Water flows at the rate of 200 I/s upwards through a tapered vertical pipe. The diameter at Marks(3) CLO5) the bottom is 240 mm and at the top 200 mm and the length is 5m. The pressure at the bottom is 8 bar, and the pressure at the topside is 7.3 bar. Determine the head loss through the pipe. Express it as a function of exit velocity head.

Answer:

35°c

Explanation:

Given data in question

heat = 35 kw

work = 35 kw

temperature = 35°c

To find out

temperature of the system after this process

Solution

we know that first law of thermodynamics is Law of Conservation of Energy

i.e  energy can neither be created nor destroyed and it can be transferred from one form to another form

first law of thermodynamics is energy (∆E) is sum of heat (q) and work (w)

here we know

35 = 35 + m Cv ( T - t )

35-35 = m Cv ( T-t )

T = t

here T = final temperature

t = initial temperature

it show final temp is equal to initial temp

so we can say temp after process is 35°c

Answer:

True, hope this helps but there no school right now its summer

Answer: The correct answer is Option d.

Explanation:

Ohm's law is defined as the law which gives us the relationship between voltage and current.

In electronics, the equation used to represent this law is:

[tex]E=IR[/tex]

where,

E = voltage of the circuit. The unit for this is Volts.

I = current of the circuit. The unit for this is Amperes

R = resistance of the circuit. The unit for this is Ohms.

Hence, the correct answer is Option d.

Answer:

explained below

Explanation:

An Abandoned well is well no longer in use or in such a state of despair that ground water can no longer be pumped out of it in useable quantity.

Following are few reasons for abandoning wells:

1. When the level ground water level falls down the well becomes redundant. And in recent times the ground water level has fallen to appreciable magnitude.

2. Wells represent potential conduits or pathways for surface contaminants to reach ground water supply.The ground water contamination at your well is likely to show up in municipal water supply.

3. Moreover, if the well is unused it  can cause physical hazard to people and animals living nearby. As these well grow vegetation around them thus hiding their hole.  

Answer:

true

Explanation:

shear strain is define as the ratio of change in deformation to the original length perpendicular to the axes of member due to shear stress.

     ε    = deformation/original length

strain is a unit less quantity but shear stain is generally expressed in radians but it can also be expressed in degree.

Answer:

Out of the four options provided, the most accurate answer is

option b) increase in the buoyancy force

Explanation:

Natural convection is a process in which thermal expansion of fluid takes place naturally due to natural buoyancy resulting in motion of fluid when it is heated.

Differences in densities result in buoyancy and natural convection depends on buoyancy force. Also higher air temperatures are found at lower densities which is found at the outlets of the channels and the larger the channel size, the larger is the buoyancy force (as the density difference will be higher).

Answer:

   [tex]\eta =\dfrac{\left (h_3-h_4\right )+(h_5-h_6)-(h_2-h_1)}{(h_3-h_2)+(h_5-h_4)}[/tex]

Explanation:

For close gas turbine:

       Gas turbine works on Brayton cycle.Gas turbine have lots of applications like ,it is use in aircraft,in land applications etc.

Reheating is the method to improve the efficiency of the gas turbine.In reheating gas is expanding in two turbine instead of one turbine alone.Two turbine like high pressure turbine and low pressure turbine are used for expansion.

In the above diagram 1-2 is a compressor,2-3 heat addition,3-4 high pressure turbine,4-5 reheating of cycle 5-6 low pressure turbine,6-1 heat rejection,

We know that    [tex]\eta =\frac{W_{net}}{Q_{s}}[/tex]

Now take [tex]h_{1},h_{2},,h_{3},h_{4},h_{5},h_{6}[/tex] represent the enthalpy of point 1,2,3,4,5,6 in the cycle respectively.

