A centrifugal pump provides a flow rate of 0.03 m/s when operating at 1750 rpm against 60 m head. Determine the pump's flow rate and developed head if the pump speed is increased to 3500 rpm.

Answer:

control volume

differential approach

Explanation:

control volume:

differential approach:

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:

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:

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:

The options a)- A blast furnace is used and d)-Coke is used to produce the heat are FALSE.

Explanation:

Aluminium is a chemical element and the most abundant metal present in the Earth's crust. An aluminium ore is called bauxite. Aluminium is extracted from its ore by the process of electrolysis, called the Hall–Héroult process. The extraction of aluminium is an expensive process as it requires large amount of electricity. The bauxite is purified to produce aluminium oxide. Then, aluminium is extracted from the aluminium oxide.

Therefore, the refining of aluminum from its ore does not involve the use of a blast furnace and coke to produce heat.

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:

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:

   [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:

True

Explanation:

An aircraft is subject to 3 primary rotations

1) About longitudinal axis known as rolling

2) About lateral axis known as known as pitching

3) About the vertical axis known as yawing

The rolling of the aircraft induces torsion in the body of the aircraft thus the fuselage structure should be capable of carrying torsion

Answer:

30.12345° can be written as : 30°7'20.42''

Or,

30 degrees 7 minutes and 20.42 seconds.

Explanation:

1 degree consists of 60 arc minutes.  

1 arc minutes consists of 60 arc seconds.  

Thus, 30.12345° can be written as:

30.12345°= 30° + 0.12345°

1° = 60'

So,

0.12345° = 0.12345*60' = 7.407'

Thus, 7.407' can be written as:

7.407' = 7' + 0.407'

1' = 60''

So,

0.407' = 0.407*60'' = 20.42''

Thus,

30.12345° can be written as : 30°7'20.42''

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:

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: 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.

Answer:

L = 46.35 m

Explanation:

GIVEN DATA

\dot m  = 0.25 kg/s

D = 40 mm

P_1 = 690 kPa

P_2 = 650 kPa

T_1 = 40° = 313 K

head loss equation

[tex][\frac{P_1}{\rho} +\alpha \frac{v_1^2}{2} +gz_1] -[\frac{P_2}{\rho} +\alpha \frac{v_2^2}{2} +gz_2] = h_l +h_m[/tex]

where[tex] h_l = \frac{ flv^2}{2D}[/tex]

[tex]h_m minor loss [/tex]

density is constant

[tex]v_1 = v_2[/tex]

head is same so,[tex] z_1 = z_2 [/tex]

curvature is constant so[tex] \alpha = constant[/tex]

neglecting minor losses

[tex]\frac{P_1}{\rho}  -\frac{P_2}{\rho} = \frac{ flv^2}{2D}[/tex]

we know[tex] \dot m[/tex] is given as[tex] = \rho VA[/tex]

[tex]\rho =\frac{P_1}{RT_1}[/tex]

[tex]\rho =\frac{690 *10^3}{287*313} = 7.68 kg/m3[/tex]

therefore

[tex]v = \frac{\dot m}{\rho A}[/tex]

[tex]V =\frac{0.25}{7.68 \frac{\pi}{4} *(40*10^{-3})^2}[/tex]

V = 25.90 m/s

[tex]Re = \frac{\rho VD}{\mu}[/tex]

for T = 40 Degree, [tex]\mu = 1.91*10^{-5}[/tex]

[tex]Re =\frac{7.68*25.90*40*10^{-3}}{1.91*10^{-5}}[/tex]

Re = 4.16*10^5 > 2300 therefore turbulent flow

for Re =4.16*10^5 , f = 0.0134

Therefore

[tex]\frac{P_1}{\rho}  -\frac{P_2}{\rho} = \frac{ flv^2}{2D}[/tex]

[tex]L = \frac{(P_1-P_2) 2D}{\rho f v^2}[/tex]

[tex]L =\frac{(690-650)*`10^3* 2*40*10^{-3}}{7.68*0.0134*25.90^2}[/tex]

L = 46.35 m

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:

(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:

Explanation:

Thermostatic expansion valve is mainly a throttling device commonly used in air conditioning systems and refrigerators.

It is an automatic valve that maintains proper flow of refrigerant in the evaporator according to  the load inside the evaporator. When the load in the evaporator is higher the valve opens and  allows the increase in flow of refrigerant and when the load reduces the valve closes a bit and  reduces the flow of refrigerant. This process leads to higher efficiency of compressor as well as the whole refrigeration system.  Thus TEV works to reduce the pressure of refrigerant from higher condenser pressure to the lower evaporator pressure. It also keeps the evaporator active.      

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:

  Rotary engine was early known by the name of internal combustion engine. It convert heat from a high pressure of combustion. The main advantage of rotary engine is that it can be operate with less number of vibration. It works on the principle of converting pressure into rotating motion. In rotary engine the expansion pressure is applied on the flank rotor.  

Answer: The rotary engine works on the same basic principle as the piston engine: combustion in the power plant releases energy to power the vehicle. However, the delivery system in the rotary engine is wholly unique. The piston engine performs four key operations: intake, compression, combustion, and exhaust.

Explanation:

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:

(a) .01

Explanation:

there is relation between friction factor and stanton number and friction factor that is stanton number is half of friction factor

stanton number =[tex]\frac{friction factor}{2}[/tex]

.005=[tex]\frac{friction factor}{2}[/tex]

friction factor =2×.005

friction factor=.01

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:

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.

B. Photons.

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

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:

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

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:

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