10 MPa and 500°C and leaves at 10 kPa with a quality of 90 percent. Neglecting the changes

Anna35 [415]

Answer:

The mass flow rate of steam m=5.4 Kg/s

Explanation:

Given:

At the inlet of turbine P=10 MPa ,T=500 C

AT the exit of turbine P=10 KPa ,x=0.9

Required power=5 MW

From steam table

<u> At 10 MPa and 500 C:</u>

h=3374 KJ/Kg ,s=6.59 KJ/Kg-K (Super heated steam table)

<u>At 10 KPa:</u>

$h_g$ =2675.1 KJ/Kg, $h_f$ =417.51 KJ/Kg

$s_g$ = 7.3 KJ/Kg-K , $s_f$ =1.3 KJ/Kg-K

So enthalpy of steam at the exit of turbine

h= $h_f+x(h_g- h_f)$

Now by putting the values

h= 417.51+0.9(2675.1- 417.51) KJ/Kg

h=2449.34 KJ/Kg

Lets take m is the mass flow rate of steam

So $5\times 10^3=m\times (3374-2449.34)$

m=5.4 Kg/s

So the mass flow rate of steam m=5.4 Kg/s

8 0

4 years ago

Part A Engineering stress and strain are calculated using the actual cross-sectional area and length of the specimen. True or fa

Galina-37 [17]

Answer: True

Explanation:

Engineering stress is the applied load divided by the original cross-sectional area of a material. It is also known as nominal stress. It can also be defined as the force per unit area of a material. Engineering Stress is usually in large numbers.

While Engineering strain is the amount that a material deforms per unit length in a tensile test. It can also be defined as extension per unit length. It has no unit as it is a ratio of lengths. Engineering Strain is in small numbers.

5 0

3 years ago

Read 2 more answers

I have a molten Ni-Cu alloy with 60 wt%Ni. (ii) I cool it down to a temperature where I have both solid and liquid phases. At th

SIZIF [17.4K]

Answer:

The percentage of the remaining alloy would become solid is 20%

Explanation:

Melting point of Cu = 1085°C

Melting point of Ni = 1455°C

At 1200°C, there is a 30% liquid and 70% solid, the weight percentage of Ni in alloy is the same that percentage of solid, then, that weight percentage is 70%.

The Ni-Cu alloy with 60% Ni and 40% Cu, and if we have the temperature of alloy > temperature of Ni > temperature of Cu, we have the follow:

60% Ni (liquid) and 40% Cu (liquid) at temperature of alloy

At solid phase with a temperature of alloy and 50% solid Cu and 50% liquid Ni, we have the follow:

40% Cu + 10% Ni in liquid phase and 50% of Ni is in solid phase.

The percentage of remaining alloy in solid is equal to

Solid = (10/50) * 100 = 20%

4 0

3 years ago

A square aluminum plate 5 mm thick and 150 mm on a side is heated while vertically suspended in quiescent air at 75°c. determine

Doss [256]

By using the boundary layer equation, the average heat transfer coefficient for the plate is equal to 4.87 W/m²k.

<u>Given the following data:</u>

Surface temperature = 15°C

Bulk temperature = 75°C

Side length of plate = 150 mm to m = 0.15 meter.

<h3>How to calculate the average heat transfer coefficient.</h3>

Since we have a quiescent room air and a uniform pole surface temperature, the film temperature is given by:

$T_f=\frac{T_{s} + T_{\infty} }{2} \\\\T_f=\frac{15 + 75 }{2} \\\\T_f = 45$

Film temperature = 45°C to K = 273 + 45 = 318 K.

For the coefficient of thermal expansion, we have:

$\beta =\frac{1}{T_f} \\\\\beta =\frac{1}{318}$

From table A-9, the properties of air at a pressure of 1 atm and temperature of 45°C are:

Kinematic viscosity, v = $1.750 \times 10^{-5}$ m²/s.

Thermal conductivity, k = 0.02699 W/mk.

Thermal diffusivity, α = $2.416 \times 10^{-5}$ m²/s.

Prandtl number, Pr = 0.7241.

