Air at 30 C, 1 bar, 50% relative humidity enters an insulated chamber operating at steady state with a mass flow rate of 3 kg/min and mixes with a saturated moist air stream entering at 5 C, 1 bar with a mass flow rate of 5 kg/min. A single mixed stream exits at 1 bar. Determine (a) the relative humidity and temperature, in C, of the exiting stream. (b) the rate of exergy destruction, in kW, for T0

GIS

The full for GIS is : Geographic Information System.

The GIS is a system of mapping that is used to create, analyze, manage and map all the types of data.

It is a method capturing, checking as well as displaying all the data that is related to the position of the surface of earth.

Remote Sensing

Remote sensing is defined as a process where the various physical characteristics of an area are constantly monitored and detected by a process of measuring the reflection of the emitted radiations at ta distance.

The snow covered areas of the Kabul basin can be monitored and analyzed by using the process of GIS and remote sensing.

They provide a continuous source of information and data about snow covered peaks, the amount of snow to the authorities.

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Answer:

3.983 MW

Explanation:

Given that:

At the inlet:

Velocity (v₁) = 30 m/s

Enthalpy (h₁) = 3348 kJ/kg

At the outlet:

Velocity (v₂) = 60 m/s

Enthalpy (h₂) = 2550 kJ/kg

Mass flow rate (m) = m₁ = m₂ = 5kg/s

According to the steady flow energy equation:

[tex]Q+ m_1 (h_1 + \dfrac{v_1^2}{2000}+ \dfrac{gz_1}{1000} )= m_2(h_2+\dfrac{v_2^2}{2000}+\dfrac{gz_2}{1000})+W_{shaft}[/tex]

Since the elevation (z) is negligible and flow via the turbine is adiabatic:

Then,

Q = 0 and z₁ = z₂

∴

[tex]W_{shaft} = (mh_1-mh_2) + (\dfrac{mv_1^2-mv_2^2}{2000})[/tex]

[tex]W_{shaft} = ((5*3348) -(5*2550)) + (\dfrac{(5*(30)^2)-(5*(60)^2)}{2000})[/tex]

[tex]W_{shaft} = (16740-12750) + (\dfrac{4500-18000}{2000})[/tex]

[tex]W_{shaft} = (16740-12750) + (-6.75)[/tex]

[tex]W_{shaft} = 3983.25 \ kW[/tex]

[tex]\mathbf{W_{shaft} = 3.983 \ MW}[/tex]

Answer:

efficiency

Explanation:

Answer:

The correct answer is Efficiency.

Explanation:

I got it right on the plato test.

Answer:

I don't know

............................

The output will be an error message since the input() function in Python always takes input as a string, even if the entered value is an integer like 12.

Here,

It cannot directly concatenate a string with an integer.

In order to fix this, we need to convert the input string into an integer using the int() function,

```

x = int(input())

x = x + "abc" # this will give an error message

```If we fix the code, the correct output will be:

`>>> x = int(input())`

`12`

`>>> x + "abc"`

`TypeError: unsupported operand type(s) for +: 'int' and 'str'`

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Answer:

b

Explanation:

what is anythin efficiency if ta

Answer:

Explanation:

f = 50.0 Hz, L = 0.650 H, π = 3.14

C = 4.80 μF, R = 301 Ω resistor. V = 120volts

XL = wL = 2πfL

= 2×3.14×50* 0.650

= 204.1 Ohm

Xc= 1/wC

Xc = 1/2πfC

Xc = 1/2×3.14×50×4.80μF

= 1/0.0015072

= 663.48Ohms

1. Total impedance, Z = sqrt (R^2 + (Xc-XL)^2)= √ 301^2+ (663.48Ohms - 204.1 Ohm)^2

√ 90601 + (459.38)^2

√ 90601+211029.98

√ 301630.9844

= 549.209

Z = 549.21Ohms

2. I=V/Z = 120/ 549.21Ohms =0.218Ampere

3. P=V×I = 120* 0.218 = 26.16Watt

Note that

I rms = Vrms/Xc

= 120/663.48Ohms

= 0.18086A

4. I(max) = I(rms) × √2

= 0.18086A × 1.4142

= 0.2557

= 0.256A

5. V=I(max) * XL

= 0.256A ×204.1

=52.2496

= 52.250volts

6. V=I(max) × Xc

= 0.256A × 663.48Ohms

= 169.85volts

7. Xc=XL

1/2πfC = 2πfL

1/2πfC = 2πf× 0.650

1/2×3.14×f×4.80μF = 2×3.14×f×0.650

1/6.28×f×4.8×10^-6 = 4.082f

1/0.000030144× f = 4.082×f

1 = 0.000030144×f×4.082×f

1 = 0.000123f^2

f^2 = 1/0.000123048

f^2 = 8126.922

f =√8126.922

f = 90.14 Hz

Answer:

True

Explanation:

There are 16 in all and 4 are not mandatory

There are some sections of the SDS that are not mandatory. Thus, option A is correct.

