Water flows in a pipe of diameter 450 mm at an average velocity of 3 m/s. Determine the volumetric flow rate and the mass flow rate. How long will it take to fill a tank with measurements 5 m x 6 m x 20 m?

Answers

Answer 1

Answer:

Volumetric flow rate = 0.4773 m³/s

Mass flow rate = 477.3 kg/s

It will take 286.38 seconds to fill a tank with measurements 5 m x 6 m x 20 m

Explanation:

Given:

Diameter of the pipe through which the water is flowing = 450 mm

Radius = Diameter/2

Thus, Radius of the pipe = 225 mm

The conversion of mm into m is shown below:

1 mm = 10⁻³ m

Radius of the pipe = 225×10⁻³ m

The area of the cross-section = π×r²

So, Area of the pipe = π×/(225×10⁻³)² m² = 0.1591 m²

Also, Given : The water flowing rate = 3 m/s

Volumetric flow rate is defined as the amount of flow of the fluid in 1 sec.

[tex]Volumetric\ flow= \frac {Volume\ passed}{Time taken}[/tex]

This, can be written as Velocity of the fluid from the cross-section area of the pipe.

Q = A×v

Where,

Q is Volumetric flow rate

A is are though which the fluid is flowing

v is the velocity of the fluid

So,

Q = 0.1591 m²×3 m/s = 0.4773 m³/s

Mass flow rate is defined as the mass of the fluid passes per unit time.

[tex]\dot {m}= \frac {Mass\ passed}{Time taken}[/tex]

The formula in terms of density can be written as:

[tex]Density=\frac{Mass}{Volume}[/tex]  

So, Mass:

[tex]Mass= Density \times {Volume}[/tex]

Dividing both side by time, we get:

[tex]\dot {m}= Density \times {Q}[/tex]

Where,

[tex]\dot {m}[/tex] is the mass flow rate

Q is Volumetric flow rate

Density of water = 1000 kg/m³

Thus, Mass flow rate:

[tex]\dot {m}= 1000 \times {0.4773} Kgs^{-1}[/tex]

Mass flow rate = 477.3 kg/s

The time taken to fill the volume of measurement 5 m× 6 m× 20 m can be calculated from the formula of volumetric flow rate as:

t= Q×V

So,

Volume of Cuboid = 600 m³

Time = 0.4773 m³/s × 600 m³ = 286.38 s


Related Questions

In a flow over a flat plate, the Stanton number is 0.005: What is the approximate friction factor for this flow a)- 0.01 b)- 0.02 c)- 0.001 d)- 0.1

Answers

Answer:

(a) .01

Explanation:

stanton number is a dimensionless quantity stanton is expressed as [tex]\frac{heat transer}{thermal capacity}[/tex]stanton number is discovered by Thomas edward stanton

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

A slider of mass 0.25 kg on a string, 0.5 m long is rotating around a pivot on a frictionless table. The velocity of the slider is initially 0.05 m/s. When the string is pulled into a radius of 0.125 m how fast is the mass spinning?

Answers

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

What is the thermal efficiency of this reheat cycle in terms of enthalpies?

Answers

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]

Which of the following is a correct formula of Ohm s Law (a) E= R/I (b) E=1+R (c)E=I/R (d) E= IR

Answers

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.

0il with a relative density of 0,8 flows in a pipe of diameter 60 mm. A venturi meter having a throat diameter of 35 mm is installed in the pipeline. The pressure difference is measured with a mercury manometer. The levels of the manometer differ by 22 mm. The venturi meter has a discharge coefficient of 0,98. Calculate the flow rate of the oil.

Answers

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

Radiation heat transfer occurs from any object that is above 0K. a) True b) False

Answers

Answer:

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

A pipe which is on a slope, transports water downwards. A doubling of cross sectional area takes place 6 above the reference level. The pressure in the smaller pipe, just before the enlargement, is 860 kPa. The flow velocity in the large pipe is 2,4 m/s. Determine the pressure in kPa at a point 1,5 m above the reference level. Ignore friction losses.