Next, we would solve for the Rayleigh number to enable us determine the heat transfer coefficient by using the boundary layer equations:

$R_{aL}=\frac{g\beta \Delta T l^3}{v\alpha } \\\\R_{aL}=\frac{9.8 \;\times \;\frac{1}{318} \;\times \;(75-15) \;\times \;0.15^3 }{1.750 \times 10^{-5}\; \times \;2.416 \times 10^{-5} } \\\\R_{aL}=\frac{9.8\; \times 0.00315 \;\times \;60\; \times\; 0.003375 }{4.228 \times 10^{-10} }\\\\R_{aL}=1.48 \times 10^{7}$

Also take note, g(Pr) is given by this equation:

$g(P_r)=\frac{0.75P_r}{[0.609 \;+\;1.221\sqrt{P_r}\; +\;1.238P_r]^\frac{1}{4} } \\\\g(P_r)=\frac{0.75(0.7241)}{[0.609 \;+\;1.221\sqrt{0.7241}\; +\;1.238(0.7241)]^\frac{1}{4} }\\\\g(P_r)=\frac{0.543075}{[0.609 \;+\;1.221\sqrt{0.7241}\; +\;1.238(0.7241)]^\frac{1}{4} }\\\\g(P_r)=\frac{0.543075}{[2.5444]^\frac{1}{4} }\\\\g(P_r)=\frac{0.543075}{1.2630 }$

g(Pr) = 0.430

For GrL, we have:

$G_{rL}=\frac{R_{aL}}{P_r} \\\\G_{rL}=\frac{1.48 \times 10^7}{0.7241} \\\\G_{rL}=1.99 \times 10^7$

Since the Rayleigh number is less than 10⁹, the flow is laminar and the condition is given by:

$N_{uL}=\frac{h_{L}L}{k} = \frac{4}{3} (\frac{G_{rL}}{4} )^\frac{1}{4} g(P_r)\\\\h_{L}=\frac{0.02699}{0.15} \times [\frac{4}{3} \times (\frac{1.99 \times 10^7}{4} )^\frac{1}{4} ]\times 0.430\\\\h_{L}= 0.1799 \times 62.9705 \times 0.430\\\\h_{L}=4.87\;W/m^2k$

Based on empirical correlation method, the average heat transfer coefficient for the plate is given by this equation:

$N_{uL}=\frac{h_{L}L}{k} =0.68 + \frac{0.670 R_{aL}^\frac{1}{4}}{[1+(\frac{0.492}{P_r})^\frac{9}{16}]^\frac{4}{19} } \\\\h_{L}=\frac{0.02699}{0.15} \times ( 0.68 + \frac{0.670 (1.48 \times 10^7)^\frac{1}{4}}{[1+(\frac{0.492}{0.7241})^\frac{9}{16}]^\frac{4}{19} })\\\\h_{L}=4.87\;W/m^2k$

Read more on heat transfer here: brainly.com/question/10119413

3 0

2 years ago

Water at 98 F flows in a pipe 4 inches in diamter and 870 feet in length at a rate of 260 gallons per minute (GPM). What is the

Basile [38]

Answer:

Reynolds number is 293068.43.

Explanation:

Step1

Given:

Water temperature is 98 °F.

Diameter of the pipe is 4 in

Length of the pipe is 870 feet.

Flow rate of water is 260 gallons per minute.

Calculation:

Step2

Water kinematic viscosity at temperature 98 F is $7.55\times10^{-6}$ ft²/s.

Find the velocity of the water as follows:

Q=AV

$Q=\frac{\pi}{4}D^{2}V$

$(260 gal/min)(\frac{0.00222801 ft^{3}/s}{1 gal/min})=\frac{\pi}{4}(\frac{4}{12})^{2}V$

V=6.638 ft/s

Step3

Expression for Reynolds number is given as follows:

$Re=\frac{vd}{\nu}$

Here, v is velocity, $\nu$ is kinematic viscosity, d is diameter and Re is Reynolds number.

Substitute the values in the above equation as follows:

$Re=\frac{vd}{\nu}$

$Re=\frac{6.638\times(4in)(\frac{1ft}{12in})}{7.55\times10^{-6}}$

Re= 293068.43

Thus, the Reynolds number is 293068.43.

7 0

4 years ago