A 3-ft-diameter duct is used to carry ventilating air ( , ) into a vehicular tunnel at a rate of 11000 ft3/min. Tests show that the pressure drop is 1.2 in. of water per 1500 ft of duct. What is (a) the value of the friction factor for this duct and (b) the approximate size of the equivalent roughness of the surface of the duct

This question is incomplete, the complete question as well as the missing diagram is uploaded below;

Consider a mixing tank with a volume of 4 m³. Glycerin flows into a mixing tank through pipe A with an average velocity of 6 m/s, and oil flow into the tank through pipe B at 3 m/s. Determine the average density of the mixture that flows out through the pipe at C. Assume uniform mixing of the fluids occurs within the 4 m³ tank.

Take [tex]p_o[/tex] = 880 kg/m³ and [tex]p_{glycerol[/tex] = 1260 kg/m³    

Answer:

the average density of the mixture that flows out through the pipe at C is 1167.8 kg/m³  

Explanation:

Given that;

Inlet velocity of Glycerin, [tex]V_A[/tex] = 6 m/s

Inlet velocity of oil, [tex]V_B[/tex] = 3 m/s  

Density velocity of glycerin, [tex]p_{glycerol[/tex] = 1260 kg/m³

Density velocity of glycerin, Take [tex]p_o[/tex] = 880 kg/m³

Volume of tank V = 4 m

from the diagram;

Diameter of glycerin pipe, [tex]d_A[/tex] = 100 mm = 0.1 m

Diameter of oil pipe, [tex]d_B[/tex] = 80 mm = 0.08 m

Diameter of outlet pipe [tex]d_C[/tex] = 120 mm = 0.12 m

Now, Appling the discharge flow equation;

[tex]Q_A + Q_B = Q_C[/tex]

[tex]A_Av_A + A_Bv_B = A_Cv_C[/tex]

π/4 × ([tex]d_A[/tex])²[tex]v_A[/tex] + π/4 × ([tex]d_B[/tex] )²[tex]v_B[/tex] = π/4 × ([tex]d_C[/tex])²[tex]v_C[/tex]

we substitute

π/4 × (0.1 )² × 6 + π/4 × (0.08 )² × 3 = π/4 × (0.12)²[tex]v_C[/tex]

0.04712 + 0.0150796 = 0.0113097[tex]v_C[/tex]

0.0621996 = 0.0113097[tex]v_C[/tex]

[tex]v_C[/tex] = 0.0621996 / 0.0113097

[tex]v_C[/tex]  = 5.5 m/s

Now we apply the mass flow rate condition

[tex]m_A + m_B = m_C[/tex]

[tex]p_{glycerin}A_Av_A + p_0A_Bv_B = pA_Cv_C[/tex]  

so we substitute

1260 × π/4 × (0.1 )² × 6 + 880 × π/4 × (0.08 )² × 3 = p × π/4 × (0.12)² × 5.5

1260 × 0.04712 + 880 × 0.0150796 = p × 0.06220335

59.3712 + 13.27 = 0.06220335p  

72.6412 = 0.06220335p    

p = 72.6412 / 0.06220335

p =  1167.8 kg/m³  

Therefore, the average density of the mixture that flows out through the pipe at C is 1167.8 kg/m³  

Explanation:

When a problem is discovered in the system design or manufacturing process, it generally gets reported to all concerned (if the company has an appropriate culture). Depending on the nature of the problem, a "quick fix" may be found by the discoverer, or by someone to whom the discovery is reported. In some situations, what seems a simple problem requires a rethinking of the entire manufacturing process, possible product recalls, possible plant retrofits, and even larger ramifications. This can happen regardless of where along the line the problem is discovered.

__

Hypothetical example:

A test technician determines that a lithium battery charger sometimes gets confused and doesn't shut down properly--continuing to charge the battery when it should not. If the proximate cause is a wire harness routing or a component improperly installed, a manufacturing engineer may be called in to address the issue. The manufacturing engineer's investigation may determine that a number of battery chargers have been delivered with the problem. Further investigation may reveal a potential for fire in situations where injury or loss of life are possible outcomes.

