A shell-and tube heat exchanger (two shells, four tube passes) is used to heat 10,000 kg/h of pressurized water from 35 to 120 oC with 5000 kg/h pressurized water entering the exchanger at 300 oC. If the overall heat transfer coefficient is 1500 W/m^2-K, determine the required heat exchanger area.

Answer:

positive sides:

negative sides:

Answer:

Explanation:

For cylinder

Diameter d = 60 inches

thickness t = 0.625 inches

circumferential (hoop) stress = 30 ksi

[tex]hoop \ \ stress =\sigma_1=\frac{P_1d}{2t}\\\\\sigma_1=30ksi\\\\30000=\frac{P_1\times 60}{2\times0.625}\\\\P_1=624psi[/tex]

[tex]longitudinal \ \ stress =\sigma_2=\frac{P_2d}{2t}\\\\\sigma_2=30ksi\\\\30000=\frac{P_2\times 60}{4\times0.625}\\\\30000=\frac{P_2\times 60}{2.5}\\\\75000=P_2\times60\\\\P_2=\frac{75000}{60} \\\\P_1=1250psi[/tex]

Therefore maximum pressure without failure is P₁ = 625 psi

ii) For Sphere

[tex]\sigma_1=\sigma_2=\frac{Pd}{4t} \\\\P=\frac{30000\times 4 \times 0.625}{60} \\\\=\frac{75000}{60}\\\\=1250\ \ psi[/tex]

Answer:

a. T-V and P-V diagram are included

b. State 1: Specific volume = 0.0811753 m³/kg

State 2: Specific volume = 0.0811753 m³/kg

State 3: Specific volume = 0.0804155 m³/kg

c. Z = 51.1

d. Quality for state 1 = 100%

Quality for state 2 = 63.47%

Quality for state 3 = 100%

Explanation:

a. T-V and P-V diagram are included

b. State 1: Water vapor

P₁ = 3.0 MPa = 30 bar

T₁ = 300°C = 573.15

Saturation temperature = 233.86°C Hence the steam is super heated

Specific volume = 0.0811753 m³/kg

State 2:

Constant volume formula is P₁/T₁ = P₂/T₂

Specific volume = 0.0811753 m³/kg

T₂ = 200°C = 473.15

Therefore, P₂ = P₁/T₁ × T₂ = 3×473.15/573.15 = 2.4766 MPa

At T₂ water is mixed water and steam and the [tex]v_f[/tex] = 0.00115651 m³/kg

[tex]v_g[/tex] = 0.127222 m³/kg

State 3:

P₃ = 2.5 MPa

T₃ = 200°C

Isothermal compression P₂V₂ = P₃V₃

V₃ = P₂V₂ ÷ P₃ = 2.4766 × 0.0811753/2.5 = 0.0804155 m³/kg

Specific volume = 0.0804155 m³/kg

2) Compressibility factor is given by the relation;

[tex]Z = \dfrac{PV}{RT} = \dfrac{3\times 10^6 \times 0.0811753 }{8.3145 \times 573.15} = 51.1[/tex]

Z = 51.1

3) Gas quality, x, is given by the relation

[tex]x = \dfrac{Mass_{saturated \, vapor}}{Total \, mass} = \dfrac{v - v_f}{v_g - v_f}[/tex]

Quality at state 1 = Saturated quality = 100%

State 2 Vapor + liquid Quality

Gas quality = (0.0811753 - 0.00115651)/ (0.127222-0.00115651) = 63.47%

State 3: Saturated vapor, quality = 100%.

Answer:

Both technician A and technician B are correct

Explanation:

A planetary gearbox consists of a gearbox with the input shaft and the output shaft that is aligned to each other. It is used to transfer the largest torque in the compact form. A planetary gearbox has a compact size and low weight and it has high power density.

One planetary gear set can provide gear reduction, overdrive, and reverse. Also, most transmissions today use compound (multiple) planetary gears set.

So, both technician A and technician B are correct.

Answer:

Explanation:

Given that,

The area of glass [tex]A_g[/tex] = [tex]0.11m^2[/tex]

The thickness of the glass [tex]t_g=4mm=4\times10^-^3m[/tex]

The area of the styrofoam [tex]A_s=11m^2[/tex]

The thickness of the styrofoam [tex]t_s=0.20m[/tex]

The thermal conductivity of the glass [tex]k_g=0.80J(s.m.C^o)[/tex]

The thermal conductivity of the styrofoam  [tex]k_s=0.010J(s.m.C^o)[/tex]

Inside and outside temperature difference is ΔT

The heat loss due to conduction in the window is

[tex]Q_g=\frac{k_gA_g\Delta T t}{t_g} \\\\=\frac{(0.8)(0.11)(\Delta T)t}{4.0\times 10^-^3}\\\\=(22\Delta Tt)j[/tex]

The heat loss due to conduction in the wall is

[tex]Q_s=\frac{k_sA_s\Delta T t}{t_g} \\\\=\frac{(0.010)(11)(\Delta T)t}{0.20}\\\\=(0.55\Delta Tt)j[/tex]

