A thin-walled pressure vessel is constructed by rolling a 6 mm thick steel sheet into a cylindrical shape, welding the seam along line A-B, and capping the ends. The vessel is subjected to an internal pressure of 1.25 MPa. What is the normal stress normal to line A-B?

Answer: Provided in the explanation section

Explanation:

A precise explanation is provided here to make it easy for understanding

(a). The transfer option shows how a filter affects the amplitude and phase of input components as a function of frequency. It is defined as the output phasor divided by the input phasor as a functoon of frequency.

H(jw) = V₀ (jw) / Vin(jw)

(b).  To determine the transfunction of a filter, we connect a sinusoidal source to be the input part, measure the amplitude and phases of both the input signal and resulting output signal using voltmeters.

Oscilloscope or other instruments, and divide the Output phasor by the input phasor. This is is repeated for all frequencies.

cheers i hope this helped !!

Answer:

Continuous insulation helps in eliminating the  necessity of applying extra materials to achieve moisture barrier demands, reducing labor and material cost.

Continuous insulation helps building last much longer over a period of time,without the need to upgrade and repair.

Inside the placement of continuous insulation in the envelope needed includes the following; the foundation wall or slab insulation, balcony interface, bond joist insulation, insulation against sub floor, on the interior of masonry wall.

Explanation:

Solution:

The analytical path to success contains a well planned and designed building exterior, which avoids the loss of energy, control cost and the maximization of technology advancement in materials.

Property installed continuous insulation on the exterior can also execute as an air barrier. the flashing of wall penetrations can form into a drainage plane. this plane can stop potentially damaging moisture from entering into the wall assembly.

By making use of continuous insulation it helps in removing necessity of applying extra materials to achieve moisture barrier demands, reducing labor and material cost.

Continuous insulation helps building to withstand the test of time without the need to upgrade and repair

Now inside the placement of continuous insulation in the envelope is needed as follows:

Answer:

No. of Machines Required = 47

Explanation:

First, we calculate the number of parts that can be manufactured by one machine during the shift. For that purpose, we use following formula:

No. of Parts Manufactured by 1 Machine = Total Operating Time/Time taken to perform Operation

where,

Total Operating Time = 8 h - 1 h = (7 h)(60 min/h) = 420 min

Time taken to perform operation = 20 min

Therefore,

No. of Parts Manufactured by 1 Machine = 420 min/20 min

No. of Parts Manufactured by 1 Machine = 21

Now, we will calculate the no. of non-defective parts manufactured by 1 machine. Since, it is given that one-fifth of the parts manufactured by machine are defective. Therefore, the non-defective parts will be: 1 - 1/5 = 4/5 (four-fifth).

No. of Non- Defective Parts Manufactured by 1 Machine = N = (4/5)(21)

N = 16.8 = 16 (Since, the 17th part will not be able to complete in time)

So, the no. of machines required to produce 750 non-defective parts is given by:

No. of Machines Required = No. of non-defective parts required/N

No. of Machines Required = 750/16

No. of Machines Required = 46.9

No. of Machines Required = 47 (Since, 46 machines will not be able complete the job)

Answer:

The answer is 0.06027 m³

Explanation:

Solution

Given that:

The first step to take is to determine the initial state of the volume for R-134a refrigerant

Now,

v₁ =V/m

V = the volume of weighted piston cylinder device at the normal state

m = the mass of the R-134a refrigerant

Thus,

We substitute the values 0.14 for V and 0.2 kg for m

Which results in

v₁ = 0.14/0.2

v₁  = 0.7 m₃/kg

The next step is to find the saturated pressure of the R-134a refrigerant from the temperature table of saturated refrigerant R-134a which is equivalent to the normal temperature of 26.4°C.

Thus, by applying the method of interpolation we have the following

P₁ =101.73 - ((101.73-92.76) * (-26-(-26.4)/-26- (-28))

P₁ = 99.936 kPa

So,

The refrigerant in the weighted piston–cylinder device is then heated until the temperature gets to a 100°C

Hence, the temperature and pressure at a state of two becomes

P₂ = 99.936 kPa

T₂ = 100°C

The next step is to determine the specific volume of the refrigerant R-134a at a final state from the super heated refrigerant R-134a which is equivalent to the pressure of 99.936 kPa

v₂ =0.30138 m³ /kg

Now,

we now calculate the final state of  the weighted piston cylinder device

V₂ = mv₂

V₂ = 0.2 * 0.30138

V₂ = 0.06027 m³

Hence ,the final volume of the weighted cylinder piston device is 0.06027 m³

The complete question is;

The wheel and the attached reel have a combined weight of 50 lb and a radius of gyration about their center of ka = 6 in. If pulley B that is attached to the motor is subjected to a torque of M = 50 lb.ft, determine the velocity of the 200lb crate after the pulley has turned 5 revolutions. Neglect the mass of the pulley.

The image of this system is attached.

Answer:

Velocity = 11.8 ft/s

Explanation:

Since the wheel at A rotates about a fixed axis, then;

v_c = ω•r_c

r_c is 4.5 in. Let's convert it to ft.

