Ductility (increases/decreases/does not change) with temperature.

Answer:

The governing ratio for thin walled cylinders is 10 if you use the radius. So if you divide the cylinder´s radius by its thickness and your result is more than 10, then you can use the thin walled cylinder stress formulas, in other words:

or using the diameter:

Answer:

37 psi

Explanation:

For ideal gases this equation applies:

p1*V1/T1 = p2*V2/T2

Since we are assuming volume remains constant:

V2 = V1

p1/T1 = p2/T2

p2 = p1*T2/T1

The temperatures must be in absolute scale.

T1 = 15 + 273  = 288 K

T2 = 60 + 273 = 333 K

Then:

p2 = 32 * 333 / 288 = 37 psi

After the emulsion is made, fuel droplet containing water droplets increase its temperature inside the engine,  because of that, water droplets explode causing water rapid evaporation that break down fuel droplet, in consequence, smaller fuel droplets are generated. This makes the temperature of the flame during the combustion decrease, and that is why the reduction of NOx is accomplished by this method .

Answer with Explanation:

The relation between power and energy is

[tex]Energy=Power\times Time[/tex]

Since the nuclear reactor operates at 1200 MW throughout the year thus the energy produced in 1 year equals

[tex]E=1200\times 10^{6}\times 3600\times 24\times 365=3.784\times 10^{16}[/tex]

Now from the energy mass equivalence we have

[tex]E=mass\times c^2[/tex]

where

'c' is the speed of light in free space

Thus equating both the above values we get

[tex]3.784\times 10^{16}=mass\times (3\times 10^{8})^{2}\\\\\therefore mass=\frac{3.784\times 10^{16}}{9\times 10^{16}}=0.42kg[/tex]

Since it is given that 1 kg of mass is 34% effective thus the mass reuired for the reactor is

[tex]mass_{req}=\frac{mass}{\eta }=\frac{0.43}{0.34}=1.235[/tex]

Thus 1.235 kg of nuclear fuel is reuired for operation.

Answer:

The given statement is correct.

Explanation:

When we load a material axially stresses are developed in the material. these stresses arise internally to keep the object in equilibrium.

When a tensile load is applied to an material since the nature of the force is to cause elongation in the material, axial strain is developed in the material or we can say that any arbitrary point in the material undergoes a displacement in the direction of the applied load. The stresses that are developed in the material tend to tear the material apart.

Strength of a material by definition is it's ability to resist deformation, thus tensile strength can be defined as the resistance of the material to tensile strains which have the tendency to tear the material apart.

Answer and Explanation:

AIR PREHEAT ER :

Answer:

c) normal

Explanation:

The pressure forces on a submersed object will be normal to the surfaces of the object. Hydrostatic pressure cannot apply tangential forces, so only the normal components exist.

Also, it should be noted that in the hydrostatic case (submerged object) all pressures depend linearly of depth. For small object we can approximate this with equal pressures everywhere.

Answer:

c)Conduction

Explanation:

As we know that conduction heat transfer is take place due to movement of electron.When we fall on the campfire then the primary heat transfer mechanism is conduction because .

Convection heat transfer taker place due to motion of fluid.

Radiation heat transfer dominates at very high temperature but in campfire the temperature is not too high.

So option C is correct.

Answer:

The correct answer is option 'a': Thermal and Electrical energy

Explanation:

We know that the solar radiation that we receive from sun provides us heat. Thus the solar radiation is a natural source of thermal energy which can be utilized in solar cooker's to cook food or to warm water to generate steam in Solar thermal power plants. The steam generated is used to drive a turbine and hence produce electricity.

As we know that solar radiation is converted into directly usable forms of energy by the solar panels. The solar panels work on the principle of photo-electric effect in which light energy is directly converted into electrical energy to run our electrical devices at home such as light bulbs, fans, e.t.c. An excellent application of this principle is the international space station that orbits the earth and is fully powered by solar energy.

