Steam in a heating system flows through tubes whose outer diameter is 5 cm and whose walls are maintained at a temperature of 198.06°C. Circular copper alloy fins (k =285 W/m · °C) of outer diameter 6 cm and constant thickness 1 mm are attached to the tube. The space between the fins is 3 mm, and thus there are 250 fins per meter length of the tube. Heat is transferred to the surrounding water at T= 43.06°C, with a heat transfer coefficient of 5300 W/m2 · °C. Determine the increase in heat transfer from the tube per meter of its length as a result of adding fins and fin effectiveness

Steam In A Heating System Flows Through Tubes Whose Outer Diameter Is 5 Cm And Whose Walls Are Maintained

Answers

Answer 1

Answer:

rf

Explanation:

attached to the tube. The space between the fins is 3 mm, and thus there are 250 fins per meter length of the tube. Heat is transferred to the surrounding water at T= 43.06°C, with a heat transfer coefficient of 5300 W/m2 · °C. Determine the increase in heat transfer from the tube per meter of its length as a resu.


Related Questions

Nate needs to replace the cable to his lamp. He is stripping it to connect it to the termils. What should he remember to do with the knife

Answers

Answer: i got you its d

Explanation:had the smae question as you

Discuss in detail the manners of interaction with opposite gender

Answers

Answer:

8 Tips on Better Communication with the Opposite Sex

Put emotions away. Ladies, this one is more aimed at us, for the most part. ...

Forget your pride. In discussions, especially these days, people always want to be the one that prevails. .

Put yourself in their shoes. .

Listen. ...

Respond. ...

Actually communicate. ...

Be detailed. ...

Don't communicate too much.

Explanation:

If it is struck by a rigid block having a weight of 550 lblb and traveling at 2 ft/sft/s , determine the maximum stress in the cylinder. Neglect the mass of the cylinder. Express your answer to three significant figures and include appropriate units.

Answers

This question is incomplete, The missing image is uploaded along this answer below;

Answer:

the maximum stress in the cylinder is 3.23 ksi

 

Explanation:

Given the data in the question and the diagram below;

First we determine the initial Kinetic Energy;

T = [tex]\frac{1}{2}[/tex]mv²

we substitute

⇒ T = [tex]\frac{1}{2}[/tex] × (550/32.2) × (2)²

T = 34.16149 lb.ft

T =  ( 34.16149 × 12 ) lb.in  

T = 409.93788 lb.in

Now, the volume will be;

V = [tex]\frac{\pi }{4}[/tex]d²L

from the diagram; d = 0.5 ft and L = 1.5 ft

so we substitute

V =  [tex]\frac{\pi }{4}[/tex] × ( 0.5 × 12 in )² × ( 1.5 × 12 in )

V = 508.938 in³

So by conservation of energy;

Initial energy per unit volume = Strain energy per volume

⇒ T/V = σ²/2E

from the image; E = 6.48(10⁶) kip

so we substitute

⇒ 409.93788 / 508.938 = σ²/2[6.48(10⁶)]

508.938σ² =  5,312,794,924.8

σ² = 10,438,982.5967  

σ = √10,438,982.5967

σ = 3230.9414  

σ = 3.2309 ksi  ≈ 3.23 ksi    { three significant figures }

Therefore, the maximum stress in the cylinder is 3.23 ksi

How should backing plates, struts, levers, and other metal brake parts be cleaned?

Answers

Answer: Cleaning of mechanical parts is necessary to remove contaminants, and to avoid clogging of wastes which could restrict the functioning of the machine.

Explanation:

There are different agents used for cleaning different machine instruments to prevent their corrosion and experience proper cleaning.

Backing plates must be dry cleaned using a cotton cloth to remove the dirt, dust or any other dry contaminant.

Struts can be wet cleaned by applying alcoholic solvent.

Levers can be cleaned using a mineral spirit.

Metallic plates can be cleaned using water based solution or water.

A tiger cub has a pattern of stripes on it for that is similar to that of his parents where are the instructions stored that provide information for a tigers for a pattern

Answers

probably in it's chromosomes

Two ball bearings from different manufacturers are being considered for a certain application. Bearing A has a catalog rating of 2.12 kN based on a catalog rating system of 3000 hours at 500 rev/min. Bearing B has a catalog rating of 7.5 kN based on a catalog that rates at 106 cycles. For a given application, determine which bearing can carry the larger load.

