A long conducting rod of rectangular cross section (20 mm 30 mm) and thermal conductivity k 20 W/m K experiences uniform heat generation at a rate q . 5 107 W/m3, while its surfaces are maintained at 300 K. Using a finite-difference method with a grid spacing of 5 mm, determine the temperature distribution in the rod.

Answer:

See explanation

Explanation:

Solution:-

- The shell and tube heat exchanger are designated by the order of tube and shell passes.

- A single tube pass: The fluid enters from inlet, exchange of heat, the fluid exits.

- A multiple tube pass: The fluid enters from inlet, exchange of heat, U bend of the fluid, exchange of heat, .... ( nth order of pass ), and then exits.

- By increasing the number of passes we have increased the "retention time" of a specific volume of tube fluid; hence, providing sufficient time for the fluid to exchange heat with the shell fluid.

- By making more U-turns we are allowing greater length for the fluid flow to develop with " constriction and turns " into turbulence. This turbulence usually at the final passes allows mixing of fluid and increases the heat transfer coefficient by:

                                U ∝ v^( 0.8 )    .... ( turbulence )

- The higher the velocity of the fluids the greater the heat transfer coefficient. The increase in the heat transfer coefficient will allow less heat energy carried by either of the fluids to be wasted ; hence, reduced losses.

Thereby, increases the thermal efficiency of the heat exchanger ( higher NTU units ).

Answer:

1) twelve

Explanation:

The dual voltage motors are used in day to day operations. The wye is connected with 9 lead motors. Maximum resistance can be obtained if the resistance are connected in series. To check resistance of dual voltage wye motor there must be twelve resistance readings of 1 ohm each.

Answer:

Attached is the complete question but the weight of the mailbox and cross bar differs from the given values which are : weight of mail box = 3.2 Ib, weight of the uniform cross member = 10.3 Ib

Answer : moment of inertia = 186.7 Ib - in

Explanation:

Given data

weight of the mailbox = 3.2 Ib

weight of the uniform cross member = 10.3 Ib

The origin is of mailbox and cross member is 0

The perpendicular distance from Y axis of centroid of the mailbox

= 4 + (25/2) = 16.5"

The centroid of the bar =( ( 1 + 25 + 4 + 4 ) / 2 ) - 4  = 13"

therefore The moment of Inertia( Mo) = (3.2 * 16.5) + ( 10.3 * 13)

                                     = 52.8 + 133.9 = 186.7 Ib-in

The combined moment about O due to the weight of the mailbox and the cross member AB is; M_o = 122.4 lb.in (ccw)

We are given;

Weight of mailbox; W_m = 3.2 lb

Weight of uniform cross member; W_c = 10.3 lb

Now, from the attached diagram, let us calculate the geometric location of the mailbox and uniform cross section from point O.

Geometric location of mailbox from point O; g_m = 3 + (19/2) = 12.5 in

Geometric location of cross member from point O;

g_c = (¹/₂(1 + 19 + 3 + 7)) - 7

g_c = 8 in

Thus. combined moment about point O is;

M_o = (W_m × g_m) + (W_c × g_c)

M_o = (3.2 × 12.5) + (10.3 × 8)

M_o = 122.4 lb.in

Since positive then it is counterclockwise. Thus;

M_o = 122.4 lb.in (ccw)

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Answer:

The speed will be "3.58 ft/s". The further explanation is given below.

Explanation:

Number of knots

= 15

For the similarity of Froude number:

⇒  [tex]\frac{V_{m}}{\sqrt{g_{m}l_{m}} }=\frac{V}{\sqrt{gl} }[/tex]

Here,

[tex]l = length[/tex]

[tex]g_{m}=g[/tex]

⇒  [tex]\frac{V_{m}}{V}=\sqrt{\frac{l_{m}}{l} }[/tex]

    [tex]V_{m}=\sqrt{\frac{1}{50} }\times number \ of \ knots[/tex]

         [tex]=\sqrt{\frac{1}{50}}\times 15[/tex]

         [tex]=2.12 \ knots[/tex]

Now,

⇒  [tex]1 \ knots=0.514\times 3.281[/tex]

                 [tex]=1.69 \ ft/s[/tex]

So that,

⇒  [tex]V_{m}=2.12\times 1.69[/tex]

          [tex]=3.58 \ ft/s[/tex]

Answer:

dx/dt = kx(9000-x) where k > 0

Explanation:

Number of students in the campus, n = 9000

Number of students who have contracted the flu = x(t) = x

Number of students who have bot yet contracted the flu = 9000 - x

Number of Interactions between those that have contracted the flu and those that are yet to contract it = x(9000 - x)

The rate of spread of the disease = dx/dt

Note: the rate at which the disease spread is proportional to the number of interactions between those that have contracted the flu and those that have not contracted it.