Assessment of the design by manufacturing, process, and design engineers may reveal more than one potential cause of the shutdown/fire-hazard issue, and that the location and nature of any fire may release toxins or cause damage to systems and equipment beyond those in the immediate vicinity of the defective battery and/or charger.

Such ramifications may require the attention of one or more systems engineers and/or a rethinking of system failure modes and effects, including fire detection and suppression, throughout the product. The product may be effectively "grounded" (taken out of service), with possible customer revenue implications, until such time as the issues can be satisfactorily resolved.

(Note: 787 Dreamliner battery problems were caused by the physics of the battery construction, not the charger. The rest of the scenario above is not a bad match.)

Answer:

[tex]\mathbf{C_{10} = 137.611 \ kN}[/tex]

Explanation:

From the information given:

Life requirement = 40 kh = 40 [tex]40 \times 10^{3} \ h[/tex]

Speed (N) = 520 rev/min

Reliability goal [tex](R_D)[/tex] = 0.9

Radial load [tex](F_D)[/tex] = 2600 lbf

To find C10 value by using the formula:

[tex]C_{10}=F_D\times \pmatrix \dfrac{x_D}{x_o +(\theta-x_o) \bigg(In(\dfrac{1}{R_o}) \bigg)^{\dfrac{1}{b}}} \end {pmatrix} ^{^{^{\dfrac{1}{a}}[/tex]

where;

[tex]x_D = \text{bearing life in million revolution} \\ \\ x_D = \dfrac{60 \times L_h \times N}{10^6} \\ \\ x_D = \dfrac{60 \times 40 \times 10^3 \times 520}{10^6}\\ \\ x_D = 1248 \text{ million revolutions}[/tex]

[tex]\text{The cyclindrical roller bearing (a)}= \dfrac{10}{3}[/tex]

The Weibull parameters include:

[tex]x_o = 0.02[/tex]

[tex](\theta - x_o) = 4.439[/tex]

[tex]b= 1.483[/tex]

∴

Using the above formula:

[tex]C_{10}=1.4\times 2600 \times \pmatrix \dfrac{1248}{0.02+(4.439) \bigg(In(\dfrac{1}{0.9}) \bigg)^{\dfrac{1}{1.483}}} \end {pmatrix} ^{^{^{\dfrac{1}{\dfrac{10}{3}}}[/tex]

[tex]C_{10}=3640 \times \pmatrix \dfrac{1248}{0.02+(4.439) \bigg(In(\dfrac{1}{0.9}) \bigg)^{\dfrac{1}{1.483}}} \end {pmatrix} ^{^{^{\dfrac{3}{10}}[/tex]

[tex]C_{10} = 3640 \times \bigg[\dfrac{1248}{0.9933481582}\bigg]^{\dfrac{3}{10}}[/tex]

[tex]C_{10} = 30962.449 \ lbf[/tex]

Recall that:

1 kN = 225 lbf

∴

[tex]C_{10} = \dfrac{30962.449}{225}[/tex]

[tex]\mathbf{C_{10} = 137.611 \ kN}[/tex]

Answer:

(a) 6.36 mg/L

(b) 5.60 mg/L

Explanation:

(a)

Using the formula below to find the required ultimate BOD of the mixture.

[tex]L_o = \dfrac{Q_wL_w+ Q_rL_r}{Q_W+Q_r}[/tex]

where;

Q_w = volumetric flow rate wastewater

Q_r = volumetric flow rate of the river just upstream of the discharge point

L_w = ultimate BOD of wastewater

Replacing the given values:

[tex]L_o = \dfrac{(1 \ m^3/L ) (40 \ mg/L) + (10 \ m^3/L) (3 \mg/L)}{(1m^3/L) +(10 \ m^3/L)} \\ \\ L_o = 6.36 \ mg/L[/tex]

(b)

The Ultimate BOD is estimated as follows:

Recall that:

[tex]time(t) = \dfrac{distance }{speed}[/tex]

replacing;

distance with 10000 m and speed with [tex]\dfrac{11 \ m^3/s}{55 \ m^2}[/tex]