So, r_c = 4.5/12 ft = 0.375 ft

Thus;

v_c = 0.375ω

Now the mass moment of inertia about of wheel A about it's mass centre is given as;

I_a = m•(k_a)²

The mass in in lb, so let's convert to slug. So, m = 50/32.2 slug = 1.5528 slug

Also, let's convert ka from inches to ft.

So, ka = 6/12 = 0.5

So,I_a = 1.5528 × 0.5²

I_a = 0.388 slug.ft²

The kinetic energy of the system would be;

T = Ta + Tc

Where; Ta = ½•I_a•ω²

And Tc = ½•m_c•(v_c)²

So, T = ½•I_a•ω² + ½•m_c•(v_c)²

Now, m_c is given as 200 lb.

Converting to slug, we have;

m_c = (200/32.2) slugs

Plugging in the relevant values, we have;

T = (½•0.388•ω²) + (½•(200/32.2)•(0.375ω)²)

This now gives;

T = 0.6307 ω²

The system is initially at rest at T1 = 0.

Resolving forces at A, we have; Ax, Ay and Wa. These 3 forces do no work.

Whereas at B, M does positive work and at C, W_c does negative work.

When pulley B rotates, it has an angle of; θ_b = 5 revs × 2π rad/revs = 10π

While the wheel rotates through an angle of;θ_a = (rb/ra) • θ_b

Where, rb = 3 in = 3/12 ft = 0.25 ft

ra = 7.5 in = 7.5/12 ft = 0.625 ft

So, θ_a = (0.25/0.625) × 10π

θ_a = 4π

Thus, we can say that the crate will have am upward displacement through a distance;

s_c = r_c × θ_a = 0.375 × 4π

s_c = 1.5π ft

So, the work done by M is;

U_m = M × θ_b

U_m = 50lb × 10π

U_m = 500π

Also,the work done by W_c is;

U_Wc = -W_c × s_c = -200lb × 1.5π

U_Wc = -300π

From principle of work and energy;

T1 + (U_m + U_Wc) = T

Since T1 is zero as stated earlier,

Thus ;

0 + 500π - 300π = 0.6307 ω²

0.6307ω² = 200π

ω² = 200π/0.6307

ω² = 996.224

ω = √996.224

ω = 31.56 rad/s

We earlier derived that;v_c = 0.375ω

Thus; v_c = 0.375 × 31.56

v_c = 11.8 ft/s

The minimum circuit ampacity for an air-conditioning condensing unit calculated by compressor amps + fan amps. The correct option is 1.

Ampacity is the term used for the greatest current conveying limit, in amperes, of a specific electrical gadget.  The current carrying capacity is generally depended upon the electrical cable and is calculated as the maximum amount of current a cable can withstand before it warms past the most extreme working temperature.

So, in air conditioning units,  the greatest current carrying capacity is  equal to the addition of compressor and fan current capacity.

Thus, the correct option is 1.

Learn more about ampacity.

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Answer:

a co mam zroic!!

Explanation:

Answer: Fossil remains of the same land-dwelling animal.

Explanation: Fossil remains which were found to belong to same land dwelling animals, in South America and Africa was used as evidence to help support the theory of Tectonics plates, what this theory simply means is that the whole continents of earth were once fused together until a tectonic plate caused it’s division. Since same remains were found in Africa and South America this shows that both continents were once fused together.

Answer:

Fossil remains of the same land-dwelling animal

Explanation:

Fossil remains tell us where the animals once lived and how by the movement of plate spearated their remaind that was burried thousands of years ago.

Answer:

thrust = ML[tex]T^{-2}[/tex]  

Explanation:

T = p[tex]V^{2}[/tex][tex]D^{2}[/tex] x [tex]\frac{ND}{V}[/tex]Re

where D is diameter

p is the density

N is the revolution per second

Re is the Reynolds number which is equal to  puD/μ

where p is the fluid density

u is the fluid velocity

μ is the fluid viscosity = kg/m.s = M[tex]L^{-1}[/tex][tex]s^{-1}[/tex]

Reynolds number is dimensionless so it cancels out

diameter is m = L

speed is in m/s = L[tex]T^{-1}[/tex]

fluid density is in kg/[tex]m^{3}[/tex] = M[tex]L^{-3}[/tex]

N is in rad/s = L[tex]L^{-1}[/tex][tex]T^{-1}[/tex] = [tex]T^{-1}[/tex]

combining these dimensions into the equation, we have

thrust = M[tex]L^{-3}[/tex][tex]( LT^{-1}) ^{2}[/tex][tex]L^{2}[/tex][tex]\frac{T^{-1}L }{LT^{-1} }[/tex]

= M[tex]L^{-3}[/tex][tex]L^{2}[/tex][tex]T^{-2}[/tex][tex]L^{2}[/tex]

thrust = ML[tex]T^{-2}[/tex]   which is the dimension for a force which indicates that thrust is a type of force

Answer:

The answer is "0.0728"

Explanation:

Given value:

[tex]P_0= 14.696\ ps\\\\\ p _{0}= 0.00238 \frac{slue}{ft^{3}}\\\\\ A= 0.05 ft^2\\\\\ T_0= 59^{\circ}f = 518.67R\\\\\ air \ k= 1\\\\ \ cirtical \ pressure ( P^*)=P_0\times \frac{2}{k+1}^{\frac{k}{k-1}}\\[/tex]

                                     [tex]= 14.696\times (\frac{2}{1.4+1})^{\frac{1.4}{1.4-1}}\\\\=7.763 Psia\\\\[/tex]

if [tex]P<P^{*} \to[/tex] flow is chocked

if [tex]P>P^{*} \to[/tex] flow is not chocked

When  P= 10 psia < [tex]P^{*}[/tex] [tex]\to[/tex] not chocked

match number:

[tex]\ for \ P= \ 10\ G= \sqrt{\frac{2}{k-1}[(\frac{\ p_{0}}{p})^{\frac{k-1}{k}}-1]}[/tex]

                       [tex]= \sqrt{\frac{2}{1.4-1}[(\frac{14.696}{10})^{\frac{1.4-1}{1.4}}-1]}[/tex]

[tex]M_0=7.625[/tex]

[tex]p=p_0(1+\frac{k-1}{2} M_0 r)^\frac{1}{1-k}[/tex]

  [tex]=0.00238(1+\frac{1.4-1}{2}0.7625`)^{\frac{1}{1-1.4}}\\\\\ p=0.001808 \frac{slug}{ft^3}[/tex]

[tex]\ T= T_0(1+\frac{k-1}{2} Ma^r)^{-1}\\\\\ T=518.67(1+\frac{1.4-1}{2} 0.7625^2)^{-1}\\\\\ T=464.6R\\\\[/tex]

[tex]\ velocity \ of \ sound \ (C)=\sqrt{KRT}\\\\[/tex]

                                    [tex]=\sqrt{1.4\times1716\times464.6}\\\\=1057 ft^3\\\\[/tex]

R= gas constant=1716

[tex]m=PAV\\\\[/tex]

    [tex]=0.001808\times0.05\times(Ma.C)\\\\=0.001808\times0.05\times0.7625\times 1057\\\\=0.0728\frac{slug}{s}[/tex]

Answer:

Cpk = 1.33

Explanation:

Given:

Mean = 50

Sd = 2

USL = 60

LSL = 42

The Cpk means process capability index. it helps decide the specification limit when the nominal value is not the central value of upoer specification limint (USL) and lower specification limit (LSL).

The Cpk can be derived using the formula:

[tex]Cpk = min [\frac{(usl - mean)}{3 * \sigma}, \frac{(mean - lsl)}{3*\sigma}] [/tex]

[tex]Cpk = min [\frac{(60 - 50)}{3*2} , \frac{(50 - 42)}{3*2}] [/tex]

Solving further,

[tex]Cpk = min [\frac{10}{6} , \frac{8}{6}] [/tex]

Cpk = min ( 1.67 , 1.33)

Cpk = 1.33

Cpk = 1.33

Answer:

Option 4 =>  only different angles for theta 4, Theta 3 should have the same values in both cases.

Explanation:

The vector loop of Section 4.6 that was made as the reference diagram has the vector loop closing itself and the vectors summing around loop zero. Also, it has one DOF mechanism and the vector length is equal to the vector link.

NOTE: Kindly check the attached file or picture for the diagram of Section 4.6 diagram.

We can see from the diagram below that the V3 and V4 can be determined or Calculated with parameters such as theta 2, all links length and length d.

Therefore, the cross and the open configurations of the circuit will produce from the diagram will be " only different angles for theta 4, Theta 3 should have the same values in both cases."( That is option 4)

Answer:

Pillar drill are small and used for woodworking while Radial arm pillar is mounted on a very large column.

Answer:

(a) The mean time to fail is 9491.22 hours

The standard deviation time to fail is 9491.22 hours

(b) 0.5905

(c) 3.915 × 10⁻¹²

(d) 2.63 × 10⁻⁵

Explanation:

(a) We put time to fail = t

∴ For an exponential distribution, we have f(t) = [tex]\lambda e^{-\lambda t}[/tex]

Where we have a failure rate = 10% for 1000 hours, we have(based on online resource);

[tex]P(t \leq 1000) = \int\limits^{1000}_0 {\lambda e^{-\lambda t}} \, dt = \dfrac{e^{1000\lambda}-1}{e^{1000\lambda}} = 0.1[/tex]

e^(1000·λ) - 0.1·e^(1000·λ) = 1

0.9·e^(1000·λ) = 1

1000·λ = ㏑(1/0.9)

λ = 1.054 × 10⁻⁴

Hence the mean time to fail, E = 1/λ = 1/(1.054 × 10⁻⁴) = 9491.22 hours

The standard deviation = √(1/λ)² = √(1/(1.054 × 10⁻⁴)²)) = 9491.22 hours

b) Here we have to integrate from 5000 to ∞ as follows;