So total heat supplied [tex]Q_S[/tex]=

[tex]\left (h_3-h_2\right )+\left (h_5-h_4\right )[/tex]

Net work out put

[tex]W_{net}[/tex]=[tex]\left (h_5-h_6\right )-\left (h_2-h_1\right )[/tex]

So efficiency   [tex]\eta =\frac{W_{net}}{Q_{s}}[/tex]

      [tex]\eta =\dfrac{\left (h_3-h_4\right )+(h_5-h_6)-(h_2-h_1)}{(h_3-h_2)+(h_5-h_4)}[/tex]

Answer:

b). False

Explanation:

A refractory material is a type of material that can withstand high temperatures without loosing its strength. They are used in reactors, furnaces, kilns, etc.

    Refractory materials are certain super alloys and ceramics materials.

Properties of refractory materials :

1. Refractory materials have high melting point.

2.They acts barriers between high heat zone and low heat zone.

3. The specific heat of refractory material is very low.

4. Refractories that have high bulk densities are better in quality.

Hence, Refractory materials have a very high melting temperature.

Answer and explanation:

Deformation means change in position plastic deformation mainly cause due to motion of dislocation

THERE ARE MAINLY TWO MECHANISM BY WHICH PLASTIC DEFORMATION TAKES PLACE

SLIP : slip is a process of sliding of blocks over one  another along the planes  these planes are called slip planes slip takes place when the shear stress exceeds than the critical value of stress distance between the slip planes are called slip lines the resistance for slip plane is very less as compared to any other planes the slip plane is the plane has very high density

Answer:

(b)False

Explanation:

[tex]I_{xy}[/tex] defined as

      [tex]I_{xy}[/tex] =[tex]\int \left (x\cdot y\right )dA[/tex]

Where x is the distance from centroidal x-axis

           y is the distance from centroidal y-axis

          dA is the elemental area.

The product of x and y can be positive or negative ,so the value of  [tex]I_{xy}[/tex] can be positive as well as negative .

So from the above expressions we can say that the product of [tex]I_{x},I_y[/tex] is different from [tex]I_{xy}[/tex] .

Answer:

0.05 J/K

Explanation:

Given data in question

heat (Q) = 10 J

temperature (T) = 200 K

to find out

the change in entropy of the system

Solution

we will solve this by the entropy change equation

i.e  ΔS = ΔQ/T           ...................1

put the value of heat Q and Temperature T in equation 1

ΔS is the enthalpy change and T is the temperature

so  ΔS = 10/200

ΔS = 0.05 J/K

Answer:

b). false

Explanation:

Lean manufacturing

Lean manufacturing, a philosophy developed by Toyota Production System are means to eliminate wastes. They are defined as the techniques or management activities in eliminating wastes and increasing the efficiency inside an organisation.

    According to the concept of lean manufacturing, mainly seven types of wastes are identified. They are :

1. Transportation waste

2. Inventory waste

3. Over production

4. Waiting

5. Defects

6. Motion waste

7. Non utilized talent

All these waste affect greatly to the efficiency of an organisation and devalue its services.  Lean manufacturing advises to prevent  all these waste in order to increase the productivity.

       All the management activities and techniques used in lean manufacturing may be different according to the business application but they are all based on the same basic principle of removing wastes and errors and increase efficiency.

The different sectors that are benefiting from lean manufacturing methodology are

Healthcare

Hospitality

Food and Beverage

Government

Manufacturing

Lean manufacturing can be used in different sectors.

Answer:

the flow rate of the oil is 2.5 m³/s

Explanation:

Given data

relative density (S) = 0.8

diameter (d1) = 60 mm = 0.06 m

diameter (d2) = 35 mm = 0.035 m

height (h) = 22 mm = 0.022 m

discharge coefficient (Cd) = 0.98

To find out

the flow rate of the oil

solution

we know the formula for rate of flow i.e.

flow rate = Cd a1 a2 [tex]\sqrt{2 g n }[/tex] /  [tex]\sqrt{a1^{2} a2^{2} }[/tex]    ...............1

here first we find area a1 and a2 i.e.

