When the transportation of natural gas in a pipeline is not feasible for economic reasons, it is first liquefied using nonconventional refrigeration techniques and then transported in super-insulated tanks. In a natural gas liquefaction plant, the liquefied natural gas (LNG) enters a cryogenic turbine at 30 bar and –160°C at a rate of 20 kg/s and leaves at 3 bar. If 120 kW power is produced by the turbine, determine the efficiency of the turbine. Take the density of LNG to be 423.8 kg/m3.

Answer:1 common law

Explanation:

It also says that if Cathy wasn’t wearing a hardhat hat she is responsible for her own injury, more than one answer is correct.

In this case, if Cathy's employer completes compliance and general duty requirements then the organization may not be held liable and again, the law can generally hold Cathy responsible for the injuries as she was not wearing the proper kits for such work.

According to OSHA, Cathy’s employer may not be held liable for her injury if it fulfilled compliance and general duty requirements.

You are entitled to a secure workplace. To stop workers from being murdered or suffering other types of harm at work, the Occupational Safety and Health Act of 1970 (OSH Act) was passed. According to the legislation, companies are required to give their workers safe working environments.

Therefore, more than one answer is correct.

Learn more about OSHA, here:

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Answer:

the  pressure drop  is 0.21159 atm

Explanation:

Given that:

length of the reactor L = 2.5 m

inside diameter of the reactor d= 0.025 m

diameter of alumina sphere [tex]dp[/tex]= 0.003 m

particle density  = 1300 kg/m³

the bed void fraction [tex]\in =[/tex]  0.38

superficial mass flux m = 4684 kg/m²hr

The Feed is  methane with pressure P = 5 bar and temperature T = 400 K

Density of the methane gas [tex]\rho[/tex] = 0.15 mol/dm ⁻³

viscosity of methane gas [tex]\mu[/tex] = 1.429  x 10⁻⁵ Pas

The objective is to determine the pressure drop.

Let first convert the Density of the methane gas from 0.15 mol/dm ⁻³  to kg/m³

SO; we have :

Density =  0.15 mol/dm ⁻³  

Molar mass of methane gas (CH₄) = (12 + (1×4) ) = 16 mol

Density =  [tex]0.1 5 *\dfrac{16}{0.1^3}[/tex]

Density =  2400

Density [tex]\rho_f[/tex] =  2.4 kg/m³

Density = mass /volume

Thus;

Volume = mass/density

Volume of the methane gas =  4684 kg/m²hr / 2.4 kg/m³

Volume of the methane gas = 1951.666 m/hr

To m/sec; we have :

Volume of the methane gas = 1951.666 * 1/3600 m/sec = 0.542130 m/sec

[tex]Re = \dfrac{dV \rho}{\mu}[/tex]

[tex]Re = \dfrac{0.025*0.5421430*2.4}{1.429*10^5}[/tex]

[tex]Re=2276.317705[/tex]

For Re > 1000

[tex]\dfrac{\Delta P}{L}=\dfrac{1.75 \rho_f(1- \in)v_o}{\phi_sdp \in^3}[/tex]

[tex]\dfrac{\Delta P}{2.5}=\dfrac{(1.75 *2.4)(1- 0.38)*0.542130}{1*0.003 (0.38)^3}[/tex]

[tex]\Delta P=8575.755212*2.5[/tex]

[tex]\Delta = 21439.38803 \ Pa[/tex]

To atm ; we have

[tex]\Delta P = \dfrac{21439.38803 }{101325}[/tex]

[tex]\Delta P =0.2115903087 \ atm[/tex]

ΔP  ≅  0.21159 atm

Thus; the  pressure drop  is 0.21159 atm

Answer:

Exit temperature = 32°C

Explanation:

We are given;

Initial Pressure;P1 = 100 KPa

Cp =1000 J/kg.K = 1 KJ/kg.k

R = 500 J/kg.K = 0.5 Kj/Kg.k

Initial temperature;T1 = 27°C = 273 + 27K = 300 K

volume flow rate;V' = 15 m³/s

W = 130 Kw

Q = 80 Kw

Using ideal gas equation,

PV' = m'RT

Where m' is mass flow rate.

Thus;making m' the subject, we have;

m' = PV'/RT

So at inlet,

m' = P1•V1'/(R•T1)

m' = (100 × 15)/(0.5 × 300)

m' = 10 kg/s

From steady flow energy equation, we know that;

m'•h1 + Q = m'h2 + W

Dividing through by m', we have;

h1 + Q/m' = h2 + W/m'

h = Cp•T

Thus,

Cp•T1 + Q/m' = Cp•T2 + W/m'

Plugging in the relevant values, we have;

(1*300) - (80/10) = (1*T2) - (130/10)

Q and M negative because heat is being lost.

