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svetoff [14.1K]
2 years ago
5

Which of the following is not an assumption for the Method of Joints?

Engineering
1 answer:
fenix001 [56]2 years ago
5 0

The option that is not an assumption for the Method of Joints is that

  • The truss is stable
  • Members are straight
  • Members are weightless

<h3>What is an assumption?</h3>

This is known to be people's believe in regard to a thing that is said  to be true or maybe probably true.

Note that The option that is not an assumption for the Method of Joints is that

  • The truss is stable
  • Members are straight
  • Members are weightless

Learn more about assumption from

brainly.com/question/2875871

#SPJ1

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11–17 A long, thin-walled double-pipe heat exchanger with tube and shell diameters of 1.0 cm and 2.5 cm, respectively, is used t
lana [24]

Answer:

the overall heat transfer coefficient of this heat exchanger is 1855.8923 W/m²°C

Explanation:

Given:

d₁ = diameter of the tube = 1 cm = 0.01 m

d₂ = diameter of the shell = 2.5 cm = 0.025 m

Refrigerant-134a

20°C is the temperature of water

h₁ = convection heat transfer coefficient = 4100 W/m² K

Water flows at a rate of 0.3 kg/s

Question: Determine the overall heat transfer coefficient of this heat exchanger, Q = ?

First at all, you need to get the properties of water at 20°C in tables:

k = 0.598 W/m°C

v = 1.004x10⁻⁶m²/s

Pr = 7.01

ρ = 998 kg/m³

Now, you need to calculate the velocity of the water that flows through the shell:

v_{w} =\frac{m}{\rho \pi (\frac{d_{2}^{2}-d_{1}^{2}  }{4} )} =\frac{0.3}{998*\pi (\frac{0.025^{2}-0.01^{2}  }{4}) } =0.729m/s

It is necessary to get the Reynold's number:

Re=\frac{v_{w}(d_{2}-d_{1}) }{v} =\frac{0.729*(0.025-0.01)}{1.004x10^{-6} } =10891.4343

Like the Reynold's number is greater than 10000, the regime is turbulent. Now, the Nusselt's number:

Nu=0.023Re^{0.8} Pr^{0.4} =0.023*(10891.4343)^{0.8} *(7.01)^{0.4} =85.0517

The overall heat transfer coefficient:

Q=\frac{1}{\frac{1}{h_{1} }+\frac{1}{h_{2} }  }

Here

h_{2} =\frac{kNu}{d_{2}-d_{1}} =\frac{0.598*85.0517}{0.025-0.01} =3390.7278W/m^{2}C

Substituting values:

Q=\frac{1}{\frac{1}{4100}+\frac{1}{3390.7278}  } =1855.8923W/m^{2} C

5 0
3 years ago
As part of a chemistry experiment, Austin ends up with a beaker containing a mixture of two liquids, which he knows have boiling
asambeis [7]

Answer:

Distillation, heat

Explanation:

Here in this question, we simply want to look at the best options that could fit in the gaps.

We have a mixture of liquids having boiling points which is far from each other.

Whenever we have a mixture of liquids with boiling points far away from each other, the best technique to use in separating them is to use distillation. That is why we have that as the best fit for the first missing gap.

Now, to get the liquids to separate from each other, we shall be needing the heating mantle for the application of heat. This ensures that the mixture is vaporized. After vaporization, the condensing tube will help to condense the vapor of each of the liquids once we reach the boiling point of either of the two.

Kindly note that the liquid with the lower temperature will evaporate first and will be first obtained. In fact after reaching a little above the boiling point of the lower boiling liquid, we can be sure that what we have left in the mixture pot is the second other liquid with the higher boiling point.

7 0
3 years ago
Whats the pros about being in the army
Cerrena [4.2K]

Answer:

Benefits that last a lifetime. The Army offers you money for education, comprehensive health care, generous vacation time, family services and support groups, special pay and cash allowances to cover the cost of living.

8 0
3 years ago
On-site oil storage containers must be marked "Used Oil."<br><br> True. <br><br> False.
Scrat [10]

Answer:

How must used oil storage containers be marked? Containers and aboveground tanks used to store used oil at generator facilities must be labeled or marked clearly with the words “Used Oil" (40 CFR Section 279.22(c)).

Explanation:

i think it will help you

7 0
3 years ago
Saturated liquid-vapor mixture of water, called wet steam, in a steam line at 1500 kPa is throttled to 50 kPa and 100°C. What is
frozen [14]

Answer:

x = 0.944

Explanation:

Steam at outlet is an superheated steam, since T > T_{sat}. From steam tables, the specific enthalpy is:

h_{out}=2682.4\,\frac{kJ}{kg}

The throttle valve is modelled after the First Law of Thermodynamics:

h_{in} = h_{out}

Hence, specific enthalpy at inlet is:

h_{in}=2682.4\,\frac{kJ}{kg}

The quality in the steam line is:

x = \frac{2682.4\,\frac{kJ}{kg}-844.55\,\frac{kJ}{kg}}{2791.0\,\frac{kJ}{kg} - 844.55\,\frac{kJ}{kg} }

x = 0.944

6 0
3 years ago
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