[tex]\frac{dx}{dt} \alpha [x(9000 -x)]\\[/tex]

Introducing a constant of proportionality, k:

dx/dt = kx(9000-x) where k > 0

Answer:

HVACR Industry Trade Groups

Explanation:

[tex]

<html>

<head>

<meta name="viewport" content="width=device-width, initial-scale=1">

<title>Time Picker</title>

</head>

<body>

<!--24 Hours format-->

<input type="time" placeholder="Enter Time" />

<input type="date">

</body>

</html>

[/tex]

Answer:

B- Fluorinated gas

Explanation:

Answer:

B.) fluorinated gas

Explanation:

The negative mark is balanced by a positive mark on the set key scale while the jaws are closed.

It is common practice to shut the jaws or faces of the system before taking some reading to guarantee a zero reading. If not, please take care of the read. This read is referred to as "zero defect."

There are two forms of zero error:

zero-mistake positive; and

Non-null mistake.

----------------------------

Hope this helps!

Brainliest would be great!

----------------------------

With all care,

07x12!

Answer:

a) T₃ = 1818.8 K

b) η = 0.614 = 61.4%

c) MEP = 660.4 kPa

Explanation:

a) According to Table A-2 of The ideal gas specific heat of gases, the properties of air are as following:

At 300K

The specific heat capacity at constant pressure = [tex]c_{p}[/tex] = 1.005 kJ/kg.K,

The specific heat capacity at constant volume = [tex]c_{v}[/tex] = 0.718 kJ/kg.K

Gas constant R for air = 0.2870 kJ/kg·K

Ratio of specific heat  k = 1.4

Isentropic Compression :

[tex]T_{2}[/tex] =  [tex]T_{1}[/tex]  [tex](v1/v2)^{k-1}[/tex]

   = 300K ([tex]16^{0.4}[/tex])

[tex]T_{2}[/tex]    = 909.4K

P = Constant heat Addition:

[tex]P_{3}v_{3} / T_{3} = P_{2} v_{2} /T_{2}[/tex]

[tex]T_{3}=v_{3}/v_{2}T_{2}[/tex]

2[tex]T_{2}[/tex] = 2(909.4K)

      = 1818.8 K

b) [tex]q_{in}[/tex] = [tex]h_{3}-h_{2}[/tex]

         =  [tex]c_{p}[/tex] ([tex]T_{3}[/tex] - [tex]T_{2}[/tex])

         = (1.005 kJ/kg.K)(1818.8 - 909.4)K

         = 913.9 kJ/kg

Isentropic Expansion:

[tex]T_{4}[/tex] =  [tex]T_{3}[/tex]  [tex](v3/v4)^{k-1}[/tex]

    =  [tex]T_{3}[/tex] [tex](2v_{2} /v_{4} )^{k-1}[/tex]

    = 1818.8 K (2 / 16[tex])^{0.4}[/tex]

    = 791.7K

v = Constant heat rejection

[tex]q_{out}[/tex] = μ₄ - μ₁

      = [tex]c_{v} ( T_{4} - T_{1} )[/tex]

      = 0.718 kJ/kg.K (791.7 - 300)K

      = 353 kJ/kg

 η[tex]_{th}[/tex] = 1 - [tex]q_{out}[/tex] / [tex]q_{in}[/tex]

       = 1 - 353 kJ/kg / 913.9 kJ/kg

       = 1 - 0.38625670

       = 0.6137

       = 0.614

      = 61.4%

c) [tex]w_{net}._{out}[/tex] = [tex]q_{in}[/tex] - [tex]q_{out}[/tex]

                = 913.9 kJ/kg - 353 kJ/kg

                = 560.9 kJ/kg

[tex]v_{1} = RT_{1} /P_{1}[/tex]

   = (0.287 kPa.m³/kg/K)*(300 K) / 95 kPa

   =  86.1 / 95

   = 0.9063 m³/kg = v[tex]_{max}[/tex]

[tex]v_{min} =v_{2} = v_{max} /r[/tex]

Mean Effective Pressure = MEP =   [tex]w_{net,out}/v_{1} -v_{2}[/tex]