300 - 8 + 13 = T2

T2 = 305 K = 305 - 273 °C = 32 °C

Answer:

Explanation:

Please

Answer:

The conversion in the real reactor is = 88%

Explanation:

conversion = 98% = 0.98

process rate = 0.03 m^3/s

length of reactor = 3 m

cross sectional area of reactor = 25 dm^2

pulse tracer test results on the reactor :

mean residence time ( tm) = 10 s and variance (∝2) = 65 s^2

note:  space time (t) =

t = [tex]\frac{A*L}{Vo}[/tex]   Vo = flow metric flow rate , L = length of reactor , A = cross sectional area of the reactor

therefore (t) = [tex]\frac{25*3*10^{-2} }{0.03}[/tex] = 25 s

since the reaction is in first order

X = 1 - [tex]e^{-kt}[/tex]

[tex]e^{-kt}[/tex] = 1 - X

kt = In [tex]\frac{1}{1-X}[/tex]

k = In [tex]\frac{1}{1-X}[/tex] / t  

X = 98% = 0.98 (conversion in PFR ) insert the value into the above equation then  

K = 0.156 [tex]s^{-1}[/tex]

Calculating Da for a closed vessel

; Da = tk

      = 25 * 0.156 = 3.9

calculate Peclet number Per using this equation

0.65 = [tex]\frac{2}{Per} - \frac{2}{Per^2} ( 1 - e^{-per})[/tex]

therefore

[tex]\frac{2}{Per} - \frac{2}{Per^2} (1 - e^{-per}) - 0.65 = 0[/tex]

solving the Non-linear equation above( Per = 1.5 )

Attached is the Remaining part of the solution

Answer:

Explanation:

The formula for critical stress is

[tex]\sigma_c=\frac{K}{Y\sqrt{\pi a} }[/tex]

[tex]\sigma_c =\texttt{critical stress}[/tex]

K is the plane strain fracture toughness

Y is dimensionless parameters

We are to Determine the Critical stress

Now replacing the critical stress with 54.8

a with 0.2mm = 0.2 x 10⁻³

Y with 1

[tex]\sigma_c=\frac{54.8}{1\sqrt{\pi \times 0.2\times10^{-3}} } \\\\=\frac{54.8}{\sqrt{6.283\times10^{-4}} } \\\\=\frac{54.8}{0.025} \\\\=2186.20Mpa[/tex]

The fracture will not occur because this material can handle a stress of 2186.20Mpa  before fracture. it is obvious that is greater than 2023Mpa

Therefore, the specimen does not failure for surface crack of 0.2mm

Answer:

Electricity cost = $9.36

Explanation:

Given:

Electric power = 400 W = 0.4 KW

Unit cost of electricity = $0.13/kWh

Overall time = 1/4 (30 days) (24 hours) = 180 hours

Find:

Electricity cost

Computation:

Electricity cost = Electric power  x Unit cost of electricity x Overall time

Electricity cost = 0.4 x $0.13 x 180

Electricity cost = $9.36

Given:

Electric power = 400 W = 0.4 KW

Over all Time  = 30(1/4) = 7.5 days

Unit cost of electricity = $0.13/kWh

Find:

Electricity cost.

Computation:

Electricity cost = Electric power x Unit cost of electricity x Over all Time

Electricity cost = 0.4 x 0.13 x 7.5

Electricity cost = $

Answer:

Fluid viscosity, [tex]\mu = 2.57 * 10^{-3} kgm^{-1}s^{-1}[/tex]

Explanation:

Container depth, D = 12 cm = 0.12 m

Tube length, l = 30 cm = 0.3 m

Specific Gravity, [tex]\rho[/tex] = 0.95

Tube diameter, d = 2 mm = 0.002 m

Rate of flow, Q = 1.9 cm³/s = 1.9 * 10⁻⁶ m³/s

Calculate the velocity at point 2 ( check the diagram attached)

Rate of flow at section 2, [tex]Q = A_2 v_2[/tex]

[tex]Area, A_2 = \pi d^{2} /4\\A_2 = \pi/4 * 0.002^2\\A_2 = 3.14159 * 10^{-6} m^2[/tex]

[tex]v_{2} = Q/A_{2} \\v_{2} =\frac{1.9 * 10^{-6}}{3.14 * 10^{-6}} \\v_{2} = 0.605 m/s[/tex]

Applying the Bernoulli (energy flow) equation between Point 1 and point 2 to calculate the head loss:

[tex]\frac{p_{1} }{\rho g} + \frac{v_{1}^2 }{2 g} + z_1 = \frac{p_{2} }{\rho g} + \frac{v_{2}^2 }{2 g} + z_2 + h_f\\ z_1 = L + l = 0.12 + 0.3\\z_1 = 0.42\\p_1 = p_{atm}\\v_1 = 0\\z_2 = 0\\\frac{p_{atm} }{\rho g} + \frac{0^2 }{2 g} + 0.42= \frac{p_{atm} }{\rho g} + \frac{0.605^2 }{2 *9.8} +0 + h_f\\h_f = 0.401 m[/tex]