Answers

Answer:

F[tex]_D[/tex] for A > F[tex]_D[/tex] for B

Hence, Bearing A can carry the larger load

Explanation:

Given the data in the question,

First lets consider an application which requires desired speed of n₀ and a desired life of L₀.

Lets start with Bearing A

so we write the relation between desired load and life catalog load and life;

[tex]F_R(L_Rn_R60)^{1/a}[/tex] = [tex]F_D(L_Dn_D60)^{1/a}[/tex]

where F[tex]_R[/tex] is the catalog rating( 2.12 kN)

L[tex]_R[/tex] is the rating life ( 3000 hours )

n[tex]_R[/tex] is the rating speed ( 500 rev/min )

F[tex]_D[/tex] is the desired load

L[tex]_D[/tex] is the desired life ( L₀ )

n[tex]_D[/tex]  is the the desired speed ( n₀ )

Now as we know, a = 3 for ball bearings

so we substitute

[tex]2.12( 3000 * 500 * 60 )^{1/3[/tex]  =  [tex]F_D( L_0n_060)^{1/3[/tex]    

950.0578 = [tex]F_D( L_0n_0)^{1/3} 3.914867[/tex]    

950.0578 / 3.914867 = [tex]F_D( L_0n_0)^{1/3}[/tex]

242.6794 =   [tex]F_D( L_0n_0)^{1/3}[/tex]

F[tex]_D[/tex] for A =  (242.6794 / [tex]( L_0n_0)^{1/3}[/tex] ) kN

Therefore the load that bearing A can carry is  (242.6794 / [tex]( L_0n_0)^{1/3}[/tex] ) kN

Next is Bearing B

[tex]F_R(L_Rn_R60)^{1/a}[/tex] = [tex]F_D(L_Dn_D60)^{1/a}[/tex]

F[tex]_R[/tex] = 7.5 kN, [tex](L_Rn_R60) = 10^6[/tex]

Also, for ball bearings, a = 3

so we substitute

[tex]7.5(10^6)^{1/3[/tex] = [tex]F_D(L_0n_060)^{1/3}[/tex]

750 =  [tex]F_D(L_0n_0)^{1/3} 3.914867[/tex]

750 / 3.914867  =  [tex]F_D(L_0n_0)^{1/3}[/tex]

191.5773 = [tex]F_D(L_0n_0)^{1/3}[/tex]

F[tex]_D[/tex] for B = ( 191.5773 / [tex](L_0n_0)^{1/3}[/tex] ) kN

Therefore, the load that bearing B can carry is  ( 191.5773 / [tex](L_0n_0)^{1/3}[/tex] ) kN

Now, comparing the Two results above,

we can say;

F[tex]_D[/tex] for A > F[tex]_D[/tex] for B

Hence, Bearing A can carry the larger load

The unit of solar radiation?

Answers

Answer: The solar irradiance is measured in watt per square metre (W/m2) in SI units. Solar irradiance is often integrated over a given time period in order to report the radiant energy emitted into the surrounding environment (joule per square metre, J/m2) during that time period.

Explanation: hope that helped!

Watts is the unit of solar radiation

The evaporator:
A. directs airflow to the condenser.
B. absorbs heat from the passenger compartment.
C. removes moisture from the refrigerant.
D. restricts refrigerant flow.

Answers

Answer:

Option B

Explanation:

An evaporator along with  cold low pressure refrigerant absorbs heat from the air within the passenger compartment thereby supplying cool air for the occupants.

Hence, option B is correct

Some project managers prefer the PERT chart over the Gantt chart because it clearly illustrates task dependencies. A PERT chart, however, can be much more difficult to interpret, especially on complex projects. Alternatively, some project managers may choose to use both techniques. If you are the project manager of a residential construction project, will you prefer PERT chart to Gantt chart? Explain why?

Answers

Answer:

PERT Chart and GANTT Chart

As the project manager of a residential construction project, I will prefer the PERT chart to the GANTT chart because a PERT chart displays task dependencies unlike a Gantt chart.  With the PERT chart, the sequence of tasks is clearly mapped out.  Dependent tasks are carried out when other tasks that they depend on have been executed.

Explanation:

By definition, a Gantt chart is like a bar chart that lays out project tasks and timelines using bars.  On the other hand, a PERT chart follows a structure in the form of flow charts or network diagrams.  It displays all the project tasks in separate boxes.  The boxes are then connected with arrows which clearly show the task dependencies.