                                                    = [tex]w_{net,out}/v_{1}(1-1)/r[/tex]

                                                    = 560.9 kJ/kg / (0.9063 m³/kg)*(1-1)/16

                                                    = (560.9 kJ / 0.8493m³) (kPa.m³/kJ)

                                                    = 660.426 kPa

Mean Effective Pressure = MEP = 660.4 kPa

The temperature after the addition process is 1724.8k, the thermal efficiency of the engine is 56.3% and the mean effective pressure is 65.87kPa

Assumptions made:

a) The temperature after the addition process:

Considering the process 1-2, Isentropic expansion

at

[tex]T_1=300k\\u_1=214.07kJ/kg\\v_o_1=621.3\\v_o_2=\frac{v_2}{v_1} *v_o_1[C.R=16]=v_2/v_1\\v_o_2=(v_2/v_1)v_o_1=1/16*621.2=38.825[/tex]

From using this value, v[tex]_o_2[/tex]=38.825, solve for state point 2;

[tex]T_2=862.4k\\h_2=890.9kJ/kg[/tex]

Considering the process 2-3 (state of constant heat addition)

[tex]\frac{p_3v_3}{t_3}=\frac{p_2v_2}{t_2} \\\\T_3=\frac{P_3V_3T_2}{V_2} \\T_3=(\frac{V_3}{V_2}) T_2\\\frac{v_3}{v_2}=2\\T_3=2(862.4)=1724.8k\\[/tex]

NB: p[tex]_3[/tex]≈p[tex]_2[/tex]

b) The thermal efficiency of the engine is

Q[tex]_i_n[/tex]=h[tex]_3-h_2[/tex] = 1910.6-890.9=1019.7kJ/kg

Considering process 3-4,

[tex]v_o_4=\frac{v_A}{v_2}\\ v_o_3 =\frac{V_a}{V_2}*\frac{v_2}{v_3}\\v_o_3=\frac{16}{2}*4.546\\v_o_3=36.37;v_4=659.7kJ/kg[/tex]

Q[tex]_o_u_t=v_4-u_1=659.7-214.07=445.3kJ/kg[/tex]

nth = [tex]1-\frac{Q_o_u_t}{Q_i_n}=1-\frac{445.63}{1019.7}=0.5629*100=56.3%[/tex]%

The thermal efficiency is 56.3%

W[tex]_n_e_t[/tex]=[tex]Q_i_n-Q_o_u_t=574.07kJ/kg[/tex]

[tex]v_1=\frac{RT_1}{p_1}=\frac{0.287*300}{95}=0.906m^3/kg\\v_2=v_1/16=0.05662m^3/kg\\[/tex]

Therefore, the mean effective pressure of the system engine is

[tex]\frac{W_n_e_t}{v_1-v_2}=675.87kPa[/tex]

The mean effective pressure is 65.87kPa as calculated above

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Answer:

The pressure drop across the pipe also reduces by half of its initial value if the viscosity of the fluid reduces by half of its original value.

Explanation:

For a fully developed laminar flow in a circular pipe, the flowrate (volumetric) is given by the Hagen-Poiseulle's equation.

Q = π(ΔPR⁴/8μL)

where Q = volumetric flowrate

ΔP = Pressure drop across the pipe

μ = fluid viscosity

L = pipe length

If all the other parameters are kept constant, the pressure drop across the circular pipe is directly proportional to the viscosity of the fluid flowing in the pipe

ΔP = μ(8QL/πR⁴)

ΔP = Kμ

K = (8QL/πR⁴) = constant (for this question)

ΔP = Kμ

K = (ΔP/μ)

So, if the viscosity is halved, the new viscosity (μ₁) will be half of the original viscosity (μ).

μ₁ = (μ/2)

The new pressure drop (ΔP₁) is then

ΔP₁ = Kμ₁ = K(μ/2)

Recall,

K = (ΔP/μ)

ΔP₁ = K(μ/2) = (ΔP/μ) × (μ/2) = (ΔP/2)

Hence, the pressure drop across the pipe also reduces by half of its initial value if the viscosity of the fluid reduces by half of its value.

Hope this Helps!!!

Answer:

The answer is 920 kJ

Explanation:

Solution

Given that:

Mass = 5kg

Pressure = 600 kPa

Temperature = 80° C

Liquid vapor mixture state (quality) = 0.3

Now we find out the amount of heat extracted in the process

Thus

Properties of  RI34a at:

P₁ = 600 kPa

T₁ = 80° C

h₁ = 320 kJ/kg

So,

P₁ = P₂ = 600 kPa

X₂ =0.3

h₂ = 136 kJ/kg

Now

The heat removed Q = m(h₁ -h₂)

Q = 5 (320 - 136)

Q= 5 (184)

Q = 920 kJ

Therefore the amount of heat extracted in the process is 920 kJ

Answer:

Tech A is correct.