For laminar flow, the head loss is given by the formula:

[tex]h_f = \frac{128 Q \mu l}{\pi \rho g d^4} \\\\0.401 = \frac{128 * 1.9 * 10^{-6} * 0.3 \mu}{\pi *0.95* 9.8* 0.002^4}\\\\\\\mu = \frac{0.401 * \pi *0.95* 9.8* 0.002^4}{128 * 1.9 * 10^{-6} * 0.3} \\\\\mu = 2.57 * 10^{-3} kgm^{-1}s^{-1}[/tex]

Answer:

0.00257 kg / m.s

Explanation:

Given:-

- The specific gravity of a liquid, S.G = 0.95

- The depth of fluid in free container, h = 12 cm

- The length of the vertical tube , L = 30 cm

- The diameter of the tube, D = 2 mm

- The flow-rate of the fluid out of the tube into atmosphere, Q = 1.9 cm^3 / s

Find:-

To determine the viscosity of a liquid

Solution:-

- We will consider the exit point of the fluid through the vertical tube of length ( L ), where the flow rate is measured to be Q = 1.9 cm^3 / s

- The exit velocity ( V2 ) is determined from the relation between flow rate ( Q ) and the velocity at that point.

                              [tex]Q = A*V_2[/tex]

Where,

             A: The cross sectional area of the tube

- The cross sectional area of the tube ( A ) is expressed as:

                                [tex]A = \pi \frac{D^2}{4} \\\\A = \pi \frac{0.002^2}{4} \\\\A = 3.14159 * 10^-^6 m^2[/tex]

- The velocity at the exit can be determined from the flow rate equation:

                              [tex]V_2 = \frac{Q}{A} \\\\V_2 = \frac{1.9*10^-^6}{3.14159*10^-^6} \\\\V_2 = 0.605 \frac{m}{s}[/tex]

- We will apply the energy balance ( head ) between the points of top-surface ( free surface ) and the exit of the vertical tube.

                    [tex]\frac{P_1}{p*g} + \frac{V^2_1}{2*g} + z_t_o_p = \frac{P_2}{p*g} + \frac{V^2_2}{2*g} + z_d_a_t_u_m + h_L[/tex]

- The free surface conditions apply at atmospheric pressure and still ( V1 = 0 ). Similarly, the exit of the fluid is also to atmospheric pressure. Where, z_top is the total change in elevation from free surface to exit of vertical tube.

- The major head losses in a circular pipe are accounted using Poiessel Law:

                           [tex]h_L = \frac{32*u*L*V}{S.G*p*g*D^2}[/tex]

Where,

                  μ: The dynamic viscosity of fluid

                  L: the length of tube

                  V: the average velocity of fluid in tube

                  ρ: The density of water

- The average velocity of the fluid in the tube remains the same as the exit velocity ( V2 ) because the cross sectional area ( A ) of the tube remains constant throughout the tube. Hence, the velocity also remains constant.

- The energy balance becomes:

                      [tex]h + L = \frac{V_2^2}{2*g} + \frac{32*u*L*V_2}{S.G*p*g*D^2} \\\\0.42 = \frac{0.605^2}{2*9.81} + \frac{32*u*(0.3)*(0.605)}{0.95*998*9.81*0.002^2} \\\\u = 0.00257 \frac{kg}{m.s}[/tex]

- Lets check the validity of the Laminar Flow assumption to calculate the major losses:

                     [tex]Re = \frac{S.G*p*V_2*D}{u} \\\\Re = \frac{0.95*998*0.605*0.002}{0.00257} \\\\Re = 446 < 2100[/tex]( Laminar Flow )

Answer:

hello the required diagram is missing attached to the answer is the required diagram

7.9954 kip.ft

Explanation:

AB = 1550-Ib ( weight acting on AB )

BCD = 190 - Ib ( weight of cage )

169-Ib = weight of man inside cage

Attached is the free hand diagram of the question

calculate distance [tex]x![/tex]

= cos 75⁰ = [tex]\frac{x^!}{10ft}[/tex]

    [tex]x! = 10 * cos 75^{o}[/tex] = 2.59 ft

calculate distance x

= cos 75⁰ = [tex]\frac{x}{30ft}[/tex]

x = 30 * cos 75⁰ = 7.765 ft

The resultant moment  produced by all the weights about point A

∑ Ma = 0

Ma = 1550 * [tex]x![/tex] + 190 ( x + 2.5 ) + 169 ( x + 2.5 + 1.75 )

Ma = 1550 * 2.59 + 190 ( 7.765 + 2.5 ) + 169 ( 7.765 + 2.5 + 1.75 )

      = 4014.5 + 1950.35 + 2030.535

      = 7995.385 ft. Ib ≈ 7.9954 kip.ft

Answer:

The answer is option # 2. (Constraints).