Assume, X Company Limited (XCL) is one of the leading 4th generation Life Insurance
Companies in Bangladesh. The Company is fully customer focused. This Life insurance company are
experimenting with analysis of consumer profiles (to determine whether a person eats healthy food,
exercises, smokes or drinks too much, has high-risk hobbies, and so on) to estimate life expectancy.
Companies might use the analysis to find populations to market policies to. From the perspective of
privacy, what are some of the key ethical or social issues raised? Evaluate some of them.

Answers

Answer:

The issues related to the privacy are:

1. Informational privacy

2. Discrimination factors

3. Biased grouping on the basis of Data mining

4. Lack of consent

5. Morally wrong

6. Illegal distribution of information risks

7. Possibility of threat to life

Let's look at some major concerns:

1. Informational privacy : The concept of privacy of the personal information is totally nullified when the information is being used for a purpose other than the intended one for which it was given. This unethical use of information even for general purposes is not correct and is a matter of concern. It is more like using the sensitive data of others for personal benefit which is purely objectionable and raises security issues. Sometimes the data is also shared with the potential employers which might have certain impacts we are unaware of.

2. Data mining issues : The process of using a certain information to arrive and understand the trend and outcomes is called data mining. In this case, the consumer's data undergoes grouping and might get placed in the wrong group rather than the actual one. Also, there can be a case of biasing towards the groups which are not be focused on, or are not a part of the intended audience. This leads to the discrimination factors if we see it from a social point of view.

3. Lack of consent : Use of information without the consent or awareness of the consumers raises concern over the business ethics followed by the company. No one deserves the right to misuse information for his personal benefits without any of its information to the consumer. It is morally wrong and againt the work ethics. Moreover, it raises trust issues between the two involved, and hence is socially unacceptable.Explanation:

A 10 cm thick slab (density 8530 kg/m3 , specific heat 380J/kg K, conductivity 110 W/m K) initially at 650C is being cooled by air at 15C (convection coefficient 220 W/m2 K) at left surface. The right surface is insulated. We want to estimate temperature in the slab. a. What will be the temperature of the slab after a very long time

Answers

Answer:

The temperature after a long time will return to 15°C

Explanation:

Determine the temperature of the slab after a very long time

First we calculate the heat flow for m^2 area normal to the surface

= q / A = 650°c - 15°C / ( 1 / h  + L / K )

           = 635°c  / ( 1 / 220 + 0.1 / 110 )  = 116.416 kw/m^2

Total heat content in the slab is calculated as

= m* c * ΔT

= 8530 * A * 0.1 * 380 * ( 650 - 15 )

= 205828.9  kJ/m^2

The temperature will return to 15°C after a long time

As part of a heat treatment process, cylindrical, 304 stainless steel rods of 100-mm diameter are cooled from an initial temperature of 500 C by suspending them in an oil bath at 30 C. If a convection coefficient of 500 W/m2 K is maintained by circulation of the oil, how long does it take for the centerline of a rod to reach a temperature of 50 C, at which point it is withdrawn from the bath

Answers

Answer:

Explanation:

Given that:

diameter = 100 mm

initial temperature = 500 ° C

Conventional coefficient = 500 W/m^2 K

length  = 1 m

We obtain the following data from the tables A-1;

For the stainless steel of the rod [tex]\overline T = 548 \ K[/tex]

[tex]\rho = 7900 \ kg/m^3[/tex]

[tex]K = 19.0 \ W/mk \\ \\ C_p = 545 \ J/kg.K[/tex]

[tex]\alpha = 4.40 \times 10^{-6} \ m^2/s \\ \\ B_i = \dfrac{h(\rho/4)}{K} \\ \\ =0.657[/tex]

Here, we can't apply the lumped capacitance method, since Bi > 0.1

[tex]\theta_o = \dfrac{T_o-T_{\infty}}{T_i -T_\infty}} \\ \\ \theta_o = \dfrac{50-30}{500 -30}} \\ \\ \theta_o = 0.0426\\[/tex]

[tex]0.0426 = c_1 \ exp (- E^2_1 F_o_)\\ \\ \\ 0.0426 = 1.1382 \ exp (-10.9287)^2 \ f_o \\ \\ = f_o = \dfrac{In(0.0374)}{0.863} \\ \\ f_o = 3.81[/tex]