Explanation:

Gears are toothed wheels that can be used to transmit power. When two or more gears are in tandem, a gear train is formed.

Gear ratio = [tex]\frac{number of teeth of the driven gear}{Number of teeth of the driving gear}[/tex]

                = [tex]\frac{27}{9}[/tex]

                = [tex]\frac{3}{1}[/tex]

Gear ratio = 3:1

The driver gear is called the input gear since it transfers its power to the driven gear. While the driven gear is called the output gear because it produces an effect due to both gears.

Tech A is correct.

Answer:

The answer to this question can be defined as follows:

Explanation:

The sky driver began his sky journey from Arkansas, drove across the Tennessee River then landed in North Carolina. He returned to both the north in the very same direction. He began with NC, traveled through Tennessee, eventually lands in Arkansas. But North Carolina has been in the third state on which skydiver was traveling over, and It's also more than 700 miles from Arkansas to the NC.

Answer:

hello the answer options are missing here are the options

A)The thickness of the heated region near the plate is increasing

B)The velocities near the plates are increasing

C)The fluid temperature near the plate are increasing

ANSWER : all of the above

Explanation:

Laminar flow  is the flow of a type of fluid across the surface of an object following regular paths and it is unlike a turbulent flow which flows in irregular paths (encountering fluctuations)

For laminar flow over a hot flat plate, the local heat transfer coefficient decreases with distance because :

Answer: (0,0)+ (1,0)= 1 lines upwards( suggesting that this is a line graph not saying it is but as an example) an (1,1) and (0,-1) all make a small square ( as this is a 2 dimensional graph that it has a negative side too,(below the positive side)) i hope this helps and is what you are looking for

Explanation:

The criteria for a guard having to be used on a machine is;

As a safety measure If the operation exposes you to an injury.

When operating a machine, there are possibilities that the operator could be injured or exposed to injury.

Due to the possible safety issues when operating a machine, the Occupational Safety and Health Administration (OSHA) in their 29 code mandated that a safeguard must be put at each machine to ensure that there is adequate safety that prevents or minimizes the risk of getting injured.

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Answer:

Hello your question lacks some values here are the values

T1 = 500⁰c,  T7 = 70⁰c, Qb = 240000 kj/s

answer : A)  56%

               B) 134400 kw ≈  134.4 Mw

Explanation:

Given values

T1 (tmax) = 500⁰c = 773 k

T7(tmin) = 70⁰c = 343 k

Qb = 240000 kj/s

A) Determine the maximum possible efficiency

[tex]n_{max}[/tex] = 1 - [tex]\frac{tmin}{tmax}[/tex] * 100

       = 1 - ( 343 / 773 )

       = 1 - 0.44 = 0.5562 * 100 ≈ 56%

B) Determine the power output for this complex steam power plant design

[tex]p_{out}[/tex] = Qb * max efficiency

      = 240000 kj/s * 56%

      = 240000 * 0.56 = 134400 kw ≈  134.4 Mw

Answer:

  radiative heat loss substantially increases as the wall temperature declines

Explanation:

The body's heat loss due to convection is ...

  (2 W/m^2·K)((32 -20)K) = 24 W/m^2

__

The body's heat loss due to radiation in the summer is ...

  [tex]\epsilon\sigma(T_b^4-T_w^4)\quad\text{where $T_b$ and $T_w$ are body and wall temperatures ($^\circ$K)}\\\\0.90\cdot 5.6703\cdot 10^{-8}(305.15^4-300.15^4)\,\text{W/m$^2$}\\\\\approx 28.3\,\text{W/m$^2$}[/tex]

The corresponding heat loss in the winter is ...

  [tex]0.90\cdot 5.6703\cdot 10^{-8}(305.15^4-287.15^4)\,\text{W/m$^2$}\\\\\approx 95.5\,\text{W/m$^2$}[/tex]

Then the total of body heat losses to surroundings from convection and radiation are ...

  summer: 24 +28.3 = 52.3 . . . W/m^2

  winter: 24 +95.5 = 119.5 . . . W/m^2

__

It is reasonable that a person would feel chilled in the winter due to the additional radiative loss to the walls in the winter time. Total heat loss is more than doubled as the wall temperature declines.