[tex]p(t>5000) = \int\limits^{\infty}_{5000} {\lambda e^{-\lambda t}} \, dt =\left [ -e^{\lambda t}\right ]_{5000}^{\infty} = e^{5000 \lambda} = 0.5905[/tex]

(c) The Poisson distribution is presented as follows;

[tex]P(x = 3) = \dfrac{\lambda ^x e^{-x}}{x!} = \frac{(1.0532 \times 10^{-4})^3 e^{-3} }{3!} = 3.915\times 10^{-12}[/tex]

p(x = 3) = 3.915 × 10⁻¹²

d) Where at least 2 components fail in one half hour, then 1 component is expected to fail in 15 minutes or 1/4 hours

The Cumulative Distribution Function is given as follows;

p( t ≤ 1/4) [tex]CDF = 1 - e^{-\lambda \times t} = 1 - e^{-1.054 \times 10 ^{-4} \times 1/4} = 2.63 \times 10 ^{-5}[/tex].

A. The mean time to fail is:

For an exponential distribution, we have f(t) = e^(1000·λ) - 0.1·e^(1000·λ)= 1

0.9·e^(1000·λ) = 1

1000·λ = ㏑(1/0.9)

λ = 1.054 × 10⁻⁴

Hence the mean time to fail, E = 1/λ = 1/(1.054 × 10⁻⁴) = 9491.22 hours

The standard deviation = √(1/λ)² = √(1/(1.054 × 10⁻⁴)²)) = 9491.22 hours

Exponential distribution is the probability the distribution of the time between events in a Poisson point process.

B. Here we have to integrate from 5000 to ∞ as follows: 0.5905

C. The Poisson distribution is presented as follows;

p(x = 3) = 3.915 × 10⁻¹²

D. Where at least 2 components fail in one half hour, then 1 component is expected to fail in 15 minutes or 1/4 hours.

The Cumulative Distribution Function is given as follows; p( t ≤ 1/4) = 2.63 × 10⁻⁵

Therefore, the correct answers are:

(a) The mean time to fail is 9491.22 hours

The standard deviation time to fail is 9491.22 hours

(b) 0.5905

(c) 3.915 × 10⁻¹²

(d) 2.63 × 10⁻⁵

Read more about exponential here:

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Answer:

The inside temperature, [tex]T_{in}[/tex] is approximately 248 °C.

Explanation:

The parameters given are;

Temperature of the air = 20°C

Carbon steel surface temperature 250°C

Area of surface = 50 cm × 75 cm = 0.5 × 0.75 = 0.375 m²

Convection heat transfer coefficient = 25 W/(m²·K)

Heat lost by radiation = 300 W

Assumption,

Air temperature = 20 °C

Hot plate temperature = 250 °C

Thermal conductivity K = 65.2 W/(m·K)

Steady state heat transfer process

One dimensional heat conduction

We have;

Newton's law of cooling;

q = h×A×([tex]T_s[/tex] - [tex]T_{\infty[/tex]) + Heat loss by radiation

= 25×0.325×(250 - 20) + 300

= 2456.25 W

The rate of energy transfer per second is given by the following relation;

[tex]P = \dfrac{K \times A \times \Delta T}{L}[/tex]

Thermal conductivity K = 65.2 W/(m·K)

Therefore;

[tex]2456.25 = \dfrac{65.2 \times 0.375 \times (250 - T_{in})}{0.02}[/tex]

[tex]T_{in} = 250 - \dfrac{2456.25 \times 0.02}{65.2 \times 0.375} = 247.99 ^{\circ}C[/tex]

The inside temperature, [tex]T_{in}[/tex] = 247.99 °C  ≈ 248 °C.

Answer:

Delta star transformer

Secondary side

[tex]V_L[/tex] = 415 V

[tex]V_P[/tex] = 239.6 V

[tex]I_P[/tex] = Line current, [tex]I_L[/tex] = 800 A

Primary side

[tex]V_L[/tex] = Phase voltage, [tex]V_P[/tex] = 3300 V

Primary phase current = 58.08 A

[tex]I_L[/tex]  = 100.6 A

Turns ratio = 83/(660·√3)

Star-star transformer

Secondary side

[tex]V_L[/tex] = 3300 V

[tex]V_P[/tex] = 1905.26 V

[tex]I_P[/tex] = Line current, [tex]I_L[/tex] =  100.6 A

Primary side

[tex]V_L[/tex] = 11000 V

[tex]V_P[/tex] = 6350.9 V

Line current, [tex]I_L[/tex]   = Phase current, [tex]I_P[/tex]   = 3.018 A

Turns ratio = 3/100.