KINDLY NOTE that there is a picture in the question. Check the picture below for the picture.

==================================

Answer:

(1). 1.2 metres.

(2). There is going to be the same pressure.

Explanation:

From the question above we can take hold of the statement Below because it is going to assist or help us in solving this particular Question or problem;

" Measurements indicate that the material of tank B will fail and the tank will burst if the air pressure in tank B exceeds 18 kPa."

=> Also, the density of oil = 930

That is if Pressure, P in B > 18kpa there will surely be a burst.

The height, h the can waste oil be poured into tank A is;

The maximum pressure  = height × acceleration due to gravity × density) + ( acceleration due to gravity × density × height, j).

18 × 10^3 = (height, h ×  10 × 930) + 10 × (2 - 1.25) × 1000.

When we make height, h the Subject of the formula then;

Approximately, Height, h = 1.2 metres.

(2). If air is accidentally trapped in the manometer line, what will be the error in the calculation of the height we will have the same pressure.

Answer:

In both cases, reynolds number is greater than 2000,so the flows can't be laminar.

Explanation:

A) For flow in a tube of uniform diameter, the reynolds number is defined as;

Re = 2ρvr/η

where;

ρ is the fluid density,

v its speed,

η is viscosity

r is the tube radius.

In this question,

We are given;

r = 0.25cm = 0.25 × 10^(-2) m

η of water has a standard value of 1.005 × 10^(-3)

ρ of water has a standard value of 1000 kg/m³

In the reynolds equation, we don't know the velocity. So let's calculate it from;

Q' = vA

Where; Q' is flow rate = 0.5 L/s = 0.0005 m³/s

Area = πr² = π × (0.25 × 10^(-2))²

Area = 1.963 × 10^(-5) m²

So, v = Q/A = 0.0005/(1.963 × 10^(-5)) = 25.5 m/s

So, Re = 2ρvr/η = (2*1000*25.5*0.25 × 10^(-2))/(1.005 × 10^(-3))

Re = 126865.67

Re > 2000 and so the flow is not laminar

B) Now,

radius = 0.9cm = 0.9 × 10^(-2) m

So, A = πr² = π × (0.9 × 10^(-2))²

A = 2.5447 × 10^(-4) m²

v = Q/A = 0.0005/(2.5447 × 10^(-4))

v = 1.965 m/s

Re = 2ρvr/η = (2*1000*1.965*0.9 × 10^(-2))/(1.005 × 10^(-3))

Re = 35194.03

Re > 2000. So flow is not laminar.

Answer:

the average normal stress in the bar is : [tex]\mathbf{\sigma = (32.5 - 20x )MPa}[/tex]

Explanation:

The free body flow of the missing diagram is attached to the answer below.

From the information given:

Let consider the sum of forces along horizontal direction to be equal to zero.

[tex]\begin{array}{l}\\\sum {{F_x}} = 0\\\\3 + 8\left( {1.25 - x} \right) - N = 0\\\\N = \left( {13 - 8x} \right){\rm{ kN}}\\\end{array}[/tex]

The average normal stress in the bar can be calculated by the formula:

[tex]\sigma = \dfrac{N}{A}[/tex]

where;

[tex]\sigma =[/tex] average normal stress in the bar

A = cross sectional area in the bar and it is given by: [tex]400*10^{-6 } m^2[/tex]

​N = (13- 8x) kN

∴ [tex]\sigma = \dfrac{(13-8)x}{400*10^{-6} m^2}[/tex]

[tex]\sigma = (32.5 - 20x )*10^3 kPa[/tex]

[tex]\sigma = (32.5 - 20x )MPa[/tex]

Thus; the average normal stress in the bar is : [tex]\mathbf{\sigma = (32.5 - 20x )MPa}[/tex]

The average normal stress in the bar as a function of x is equal to 32.5 - 20x MPa.

Given the following data:

To calculate the average normal stress in the bar as a function of x.

First of all, we would determine the sum of the forces acting on the bar in the horizontal direction;

[tex]\sum F_x=0\\\\3+8(1.25-x)-N=0\\\\3+10-8x-N=0\\\\13-8x-N=0\\\\N=(13-8x)\; kN[/tex]

For the average normal stress:

Mathematically, the average normal stress is given by this formula:

[tex]\sigma = \frac{N}{A}[/tex]

Where:

Substituting the parameters into the formula, we have;

[tex]\sigma = \frac{(13-8x) \times 10^3}{400 \times 10^{-6}}\\\\\sigma = \frac{(13-8x) \times 10^{3+6}}{400}\\\\\sigma =(32.5-20x) \times 10^{9}[/tex]

Note: 1 MPa = [tex]1\times 10^9\;Pa[/tex]

Average normal stress = 32.5 - 20x MPa.