A Safety Data Sheet, previously known as a Material Safety Data Sheet, is a thorough instructional report written by a dangerous product's producer or exporter. It gives a description of something like the product's chemical and physical characteristics.

Sections 1 through 11 and 16 must be used; sections 12 through 15 are optional but could be included. the section which will are not compulsory are:

Section 12: Environmental data

Section 13: Guidelines for Recycling

Section 14: Documentation regarding mobility

Section 15: Relevant data about regulations

In Safety Data Sheet, all the sections are not compulsory or mandatory. this statement is true. Therefore, option A is the correct option.

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Answer:

Explanation:

The only vertical force that can be applied at joint D is that of link CD. Since joint D is stationary, there must be no vertical force. Hence the force in link CD must be zero, as must the force in link DE.

At joint E, the only horizontal force is that applied by link EF, so it, too, must be zero.

Then link CE has 15 N tension.

The downward force in CE must be balanced by an upward force in CF. Of that force, only 1/√2 of it will be vertical, so the force in CF is a compression of 15√2 N.

In order for the horizontal forces at C to be balanced the 15 N horizontal compression in CF must be balanced by a 15 N tension in BC.

At joint F, the 15 N horizontal compression in CF must be balanced by a 15 N compression in FA. CF contributes a downward force of 15 N at joint F. Together with the external load of 10 N, the total downward force at F is 25 N. Then the tension in BF must be 25 N to balance that.

At joint B, the 25 N downward vertical force in BF must be balanced by the vertical component of the compressive force in AB. That component is 2/√5 of the total force in AB, which must be a compression of 25√5/2 N.

The horizontal forces at joint B include the 15 N tension in BC and the 25/2 N compression in AB. These are balanced by a (25/2+15) N = 55/2 N tension in BG.

In summary, the link forces are ...

_____

Note that the forces at the pins of G and A are in accordance with those that give a net torque about those point of 0, serving as a check on the above calculations.

Answer:

1. Yes, they are all necessary.

2. Both written and verbal communication skills are of the utmost importance in business, especially in engineering. Communication skills boost you or your teams' performance because they provide clear information and expectations to help manage and deliver excellent work.

Answer:

thread depth = 2.5 mm

thread width = 2.5 mm

mean diameter = 27.5 mm

root diameter = 25 mm

lead of screw = 5 mm

Explanation:

given data

power screw diameter D = 30 mm

thread pitch  P = 5 mm

solution

First, we get here thread depth fr square thread

thread depth = [tex]\frac{P}{2}[/tex]   ......................1

thread depth = [tex]\frac{5}{2}[/tex]

thread depth = 2.5 mm

and

thread width for square thread

thread width = [tex]\frac{P}{2}[/tex]   ......................2

thread width = [tex]\frac{5}{2}[/tex]

thread width = 2.5 mm

and

mean diameter is

mean diameter = D - [tex]\frac{P}{2}[/tex]    ................3

mean diameter = 30 - [tex]\frac{5}{2}[/tex]

mean diameter = 27.5 mm

and

root diameter is

root diameter = D - P   ....................4

root diameter = 30 - 5

root diameter = 25 mm

and

lead of screw for single thread so n = 1

so lead of screw = 1 × 5

lead of screw = 5 mm

Answer:

30 feet

Explanation:

Given data :

design speed = 30 miles/h

super elevation = 0.08

determine the width of the turning roadway

calculate the value of R = V^2 / 15( e + p)

e = 0.08 , p = 0.2 , v = 30

R = (30)^2 / 15 ( 0.08 + 0.2 )

  = 900 / 15 ( 0.28 )

  ≈ 215 ft

pavement width from the calculation above = 28 ft

width of the turning roadway = pavement width + 2 = 30 feet ( because there are two vertical widths joining up the main road at the T junction )

I inferred you want literal editing of the text above.