Explanation:

Delta star transformer

For the star Connection to the 3-Phase load,

Secondary line voltage, [tex]V_L[/tex] = 415 V

Secondary phase voltage, [tex]V_P[/tex] =  [tex]V_L[/tex] ÷ √3 = 415 ÷ √3 = 239.6 V

Secondary phase current, [tex]I_P[/tex] = Line current, [tex]I_L[/tex] = 800 A

For the primary side of the delta star transformer

Line voltage, [tex]V_L[/tex] = Phase voltage, [tex]V_P[/tex] = 3300 V

K =  415/(3300×√3) = 83/(660·√3)

Primary phase current = 83/(660·√3)× 800 A = 58.08 A

Line current, [tex]I_L[/tex]   = Phase current, [tex]I_P[/tex] × √3  = 100.6 A

The turns ratio = 83/(660·√3)

For the star-star transformer, we have;

Secondary line voltage, [tex]V_L[/tex] = 3300 V

Secondary phase voltage, [tex]V_P[/tex] =  [tex]V_L[/tex] ÷ √3 = 3300 ÷ √3 = 1905.26 V

Secondary phase current, [tex]I_P[/tex] = Line current, [tex]I_L[/tex] = 100.6 A

For the primary side of the star-star transformer

Line voltage, [tex]V_L[/tex] = 11000 V

Phase voltage, [tex]V_P[/tex] = 11000 V ÷ √3 = 6350.9 V

K =  3300/11000 = 3/100

Primary phase current = 3/100× 100.6 = 3.018 A

Line current, [tex]I_L[/tex]   = Phase current, [tex]I_P[/tex]   = 3.018 A

The turns ratio = k = 3/100.

Answer:

the rate of heat transfer into the wall is [tex]\mathbf{q__{in}} \mathbf{ = 200 W/m^2}[/tex]

the rate of heat output is [tex]\mathbf{q_{out} =182 \ W/m^2}[/tex]

the rate of change of energy stored by the wall is [tex]\mathbf{ \Delta E_{stored} = 18 \ W/m^2 }[/tex]

the convection coefficient is h = 4.26 W/m².K

Explanation:

From the question:

The temperature distribution across the wall is given by :

[tex]T(x) = ax+bx+cx^2[/tex]

where;

T = temperature in ° C

and a, b, & c are constants.

replacing 200° C for a, - 200° C/m for b and  30° C/m² for c ; we have :

[tex]T(x) = 200x-200x+30x^2[/tex]

According to the application of Fourier's Law of heat conduction.

[tex]q_x = -k \dfrac{dT}{dx}[/tex]

where the rate of heat input [tex]q_{in} = q_k[/tex] ; Then x= 0

So:

[tex]q_{in}= -k (\dfrac{d( 200x-200x+30x^2)}{dx})_{x=0}[/tex]

[tex]q_{in}= -1 (-200+60x)_{x=0}[/tex]

[tex]\mathbf{q__{in}} \mathbf{ = 200 W/m^2}[/tex]

Thus , the rate of heat transfer into the wall is [tex]\mathbf{q__{in}} \mathbf{ = 200 W/m^2}[/tex]

The rate of heat output is:

[tex]q_{out} = q_{x=L}[/tex]; where x = 0.3

[tex]q_{out} = -k (\dfrac{dT}{dx})_{x=0.3}[/tex]

replacing T with [tex]200x-200x+30x^2[/tex] and k with 1 W/m.K

[tex]q_{out} = -1 (\dfrac{d(200x-200x+30x^2)}{dx})_{x=0.3}[/tex]

[tex]q_{out} = -1 (-200+60x)_{x=0.3}[/tex]

[tex]q_{out} = 200-60*0.3[/tex]

[tex]\mathbf{q_{out} =182 \ W/m^2}[/tex]

Therefore , the rate of heat output is [tex]\mathbf{q_{out} =182 \ W/m^2}[/tex]

Using energy balance to determine the change of energy(internal energy) stored by the wall.

[tex]\Delta E_{stored} = E_{in}-E_{out} \\ \\ \Delta E_{stored} = q_{in}- q_{out} \\ \\ \Delta E_{stored} = (200 - 182 ) W/m^2 \\ \\[/tex]

[tex]\mathbf{ \Delta E_{stored} = 18 \ W/m^2 }[/tex]

Thus; the rate of change of energy stored by the wall is [tex]\mathbf{ \Delta E_{stored} = 18 \ W/m^2 }[/tex]

We all know that for a steady state, the heat conducted to the end of the plate must be convected to the surrounding fluid.

So:

[tex]q_{x=L} = q_{convected}[/tex]

[tex]q_{x=L} = h(T(L) - T _ \infty)[/tex]

where;

h is the convective heat transfer coefficient.