The net heat loss of the wall and the window is

[tex]Q=Q_g+Q_s\\\\=\frac{k_gA_g\Delta T t}{t_g}+\frac{k_sA_s\Delta T t}{t_g}\\\\=(22\Delta Tt)j +(0.55\Delta Tt)j \\\\=(22.55\Delta Tt)j[/tex]

The percentage of heat lost by the window is

[tex]=\frac{Q_g}{Q}\times 100\\\\=\frac{22\Delta T t}{22.55\Delta T t}\times 100\\\\=97.6 \%[/tex]

Answer:

The answer is VN =37.416 m/s

Explanation:

Recall that:

Pressure (atmospheric) = 100 kPa

So. we solve for the maximum velocity (m/s) to which water can be accelerated by the nozzles

Now,

Pabs =Patm + Pgauge = 800 KN/m²

Thus

PT/9.81 + VT²/2g =PN/9.81  + VN²/2g

Here

Acceleration due to gravity = 9.81 m/s

800/9.81 + 0

= 100/9.81 + VN²/19.62

Here,

9.81 * 2= 19.62

Thus,

VN²/19.62 = 700/9.81

So,

VN² =1400

VN =37.416 m/s

Note: (800 - 100) = 700

Answer:

[tex]V2 = 37.417ms^{-1}[/tex]

Explanation:

Given the following data;

Water enters the turbine nozzles (inlet) = 800kPa = 800000pa.

Nozzle outlets = 100kPa = 100000pa.

Density of water = 1000kg/m³.

We would apply, the Bernoulli equation between the inlet and outlet;

[tex]\frac{P_{1} }{d}+\frac{V1^{2} }{2} +gz_{1} = \frac{P_{2} }{d}+\frac{V2^{2} }{2} +gz_{2}[/tex]

Where, V1 is approximately equal to zero(0).

Z[tex]z_{1} = z_{2}[/tex]

Therefore, to find the maximum velocity, V2;

[tex]V2 = \sqrt{2(\frac{P_{1} }{d}-\frac{P_{2} }{d}) }[/tex]

[tex]V2 = \sqrt{2(\frac{800000}{1000}-\frac{100000}{1000}) }[/tex]

[tex]V2 = \sqrt{2(800-100)}[/tex]

[tex]V2 = \sqrt{2(700)}[/tex]

[tex]V2 = \sqrt{1400}[/tex]

[tex]V2 = 37.417ms^{-1}[/tex]

Hence, the maximum velocity, V2 is 37.417m/s

Answer:

the overall elongation δ of the bar is  1.2337 mm

Explanation:

From the information given :

According to the principle of superposition being applied to the axial load P of the system; we have:

[tex]\delta = \delta_{AB} +\delta_{BC} + \delta_{CD}[/tex]    

where;

δ = overall elongation

[tex]\delta _{AB}[/tex] = elongation of bar AB

[tex]\delta _{BC}[/tex] = elongation of  bar BC

[tex]\delta _{CD} =[/tex]  elongation of bar CD]

If we replace; [tex]\dfrac{PL}{AE}[/tex] for  δ  and bt for area;

we have:

[tex]\delta = \dfrac{P_{AB}L_{AB}}{(b_{AB}t)E} +\dfrac{P_{BC}L_{BC}}{(b_{BC}t)E}+\dfrac{P_{CD}L_{CD}}{(b_{CD}t)E}[/tex]

where ;

P = load

L = length of the bar

A = area of the cross-section

E = young modulus of elasticity

Let once again replace:

P for [tex]P_{AB}, P_{BC} , P_{CD}[/tex]  (since load in all member of AB, BC and CD will remain the same )

[tex]\dfrac{L}{4}[/tex] for [tex]L_{AB}[/tex],  

[tex]\dfrac{L}{2}[/tex] for [tex]L_{BC}[/tex] and

[tex]\dfrac{L}{4}[/tex] for [tex]L_{CD}[/tex]

[tex]2\dfrac{b}{3}[/tex] for  [tex]b_{BC}[/tex]

b for  [tex]b_{CD}[/tex]

[tex]\delta = \dfrac{P (\dfrac{L}{4})}{btE}+ \dfrac{P (\dfrac{L}{2})}{2 \dfrac{b}{3}tE}+\dfrac{P (\dfrac{L}{4})}{btE}[/tex]

[tex]\delta = \dfrac{PL}{btE}[\dfrac{1}{4}+ \dfrac{1}{2}*\dfrac{3}{2}+ \dfrac{1}{4}][/tex]

[tex]\delta = \dfrac{5}{4}\dfrac{PL}{btE} --- \ (1)[/tex]

The stress in the central portion can be calculated as:

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

[tex]\sigma = \dfrac{P}{\dfrac{2}{3}bt}[/tex]

[tex]\sigma = \dfrac{3P}{2bt}[/tex]

So; Now:

[tex]\delta = \dfrac{5}{4}* \dfrac{2 * \sigma}{3}*\dfrac{L}{E}[/tex]

[tex]\delta= \dfrac{5}{4}* \dfrac{2 * 200}{3}*\dfrac{570}{77*10^3 \ MPa}[/tex]

δ = 1.2337 mm

Therefore, the overall elongation δ of the bar is  1.2337 mm

Answer:

Technician B is correct

Explanation:

O- rings are used with oil transmission filters to avoid transmission failures but some people use  lip seals as well. either of them is  inserted onto the outer part of the transmission system i.e it is inserted/found in-between Transmission filters and the transmission systems and it main purpose is to avoid leaks and transmission failure in the short and long term.