[tex]time =\dfrac{10000 \ m}{\Bigg(\dfrac{11 \ m^3/s}{55 \ m^2}\Bigg)}\Bigg(\dfrac{1 \ hr}{3600 \ s}\Bigg) \Bigg(\dfrac{1 \ day}{24 hr}\Bigg)[/tex]

time (t) = 0.578 days

Finally; [tex]L_t = L_oe^{-kt}[/tex]

here;

k = rate of coefficient reaction

[tex]L_ t= (6.36) \times e^{-(0.22/day)(0.5758 \ days)}\\ \\ \mathbf{L_t =5.60 \ mg/L}[/tex]

Thus, the ultimate BOD = 5.60 mg/L

Answer:

the required time for the specimen is  1109.4 min

Explanation:

Given that;

diameter of metal alloy d₀ = 1.7 × 10² mm

Temperature of heat treatment T = 450°C = 450 + 273 = 723 K

Time period of heat treatment t = 250 min

Increased grain diameter 4.5 × 10² mm

grain diameter exponent n = 2.1

First we calculate the time independent constant K

dⁿ - d₀ⁿ = Kt

K = (dⁿ - d₀ⁿ) / t

we substitute

K = (( 4.5 × 10² )²'¹ - ( 1.7 × 10² )²'¹) / 250

K = (373032.163378 - 48299.511117) / 250

K = 1298.9306 mm²/min

Now, we calculate the time required for the specimen to achieve the given grain diameter ( 8.7 × 10² mm )

dⁿ - d₀ⁿ = Kt

t = (dⁿ - d₀ⁿ) / K

t = (( 8.7 × 10² )²'¹ - ( 1.7 × 10² )²'¹) / 1298.9306

t = ( 1489328.26061158 - 48299.511117) / 1298.9306

t = 1441028.74949458 / 1298.9306

t = 1109.4 min

Therefore, the required time for the specimen is  1109.4 min

Answer:

the disadvantages can be people's thoughts about them and also a rival resort nearby which looks more posh pls mark brainliest i need 5 more for next rank thank you

Explanation:

Answer:

What is the rms value Vrms of the voltage plotted in the graph?

Express your answer in volts.

Vrms = ? V

2.

When a lamp is connected to a wall plug, theresulting circuit can be represented by a simplified AC circuit, asshown in the figure. (Part B figure) Here the lamp has been replaced by a resistor with an equivalentresistance Part B figure) Here the lamp has been replaced bya resistor with an equivalent resistance R = 120 . What is the rms value Irms of the current flowing through the circuit?

Express your answer in amperes.

Irms = ? A

3.

What is the average power Pavg dissipated in the resistor?

Express your answer in watts.

Pavg = ? W

Technicians A is correct in the given scenario. The correct option is C.

Exhaust gas is a byproduct of combustion that exits the tailpipe of an internal combustion engine.

It consists of a gas mixture that includes carbon dioxide (CO2), carbon monoxide (CO), nitrogen oxides (NOx), hydrocarbons (HC), and particulate matter (PM).

Technician B is mistaken. CO2 levels in the exhaust should be less than 15%, preferably between 13% and 14.5% for petrol engines and 11% to 13% for diesel engines.

High CO2 levels can actually indicate inefficient engine operation, as it means that not all of the fuel in the engine is being burned and is being wasted as exhaust.

Thus, C is the correct answer. A technician is correct.

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Answer:

False ( B )

Explanation:

considering that the wind turbine is a horizontal axis turbine

Power generated/extracted by the turbine can be calculated as

P =  n * 1/2 * p *Av^3

where: n = turbine efficiency

           p = air density

           A = πd^2 / 4

           v =  speed

From the above equation it can seen that increasing the Blade radius by 10% will increase the Blade Area which will in turn increase the value of the power extracted by the wind turbine

Answer:

Wind energy is converted to Mechanical energy  which is then converted in to  electrical energy

Explanation:

In a wind mill the following energy conversions take place

a) Wind energy is converted into Mechanical energy (rotation of rotor blades)

b) Mechanical energy is converted into electrical energy (by using electric motor)

This electrical energy is then used for transmission through electric lines.