Answer:

1) Height of oil in right limb = 67.31 cm

2) Height of water in the right limb = 16.83 cm

Explanation:

The U-tube manometer is shown in the attached figure

Foe equilibrium The pressure at the bottom of the U tube should be same

Let the height of the water in the left limb of the manometer be [tex]h_L[/tex]

Thus the pressure at the bottom is found using the equation of pressure statics as

[tex]P_{bottom}=P_{atm}+\rho _{water}\times g\times h_{L}.............(i)[/tex]

Similarly for the liquid in the right limb the pressure at the bottom is the sum of the oil column and the water column

Thus we can write

[tex]P_{bottom}=P_{atm}+\rho _{water}\times g\times h_{}+\rho _{oil}\times g\times 4h_{}...........(ii)[/tex]

Equating the equations 'i' and 'ii' we get

[tex]P_{atm}+\rho _{water}\times g\times h_{L}=P_{atm}+\rho _{water}\times g\times h_{}+\rho _{oil}\times g\times 4h_{}\\\\\rho _{water}\times g\times 0.7=\rho _{oil}\times 4h\times g+\rho _{water}\times h\times g\\\\\therefore h_{}=\frac{0.7\times \rho _{water}}{4\times \rho _{oil}+\rho _{water}}\\\\h_{}=\frac{0.7\times 1000}{4\times 790+1000}=16.83cm[/tex]

Thus the height of oil is [tex]4\times 16.83=67.31cm[/tex] andthe height of water in the right limb is 16.83 cm.

Answer:

a) 1253 kJ

b) 714 kJ

c) 946 C

Explanation:

The thermal efficiency is given by this equation

η = L/Q1

Where

η: thermal efficiency

L: useful work

Q1: heat taken from the heat source

Rearranging:

Q1 = L/η

Replacing

Q1 = 539 / 0.43 = 1253 kJ

The first law of thermodynamics states that:

Q = L + ΔU

For a machine working in cycles ΔU is zero between homologous parts of the cycle.

Also we must remember that we count heat entering the system as positiv and heat leaving as negative.

We split the heat on the part that enters and the part that leaves.

Q1 + Q2 = L + 0

Q2 = L - Q1

Q2 = 539 - 1253 = -714 kJ

TO calculate a temperature for the heat sink we must consider this cycle as a Carnot cycle. Then we can use the thermal efficiency equation for the Carnot cycle, this one uses temperatures:

η = 1 - T2/T1

T2/T1 = 1 - η

T2 = (1 - η) * T1

The temperatures must be given in absolute scale (1453 C = 1180 K)

T2 = (1 - 0.43) * 1180 = 673 K

673 K = 946 C

Answer:

d. to supply heat to a warmer space

Explanation:

Let's analyze the options

(a) to remove heat from a colder space is the purpose of a refrigerator. Note that heat pump also do this  

(b) to remove heat from a warmer space. This option has no sense, both refrigerators and heat pumps supply heat to a warmer space

(c) to supply heat to a colder space. This option has no sense, both refrigerators and heat pumps remove heat from a colder space

(d) to supply heat to a warmer space. This is the purpose of a heat pump

Answer:

The lamp will remain off.

Explanation:

I assume this is a simple series circuit with a battery, a switch, an incandescent lamp and a capacitor.

Since the circuit is of continuous current (because it uses a battery), after the switch has remained closed for a long time the capacitor will be fully charged. Being fully charged, no continuous current will flow through the capacitor, and since it is in series with the rest of the circuit there will be no current anywhere, so the lamp will remain off.

Answer:

(a)[tex]1.308\times 10^{-4}m/sec[/tex]

(b)57.33831 pound/cubic feet

(c)120.1095 hp

Explanation:

We have

(a) 760 miles / hour

We know that [tex]1\ mile\ =0.00062m[/tex]

And 1 hour = 60×60=3600 sec

So [tex]760miles/hour=\frac{760\times 0.00062meter}{3600sec}=1.308\times 10^{-4}m/sec[/tex]

(b) 927 kg/cubic meter to mass/cubic foot

We know that 1 kg = 2.20 pound

So 921 kg = 921×2.20=2026.2 pound

We know that 1 cubic meter = 35.31 cubic feet

So 57.33831 pound / cubic feet

(c) We have to convert to hp

[tex]5.37\times 10^3kj/min=\frac{5.37\times 1000kj}{60sec}=89.5kj/sec[/tex]

We know that 1 kj /sec = 1.341 hp

So 89.5 kj/sec = 89.5××1.341=120.1095 hp

The final gage pressure of the gas after being heated from 27°C to 77°C in a closed, rigid tank can be found using Gay-Lussac's Law. After adjusting the initial and final temperatures to Kelvin and converting the gage pressure to absolute pressure, calculate the new absolute pressure and then convert back to gage pressure.