a1 = ( [tex]\pi[/tex] /4 ) × d² = ( [tex]\pi[/tex] /4 ) × 0.06² = 0.002827 m²

a2 = ( [tex]\pi[/tex] /4 ) × d² = ( [tex]\pi[/tex] /4 ) × 0.035² = 0.000962 m²

and now we find n = (density of mercury / density of oil)  - 1 × h

n = ((13.56 / 0.8)  - 1) × 0.022 = 0.3509

put all these value in equation 1

flow rate = Cd a1 a2 [tex]\sqrt{2 g n }[/tex] / [tex]\sqrt{a1^{2} a2^{2} }[/tex] 

flow rate = 0.98× 0.002827× 0.000962 [tex]\sqrt{2*9.81*0.3509}[/tex] / [tex]\sqrt{0.002827^{2} 0.000962^{2} }[/tex]

flow rate = 2.571386 m³/s

Answer: a)True

Explanation: Takt time is defined as the average time difference between  the production of the two consecutive unit of goods by the manufacturer and this rate is matched with the demand of the customer. This is the time which is calculated to find the acceptable time for which the goods unit must be produced  by the factory to meet the needs of the customer. Therefore , the statement is true that takt time is the rate at which a factory must produce to satisfy the customer's demand.

Answer:

Velocity component in x-direction [tex]u=-\frac{3}{2}x^2-\frac{1}{3}x^3[/tex].

Explanation:

   v=3xy+[tex]x^{2}[/tex]y

We know that for incompressible flow

   [tex]\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=0[/tex]

[tex]\frac{\partial v}{\partial y}=3x+x^{2}[/tex]

So   [tex]\frac{\partial u}{\partial x}+3x+x^{2}=0[/tex]

[tex]\frac{\partial u}{\partial x}= -3x-x^{2}[/tex]

By integrate with respect to x,we will find

[tex]u=-\frac{3}{2}x^2-\frac{1}{3}x^3[/tex]+C

So the velocity component in x-direction [tex]u=-\frac{3}{2}x^2-\frac{1}{3}x^3[/tex].

Answer:

0.45516

Explanation:

ENTROPY : Entropy is a measure of molecular disorder it is denoted by S. Entropy is also measured in terms of thermal energy and temperature it is equal to thermal energy per unit temperature.

from the table S₁=1.99194 KJ/kg.k (at 400k)

from the table S₂=2.21952 KJ/kg.k (at 500k)

so total entropy change is given by =m (S₂-S₁)

=2(2.21952-1.99194)

=0.45516

Answer: d) All of the above

Explanation: Polymers are the substances that have molecular structure with having same bonds in the entire molecule together.There are light weight substance which occur natural as well as artificially. They are also categorized  as thermoplastics ,thermosets, and elastomers. They also have the property of being stretching and bending under the pressure or load they are also instable. Therefore, all the options are correct statement about polymers.

B. Photons.

In a photonic material, signal transmission occurs by photons which are light particles.

Answer:

Tolerance is important for the very fact that providing proper tolerances ensures proper fittings of different parts.                                            

Explanation:

Tolerance and dimensioning is an important link between manufacturing and engineering.

Tolerance optimization leads to high cost of machined part to be produced and also provides good quality product. Whereas loose tolerance means reduction in cost but poor quality product. hence it is very important and critical to provide the right tolerance while designing a product.

                               Tolerance also influence what type of production processes to be selected by the process planners. The optimization of the tolerances during the design phase has a positive impact on the results coming out of the manufacturing processes.

                                Providing proper tolerances ensures that the parts will fit properly.

Therefore, providing proper tolerances, the engineers shares the responsibility to manufacture the parts correctly.