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Answer:

Waterfall model

Explanation:

The waterfall model is amenable to the projects. It focused on the data structure. The software architecture and detail about the procedure. It will interfere with the procedure. It interfaces with the characterization of the objects. The waterfall model is the first model that is introduced first. This model also called a linear sequential life cycle model.

The waterfall model is very easy to use. This is the earliest approach of the SDLC.

There are different phase of the waterfall:

Answer:

(a)Tb = 330.12 K (b)Tc =304.73 K (c)19.81 K/m (d) h =60.65 W/m². K

Explanation:

Solution

Given that:

The mass flow rate of engine oil m = 1.81 x 10^-3 kg/s

Diameter of the tube, D = 1cm =0.01 m

Electrical heat rate, q =76 W/m

Wall Temperature, Ts = 370 K

Now,

From the properties table of engine oil we can deduce as follows:

thermal conductivity, k =0.139 W/m .K

Density, ρ = 854 kg/m³

Specific heat, cp = 2120 J/kg.K

(a) Thus

The wall heat flux is given as follows:

qs = q/πD

=76/π *0.01

= 2419.16 W/m²

Now

The oil mean temperature is given as follows:

Tb =Ts -11/24 (q.R/k) (R =D/2=0.01/2 = 0.005 m)

Tb =370 - 11/24 * (2419.16 * 0.005/0.139)

Tb = 330.12 K

(b) The center line temperature is given below:

Tc =Ts - 3/4 (qs.R/k)= 370 - 3/4 * ( 2419.16 * 0.005/0.139)

Tc =304.73 K

(c) The flow velocity is given as follows:

V = m/ρ (πR²)

Now,

The The axial gradient of the mean temperature is given below:

dTb/dx = 2 *qs/ρ *V*cp * R

=2 *qs/ρ*[m/ρ (πR²) *cp * R

=2 *qs/[m/(πR)*cp

dTb/dx = 2 * 2419.16/[1.81 x 10^-3/(π * 0.005)]* 2120

dTb/dx = 19.81 K/m

(d) The heat transfer coefficient is given below:

h =48/11 (k/D)

=48/11 (0.139/0.01)

h =60.65 W/m². K

Answer:

a. Compromise the fire rating by one hour.

Explanation:

One hour fire rating is given to materials that can resist the fire exposure. The Acoustical panels controls reverberation and they are used for echo controls. The Fire rating is the passive fire protection which can resist standard fire. The test for fire rating also consider normal functioning of the material.

Answer:

The viscosities of the oils are 0.967 Pa.s and 1.933 Pa.s

Explanation:

Assuming the two oils are Newtonian fluids.

From Newton's law of viscosity for Newtonian fluids, we know that the shear stress is proportional to the velocity gradient with the viscosity serving as the constant of proportionality.

τ = μ (∂v/∂y)

There are oils above and below the plate, so we can write this expression for the both cases.

τ₁ = μ₁ (∂v/∂y)

τ₂ = μ₂ (∂v/∂y)

dv = 0.3 m/s

dy = (0.06/2) = 0.03 m (the plate is centered in a gap of width 0.06 m)

τ₁ = μ₁ (0.3/0.03) = 10μ₁

τ₂ = μ₂ (0.3/0.03) = 10μ₂

But the shear stress on the plate is given as 29 N per square meter.

τ = 29 N/m²

But this stress is a sum of stress due to both shear stress above and below the plate

τ = τ₁ + τ₂ = 10μ₁ + 10μ₂ = 29

But it is also given that one viscosity is twice the other

μ₁ = 2μ₂

10μ₁ + 10μ₂ = 29

10(2μ₂) + 10μ₂ = 29

30μ₂ = 29

μ₂ = (29/30) = 0.967 Pa.s

μ₁ = 2μ₂ = 2 × 0.967 = 1.933 Pa.s

Hope this Helps!!!