Answer:

No, the velocity profile does not change in the flow direction.

Explanation:

In a fluid flow in a circular pipe, the boundary layer thickness increases in the direction of flow, until it reaches the center of the pipe, and fill the whole pipe. If the density, and other properties of the fluid does not change either by heating or cooling of the pipe, then the velocity profile downstream becomes fully developed, and constant, and does not change in the direction of flow.

Answer:

Explanation:

i) In the attached truth table, TRUE means the respective key owner is present and their keys are inserted at the same time. The "Opens" column is TRUE when two owners are present, not including the case where the only two owners present are B and C.

__

ii) The second attachment is a Karnaugh map of the truth table. The circled terms are ...

  Opens = AB +AC

Answer:

863 K

Explanation:

See the attachment

Answer:

A TVS screen works when the pixels are switched on electronically using liquid crystals to rotate polarized light.

Explanation:

Answer:

zsxdcffffusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernameusernamev

Explanation:

Answer:

This doesn't represent an equilibrium state of stress

Explanation:

∝ = 1 , β = 1 ,  y = 1

x = 0 , y = 0 , z = 0 ( body forces given as 0 )

Attached is the detailed solution is and also the conditions for equilibrium

for a stress state to be equilibrium all three conditions has to meet the equilibrum condition as explained in the attached solution

Answer:

the answer is

Explanation:

Answer:

The critical buckling load is [tex]\mathbf{P_o = 141.61 \ kN}[/tex]

Explanation:

Given that:

the width of the rectangular steel = 37.5 mm = 0.0375 m

the thickness = 50 mm  = 0.05 m

the length = 1.75 m

modulus of elasticity = 200 Gpa = 200 10⁹ × Mpa

We are to calculate the critical buckling load  [tex]P_o[/tex]

Using the formula:

[tex]P_o = \dfrac{\pi ^2 E I}{L^2}[/tex]

where;

[tex]I = \dfrac{0.0375^3*0.05}{12}[/tex]

[tex]I = 2.197 * 10^{-7}[/tex]

[tex]P_o = \dfrac{\pi ^2 *200*10^9 * 2.197*10^{-7}}{1.75^2}[/tex]

[tex]P_o = 141606.66 \ N[/tex]

[tex]\mathbf{P_o = 141.61 \ kN}[/tex]

The critical buckling load is [tex]\mathbf{P_o = 141.61 \ kN}[/tex]

Answer:

- 1.19 lb/ft^3

Explanation:

You are given the following information;

Radius r = 20 ft

Speed V = 100 ft/s

You should use Bernoulli equation pertaining to streamline. That is, normal to streamline.

The pressure gradient = dp/dn

Where air density rho = 0.00238 slugs per cubic foot.

Please find the attached files for the solution and diagram.

Answer:

i) Truth Table:

A      |     B     |     C     |     O

0      |     0     |     0     |      0

0      |     0     |     1      |      0

0      |     1      |     0     |      0

0      |     1      |     1      |      0    (condition 2 not satisfied)

1       |     0     |     0     |      0

1       |     0     |     1      |      1    (both conditions satisfied)

1       |     1      |     0     |      1    (both conditions satisfied)

1       |     1      |     1      |      1    (both conditions satisfied)

ii) The minimized sum of products (SOP) expression is

O = AC + AB

Explanation:

We have three inputs A, B and C

Let O is the output.

We are given two conditions to open the vault door:

1. At  least two people must insert their keys into the assigned locks at the same  time.

2. The trainee bank teller (C) can only open the vault when the bank  manager (A) is present in the opening.

i) Construct the Truth Table

A      |     B     |     C     |     O

0      |     0     |     0     |      0

0      |     0     |     1      |      0

0      |     1      |     0     |      0

0      |     1      |     1      |      0    (condition 2 not satisfied)

1       |     0     |     0     |      0

1       |     0     |     1      |      1    (both conditions satisfied)

1       |     1      |     0     |      1    (both conditions satisfied)

1       |     1      |     1      |      1    (both conditions satisfied)

ii) SOP Expression using Karnaugh-Map:

A 3 variable Karnaugh-map is attached.

The minimized sum of products (SOP) expression is

O = AC + AB

The orange pair corresponds to "AC" and the purple pair corresponds to "AB"

Bonus:

The above expression may be realized by using two AND gates and one OR gate.  

Please refer to the attached logic circuit diagram.