Then:

[tex]Replacing \ 182 W/m^2 \ for \ q_{x=L} , (200-200x +30x \ for \ T(x) \ , 0.3 m \ for \ x \ and \ 100^0 C for \ T[/tex] We have:

182 = h(200-200×0.3 + 30 ×0.3² - 100 )

182 = h (42.7)

h = 4.26 W/m².K

Thus, the convection coefficient is h = 4.26 W/m².K

Answer:

The inlet velocity v₁ = 6.66 m/s, the exit velocity v₂ = 7.32 m/s and the volume flow rate Q = 4.72 m³/s

Explanation:

Using Bernoulli's equation

P₁ + ρgh₁ + 1/2ρv₁² = P₂ + ρgh₂ + 1/2ρv₂²

P₁ - P₂ + ρgh₁ -  ρgh₂ = 1/2ρv₂² - 1/2ρv₁²

ΔP + ρgΔh = 1/2ρ(v₂² - v₁²)  (1)

where ΔP = pressure difference = 12.35 kPa = 12350 Pa

Δh = height difference = 1.35 m

From the flow rate equation Q = A₁v₁ = A₂v₂ and v₁ = A₂v₂/A₁ = d₂²v₂/d₁² where v₁ and v₂ represent the inlet and exit velocities from the pipe and d₁ = 0.95 m and d₂ = 0.44 m represent the diameters at the top end and lower end of the pipe respectively.

Substituting v₁ into (1), we have

ΔP + ρgΔh = 1/2ρ(v₂² - (d₂²v₂/d₁² )²)

ΔP + ρgΔh = 1/2ρ(v₂² - (d₂/d₁)⁴v₂²)

v₂ = √[2(ΔP + ρgΔh)/ρ(1 - (d₂/d₁)⁴)}

substituting the values of the variables, we have

v₂ = √[2(12350 Pa + 1000 kg/m³ × 9.8 m/s² × 1.35 m)/(1000 kg/m³ (1 - (0.44 m/0.95 m)⁴))}

= √[2(12350 Pa + 13230 Pa)/(1000 kg/m³ × 0.954)]

= √[2(25580 Pa)/954 kg/m³]

= √[51160 Pa/954 kg/m³]

= √53.627

= 7.32 m/s

v₁ = d₂²v₂/d₁²

  = (0.44 m/0.95 m)² × 7.32 m/s

  = (0.954)² × 7.32 m/s

  = 6.66 m/s

The volume flow rate Q = A₁v₁

= πd₁²v₁/4

= π(0.95 m)² × 6.66 m/s ÷ 4

= 18.883 m³/s ÷ 4

= 4.72 m³/s

So, the inlet velocity v₁ = 6.66 m/s, the exit velocity v₂ = 7.32 m/s and the volume flow rate Q = 4.72 m³/s

Answer:

The volume of sludge after filtration is 0.914 m

Explanation:

Solution

Given that:

We have to find the Volume of sludge after filtration

Now,

The  Sludge concentration is = 32%,

The sludge volume = 100 m,

The sludge concentration after filtration = 35%

Then,

The mass balance equation is stated below

Cin∀in = Cout∀out

Now,

We Solve for ∀out

∀out =Cin∀in/Cout

By substituting the values

∀out = (0.32)(100 m)/(0.35) = 0.914 m

Answer:

1.50 miles/hour is the correct answer to the given question  

Explanation:

As mention in the question engine power = 190 horsepower engine.......eq(1)

we have to converted into the watt as we know that 1 horse power=746 watt

putting this value in eq(1) we get

[tex]Engine \ power\ =190 *746\\\ Engine \ power =146216\ NM/S[/tex]

Drag coefficient = 0.33

Projected area = 26.36 [tex]ft^{2}[/tex]............eq(2)

we have to converted into the [tex]m^{2}[/tex]  as we know that [tex]1m^{2}\ =\ 10.7639\ ft^{2}[/tex]

putting this value in eq(2) we get

Projected area =  [tex]2.448921\ m^{2}[/tex]

Now we to calculated the Drag Force

[tex]F\ =\frac{\ Drag \ coefficient\ *P*\ V^{2}\ *A }{2}[/tex]....................eq(3)

Putting the above value in eq(3) we get

[tex]F=\ \frac{0.33\ * 1.2 * \ V^{2 \ * 2.44821\ *N} }{2} \\F=\ 0.4848V^{2}[/tex]

As we know that

[tex]F=0.4848\ V^{2}\ *V \\F= \frac{\ 0.4848V^{3} NM}{S}[/tex]

As mention in the question

Power of Engine =Drag free force

[tex]146216\ =\ 0.4848\ V^{3}\\ V^{3} =301550.889\\V=67.05M/S[/tex]

We have to converted this value into the miles/hour

1 M/S =2.22369 M/H

Putting this value into the V we get

V=67.05 * 2.2234 M/H

V=150 M/H

ANSWER:

1) The car is not accelerating, because for the the car to turn while accelerating down the street, it has to deaccelerate (reduce speed) due to CENTRIPETAL FORCE that is acting on the car during the turn. This force acts to bring the car in the direction where it has turned to.

2) You will feel the movement as everyone in the car will tend to fall to the left, while the car makes a right turn. This is because of CENTRIFUGAL FORCE, which is acting to take the car to a left direction, while the car is turning to the right direction.

Answer is given below

Explanation:

we know that some common types of throttling devices are

so here throttling devices commonly used in refrigeration and air-conditioning because

Answer:

ESTABLISHED

Explanation:

What is TCP?

A Transmission Control Protocol (TCP) is a communication protocol which allows the exchange of data between computers in a network.  