0-rings should be lubricated before installation this is because the o-rings are usually super tight when installing and would require lubrication to ease the installation process else the rubber might get ruptured and this would lead to instant transmission failure.

Answer:

South Pole and South Pole or North Pole and North Pole.

Answer:

Settling Velocity (Up)= 2.048*10^-5 m/s

Reynolds number Re = 2.159*10^-3

Explanation:

We proceed as follows;

Diameter of Particle = 0.09 mm = 0.09*10^-3 m

Solid Particle Density = 2002 kg/m3

Solid Fraction, θ= 0.45

Temperature = 26.7°C

Viscosity of water = 0.8509*10^-3 kg/ms

Density of water at 26.7 °C = 996.67 kg/m3

The velocity between the interface, i.e between the suspension and clear water is given by,

U = [ ((nf/ρf)/d)D^3] [18+(1/3)D^3)(1/2)]

D = d[(ρp/ρf)-1)g*(ρf/nf)^2]^(1/3)

D = 2.147

U = 0.0003m/s (n = 4.49)

Up = 0.0003 * (1-0.45)^4.49 = 2.048*10^-5 m/s

Re=0.09*10^-3*2.048*10^-5*996.67/0.0008509 = 2.159*10^-3

Answer:

The final temperature and pressure in the insulated rigid tank are [tex]48\,^{\circ}C[/tex] and [tex]375\,kPa[/tex].

Explanation:

An ideal gas is represented by the following model:

[tex]P\cdot V = \frac{m}{M}\cdot R_{u} \cdot T[/tex]

Where:

[tex]P[/tex] - Pressure, measured in kilopascals.

[tex]V[/tex] - Volume, measured in cubic meters.

[tex]m[/tex] - Mass of the ideal gas, measured in kilograms.

[tex]M[/tex] - Molar mass, measured in kilograms per kilomole.

[tex]T[/tex] - Temperature, measured in Kelvin.

[tex]R_{u}[/tex] - Universal constant of ideal gases, equal to [tex]8.314\,\frac{kPa\cdot m^{3}}{kmol\cdot K}[/tex]

As tank is rigid and insulated, it means that no volume deformations in tank, heat and mass interactions with surroundings occur during expansion process. Hence, final pressure is less that initial one, volume is doubled (due to equal partitioning) and temperature remains constant. Hence, the following relationship can be derived from model for ideal gases:

[tex]\frac{P_{1}\cdot V_{1}}{T_{1}} = \frac{P_{2}\cdot V_{2}}{T_{2}}[/tex]

Now, final pressure is cleared:

[tex]P_{2} = P_{1}\cdot \frac{T_{2}}{T_{1}}\cdot \frac{V_{1}}{V_{2}}[/tex]

[tex]P_{2} = (750\,kPa)\cdot 1 \cdot \frac{1}{2}[/tex]

[tex]P_{2} = 375\,kPa[/tex]

The final temperature and pressure in the insulated rigid tank are [tex]48\,^{\circ}C[/tex] and [tex]375\,kPa[/tex].

Answer:

(a) 2.39 MPa (b) 3.03 kJ (c) 3.035 kJ

Explanation:

Solution

Recall that:

A 10 gr of air is compressed isentropically

The initial air is at = 27 °C, 110 kPa

After compression air is at = a450 °C

For air,  R=287 J/kg.K

cv = 716.5 J/kg.K

y = 1.4

Now,

(a) W efind the pressure on [MPa]

Thus,

T₂/T₁ = (p₂/p₁)^r-1/r

=(450 + 273)/27 + 273) =

=(p₂/110) ^0.4/1.4

p₂ becomes  2390.3 kPa

So, p₂ = 2.39 MPa

(b) For the increase in total internal energy, is given below:

ΔU = mCv (T₂ - T₁)

=(10/100) (716.5) (450 -27)

ΔU =3030 J

ΔU =3.03 kJ

(c) The next step is to find the total work needed in kJ

ΔW = mR ( (T₂ - T₁) / k- 1

(10/100) (287) (450 -27)/1.4 -1

ΔW = 3035 J

Hence, the total work required is = 3.035 kJ

Answer:

s = v√[2(H - h)/g]

This formula describes the distance from the building that Prof Murthy must be when you release the egg

Explanation:

First, we need to find the time required by the egg to reach the head of Professor. For that purpose, we use 1st equation of motion in vertical form:

Vf = Vi + gt

where,

Vf = Velocity of egg at the time of hitting head of the Professor

Vi = initial velocity of egg = 0 m/s  (Since, egg is initially at rest)

g = acceleration due to gravity

t = time taken by egg to come down

Therefore,

Vf = 0 + gt

t = Vf/g   ----- equation (1)

Now, we use 3rd euation of motion for Vf:

2gs = Vf² - Vi²

where,

s = height dropped = H - h

Therefore,

2g(H - h) = Vf²

Vf = √[2g(H - h)]

Therefore, equation (1) becomes:

t = √[2g(H - h)]/g

t = √[2(H - h)/g]

Now, consider the horizontal motion of professor. So, the minimum distance of professor from building can be found out by finding the distance covered by the professor in time t. Since, the professor is running at constant speed of v m/s. Therefore:

s = vt

s = v√[2(H - h)/g]

This formula describes the distance from the building that Prof Murthy must be when you release the egg

Answer:

Roll force, F = 5.6 MN

Explanation:

Sheet width, b = 0.5 m

Roll contact length, [tex]l_{p} = 40 mm[/tex]

Strength coefficient, [tex]\sigma_{0} = 200 MPa[/tex]

Thickness, h = 11 mm

Since the sheet of metal is cold worked by 40%, the reduction in thickness will be:

Δh = 40% * 11 = 0.4 * 11 = 4.4 mm

Strain, e = (Δh)/h

e = 4.4/11 = 0.4

The roll force is calculated by the formula:

[tex]F = \sigma_{f} l_{p} b[/tex]

[tex]\sigma_{f} = \sigma_{0} (e+1)\\\sigma_{f} = 200 (0.4+1)\\\sigma_{f} = 200 *1.4\\\sigma_{f} = 280 MPa[/tex]

Substituting the value of [tex]\sigma_{f}[/tex], [tex]l_{p}[/tex], and b into the formula for the roll force:

[tex]F = \sigma_{f} l_{p} b\\F = 280 * 0.04 * 0.5\\F = 5.6 MN[/tex]

Answer:

See explanation

Explanation:

See the document for the complete FBD and the introductory part of the solution.

Static Balance ( Sum of Forces = 0 ) in all three directions

                 ∑[tex]F_G_X = W - G_x = 0[/tex]

                 [tex]G_X = W = 1500 lb[/tex]

                 ∑[tex]F_G_Y = P - G_Y = 0[/tex]

                 [tex]G_Y = P = -10,800 lb[/tex]

                ∑[tex]F_G_Z = - G_Z = 0[/tex]

Where, ( [tex]G_X, G_Y, G_Z[/tex] ) are internal forces at section ( G ) along the defined coordinate axes.

Static Balance ( Sum of Moments about G = 0 ) in all three directions

              [tex]M_G = r_O_G x F_O[/tex]

Where,

              r_OG: The vector from point O to point G

              F_OG: The force vector at point O

- The vector ( r_OG ) and ( F_OG ) can be written as follows:

              [tex]r_O_G = [ -( 3 + \frac{H}{2} ) i + (\frac{r_o}{12})j - ( \frac{r_o}{12} + \frac{L}{2})k ] \\\\r_O_G = [ -( 6 ) i + (0.25)j - (6)k ] \\[/tex]

              [tex]F_O_G = [ ( W ) i + ( P ) k ]\\\\F_O_G = [ (1500) i - ( 10,800 ) k ] lb[/tex]

- Then perform the cross product of the two vectors ( r_OG ) and ( F_OG ):

     [tex]( M_G_X )i + (M_G_Y)j+ (M_G_Z)k = \left[\begin{array}{ccc}i&j&k\\-6&0.25&-6\\1500&-10,800&0\end{array}\right] \\\\\\( M_G_X )i + (M_G_Y)j+ (M_G_Z)k = -( 6*10,800 ) i - ( 6*1500 ) j + [ ( 10,800*6) - ( 0.25*1500) ] k\\\\( M_G_X )i + (M_G_Y)j+ (M_G_Z)k = - (64,800)i - (9,000)j + (64,425)k[/tex]

- The internal torque ( T ) and shear force ( V ) that act on slice ( G ) are due to pressure force ( P ) as follows:

             [tex]T = P*[\frac{L}{2}] = (10,800)*(6) = 64,800 lb.ft[/tex]

             [tex]V = P = -10,800 lb[/tex]

- For the state of stress at point "C" we need to determine the the normal stress along x direction ( σ_x ) and planar stress ( τ_xy ) as follows:-

             σ_x = [tex]-\frac{G_x}{A} - \frac{M_G_Y. z*}{I_Y_Y} + \frac{M_G_Z. y*}{I_Z_Z}[/tex]

Where,

          A: The area of pipe cross section

          [tex]A = \pi * [ ( \frac{r_o}{12})^2 - ( \frac{r_i}{12})^2 ] = \pi * [ ( \frac{3}{12})^2 - ( \frac{2.75}{12})^2 ] = 0.03136 ft^2[/tex]

          z*: The distance of point "C" along z-direction from central axis ( x )

          [tex]z*= [\frac{r_i}{12} ] = [\frac{2.75}{12} ] = 0.22916 ft[/tex]

         I_YY: The second area moment of pipe along and about "y" axis:

         [tex]I_Y_Y = \frac{\pi }{4} * [ (\frac{r_o}{12})^4 - (\frac{r_i}{12})^4 ]=\frac{\pi }{4} * [ (\frac{3}{12})^4 - (\frac{2.75}{12})^4 ] \\\\I_Y_Y = 0.00090 ft^4[/tex]

         y*: The distance of point "C" along y-direction from central axis ( x )

         [tex]y* = 0[/tex]

- The normal stress ( σ_x ) becomes:

          σ_x = [tex][-\frac{1500}{0.03136} - \frac{-9,000*0.22916}{0.00090} + \frac{64,425*0}{0.00090} ] * (\frac{1}{12})^2 = 15.5 ksi[/tex]

- The planar stress is ( τ_xy ) is a contribution of torsion ( T ) and shear force ( V ):

           τ_xy = [tex]- \frac{T.c}{J} + \frac{V.Q}{I.t}[/tex]

Where,

           c: The radial distance from central axis ( x ) and point "C".