To determine the final gage pressure of a gas modeled as an ideal gas in a closed, rigid tank after it is heated from an initial temperature of 27°C to a final temperature of 77°C, we can use the ideal gas law in a form that relates pressure and temperature while keeping the volume and number of moles constant (Gay-Lussac's Law). We're told that the initial gage pressure is 300 kPa, and we need to find the final gage pressure. The local atmospheric pressure is given as 1 atm, which is equivalent to approximately 101.325 kPa. The formula for the final pressure, assuming no changes in the amount of gas or volume, is P2 = P1 * (T2/T1), where pressures are absolute pressures.

The steps to solve the problem are:

The answer explains how to calculate the initial pressure of a gas using the combined gas law equation.

The initial pressure of the gas in the closed system can be found using the combined gas law equation:

P1V1/T1 = P2V2/T2

Substitute the given values to find the initial pressure, which is calculated to be 1.6 atm.

Therefore, the initial pressure of the gas was 1.6 atm.

Answer:

a) 24 kg

b) 32 kg

Explanation:

The gauge pressure is of the gas is equal to the weight of the piston divided by its area:

p = P / A

p = m * g / (π/4 * d^2)

Rearranging

p * (π/4 * d^2) = m * g

m = p * (π/4 * d^2) / g

m = 1200 * (π/4 * 0.5^2) / 9.81 = 24 kg

After the weight is added the gauge pressure is 2.8kPa

The mass of piston plus addded weight is

m2 = 2800 * (π/4 * 0.5^2) / 9.81 = 56 kg

56 - 24 = 32 kg

The mass of the added weight is 32 kg.

Answer:

a) 333,33 b)291,67

Explanation:

you have both equations :

C=50+0.5V and V=2500-10C

and you want to know the variable V so, you can calculate V in function of C, and you have already clear the variable "C", then you replace for (a):

for b) they tell you that you increase your constant of the equiation C=50+0.5V (remember that the constant is the one alone, wihout any variable, in this case "50") increase in 25 so, your equiation for this point of C is C=75+0.5V, then you do the same:

It's a system of two equations and two variables, wich gives you a compatible define system, it gives you only one solution.

I hope it helps you.

To determine the volume of traffic across the bridge, we can solve an equation relating cost and traffic volume. The volume of traffic across the bridge is approximately 333 vehicles per day. If a toll of 25 cents is added, the volume of traffic across the bridge is approximately 292 vehicles per day.

To determine the volume of traffic across the bridge, we need to solve the equation for V. The equation relating cost and volume of traffic is given by V = 2500 - 10C. Substituting the cost function C = 50 + 0.5V into the equation, we get V = 2500 - 10(50 + 0.5V). Simplifying the equation, we find V = 2500 - 500 - 5V. Combining like terms, we get 6V = 2000, which yields V = 333.33. Therefore, the volume of traffic across the bridge is approximately 333 vehicles per day.

To find the volume across the bridge if a toll of 25 cents is added, we can modify the cost function to be C = 50 + 0.5V + 25 cents. We again substitute this into the equation V = 2500 - 10C and solve for V. Substituting in the new cost function, we get V = 2500 - 10(50 + 0.5V + 25). Simplifying the equation, we find V = 2500 - 10(75 + 0.5V). Continuing to simplify, we have V = 2500 - 750 - 5V. Combining like terms, we get 6V = 1750, which yields V = 291.67. Therefore, the volume across the bridge with the toll of 25 cents is approximately 292 vehicles per day.

Answer:

Stress is a force that acts on a unit area of a material. The strength of a material is how much stress it can bear without permanently deforming or breaking.

Is the beam design acceptable for a SF of 2? YES

Explanation:

Your factor of safety is 2, this means your stress allowed is:

Where:

Now we are going to calculate the shear stress and bending stresses of the proposed scenario. If the calculated stresses are less than the allowed stress, that means the design is adequate for a factor of safety of 2.

First off we calculate the reaction force on your beam. And for this you do sum of forces in the Y direction and equal to 0 because your system is in equilibrium:

Knowing this reaction force you can already calculate the shear stress on the cantilever beam:

Now, you do a sum of moments at the fixed end of your cantilever beam, so you can cancel off any bending moment associated with the reaction forces on the fixed end, and again equal to 0 because your system is in equilibrium.