Answer:

3.42 kW

Explanation:

calculate the mass flow rate of the water = density * velocity * area

= 13.3 kg/s

where Area of pipe  =  π/4 * d^2 =  π/4 * 0.075^2 m  =  0.0044 m^2

given that : 1 Kg = 1 N s^2 / m,      1 Watt = 1 N m /s

calculate : power produced by discharge pipe

Discharge Pressure x Volume flow =

Pressure  * Area * velocity

P  * 0.0044 * 3  = 2253 N m /s

calculate: power due the mass of water  

mass of water * 2  = 258 N m/s

where:  mass of water = density * volume

Calculate power produced due the velocity of exiting water

= m * V2/2 = 59.4 N m/s

hence The  power added to fluid = 2253 watts

power added to the fluid by pump = 2253 / 0.75 = 3.42 kW

Solution :

Given :

Velocity, [tex]$V_1 =100$[/tex] cm/sec

Pressure,  [tex]$P_1 = 120 $[/tex] mm Hg

Then, [tex]$P_1 = \rho_1 g h$[/tex]

[tex]$P_1 = 0.120 \times 13.6 \times 1000 \times 9.81$[/tex]

    = 16.0092 kPa

[tex]$P_2 = 110 $[/tex] mm Hg

[tex]$P_2 = \rho_2 g h$[/tex]

    [tex]$= 0.110 \times 13.6 \times 1000 \times 9.81$[/tex]

    = 14.675 kPa

Then blood is incompressible,

[tex]$A_1v_1=A_2v_2$[/tex]

[tex]$\frac{\pi}{4}(25)^2\times 100=\frac{\pi}{4}(21)^2\times v_2$[/tex]

[tex]$v_2=141.72 \ cm/s$[/tex]

Then the linear momentum conservation fluid :

(Blood ) in y - direction

[tex]$P_1A_1+ P_2A_2-F_g = m_2v_2-m_1v_1$[/tex]

[tex]$m_1=m_2=P_1A_1v_1$[/tex]

              [tex]$=1.50 \times \frac{\pi}{4}\times (0.025)^2 \times 1.00$[/tex]

              = 0.515 kg/ sec

Then the linear conservation of momentum of blood in y direction.

[tex]$P_1A_1+ P_2A_2-F_g = m_2v_2-m_1v_1$[/tex]

[tex]$16.0092 \times 1000 \times \frac{\pi}{4} \times (0.025)^2+14.675 \times 1000\times \frac{\pi}{4}\times (0.021)^2$[/tex]

[tex]$-F_y=0.515(-1.4172-1)$[/tex]

7.858+5.0828- Fy = 0.515(-2.4172)

Fy = 14.1856 N

Here are the eight stages of the recruitment process and what Garth might expect to occur or carry out at each stage in the explanation part.

The process of identifying, attracting, and selecting qualified candidates for a job opening in an organisation is known as recruitment.

Here are the eight stages of the recruitment process, as well as what Garth might expect to happen or do at each stage:

Thus, these are the stages of recruitment.

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Answer:

A

Explanation:

Actually I don't know anything about American history, I chose it because South Africa is not in the least

Answer:

what?????

Explanation:

Answer:

In C++:

#include <iostream>

#include <cmath>

using namespace std;

int main(){

 double total = 0;

 for(int i = -170; i<=-130;i++){

     if(i%2 == 0 && i%7==0){

         double angle = i*3.14159/180;

         total+=sin(angle);      }  }

 cout<<"Total: "<<total;

 return 0; }

Explanation:

This initializes the total to 0

 double total = 0;

This iterates from -170 to - 130

 for(int i = -170; i<=-130;i++){

This checks if the current number is even and is divided by 7

     if(i%2 == 0 && i%7==0){

This converts the number from degrees to radians

         double angle = i*3.14159/180;

This adds the sine of the number

         total+=sin(angle);      }  }

This prints the calculated total

 cout<<"Total: "<<total;