           [tex]c = \frac{r_i}{12} = \frac{2.75}{12} = 0.22916 ft[/tex]

          J: The polar moment of inertia of the annular cross section of pipe:

          [tex]J = \frac{\pi }{2}* [ ( \frac{r_o}{12})^4 - ( \frac{r_i}{12})^4 ] = \frac{\pi }{2}* [ ( \frac{3}{12})^4 - ( \frac{2.75}{12})^4 ] = 0.00180 ft^4[/tex]

          Q: The first moment of area for point "C" = semi-circle

          [tex]Q = Y_c*A_c = \frac{4*( r_m)}{3\pi } * \frac{\pi*( r_m)^2 }{2} = \frac{2. ( r_m)^3}{3} \\\\Q = \frac{2. [ ( \frac{r_o}{12})^3 - ( \frac{r_i}{12})^3] }{3} = \frac{2. [ ( \frac{3}{12})^3 - ( \frac{2.75}{12})^3] }{3} = 0.00239ft^3[/tex]

          I: The second area moment of pipe along and about "y" axis:

         [tex]I_Y_Y = \frac{\pi }{4} * [ (\frac{r_o}{12})^4 - (\frac{r_i}{12})^4 ]=\frac{\pi }{4} * [ (\frac{3}{12})^4 - (\frac{2.75}{12})^4 ] \\\\I_Y_Y = 0.00090 ft^4[/tex]    

         t: The effective thickness of thin walled pipe:

         [tex]t = 2* [ \frac{r_o}{12} - \frac{r_i}{12} ] = 2* [ \frac{3}{12} - \frac{2.75}{12} ] = 0.04166 ft[/tex]

- The planar stress is ( τ_xy ) becomes:

        τ_xy =  [tex][ - \frac{-64,800*0.22916}{0.0018} + \frac{-10,800*0.00239}{0.0009*0.04166} ] * [ \frac{1}{12}]^2 = 52.4 ksi[/tex]

- The principal stresses at point "C" can be determined from the following formula:-

       σ_x = 15.55 ksi,  σ_y = 0 ksi , τ_xy = 52.4 ksi

       σ_1 =[tex]\frac{sigma_x+sigma_y}{2} + \sqrt{(\frac{sigma_x+sigma_y}{2})^2 + (tow_x_y)^2 }[/tex]

       σ_2 = [tex]\frac{sigma_x+sigma_y}{2} - \sqrt{(\frac{sigma_x+sigma_y}{2})^2 + (tow_x_y)^2 }[/tex]

        σ_1 = [tex]\frac{15.55+0}{2} + \sqrt{(\frac{15.55+0}{2})^2 + (52.4)^2 } = 60.75 ksi[/tex]

        σ_2 =[tex]-\sqrt{\left(\frac{15.55+0}{2}\right)^2\:+\:\left(52.4\right)^2\:}+\frac{15.55+0}{2} = -45.20 ksi[/tex]

- The angle of maximum plane stress ( θ ):

       θ = [tex]0.5*arctan ( \frac{tow_x_y}{\frac{sigma_x-sigma_y}{2} } )= 0.5*arctan*( \frac{52.4}{7.8} ) = 40.8 deg[/tex]

Note: The plane stresses at point D are evaluated using the following procedure given above. Due to 5,000 character limit at Brainly, i'm unable to post here.

Answer:

A periodic function is a function that returns to its value over a certain period at regular intervals an example is the wave form of flux density (B) = sin wt

Explanation:

A periodic function is a function that returns to its value over a certain period at regular intervals an example is the wave form of flux density (B) = sin wt

attached to the answer is a free plot of the output starting with zero degree for one coil rotating in a uniform magnetic field

B ( wave flux density ) = Bm sinwt  and w = 2[tex]\pi[/tex]f = [tex]\frac{2\pi }{T}[/tex] rad/sec

Answer:

None of these

Explanation:

There are different types of amplifiers, and each has different characteristics.

Therefore, the correct answer is "None of these "

Answer:

a = 0.07m or 70mm

b = 0.205m or 205mm

Explanation:

Given the following data;

Modulus of rigidity, G = 14MPa=14000000Pa.

c = 80mm = 0.08m.

P = 46kN=46000N.

Shearing stress (r) in the rubber shouldn't exceed 1.4MPa=1400000Pa.

Deflection (d) of the plate is to be at least 7mm = 0.007m.