Explanation:

Here's a correction of the sentences:

1. The author, an expert in cybersecurity will speak via Zoom on Wednesday.

In this sentence, punctuation mark ( , ) was added and the word  'this' was replaced with 'on'.

2. Williams' book contains many illustrations, which makes it easy to read.

Added punctuation and made a revision of the sentence.

3. Based on the available evidence, the university administrators have opted for a hybrid format for the fall, which begins September 20.

Mainly added punctuations to make the senstence clarer.

4. (Thesis statement) I believe Free laptops should be offered to all students who need them.

Made a few additions.

Answer: answer provided in the explanation section.

Explanation:

Weather phenomenons that would impart Aviation Operations in Santa Barbara -

1. Although winters are cold, wet, and partly cloudy here. It is in general favorable for flying. But sometimes strong winds damage this pleasant weather.

2.  The Sundowner winds cause rapid warming and a decrease in relative humidity. The wind speed is very high surrounding this area for this type of wind.  

3. Cloud is an important factor that affects aviation operations. Starting from April, here the sky is clouded up to November. The sky is overcast (80 to 100 percent cloud cover) or mostly cloudy (60 to 80 percent) 44% on a yearly basis. Thus extra cloud cover can trouble aviation operations.

4. The average hourly wind speed can also be a factor. This also experiences seasonal variations, these variations are studied carefully in the aviation industry. The windier part of the year starts in January and ends in June. In April, the wind speed can reach 9.5 miles per hour.

This and more are some factors to look into when considering wheather conditions that would affect aviation operations.

I hope this was a bit helpful. cheers

Answer: hello some aspects of your question is missing below is the missing information

The gas tank is made from A-36 steel and has an inner diameter of 1.50 m.

answer:

≈ 22.5 mm

Explanation:

Given data:

Inner diameter = 1.5 m

pressure = 5 MPa

factor of safety = 1.5

Calculate the required minimum wall thickness

maximum-shear-stress theory ( σ allow ) = σγ / FS

                                                  = 250(10)^6 / 1.5  = 166.67 (10^6) Pa

given that |σ| = σ allow  

3.75 (10^6) / t = 166.67 (10^6)

∴ t ( wall thickness ) = 0.0225 m   ≈ 22.5 mm

Answer:

Hello bro

Explanation:

I think i can help you something but can you translate it on english plzz

Answer:

business

Explanation:

because shopping has great income

Answer:

is ur question right

Explanation:

is ur question right

is ur question right

The amount of heat rejected during the process in kJ/kg is 2133.4 kJ/kg.

Heat is defined as the process of kinetic energy being transferred from an energy source to a medium or an item. These three mechanisms radiation, conduction, and convection can all transmit energy. These three modes of heat transfer are typically combined, and it often happens on its own. The heat energy can be transformed into other types of energy, such as mechanical and kinetic energy. Thermal (heat) energy is a type of energy that is produced by the movement of particles and is a component of all matter.

The process rejects a certain amount of heat, which is

-q = w + Δu

q = - w - Δu

q = - 185.5 kJ/kg - 1948.9 kJ/kg

q = 2133.4 kJ/kg

Thus, the amount of heat rejected during the process in kJ/kg is 2133.4 kJ/kg.

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Answer:

b. What is the operator's efficiency for the day?

                                                      AND

e. What would be the proper piece rate (rate expressed in money) for this job, assuming that the above time standard is correct?

Explanation:

Answer:

The answers to your questions are given below.

Explanation:

The following data were obtained from the question:

Red (R) = 4

White (W) = 7

1. Determination of the sample space, S.

The box contains 4 red balls and 7 white balls. Therefore, the sample space (S) can be written as follow:

S = {R, R, R, R, W, W, W, W, W, W, W}

nS = 11

2. Determination of the probability of all results that appeared in the sample space.

From the question, we were told that the two balls was drawn with return. There, the probability of all results that appeared in the sample space can be given as follow:

i. Probability that the first draw is red and the second is also red.

P(R1) = nR/nS

Red (R) = 4

Space space (S) = 11

P(R1) = nR/nS

P(R1) = 4/11

P(R2) = nR/nS

P(R2) = 4/11

P(R1R2) = P(R1) x P(R2)

P(R1R2) = 4/11 x 4/11

P(R1R2) = 16/121

Therefore, the Probability that the first draw is red and the second is also red is 16/121.

ii. Probability that the first draw is red and the second is white.