Answer: A

Original length = 1.60 m

Explanation: given that due to loading, a line segment has length 2 with constant normal strain 0.25

Strain is the ratio of extension to original length. That is,

Strain = e/L

If a line segment has length 2, that means:

e + L = 2

e = 2 - L

And given that the strain = 0.25

Substitute all the parameters into the formula

0.25 = ( 2 - L ) / L

Cross multiply

0.25L = 2 - L

Collect the like terms

0.25L + L = 2

1.25L = 2

L = 2/ 1.25

L = 1.6 m

Therefore, original length is 1.6 metres

Answer:

hello attached is the free body diagram of the missing figure

Fr = [tex]\frac{\pi }{4} D^2 [ ( P1 - P2) - pV^2 ][/tex]

Explanation:

Average velocity is constant i.e V1 = V2 = V

The momentum equation for the flow in the Z - direction can be expressed as

-Fr + P1 Ac - P2 Ac = mB2V2 - mB1V1 ------- equation 1

Fr = horizontal force on the bolts

P1 = pressure of fluid at entrance

V1 = velocity of fluid at entrance

Ac = cross section area of the pipe

P2 and V2 = pressure and velocity of fluid at some distance

m = mass flow rate of fluid

B1 = momentum flux at entrance , B2 = momentum flux correction factor

Note; average velocity is constant hence substitute V for V1 and V2

equation 1 becomes

Fr = ( P1 - P2 ) Ac + mV ( 1 - 2 )

Fr = ( P1 - P2 ) Ac - mV ---------------- equation 2

equation for mass flow rate

m = pAcV  

p = density of the fluid

insert this into equation 2 EQUATION 2 BECOMES

Fr = ( P1 - P2) Ac - pAcV^2

    = Ac [ (P1 - P2) - pV^2 ]  ---------- equation 3

Note Ac = [tex]\frac{\pi }{4} D^2[/tex]

Equation 3 becomes

Fr = [tex]\frac{\pi }{4} D^2[/tex] [ (P1 -P2 ) - pV^2 ] ------- relation for the horizontal force acting on the bolts      

Answer:

I have attached the diagram for this question below. Consult it for better understanding.

Find the cross sectional area AB:

A = (1.6mm)(12mm) = 19.2 mm² = 19.2 × 10⁻⁶m

Forces is given by:

F = 2.75 × 10³ N

Horizontal Stress can be found by:

σ (x) = F/A

σ (x) = 2.75 × 10³ / 19.2 × 10⁻⁶m

σ (x) = 143.23 × 10⁶ Pa

Horizontal Strain can be found by:

ε (x) = σ (x)/ E

ε (x) = 143.23 × 10⁶ / 200 × 10⁹

ε (x) = 716.15 × 10⁻⁶

Find Vertical Strain:

ε (y) = -v · ε (y)

ε (y) = -(0.3)(716.15 × 10⁻⁶)

ε (y) = -214.84 × 10⁻⁶

For L = 0.05m

Change (x) = L · ε (x)

Change (x) = 35.808 × 10⁻⁶m

For W = 0.012m

Change (y) = W · ε (y)

Change (y) = -2.5781 × 10⁻⁶m

For t= 0.0016m

Change (z) = t · ε (z)

where

ε (z) = ε (y) ,so

Change (z) = t · ε (y)

Change (z) = -343.74 × 10⁻⁹m

A = A(final) - A(initial)

A = -8.25 × 10⁻⁹m²

(Consult second picture given below for understanding how to calculate area)

The resulting change in the 50-mm gauge length; the width of portion AB of the test coupon; the thickness of portion AB; the cross- sectional area of portion AB are respectively; Δx = 35.808 × 10⁻⁶ m; Δy = -2.5781 × 10⁻⁶m; Δ_z = -343.74 × 10⁻⁹m; A = -8.25 × 10⁻⁹m²

Let us first find the cross sectional area of AB from the image attached;

A = (1.6mm)(12mm) = 19.2 mm² = 19.2 × 10⁻⁶m

We are given;

Tensile Load; F = 2.75 kN = 2.75 × 10³ N

Horizontal Stress is calculated from the formula;

σₓ = F/A

σₓ = (2.75 × 10³)/(19.2 × 10⁻⁶)m

σₓ = 143.23 × 10⁶ Pa

Horizontal Strain is calculated from;

εₓ = σₓ/E

We are given E = 200 GPa = 200 × 10⁹ Pa

Thus;

εₓ = (143.23 × 10⁶)/(200 × 10⁹)

εₓ = 716.15 × 10⁻⁶

Formula for Vertical Strain is;

ε_y = -ν * εₓ

We are given ν = 0.30. Thus;

ε_y  = -(0.3) * (716.15 × 10⁻⁶)

ε_y  = -214.84 × 10⁻⁶

A) We are given;

Gauge Length; L = 0.05m

Change in gauge length is gotten from;

Δx = L * εₓ

Δx = 0.05 × 716.15 × 10⁻⁶

Δx = 35.808 × 10⁻⁶ m

B) From the attached diagram, the width is;

W = 0.012m

Change in width is;

Δy = W * ε_y

Δy = 0.012 * -214.84 × 10⁻⁶

Δy = -2.5781 × 10⁻⁶m

C) We are given;

Thickness of plate; t = 1.6 mm = 0.0016m

Change in thickness;

Δ_z = t * ε_z

where;

ε_z = ε_y

Thus;

Δ_z = t * ε_y

Δ_z = 0.0016 * -214.84 × 10⁻⁶

Δ_z = -343.74 × 10⁻⁹m

D) The change in cross sectional area is gotten from;

ΔA = A_final - A_initial

From calculating the areas, we have;

A = -8.25 × 10⁻⁹ m²

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Answer:

A.) 7.13 degree north east

B.) 806.23 km/h

C.) 7.44 hours

D.) 0.06 hours

Explanation:

Assume Washington is due east of San Francisco and Francisco with Washington in the positive x-direction

Also, the cross wind is blowing from north to south of 100 km/hr in y coordinate direction.