[tex]t_f = \dfrac{f_o r^2}{\alpha} \\ \\ t_f = \dfrac{3.81 \times (0.05)^2}{4.40 \times 10^{-6}} \\ \\ t_f= 2162.5 \\ \\ t_f = 36 mins[/tex]

However, on a single rod, the energy extracted is:

[tex]\theta = pcv (T_i - T_{\infty} )(1 - \dfrac{2 \theta}{c} J_1 (\zeta) ) \\ \\ = 7900 \\times 546 \times 0.007854 \times (500 -300) (1 - \dfrac{2 \times 0.0426}{1.3643}) \\ \\ \theta = 1.54 \times 10^7 \ J[/tex]

Hence, for centerline temperature at 50 °C;

The surface temperature is:

[tex]T(r_o,t) = T_{\infty} +(T_1 -T_{\infty}) \theta_o \ J_o(\zeta_1) \\ \\ = 30 + (500-30) \times 0.0426 \times 0.5386 \\ \\ \mathbf{T(r_o,t) = 41.69 ^0 \ C}[/tex]

Help meeeeeeeee plzzzzz need explanation

Answers

the picture is blank for me what does it say i can comment the answer plz mark brainlyist

In the context of mechanical systems, what does the term efficiency mean? OA the factor by which a machine multiplies a force B. the ratio of a machine's power to the force of its input Ос. the rate at which a machine performs work D. the rate at which a machine consumes energy E. the ratio of the work output of a machine to the work input​

Answers

Answer:

E

Explanation:

I have a big brain and I just took the test and got it correct.

A noisy transmission channel has a per-digit error probability p = 0.01.
(a) Calculate the probability of more than one error in 10 received digits?

Answers

Answer:

The appropriate answer is "0.0043".

Explanation:

The given values is:

Error probability,

p = 0.01

Received digits,

n = 10

and,

[tex]x\sim Binomial[/tex]

As we know,

⇒  [tex]P(x)=\binom{n}{x}p^xq^{n-x}[/tex]

Now,

⇒  [tex]P(x >1) =1- \left \{ P(x=0)+P(x=1) \right \}[/tex]

⇒                 [tex]=1-\left \{\binom{10}{0}(0.01)^0(0.99)^{10-0}+\binom{10}{0}(0.01)^1(0.99)^{10-1} \right \}[/tex]

⇒                 [tex]=1-0.9957[/tex]

⇒                 [tex]=0.0043[/tex]

Which option identifies the step of the implementation phase represented in the following scenario?

A company in the gaming industry decides to integrate movement controls in its newest hardware release. It sends an online survey to interactive gamers to identify their highest priorities.

establishing a process and budget

using communication tools

building and assembling a team

setting up a change order process

Answers

Answer:

Which option identifies the step of the implementation phase represented in the following scenario?

A company in the gaming industry decides to integrate movement controls in its newest hardware release. It sends an online survey to interactive gamers to identify their highest priorities.

establishing a process and budget

using communication tools

building and assembling a team

setting up a change order process

Explanation:

#carryonml

Answer:

using communication tools

Explanation:

The correct answer is using communication tools. Communication tools such as online surveys help project teams identify customers’ wants and needs.

A 4-L pressure cooker has an operating pressure of 175 kPa. Initially, one-half of the volume is filled with liquid and the other half with vapor. If it is desired that the pressure cooker not run out of liquid water for 75 min, determine the highest rate of heat transfer allowed.

Answers

Answer:

the highest rate of heat transfer allowed is 0.9306 kW

Explanation:

Given the data in the question;

Volume = 4L = 0.004 m³

V[tex]_f[/tex] = V[tex]_g[/tex] = 0.002 m³

Using Table ( saturated water - pressure table);

at pressure p = 175 kPa;

v[tex]_f[/tex] = 0.001057 m³/kg

v[tex]_g[/tex] = 1.0037 m³/kg

u[tex]_f[/tex] = 486.82 kJ/kg

u[tex]_g[/tex] 2524.5 kJ/kg

h[tex]_g[/tex] = 2700.2 kJ/kg

So the initial mass of the water;

m₁ = V[tex]_f[/tex]/v[tex]_f[/tex] + V[tex]_g[/tex]/v[tex]_g[/tex]

we substitute

m₁ = 0.002/0.001057  + 0.002/1.0037

m₁ = 1.89414 kg

Now, the final mass will be;

m₂ = V/v[tex]_g[/tex]

m₂ = 0.004 / 1.0037

m₂ = 0.003985 kg

Now, mass leaving the pressure cooker is;

m[tex]_{out[/tex] = m₁ - m₂

m[tex]_{out[/tex] = 1.89414  - 0.003985

m[tex]_{out[/tex] = 1.890155 kg

so, Initial internal energy will be;