Red (R) = 4

White (W) = 7

Space space (S) = 11

P(R) = nR/nS

P(R) = 4/11

P(W) = nW/nS

P(W) = 7/11

P(RW) = P(R) x P(W)

P(RW) = 4/11 x 7/11

P(RW) = 28/121

Therefore, the probability that the first draw is red and the second is white is 28/121.

iii. Probability that the first draw is white and the second is also white.

White (W) = 7

Space space (S) = 11

P(W1) = nW/nS

P(W1) = 7/11

P(W2) = nW/n/S

P(W2) = 7/11

P(W1W2) = P(W1) x P(W2)

P(W1W2) = 7/11 x 7/11

P(W1W2) = 49/121

Therefore, the probability that the first draw is white and the second is also white is 49/121.

iv. Probability that the first draw is white and the second is red.

Red (R) = 4

White (W) = 7

Space space (S) = 11

P(W) = nW/nS

P(W) = 7/11

P(R) = nR/nS

P(R) = 4/11

P(WR) = P(W) x P(R)

P(WR) = 7/11 x 4/11

P(WR) = 28/121

Therefore, the probability that the first draw is white and the second is red is 28/121.

Answer:

b. abc

Explanation:

The phase rotation and its sequence is order in which voltage is reached to their respective sources. The wave forms of a polyphase flow through AC source to the panel. The three phase system has two possible sequence which can be a-b-c or c-b-a. The correct answer is therefore b.

Answer:

D.  Resistance = 30 ohms

Explanation:

Using Ohm's law

V = I times R

Given:

V = 60 V

I = 2 A

Resistance = V / I = 60 V / 20 A

Resistance = 30 ohms

Answer:

[tex]w_{out}=319.1\frac{BTU}{lbm}[/tex]

Explanation:

Hello,

In this case, for the inlet stream, from the steam table, the specific enthalpy and entropy are:

[tex]h_1=1456.0\frac{BTU}{lbm} \ \ \ s_1=1.6413\frac{BTU}{lbm*R}[/tex]

Next, for the liquid-vapor mixture at the outlet stream we need to compute its quality by taking into account that since the turbine is adiabatic, the entropy remains the same:

[tex]s_2=s_1[/tex]

Thus, the liquid and liquid-vapor entropies are included to compute the quality:

[tex]x_2=\frac{s_2-s_f}{s_{fg}}=\frac{1.6313-0.39213}{1.28448}=0.965[/tex]

Next, we compute the outlet enthalpy by considering the liquid and liquid-vapor enthalpies:

[tex]h_2=h_f+x_2h_f_g=236.14+0.965*933.69=1136.9\frac{BTU}{lbm}[/tex]

Then, by using the first law of thermodynamics, the maximum specific work is computed via:

[tex]h_1=w_{out}+h_2\\\\w_{out}=h_1-h_2=1456.0\frac{BTU}{lbm}-1136.9\frac{BTU}{lbm}\\\\w_{out}=319.1\frac{BTU}{lbm}[/tex]

Best regards.

Answer:

uhhhhhhh lemme think

Explanation:

Answer:

A) Flow depth = 2.46 m, Froude number after jump = 0.464

B) head loss = 0.572 m

C) dissipation ratio = 0.173

Explanation:

Given data :

Velocity before jump ( v1 ) = 7 m/s

flow depth before jump ( y1 ) = 0.8 m

g = 9.81 m/s

Esi = 3.3 m ( calculated )

attached below is a detailed solution of the problem

Answer:

C) Decreases with an increase in temperature

Explanation:

As the temperature of a transistor increases, the thermal runaway property of the transistor becomes more significant and the transistors, conducting more freely as a result of the rise in temperature, causes an increase in the collector current or leakage current. The transistor base-emitter voltage decreases as a result.

With increased heating due to heavy current flow, the transistor is damaged.

Answer:

  Put a 10.0-ohm resistor in the circuit. Measure the current in the circuit. Replace the 10.0-ohm resistor with a 20.0-ohm resistor. Measure the new current. Continue replacing the resistor with a different resistor of known resistance. Measure the current for each resistor. Record all data.

Explanation:

The only design that has resistance varying with everything else remaining the same is the first design. That would be what you'd want to do if you're exploring the effect of resistance on current.

Answer:

"Process capability" is the correct answer.