From shearing strain;

[

[tex]Modulus Of Elasticity, E = \frac{d}{a} =\frac{r}{G}[/tex]

Making a the subject formula;

[tex]a = \frac{Gd}{r}[/tex]

Substituting into the above formula;

[tex]a = \frac{14000000*0.007}{1400000}[/tex]

[tex]a = \frac{98000}{1400000}[/tex]

[tex]a = 0.07m or 70mm[/tex]

a = 0.07m or 70mm.

Also, shearing stress;

[tex]r = \frac{P}{2bc}[/tex]

Making b the subject formula;

[tex]b = \frac{P}{2cr}[/tex]

Substituting into the above equation;

[tex]b = \frac{46000}{2*0.08*1400000}[/tex]

[tex]b = \frac{46000}{224000}[/tex]

[tex]b = 0.205m or 205mm[/tex]

b = 0.205m or 205mm

Answer:

Explanation:

Given that:

V = 12.5m/s

L= 2.70m

b= 0.65m

[tex]T_{ \infty} = 27^0C= 273+27 = 300K[/tex]

[tex]T_s= 127^0C = (127+273)= 400K[/tex]

P = 1atm

Film temperature

[tex]T_f = \frac{T_s + T_{\infty}}{2} \\\\=\frac{400+300}{2} \\\\=350K[/tex]

dynamic viscosity =

[tex]\mu =20.9096\times 10^{-6} m^2/sec[/tex]

density = 0.9946kg/m³

Pr = 0.708564

K= 229.7984 * 10⁻³w/mk

Reynolds number,

[tex]Re = \frac{SUD}{\mu} =\frac{\ SUl}{\mu}[/tex]

[tex]=\frac{0.9946 \times 12.5\times 2.7}{20.9096\times 10^-^6} \\\\Re=1605375.043[/tex]

we have,

[tex]Nu=\frac{hL}{k} =0.037Re^{4/5}Pr^{1/3}\\\\\frac{h\times2.7}{29.79\times 10^-63} =0.037(1605375.043)^{4/5}(0.7085)^{1/3}\\\\h=33.53w/m^2k[/tex]

we have,

heat transfer rate from top plate

[tex]\theta _1 =hA(T_s-T_{\infty})\\\\A=Lb\\\\=2.7*0.655\\\\ \theta_1=33.53*2.7*0.65(127/27)\\\\ \theta_1=5884.51w[/tex]

Answer:

a) 4160 V

b) 12 kW and 81 kVAR

c)  54 kW and 477 kVAR

Explanation:

1) The phase voltage is given as:

[tex]V_p=\frac{3810.5}{\sqrt{3} }=2200 V[/tex]

The complex power S is given as:

[tex]S=560.1(0.707 +j0.707)+132=660\angle 36.87^o \ KVA[/tex]

[tex]where\ S^*\ is \ the \ conjugate\ of \ S\\Therefore\ S^*=660\angle -36.87^oKVA[/tex]

The line current I is given as:

[tex]I=\frac{S^*}{3V}=\frac{660000\angle -36.87}{3(2200)} =100\angle -36.87^o\ A[/tex]

The phase voltage at the sending end is:

[tex]V_s=2200\angle 0+100\angle -36.87(0.4+j2.7)=2401.7\angle 4.58^oV[/tex]

The magnitude of the line voltage at the source end of the line ([tex]V_{sL}=\sqrt{3} |V_s|=\sqrt{3} *2401.7=4160V[/tex]

b) The Total real and reactive power loss in the line is:

[tex]S_l=3|I|^2(R+jX)=3|100|^2(0.4+j2.7)=12000+j81000[/tex]

The real power loss is 12000 W = 12 kW

The reactive power loss is 81000 kVAR = 81 kVAR

c) The sending power is:

[tex]S_s=3V_sI^*=3(2401.7\angle 4.58)(100\angle 36.87)=54000+j477000[/tex]

The Real power delivered by the supply = 54000 W = 54 kW

The Reactive power delivered by the supply = 477000 VAR = 477 kVAR

Answer:

a) [tex]\ddot n = 50400\,\frac{rev}{min^{2}}[/tex], b) [tex]n = 2450\,rev[/tex]

Explanation:

a) The acceleration experimented by the grinding wheel is:

[tex]\ddot n = \frac{4200\,\frac{rev}{min} - 0 \,\frac{rev}{min} }{\frac{5}{60}\,min }[/tex]

[tex]\ddot n = 50400\,\frac{rev}{min^{2}}[/tex]

Now, the number of revolutions done by the grinding wheel in that period of time is:

[tex]n = \frac{(4200\,\frac{rev}{min} )^{2}-(0\,\frac{rev}{min} )^{2}}{2\cdot \left(50400\,\frac{rev}{min^{2}} \right)}[/tex]

[tex]n = 175\,rev[/tex]

b) The acceleration experimented by the grinding wheel is:

[tex]\ddot n = \frac{0 \,\frac{rev}{min} - 4200\,\frac{rev}{min} }{\frac{70}{60}\,min }[/tex]

[tex]\ddot n = -3600\,\frac{rev}{min^{2}}[/tex]

Now, the number of revolutions done by the grinding wheel in that period of time is:

[tex]n = \frac{(0\,\frac{rev}{min} )^{2} - (4200\,\frac{rev}{min} )^{2}}{2\cdot \left(-3600\,\frac{rev}{min^{2}} \right)}[/tex]

[tex]n = 2450\,rev[/tex]

Answer:

[tex]W_{net} = - 1223 kJ[/tex]

Explanation:

State 1:

[tex]P_1 = 20 bar\\T_1 = 360^{0}C\\ h_1 = 3159.3 kJ/kg\\S_1 = 6.9917 kJ/kg[/tex]

State 2:

[tex]P_2 = 20 bar\\x_2 = 1 \\ h_2 = 2799.5 kJ/kg\\u_2 = 2600.3 kJ/kg\\v_2 = 0.09963m^3/kg[/tex]

State 3:

[tex]P_2 = 5 bar\\v_2 = v_3 \\v_3 = v_f + x_3 (v_g - v_f)\\0.09963 = (1.0926 * 10^{-3}) +x_3 (0.3749 - (1.0926 * 10^{-3}))\\x_3 = 0.263[/tex]

[tex]u_{3} = u_f + x_3 ( u_g - u_f)\\u_{3} = 639.68 + 0.263 (2561.2 - 639.68)\\u_{3} = 1146.2 kJ/kg[/tex]

State 4:

[tex]P_{4} = 5 bar\\T_4 = 360^0 C\\h_4 = 3188.4 kJ/kg\\S_4 = 7.660 kJ/kg-K\\Q_{12} = h_2 - h_1 = 2799.5-3159.3 = -359 kJ/kg\\Q_{23} = u_3 - h_2 =1146.2-2006.3 = -1454.1 kJ/kg\\Q_{34} = h_4 - h_3 = 3188.4-1196.04 = 1992.36 kJ/kg\\Q_{41} = T(S_1 - S_4) = (360 + 273) (6.9917 - 7.660) = -423.04 kJ/kg[/tex]

Calculate the network done for the cycle

[tex]W_{net} = m( Q_{12} + Q_{23} + Q_{34} + Q_{41})\\W_{net} = 5( -359.8 - 1454.1 + 1992.36 - 423.04)\\W_{net} = -1223 kJ[/tex]

Answer:

Heat rejected to cold body = 3.81 kJ

Explanation:

Temperature of hot thermal reservoir Th = 1600 K

Temperature of cold thermal reservoir Tc = 400 K

efficiency of the Carnot's engine = 1 - [tex]\frac{Tc}{Th}[/tex]

eff. of the Carnot's engine = 1 - [tex]\frac{400}{600}[/tex]

eff = 1 - 0.25 = 0.75

efficiency of the heat engine = 70% of 0.75 = 0.525

work done by heat engine = 2 kJ

eff. of heat engine is gotten as = W/Q

where W = work done by heat engine

Q = heat rejected by heat engine to lower temperature reservoir

from the equation, we can derive that

heat rejected Q = W/eff = 2/0.525 = 3.81 kJ

Answer:

(1). False, (2). True, (3). False, (4). False, (5). True.

Explanation:

The term ''contouring'' in this question does not have to do with makeup but it has to deal with the measurement of all surfaces in planes. It is a measurement in which the rough and the contours are being measured. So, let us check each questions again.

(1). In contouring, it is necessary to measure position and not velocity for feedback.

ANSWER : b =>False. IT IS NECESSARY TO MEASURE BOTH FOR FEEDBACK.

(2). In contouring during 2-axis NC machining, the two axes are moved at the same speed to achieve the desired contour.

ANSWER: a=> True.

(3). Job shop is another term for process layout.

ANSWER: b => False

JOB SHOP IS A FLEXIBLE PROCESS THAT IS BEING USED during manufacturing process and are meant for job Production. PROCESS LAYOUT is used in increasing Efficiency.

(4). Airplanes are normally produced using group technology or cellular layout.

ANSWER: b => False.

(5). In manufacturing, value-creating time is greater than takt time.

ANSWER: a => True.

Answer:

The Poisson's Ratio of the bar is 0.247

Explanation:

The Poisson's ratio is got by using the formula

Lateral strain / longitudinal strain

Lateral strain = elongation / original width (since we are given the change in width as a result of compession)

Lateral strain = 0.15mm / 40 mm =0.00375

Please note that strain is a dimensionless quantity, hence it has no unit.

The Longitudinal strain is the ratio of the elongation to the original length in the longitudinal direction.

Longitudinal strain = 4.1 mm / 270 mm = 0.015185

Hence, the Poisson's ratio of the bar is 0.00375/0.015185 = 0.247

The Poisson's Ratio of the bar is 0.247

Please note also that this quantity also does not have a dimension

Answer:

The total force resisted by the flange bolts is  163.98 N

Explanation:

Solution

The first step is to find  the pipe cross section at the inlet section

Now,

A₁ = π /4 D₁²

D₁ =  diameter of the pipe at the inlet section

Now we insert 8 cm for D₁ which gives us A₁ = π /4 D (8)²

=50.265 cm² * ( 1 m²/100² cm²)

= 5.0265 * 10^⁻³ m²

Secondly, we find cross section area of  the pipe at the inlet section

A₂ = π /4 D₂²

D₂ =  diameter of the pipe at the inlet section

Now we insert 5 cm for D₁ which gives us A₁ = π /4 D (5)²

= 19.63 cm² * ( 1 m²/100² cm²)

= 1.963 * 10^⁻³ m²

Now,

we write down the conversation mass relation which is stated as follows:

Q₁ = Q₂

Where Q₁ and Q₂ are both the flow rate at the exist and inlet.