Answer:

0.025 m/sec

Explanation:

we have given m =0.25 kg

velocity=0.05 m/sec

radius =0.5 meter

the centrifugal force produced due to rotational motion

[tex]F_c=\frac{mv^2}{r}=\frac{0.25\times 0.05^2}{0.5}=0.00125 N[/tex]

now again using this equation for finding the final velocity

[tex]0.00125=\frac{mv_{final}^2}{r}=\frac{0.25v_{final}^2}{0.125}[/tex]

[tex]v_{final}=\sqrt{\frac{0.00125\times 0.125}{0.25}}=0.025\ m/sec[/tex]

so the final speed of mass spring will be 0.025 m/sec

Answer:

T = 212.8125°C

Explanation:

Given

radius of the wire, [tex]r_{0}[/tex] = 5 mm 0.005 m

heat generated, g = 5 x [tex]10^{7}[/tex] W/[tex]m^{3}[/tex]

outer surface temperature, [tex]T_{S}[/tex] = 180°C

Thermal conductivity, k = 8 W / m-k

Now maximum temperature occurs at the center of the wire

that is at r=0,

Therefore, [tex]T_{o}=T_{S}+\frac{g\times r_{o}^{2}}{4\times k}[/tex]

                  [tex]T_{o}=180+\frac{5\times 10^{7}\times 0.005^{2}}{4\times 8}[/tex]

                 [tex]T_{o}=219.0625[/tex]°C

Therefore, temperature at r = 2 mm

[tex]\frac{T-T_{S}}{T_{O}-T_{S}}= 1-\left (\frac{r}{r_{O}}  \right )^{2}[/tex]

[tex]\frac{T-180}{219.0625-180}= 1-\left (\frac{2}{5}  \right )^{2}[/tex]

Therefore, T = 212.8125°C

Answer:

The work required to compress the saturated water vapor to 895 kPa pressure is 130.9540 k J/Kg

Explanation:

Given data in question

temperature = 140°C

pressure  (P2) = 895 kPa

To find out

work required for compress saturated water

Solution  

We know the equation for reversible work for compress saturated water vapor

i.e.  

W =  [tex]-\int_{1}^{2}vdP-\Delta ke - \Delta pe[/tex]

w is  reversible work, v is specific volume, P is water vapor pressure and

ke is kinetic energy and pe is potential energy

and in question we have given v is constant so ke and pe will be zero

so  

W =  [tex]-\int_{1}^{2}vdP[/tex]

W =  -v( P2 - P1 )

we can given in question temperature = 140°C and use steam table "A-4 saturated water - temperature table"

at this water property P1 will be 361.53 kPa and v will be 0.50850 m³/kg

so put these value in above equation

W =  -0.50850( 104 - 361.53 )

W = 130.9540 kJ/Kg  

Answer:

strains for the respective cases are

0.287

0.318

0.127

and for the entire process 0.733

Explanation:

The formula for the true strain is given as:

[tex]\epsilon =\ln \frac{l}{l_{o}}[/tex]

Where

[tex]\epsilon =[/tex] True strain

l= length of the member after deformation

[tex]l_{o} = [/tex] original length of the member

Now for the first case we have

l= 1.6m

[tex]l_{o} = 1.2m[/tex]

thus,

[tex]\epsilon =\ln \frac{1.6}{1.2}[/tex]

[tex]\epsilon =0.287[/tex]

similarly for the second case we have

l= 2.2m

[tex]l_{o} = 1.6m[/tex]   (as the length is changing from 1.6m in this case)

thus,

[tex]\epsilon =\ln \frac{2.2}{1.6}[/tex]

[tex]\epsilon =0.318[/tex]

Now for the third case

l= 2.5m

[tex]l_{o} = 2.2m[/tex]

thus,

[tex]\epsilon =\ln \frac{2.5}{2.2}[/tex]

[tex]\epsilon =0.127[/tex]

Now the true strain for the entire process

l=2.5m

[tex]l_{o} = 1.2m[/tex]

thus,

[tex]\epsilon =\ln \frac{2.5}{1.2}[/tex]

[tex]\epsilon =0.733[/tex]

Answer:

Given:

laminar flow

and since velocity of flow is doubled, we consider [tex]v_{n}[/tex] as new velocity and [tex]v_{o}[/tex] as original velocity

Explanation:

As per laminar flow, thickness, t is given by

t = [tex]\frac{4.91x}{\sqrt( R_{ex}) }[/tex]

t =  [tex]\frac{4.91x}{\sqrt{\frac{\rho vx}{\mu }}}[/tex]

t = [tex]\frac{4.91x\mu }{\sqrt{\rho vx}}[/tex]

where,

[tex]R_{ex}[/tex] = Reynold's no.

therefore,

t ∝ [tex]\frac{1}{\sqrt{v} }[/tex]

Now,

[tex]\frac{t_{n} }{t_{o} }[/tex] = [tex]\sqrt{(\frac{v_{o} }{v_{n} })}[/tex]

[tex]\frac{t_{n} }{t_{o} }[/tex] = [tex]\sqrt{(\frac{v_{o} }{2v_{o} } )} =\frac{1}{\sqrt{2} }[/tex]

therefore,

[tex]t_{n}:t_{o} = 1:\sqrt{2}[/tex]

Answer:

a). absorber, generator and liquid pump

Explanation:

The Vapour absorption system consists of compression, expansion, condensation and evapouration processes. This system uses ammonia, lithium bromide or water as refrigerant.

                  An absorber, pump and generator is used in place of a compressor in the vapour compression refrigeration system. The operation is smooth in vapour absorption system since all the moving elements are in the pump only. This system make use of low energy like heat and can work on lower evapourator pressure. It has low Coefficient of performance.

Answer:500,551.02

Explanation:

Given

Initial enthaly of pump \left ( h_1\right )=500KJ/kg

Final  enthaly of pump \left ( h_2\right )=550KJ/kg

Final  enthaly of pump when efficiency is 100%=[tex]h_2^{'}[/tex]

Now pump efficiency is 98%

[tex]\eta [/tex]=[tex]\frac{h_2-h_1}{h_2^{'}-h_1}[/tex]

0.98=[tex]\frac{550-500}{h_2-500}[/tex]

[tex]h_2=551.02KJ/kg[/tex]

therefore initial and final enthalpy of pump for 100 % efficiency

initial=500KJ/kg

Final=551.02KJ/kg

Answer:

work done= 2.12 kJ

Explanation:

Given

N=2500 rpm

T=4.5 N.m

Period ,t= 30 s

[tex]torque =\frac{power}{2\pi N}[/tex]

[tex]power=2\pi N\times T[/tex]

P=[tex]2\times \pi \times2500 \times 4.5[/tex]

P=70,685W

P=70.685 KW

power=[tex]\frac{work done}{time}[/tex]

work done = power * time

                  = 70.685*30=2120.55J

                  = 2.12 kJ

Explanation:

A solid state laser contains a cavity like structure fitted with spherical mirrors or plane mirrors at the end filled with a rigidly bonded crystal. It uses solid as the medium. It uses glass or crystalline materials.

    It is known that active medium used for this type of laser is a solid material. This lasers are pumped optically by means of a light source which is used as a source of energy for the laser. The solid materials gets excited by absorbing energy in the form of light from the light source. Here the pumping source is light energy.  

Answer:

control volume

differential approach

Explanation:

control volume:

differential approach:

Answer:

P₂ = 830.75 kPa

Explanation:

Given:

Pressure in the smaller pipe,P₁  = 860 kPa

Velocity in the larger pipe, v₂ = 2.4 m/s

Therefore velocity in the smaller pipe, v₁ = 4.8 m/s ( velocity gets doubled since area is reduced to half )

Height at section where the area is doubled, z₁ = 6 m

Height at the section where pressure is to be calculated, z₂ = 1.5 m

Now apply Bernouli Equation between the section of enlargement and at section where pressure is to be calculated,

[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}[/tex]

[tex]\frac{860}{1000 \times 9.81}+\frac{4.8^{2}}{2\times 9.81}+6 = \frac{P_{2}}{1000 \times 9.81}+\frac{2.4^{2}}{2\times 9.81}+1.5[/tex]

P₂ = 830.75 kPa

Therefore, pressure at the section 1.5 m above datum is 830.75 kPa