Explanation:

Answer:

the rate of heat loss from the duct to the attic space = 1315.44 W

the pressure difference between the inlet and outlet sections of the duct  = 7.0045 N/m²

Explanation:

We know that properties of air 80⁰C  and 1atm  (from appendix table) are;

density p = 0.9994 kg/m³, Specifice heat Cp = 1008 J/kg.⁰C

Thermal conductivity k = 0.02953 W/m.⁰C, Prandtl number Pr = 0.7154,

Kinematic viscosity v = 2.097 × 10⁻⁵ m²/s

haven gotten that, we calculate the hydraulic diameter of square duct

Dh = 4Ac / P      { Ac = is cross sectional area of duct and P = perimeter}

now we substitute a² for Ac and 4a for P ( we know from the question that a = 0.2 m)

Dh = 4a² / 4a

Dh = 4(0.2)² / 4(0.2)

Dh = 0.2 m

Now we calculate the average velocity of air

Vₐ = Vˣ / Ac        { vˣ = volume flow rate of air}

Vₐ = Vˣ / a²      { Ac = a² }, we know that a = 0.2m₂, Vˣ = 0.15 m³

Vₐ = 0.15 / (0.2)²

Vₐ = 3.75 m/s

Next we calculate the Reynolds number

Re = Vₐ Dh / V

Re = (3.75 × 0.2) / 2.097× 10⁻⁵

Re = 35765.379

The  Reynolds number IS GREATER than 10,000

so the flow is turbulent and entry length in this case is nearly 10 times the hydraulic diameter

Lh ≈ Lt ≈ 10D

= 10 × 0.2

= 2m

As this length is quite small when compared to the total of tube, we assume fully developed flow for the entire tube length.

Now we calculate the Nusselt number from this relation;

Nu =  0.023 Re⁰'⁸ Pr⁰'³

so we substitute for Re and Pr

Nu = 0.023(35765.379)⁰'⁸ (0.7154)⁰'³

Nu = 91.4

Now calculate the convective heat transfer coefficient

h = Nu × K/ Dh

we substitute

h = 91.4 × 0.02953 W/m.°C / 0.2 m

h = 13.5 W/m².°C

We calculate the surface area of the square duct

Aₓ = 4aL       { L= length of duct}

we substitute

Aₓ = 4 × 0.2 × 8

Aₓ = 6.4 m²

Mass flow rate of air

m = pVˣ

we substitute again ( from our initials)

m = 0.9994 kg/m₃ × 0.15 m³/s

m= 0.150 kg/s

We calculate the exit temperature of the air from the duct

Te = Ts - (Ts -Ti) exp ( - hAₓ / mCp)

we know that

Ts = 60°C , Ti = 80°C, h = 13.5 W/m².°C , Aₓ = 6.4m², m = 0.150 kg/s , Cp = 1008 J/kg.°C

we substitute

Te = 60 - (60-80) exp(- ((13.5 × 6.4)/(0.15 × 1008))

Te = 71.3°

Now we calculate the rate of heat loss from the duct.

Q = mCp ( Ti -Te )

we substitute again

Q = 0.150 × 1008 × ( 80 - 71.3 )

Q = 1315.44 W

Next we calculate the estimated friction factors by using Haaland equation

1/√f = - 1.8log₁₀ [ 6.9/Re + (E/D)/3.7)¹'¹¹]

we know that E/D = relative roughness = 10⁻³

we substitute

so

1/√f = - 1.8log₁₀ [ (6.9/35765.379) + ( 10⁻³/3.7)¹'¹¹]

1/√f = - 1.8log₁₀ { 0.000192924 + 0.00010947}

1/√f = - 1.8log₁₀ 0.000302324  

√f =   1/6.334

f = (1/6.334)²

f = 0.02492

We calculate the pressure difference between inlet and outlet sections of the duct

ΔPl = fLPVa² / Dh × 2

ΔPl = {0.02492 × 8 × 0.9994 × (3.75)²} / 0.2 × 2

ΔPl = 2.8018 / 0.4

ΔPl = 7.0045 N/m²

Therefore pressure deference is 7.0045 N/m²

Answer:both

Explanation:

Answer: all of the above

Explanation:

To drive safely in the station, the correct option is D. All of the above.

It should be noted that when a person is driving, it's essential for the person to drive carefully in a station.

One should not smoke in a station. Also, it's important that the driver should turn on their hazard lights

Furthermore, it's important for one to roll their windows down, turn off the radio and also maintain idle speed.

In conclusion, the correct option is D.

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