A.) Using Cartesian system of coordinates, the direction of flight for the plane to actually arrive in Washington can be calculated by using the formula

Tan Ø = y/x

Substitute y = 100 km/h and x = 800km/h

Tan Ø = 100/800

Tan Ø = 0.125

Ø = Tan^-1(0. 125)

Ø = 7.13 degrees north east.

Therefore, the direction of flight for the plane to actually arrive in Washington is 7.13 degree north east

B.) The speed in the San Francisco to Washington direction can be achieved by using pythagorean theorem

Speed = sqrt ( 800^2 + 100^2)

Speed = sqrt (650000)

Speed = 806.23 km/h

C.) Let us use the speed formula

Speed = distance / time

Substitute the speed and distance into the formula

806.23 = 6000/ time

Make Time the subject of formula

Time = 6000/806.23

Time = 7.44 hours

D.) If there is no cross wind,

Time = 6000/800

Time = 7.5 hour

Time difference = 7.5 - 7.44

Time difference = 0.06 hours

Answer:

54.52 MPa

Explanation:

power P = 46 kW = 46000 W

speed  N = 760 rpm

outer diameter of shaft D = 40 mm = 0.04 m

inner diameter of shaft d = 25 mm = 0.025 m

torque T = P/Ω

where Ω = angular speed in rad/s

Ω = 2πN/60 = (2 x 3.142 x 760)/60

Ω = 79.59 rad/s

from this,

torque T = 46000/79.59 = 577.96 N-m

the relationship between torque T, maximum shear stress τmax, and shaft diameters D and d is stated as

T = (π / 16) τmax ([tex]D^{4}[/tex] - [tex]d^{4}[/tex])/D

imputing the values, we have

577.96 = (3.142/16) x τmax x ([tex]0.04^{4}[/tex] - [tex]0.025^{4}[/tex])/0.04

577.96 = 0.196 x τmax x (5.4 x [tex]10^{-5}[/tex])

577.96 = 1.06 x [tex]10^{-5}[/tex] x τmax

τmax  ≅ 54.52 MPa

Answer:

The correct answer to the following question will be "[tex]a_{x}[/tex] or [tex]a_{y}[/tex]".

Explanation:

So that the above is the appropriate choice.

Answer:

In this example, the delay procedure is given below in the explanation section

Explanation:

Solution

The delay procedure is given below:

LDS # $4000 // load initial memory

LDAA #$FF

STAA  DDRA

LDAA #$00 //load address

STAA DDRB

THERE LDAA PORT B

           ANDA   #%00000011// port A and port B

           CMPA   #%0000011

           BNE     THERE

           LDAA   #$FF

           STAA    PORT A

           JSR       DELAY

           LDAA    #$00

           STAA     PORT A

           BACK     BRA BACK

Answer:

The power that fluid supplies to the turbine is 1752.825 kilowatts.

Explanation:

A turbine is a device that works usually at steady state. Given that heat losses exists and changes in kinetic energy are not negligible, the following expression allows us to determine the power supplied by the fluid to the turbine by the First Law of Thermodynamics:

[tex]-\dot Q_{loss} - \dot W_{out} + \dot m \cdot \left[(h_{in}-h_{out}) + \frac{1}{2}\cdot (v_{in}^{2}-v_{out}^{2}) \right] = 0[/tex]

Output power is cleared:

[tex]\dot W_{out} = -\dot Q_{loss} + \dot m \cdot \left[(h_{in}-h_{out})+\frac{1}{2}\cdot (v_{in}^{2}-v_{out}^{2}) \right][/tex]

If [tex]\dot Q_{loss} = 0.3\,kW[/tex], [tex]\dot m = 0.85\,\frac{kg}{s}[/tex], [tex]h_{in} = 2800\,\frac{kJ}{kg}[/tex], [tex]h_{out} = 1550\,\frac{kJ}{kg}[/tex], [tex]v_{in} = 45\,\frac{m}{s}[/tex] and [tex]v_{out} = 20\,\frac{m}{s}[/tex], then:

[tex]\dot W_{out} = -0.3\,kW + \left(0.85\,\frac{kg}{s} \right)\cdot \left\{\left(2800\,\frac{kJ}{kg}-1550\,\frac{kJ}{kg} \right)+\frac{1}{2}\cdot \left[\left(45\,\frac{m}{s} \right)^{2}-\left(20\,\frac{m}{s} \right)^{2}\right] \right\}[/tex]

[tex]\dot W_{out} = 1752.825\,kW[/tex]

The power that fluid supplies to the turbine is 1752.825 kilowatts.