Answers

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

A compressed-air drill requires an air supply of 0.25 kg/s at gauge pressure of 650 kPa at the drill. The hose from the air compressor to the drill has a 40 mm diameter and is smooth. The maximum compressor discharge gauge pressure is 690 kPa. Neglect changes in air density and any effects of hose curvature. Air leaves the compressor at 40° C. What is the longest hose that can be used?

Answers

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

Explain with schematics the operating principle of solid state lasers.

Answers

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.  

Determine the work done by an engine shaft rotating at 2500 rpm delivering an output torque of 4.5 N.m over a period of 30 seconds.

Answers

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

Saturated water vapor at 140°C is compressed in a reversible, steady-flow device to 895 kPa while its specific volume remains constant. Determine the work required.

Answers

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  

In a photonic material, signal transmission occurs by which of the following? a)- Electrons b)- Photons

Answers

B. Photons.

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

Describe how the Rotary Engine works.

Answers

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:

What different between 'flow analysis using control volume method' and 'flow analysis using differential method'?

Answers

Answer:

control volume

control volume is used to determine the flow characteristics of  complex shape like turbine and compressors

differential approach

it is carried out by considering infintely small region for fluid analysis.

Explanation:

control volume:

control volume is used to determine the flow characteristics of  complex shape like turbine and compressorsit is used to determine the flow velocity within in the boundaries of control volume. it can also used for force analysis for flow. one main disadvantage of control volume is that it doesn't provide detail information about stress and pressure variation.

differential approach:

it is carried out by considering infintely small region for fluid analysis.solution of the fluid analysis is in the form of differential equationit provide detail information about the flow.

Which of the following is/are FALSE about refining aluminum from the ore state (mark all that apply) a)- A blast furnace is used b)-The ore is called bauxite c)-The process uses a lot of electricity d)-Coke is used to produce the heat

Answers

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.

A closed system contains propane at 35°c. It produces 35 kW of work while absorbing 35 kW of heat. What is process? the temperature of the system after this process.

Answers

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

It is true about polymers: a)-They are light-weight materials b)-There are three general classes: thermosets, thermoplastics and thermoset c)-They present long term instability under load d)-All the above

Answers

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.

Explain the reasons for abandoning a well.

Answers

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.  

Describe the slip mechanism that enables a metal to be plastically deformed without fracture.

Answers

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

SLIPTWINNING

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

Shear strain can be expressed in units of either degrees or radians. a)True b)- False

Answers

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.

A typical aircraft fuselage structure would be capable of carrying torsion moment. a)True b)- False

Answers

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

Calculate the change of entropy of 2 kg of air when its temperature increases from 400 K to 500 K at constant pressure equal to 300 kPa.

Answers

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

                       

Takt time is the rate at which a factory must produce to satisfy the customer's demand. a)- True b)- False

Answers

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.

The velocity of flow over a flat plate is doubled. Assuming the flow remains laminar over the entire plate, what is the ratio of the new thermal boundary layer thickness to the original boundary layer thickness?

Answers

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]

The product second moment of area Ixy is found by multiplying Ix and Iy. a)True b)- False

Answers

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] .

A long homogeneous resistance wire of radius ro = 5 mm is being used to heat the air in a room by the passage of electric current. Heat is generated in the wire uniformly at a rate of g=5'107 W/m as a result of resistance heating. If the temperature of the outer surface of the wire remains at 180°C, determine the temperature at r = 2 mm after steady operation conditions are reached. Take the thermal conductivity of the wire to be k = 8 W/m x °C.

Answers

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

Refectories are one of the types of ceramics that have low melting temperature. a)-True b)-False

Answers

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.

In a vapour absorption refrigeration system, the compressor of the vapour compression system is replaced by a a)- absorber, generator and liquid pump. b)-absorber and generator. c)- liquid pump. d)-generator.

Answers

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.

A strip ofmetal is originally 1.2m long. Itis stretched in three steps: first to a length of 1.6m, then to 2.2 m, and finally to 2.5 m. Compute the true strain after each step, and the true strain for the entire process (i.e. for stretching from 1.2 m to 2.5 m).

Answers

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]

If you add 10 J of heat to a system so that the final temperature of the system is 200K, what is the change in entropy of the system? a)-0.05 J/K b)-0.30 J/k c)-1 J/K d)-9 J/K e)-2000 J/K

Answers

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

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