4.54 Saturated liquid nitrogen at 600 kPa enters a boiler at a rate of 0.008 kg/s and exits as saturated vapor (see Fig. P4.54). It then flows into a superheater also at 600 kPa, where it exits at 600 kPa, 280 K. Find the rate of heat transfer in the boiler and the superheater.

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]

Answer:

Valvular stenosis , Valvular prolapse , Regurgitation,

Explanation:

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:

Answer:

45.32%

Explanation:

Given data:

pressure of high temperature steam = 6 MPa

low temperature heat rejection process ( Tr )  = 300 k

A) plot of cycle in t-s coordinates showing steam dome

attached below

B) Calculate thermal efficiency

thermal efficiency = 1 - (Tr / Tsat )

Tsat = 275.59°C  ≈  548.59 K  ( from steam table at Pa = 6 MPa )

back to equation 1

1 - (300 / 548.59 )

1 - 0.5468 = 0.4532 = 45.32%

Answer:

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

Explanation:

Answer:

Coolant

Explanation:

Answer:

0.05%

Explanation:

From the question, we have;

The yield strength of the mild steel, [tex]\sigma _c[/tex] = 43.5 ksi

Young's modulus of elasticity, ∈ = 29,000 ksi

The total strain, [tex]\epsilon _c[/tex] = 0.2% = 0.002

The inelatic strain [tex]\epsilon_c^{in}[/tex] is given as follows;

[tex]\epsilon_c^{in}[/tex] = [tex]\epsilon _c[/tex] - [tex]\sigma _c[/tex]/∈

Therefore, we have;

[tex]\epsilon_c^{in}[/tex] = 0.002 - 43.5/(29,000) = 0.0005

Therefore, the inelastic strain, [tex]\epsilon_c^{in}[/tex] = 0.0005 = 0.05%

Taking the inelastic strain as the residual strain, we have;

The residual strain = 0.05%

pelo o que diz na database é que você n é ser humano normal por perguntar isso!!

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.

Answer:

CAD and a database

Explanation:

The correct answer is CAD and a database. When American Motors Corportation introduced the Jeep Cherokee, it implemented CAD to increase engineering productivity and combined that with a new communications system.

Answer:

the first cleaning be scheduled 1.006 years after installation

Explanation:

 Given the data in the question;

U[tex]_{clean[/tex] = 300 W/m².K

first we determine the heat coefficient of the dirt surface;

overall heat transfer coefficient is reduced from its initial value by 25%

U[tex]_{dirt[/tex] = ( 1 - 25%) × U[tex]_{clean[/tex]

U[tex]_{dirt[/tex] = ( 1 - 0.25) × 300

U[tex]_{dirt[/tex] = 0.75 × 300

U[tex]_{dirt[/tex] = 225 W/m².K

next we find the inner fouling factor

[tex]R"_{f ,i[/tex] = [tex]a_it[/tex]

[tex]R"_{f ,o[/tex] = (2.5 × 10⁻¹¹)t

for the outer fouling water;

[tex]R"_{f ,o[/tex] = [tex]a_ot[/tex]

[tex]R"_{f ,o[/tex] = ( 1.0 × 10⁻¹¹ )t

now, we determine the total heat transfer coefficient

[tex]\frac{1}{U}[/tex] = [tex]R"_{f ,i[/tex] + [tex]R"_{f ,o[/tex]

we substitute

[tex]\frac{1}{U}[/tex] =  (3.5 × 10⁻¹¹)t

so the first cleaning duration after insulation will be;

[tex]\frac{1}{U} = \frac{1}{U_{dirt}} - \frac{1}{U{clean}}[/tex]

we substitute

(3.5 × 10⁻¹¹)t = [tex]\frac{1}{225} - \frac{1}{300}[/tex]

(3.5 × 10⁻¹¹)t = 0.001111

t = 0.001111 / (3.5 × 10⁻¹¹)

t = 31742857.142857 seconds

t = 31742857.142857 / 3.154 × 10⁷

t = 1.006 years

Therefore, the first cleaning be scheduled 1.006 years after installation

Answer:

Vgr = 0.122 = 12.2 vol %

Explanation:

Density of ferrite = 7.9 g/cm^3

Density of graphite = 2.3 g/cm^3

compute the volume percent of graphite

for a 3.9 wt% cast Iron

W∝ =  (100 - 3.9) / ( 100 -0 ) = 0.961

Wgr = ( 3.9 - 0 ) / ( 100 - 0 ) = 0.039

Next convert the weight fraction to volume fraction using the equation attached below

Vgr = 0.122 = 12.2 vol %

Answer:

true

Explanation:

the answer is true because if u don't have a valid license when operating a vehicle and you get pulled over you will get in trouble i know this because my parents got in trouble for it once