When a Transmission Control Protocol connection is up and running meaning that both sides can send and receive data then the corresponding TCP socket states is known as "ESTABLISHED".

The most common socket states are:

LISTEN:

Before a TCP connection is made, there needs to be a server with a listener that will listen on incoming connection request.

ESTABLISHED:

When a TCP connection is up and running meaning that both sides can send and receive data.

CLOSED:

The CLOSED state means that there is no TCP connection.

There are a total of 11 TCP socket states:

1. LISTEN

2. SYN-SENT

3. SYN-RECEIVED

4. ESTABLISHED

5. FIN-WAIT-1

6. FIN-WAIT-2

7. CLOSE-WAIT

8. CLOSING

9. LAST-ACK

10. TIME-WAIT

11. CLOSED

Answer:

(a) 6.91 mm (b) 160 MPa

Explanation:

Solution

Given that:

E = 200 GPa

The rod length = 48 mm

P =P¹ = 6 kN

Recall that,

1 kN = 10^3 N

1 m =10^3 mm

I GPa = 10^9 N/m²

Thus

The rod deformation is stated as follows:

δ = PL/AE-------(1)

σ = P/A----------(2)

Now,

(a) We substitute the values in equation and obtain the following:

48 * 10 ^⁻3 m =  (6 * 10³ N) (60 m)/A[ 200 * 10^9 N/m^2]

Thus, we simplify

A = (6 * 10³) (60)/ ( 200 * 10^9) (48 * 10 ^⁻3)m²

A =0.0375 * 10 ^⁻3 m²

A =37.5 mm²

A = π/4 d²

Thus,

d² = 4A /π

After inserting the values we have,

d = √37.5 * 4/3.14 mm

= 6.9116 mm

or d = 6.91 mm

Therefore, the smallest that should be used is 6.91 mm

(B) To determine the corresponding normal stress that is caused by the tensile force, we input the values in equation (2)

Thus,

σ = P/A

σ= 6 * 10 ^ 3 N/ 37. 5 * 10 ^⁻6 m²

σ= 160 MPa

Note: I MPa = 10^6 N/m²

Hence the the corresponding normal stress is σ= 160 MPa

Answer:

The minimum particle diameter that is removed at 85% is 1.474 * 10 ^⁻4 meters.

Solution

Given:

Length = 48 m

Width = 12 m

Depth = 3m

Flow rate = 4 m 3 /s

Water density = 10 3 kg/m 3

Dynamic viscosity = 1.30710 -3 N.sec/m

Now,

At the minimum particular diameter it is stated as follows:

The Reynolds number= 0.1

Thus,

0.1 =ρVTD/μ

VT = Dp² ( ρp- ρ) g/ 10μ²

Where

gn = The case/issue of sedimentation

VT = Terminal velocity

So,

0.1 = Dp³ ( ρp- ρ) g/ 10μ²

This becomes,

0.1 = 1000 * dp³ (1100-1000) g 0.1/ 10 *(1.307 * 10 ^⁻3)²

= 3.074 * 10 ^⁻6 = dp³ (.g01 * 10^6)

dp³=3.1343 * 10 ^⁻12

Dp minimum= 1.474 * 10 ^⁻4 meters.

Answer:

A) ΔL = 0.503 mm

B) Δd = -0.016 mm

Explanation:

A) From Hooke's law; σ = Eε

Where,

σ is stress

ε is strain

E is elastic modulus

Now, σ is simply Force/Area

So, with the initial area; σ = F/A_o

A_o = (π(d_o)²)/4

σ = 4F/(π(d_o)²)

Strain is simply; change in length/original length

So for initial length, ε = ΔL/L_o

So, combining the formulas for stress and strain into Hooke's law, we now have;

4F/(π(d_o)²) = E(ΔL/L_o)

Making ΔL the subject, we now have;

ΔL = (4F•L_o)/(E•π(d_o)²)

We are given;

F = 50500 N

L_o = 209mm = 0.209m

E = 65.5 GPa = 65.5 × 10^(9) N/m²

d_o = 20.2 mm = 0.0202 m

Plugging in these values, we have;

ΔL = (4 × 50500 × 0.209)/(65.5 × 10^(9) × π × (0.0202)²)

ΔL = 0.503 × 10^(-3) m = 0.503 mm

B) The formula for Poisson's ratio is;

v = -(ε_x/ε_z)

Where; ε_x is transverse strain and ε_z is longitudinal strain.

So,

ε_x = Δd/d_o

ε_z = ΔL/L_o

Thus;

v = - [(Δd/d_o)/(ΔL/L_o)]

v = - [(Δd•L_o)/(ΔL•d_o)]

Making Δd the subject, we have;

Δd = -[(v•ΔL•d_o)/L_o]

We are given v = 0.33; d_o = 20.2mm

So,

Δd = -[(0.33 × 0.503 × 20.2)/209]

Δd = -0.016 mm

Answer: Provided in the explanation section

Explanation:

Transactional Leadership - This leadership style is mainly focused on the transactions between the leader and employees. If the employees work hard, achieve the objectives and deliver the results, they are rewarded through bonuses, hikes, promotions etc. If the employees fail to achieve the desired results, they are punished by awarding lower ratings in the performance appraisal, denying opportunities etc.