We now insert A₁V₁ for Q₁ and A₂V₂ for Q₂

So,

V₁ and V₂ are defined as the velocities at the inlet and exit

We now insert 5.0265 * 10^⁻³ m² for A₁ 5 m/s for V₁ and 1.963 * 10^⁻³ m² for A₂

= 5.0265  * 5 = 1.963 * V₂

V₂ = 12.8 m/s

Note: Kindly find an attached copy of the part of the solution to the given question below

Find the given attachments for complete explanation

Answer:

  9.29 N . . . . weight of 0.948 kg sphere

Explanation:

The sum of torques on the link BOA is zero, so we have ...

  (right force at A)(OA) = (up force at B)(OB·sin(135°))

Solving for the force at B, we have ...

  up force at B = (10.53 N)(253 mm)/((553 mm)/√2) ≈ 6.81301 N

This force is due to the difference between the buoyant force on the float sphere and the weight of the float sphere. Dividing by the acceleration due to gravity, it translates to the difference in mass between the water displaced and the mass of the sphere.

  ∆mass = (6.81301 N)/(9.8 m/s^2) = 0.695205 kg

__

The center of the sphere of diameter 0.1553 m is below the waterline by ...

  (553 mm)cos(45°) -(353 mm) = 38.0300 mm

The volume of the spherical cap of radius 155.3/2 = 77.65 mm and height 77.65+38.0300 = 115.680 mm can be found from the formula ...

  V = (π/3)h^3(3r -h) = (π/3)(115.680^2)(3·77.65 -115.68) mm^3 ≈ 1.64336 L

So the mass of water contributing to the buoyant force is 1.64336 kg. For the net upward force to correspond to a mass of 0.695305 kg, the mass of the float sphere must be ...

  1.64336 kg -0.695205 kg ≈ 0.948 kg

The weight of the float sphere is then (9.8 m/s^2)·(0.948 kg) = 9.29 N

The weight of the 0.948 kg float sphere is about 9.29 N.

Answer:

Explanation:

The frequency components in the message signal are

f1 = 100Hz, f2 = 200Hz and f3 = 400Hz

When amplitude modulated with a carrier signal of frequency fc = 100kHz

Generates the following frequency components

Lower side band

[tex]100k - 100 = 99.9kHz\\\\100k - 200 = 99.8kHz\\\\100k - 400 = 99.6kHz\\\\[/tex]

Carrier frequency 100kHz

Upper side band

[tex]100k + 100 = 100.1kHz\\\\100k + 200 = 100.2kHz\\\\100k + 400 = 100.4kHz[/tex]

After passing through the SSB filter that filters the lower side band, the transmitted frequency component will be

[tex]100k, 100.1k, 100.2k\ \texttt {and}\ 100.4kHz[/tex]

At the receive these are mixed (superheterodyned) with local ocillator frequency whichh is 100.02KHz, the output frequencies will be

[tex]100.02 - 100.1k = 0.08k = 80Hz\\\\100.02 - 100.2k = 0.18k = 180Hz\\\\100.02 - 100.4 = 0.38k = 380Hz[/tex]

After passing through the SSB filter that filters the higher side band, the transmitted frequency component will be

[tex]100k, 99.9k, 99.8k\ \ and \ \99.6kHz[/tex]

At the receive these are mixed (superheterodyned) with local oscillator frequency which is 100.02KHz, and then fed to the detector whose output frequencies will be

[tex]100.02 - 99.9k = 0.12k = 120Hz\\\\100.02 - 99.8k = 0.22k = 220Hz\\\\100.02 - 99.6k = 0.42k = 420Hz[/tex]

A) The frequency Components of the Detector Output are;

80 Hz, 120 Hz and 380 Hz

B) The frequency Components if only the lower sideband is transmitted are; 120 Hz, 220 Hz and 420 Hz

A) We are given the frequency components in the message signal as;

We are told that the carrier signal has a frequency; fc = 100kHz

Thus, the frequency components generated are;

Lower side band:

Upper side band:

We are told that the local oscillator now supplies a sinusoidal wave of frequency 100.02 kHz.

Thus, the output frequencies are;

100.02 kHz - 100.1 kHz = 80 Hz

100.02 kHz - 100.2 kHz = 180 Hz

100.02 kHz - 100.4 kHz = 380 Hz

B) Repeating the analysis assuming only the lower sideband is repeated gives us the frequencies as;

100.02 kHz - 99.9 kHz = 120 Hz

100.02 kHz - 99.8 kHz = 220 Hz

100.02 kHz - 99.6 kHz = 420 Hz

Read more about Message Signals at; https://brainly.com/question/25904079