Answer:

a) 2.45 KN

b) 0.0375 m

Explanation:

[tex](a) \quad \sigma_{v}=400 \times 10^{6} \mathrm{Pa} \quad A=\frac{\pi}{4} d^{2}=\frac{\pi}{4}(5)^{2}=19.635 \mathrm{mm}^{2}=19.635 \times 10^{-6} \mathrm{m}^{2}[/tex]

[tex]P_{U}=\sigma_{U} A=\left(400 \times 10^{6}\right)\left(19.635 \times 10^{-6}\right)=7854 \mathrm{N}[/tex]

[tex]P_{\text {al }}=\frac{P_{U}}{F S}=\frac{7854}{3.2}=2454 \mathrm{N}[/tex]

(b) [tex]\quad \delta=\frac{P L}{A E}=\frac{(2454)(60)}{\left(19.635 \times 10^{-6}\right)\left(200 \times 10^{9}\right)}=37.5 \times 10^{-3} \mathrm{m}[/tex]

Answer:

i took the test and its budget

Explanation:

Budget, of the following is normally included in the criteria of a design. Thus, option (c) is correct.

The term design refers to the visual representation of image and shapes. The image is based on vector raster and bitmap. Who design the graphic is called graphic designer. The design main purpose to design of logo, word art, and advertisement design. The design mostly software as Coral Draw, Photoshop, Canva etc.

The criteria of a design are the budget. The budget was the financial as well as the monetary terminology. The criteria of the budget are the decided to how much they spend, and they save. The design on the country budget and the five-year plan.

As a result, the budget following is normally included in the criteria of a design. Therefore, option (c) is correct.

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#SPJ6

Answer:

219kJ

Explanation:

The work done (W) on a gas in an isothermal process is given by;

W = -P₁V₁ ln[tex]\frac{V_{2}}{V_1}[/tex]      -----------------(i)

Where;

P₁ = initial pressure of the gas

V₁ = initial volume of the gas

V₂ = final volume of the gas

From the question;

P₁ = 600kPa = 6 x 10⁵Pa

V₁ = 0.45m³

V₂ = 0.2m³

Substitute these values into equation (i) as follows;

W = -6 x 10⁵ x 0.45 x ln [tex]\frac{0.2}{0.45}[/tex]

W = -6 x 10⁵ x 0.45 x ln (0.444)

W = -6 x 10⁵ x 0.45 x -0.811

W = 2.19 x 10⁵

W = 219 x 10³

W = 219kJ

Therefore, the work done on the gas during the compression process is 219kJ

A) The linear density expressions for BCC [110] and [111] directions in terms of the atomic radius R are;

i) LD_110 = √3/(4R√2)

ii) LD_111 = 1/(2R)

B) The linear density values for these same two directions for iron (Fe) are;

i) LD_110 = 2.4 × 10^(9) m^(-1)

ii) LD_111 = 4 × 10^(9) m^(-1)

A) i) To find linear density expression for BCC 110, first of all we will calculate the length of the vector using the length of the unit cell which is 4R/√3 and the cell edge length which is 4R. Thus, the vector length can now be calculated from this expression;

√((4R)² - (4R/√3)²)

This reduces to; 4R√(1 - 1/3) = 4R√(2/3)

Now, the expression for the linear density of this direction is;

LD_110 =

Number of atoms centered on (110) direction/vector length of 110 direction

In this case, there is only one atom centered on the 110 direction. Thus;

LD_110 = 1/(4R√(2/3))

LD_110 = √3/(4R√2)

ii)  The length of the vector for the direction 111 is equal to 4R, since

all of the atoms whose centers the vector passes through touch one another. In addition, the vector passes through an equivalent of 2 complete atoms. Thus, the linear density is;

LD_111 = 2/(4R) = 1/(2R)

B)i) From tables, the atomic radius for iron is 0.124 nm or 0.124 x 10^(-9) m. Therefore, the linear density for the [110] direction is;

LD_110 = √3/(4R√2) = √3/(4*0.124*10^(-9)(√2))

LD_110 = 2.4 × 10^(9) m^(-1)

ii) for the 111 direction, we have;

LD_111 = 1/(2R) = 1/(2*0.124*10^(-9))

LD_111 = 4 × 10^(9) m^(-1)

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