U₁ = m[tex]_f[/tex]u[tex]_f[/tex] + m[tex]_g[/tex]u[tex]_g[/tex]

U₁ = (V[tex]_f[/tex]/v[tex]_f[/tex])u[tex]_f[/tex]  + (V[tex]_g[/tex]/v[tex]_g[/tex])u[tex]_g[/tex]

we substitute

U₁ = (0.002/0.001057)(486.82)  + (0.002/1.0037)(2524.5)

U₁ = 921.135288 + 5.030387

U₁ = 926.165675 kJ

Now, using Energy balance;

E[tex]_{in[/tex] -  E[tex]_{out[/tex] = ΔE[tex]_{sys[/tex]

QΔt - m[tex]_{out[/tex]h[tex]_{out[/tex] = m₂u₂ - U₁

QΔt - m[tex]_{out[/tex]h[tex]_g[/tex] = m₂u[tex]_g[/tex] - U₁

given that time = 75 min = 75 × 60s = 4500 sec

so we substitute

Q(4500) - ( 1.890155 × 2700.2 ) = ( 0.003985 × 2524.5 ) - 926.165675

Q(4500) - 5103.7965 = 10.06013 - 926.165675

Q(4500) = 10.06013 - 926.165675 + 5103.7965

Q(4500) = 4187.690955

Q = 4187.690955 / 4500

Q = 0.9306 kW

Therefore, the highest rate of heat transfer allowed is 0.9306 kW

MAPP gas, natural gas, propane, and acetylene can be used with oxygen to cut metal.
True
False

Answers

The answer is true.

What regulates the car engines temperature

Answers

Answer:

car’s thermostat is used to regulate the temperature of the engine to help the engine stay cool.

Explanation:

Answer:

Coolant

Explanation:

A cylindrical bar of metal having a diameter of 19.9 mm and a length of 186 mm is deformed elastically in tension with a force of 42600 N. Given that the elastic modulus and Poisson's ratio of the metal are 67.1 GPa and 0.34, respectively, determine the following: (a) The amount by which this specimen will elongate (in mm) in the direction of the applied stress. (b) The change in diameter of the specimen (in mm). Indicate an increase in diameter with a positive number and a decrease with a negative number.

Answers

Answer:

a) 0.347 mm

b)  - 0.0126 mm

Explanation:

Diameter of metal bar = 19.9 mm

length = 186 mm = 0.186 m

Tension force = 42600 N

elastic modulus = 67.1 GPa

Poisson's ratio = 0.34

a) calculate the amount by which the specimen will elongate

first calculate the area of metal bar

Area = πd^2 / 4 = π/4 ( 19.9 )^2  = 3.11 * 10^-4 m^2

Elongation ( E ) = б / e = P/A * L / ΔL

and ΔL = PL / AE

hence the elongation ( ΔL) = [ (42600 * 0.186 ) / ( 3.4*10^-4 *  67.1 * 10^9 ) ]

                                            = 3.47 * 10^-4 m ≈ 0.347 mm

b) Calculate the change in diameter of specimen

μ = -( Δd / d) / ( ΔL/L )

0.34 = - (Δd / 19.9 ) / ( 0.347 / 186 )

∴ Δd = 0.34 * 0.00186 * 19.9  = - 0.0126 mm

Which of these parts reduces the amount of pollutants in the engine exhaust?
A. Transmission
B. Catalytic converter
C. Tailpipe
D. Muffler

Answers

Answer:

Option B

Explanation:

Catalytic convertor with in the exhaust pipes of any vehicle converts harmful gases (such as hydrocarbons (HC), carbon monoxide (CO) and nitrogen oxides (NOx)) produced by combustion of fossil fuels (petrol, diesel etc.) into less harmful/toxic pollutants by acting as a catalyst for redox reaction. A catalytic convertor releases some mild gases from the exhaust like less harmful CO2, nitrogen and water vapour (steam)

Hence, option B is correct

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