In this style, leader lays emphasis on the relation with the followers.

It is a reactive style where the growth of the employee in the organization completely depends on the performance with respect to the activities and deliverables.

It is best suited for regular operations and for a settled environment by developing the existing organizational culture which is not too challenging.

It is a bureaucratic style of leadership where the leader concentrates on planning and execution rather than innovation and creation.

A transactional leader is short-term focused and result oriented. He/she doesn't consider long-term strategic objectives regarding the organization's future.

cheers i hope this helped !!

Answer:

Check below for answers

Explanation:

Matlab code:

function[Pa] = Psi-ToPa(psi)

        Pa = psi * 6894.75728;

end

a) To convert 120 psi to units of Pa, just call the function Pa using the command:

          Psi-ToPa(120)

ans =

       8.2737e+05

b) To convert 3000 psi to units of Pa, just call the function Pa using the command:

          Psi-ToPa(3000)

ans =

       2.0684e+07

Answer:

a) P* = 0.5283 MPa , T* = 624.75 K , ρ* = 2.945 kg/m^3 , V* = 501.023 m/s

b) Pe = 0.1278 MPa , Te = 416.7 K , ρe = 1.069 kg/m^3 , Ve = 818.36 m/s, Ae = 33.75 cm^2

c) m' = 2.915 kg/s

Explanation:

Given:-

- The inlet pressure, Pi = 1.0 MPa

- The inlet temperature, Ti = 750 K

- Inlet velocity is negligible

- Steady, Isentropic Flow

- The specific heat ratio of air, k = 1.4

- Exit Mach number, Mae = 2

- The throat area, Ath = 20 cm^2

- Gas constant of air, R = 0.287 KJ / kg.K

Find:-

(a) the throat conditions (static pressure, temperature, density, and mach number)

b) the exit plane conditions

c) the mass flow rate

Solution:-

- For this problem we will assume air to behave like an ideal gas with constant specific heat at RTP. Also the flow of air through the nozzle is assumed to be steady, one dimensional, and Isentropic with constant specific heat ratio ( k ).

- First we will scrutinize on the exit conditions. We have a Mach number of 2 at the exit. The flow at the exit of converging-diverging nozzle is in super-sonic region this is only possible only if sonic ( Ma = 1 ) conditions are achieved by the flow at the throat area ( minimum cross-sectional area ).

- Moreover, the flow is almost still at the inlet. Hence, we can assume that the flow has negligible velocity ( vi = 0 m/s ) at the inlet and the reservoir temperature and pressure can be assumed to be stagnation temperature and pressures as follows:

                             [tex]P_o = 1.0 MPa\\\\T_o = 750 K[/tex]

- Using the ideal gas law we can determine the stagnation density ( ρo ) as follows:

                             [tex]p_o = \frac{P_o}{RT_o} = \frac{1000}{0.287*750} = 4.64576\frac{kg}{m^3}[/tex]

- We will use the already developed results for flow which has reached sonic velocity ( Ma = 1 ) at the throat region. Use Table A - 13, to determine the critical static values at the throat region:

                            [tex]\frac{P^*}{P_o} = 0.5283\\\\P^* = 0.5283*1 = 0.5283 MPa\\\\\frac{T^*}{T_o} = 0.8333\\\\T^* = 0.8333*750 = 624.75 K\\\\ \frac{p^*}{p_o} = 0.6339\\\\p^* = 0.6339*4.64576 = 2.945 \frac{kg}{m^3} \\\\[/tex]

                            [tex]V^* = \sqrt{kRT^*} =\sqrt{1.4*287*624.75} = 501.023 \frac{m}{s}[/tex]

- Similarly, we will again employ the table A - 13 to determine the exit plane conditions for ( Ma = 2 ) as follows:

                           [tex]\frac{P_e}{P_o} = 0.1278 \\\\P_e = 0.1278*1.0 = 0.1278 MPa\\\\\frac{T_e}{T_o} = 0.5556 \\\\T_e = 0.5556*750 = 416.7 K\\\\\frac{p_e}{p_o} = 0.23 \\\\p_e = 0.23*4.64576 = 1.069 \frac{kg}{m^3} \\\\\frac{A_e}{A_t_h} = 1.6875 \\\\A_e =1.6875*20 = 33.75 cm^2\\[/tex]

- The velocity at the exit plane ( Ve ) can be determined from the exit conditions as follows:

                        [tex]V_e = Ma_e*\sqrt{kRT_e} = 2*\sqrt{1.4*287*416.7} = 818.36 \frac{m}{s}[/tex]  

- For steady flows the mass flow rate ( m' ) is constant at any section of the nozzle. We will use the properties at the throat section to determine the mass flow rate as follows:

                         [tex]m' = p^* A_t_h V^*\\\\m' = 2.945*20*10^-^4*501.023\\\\m' = 2.951 \frac{kg}{s}[/tex]