Knowing the maximum bending moment you can now calculate your bending stress as follows:

Where:

Out of the 3 values needed, we already know M. But we still need to figure out c and Ix. Getting c is very straight forward, since you have a rectangle with base (b) 2 and height (h) 5, you know the centroid is right at the center of the rectangle, meaning that the distance from the centroid to the outermost fibre would be 5in/2=2.5in

To calculate the moment of Inertia, you need to use the formula for the second moment of Inertia of a rectangle and knowing that you will use Ix since you are bending over the x axis:

Now you can use this numbers in your bending stress formula:

The shear stress is 10psi and the bending stress is 120psi, this means you are way below the stress allowed which is 50,000 psi, thus the beam design is acceptable. You could actually use a different geometry to optimize your design.

Answer:

Time needed to empty the pool is 401.35 seconds.

Explanation:

The exit velocity of the water from the orifice is obtained from the Torricelli's law as

[tex]V_{exit}=\sqrt{2gh}[/tex]

where

'h' is the head under which the flow of water occurs

Thus the theoretical discharge through the orifice equals

[tex]Q_{th}=A_{orifice}\times \sqrt{2gh}[/tex]

Now we know that

[tex]C_{d}=\frac{Q_{act}}{Q_{th}}[/tex]

Thus using this relation we obtain

[tex]Q_{act}=C_{d}\times A_{orifice}\times \sqrt{2gh}[/tex]

Now we know by definition of discharge

[tex]Q_{act}=\frac{d}{dt}(volume)=\frac{d(lbh)}{dt}=Lb\cdot \frac{dh}{dt}[/tex]

Using the above relations we obtain

[tex]Lb\times \frac{dh}{dt}=AC_{d}\times \sqrt{2gh}\\\\\frac{dh}{\sqrt{h}}=\frac{AC_{d}}{Lb}\times \sqrt{2g}dt\\\\\int_{1.5}^{0}\frac{dh}{\sqrt{h}}=\int_{0}^{t}\frac{0.62\times 0.3}{15\times 9}\times \sqrt{2\times 9.81}\cdot dt\\\\[/tex]

The limits are put that at time t = 0 height in pool = 1.5 m and at time 't' the height in pool = 0

Solving for 't' we get

[tex]\sqrt{6}=6.103\times 10^{-3}\times t\\\\\therefore t=\frac{\sqrt{6}}{6.103\times 10^{3}}=401.35seconds.[/tex]

Answer:

P=25000lbf

Explanation:

For this problem we will use the equation that relates, the effort, the area and the force for an element under normal stress.

σ=P/A

σ=stress=20kSI=20 000 lbf/in ^2

P=force

A=area

solving for P

P=Aσ

P=(20 000 lbf/in ^2)(1.25in^2)

P=25000lbf

Answer:

The maximum magnitude P of the loads that can be applied to the truss = 25,000 Pounds or 111,205.5 Newtons.

Explanation:

In order to calculate the maximum load P, we will make use of the formula: Maximum average stress (20 ksi) = maximum load P ÷ cross-sectional area (1.25 in²)

Make P (the maximum load) the subject of the formula: P = 20 ksi × 1.25 in².

Before moving further, we have to convert the average normal stress (in ksi) to an appropriate unit: The average normal stress = 20 ksi = 20 kip per square inch (kip/in²)

But 1 kip = 1000 Pounds (i.e., 1000 lb)

Therefore, 20 ksi = 20,000 Pounds/in².

Therefore, P (maximum load) = 20,000 pounds/in² × 1.25 in² = 25,000 Pounds = 111,205.5 Newtons (because 1 Pound = 4.44822 Newtons).

Answer:

Option c is True

Explanation:

a)- Machines are structures made of one or more movable parts. is false statement because Machines can be made without any movable part as in simple machines like inclined plains, wedges, screw etc.

b) Machines have a least one multi force member as part. is false statement since it is not a compulsory condition.

c) Machines transmit force or alter the effect. This is a true statement as machine are made in  the first place to transmit forces and alter the effect.  

d) Machines can be Contrained.

Answer:

21.456 kJ/h

Explanation:

See the figure attached. In this case  

[tex]W_{cycle} = 4 kW [/tex]

[tex]Q_{out} = \text{heating effect}[/tex]

Coefficient of performance in heat pump is defined by

[tex]COP = \frac{Q_{out}}{W_{cycle}} [/tex]

[tex]Q_{out} =COP*W_{cycle} [/tex]

[tex]Q_{out} =1.49*4 \, W [/tex]

[tex]Q_{out} = 5.96 \, W [/tex]

Now it is necessary to change units, remember that Watt (W) is defined as J/s

[tex]Q_{out} = 5.96 \frac{J}{s} \frac{3600s}{1 h} \frac{1 kJ}{1000 J}[/tex]

[tex]Q_{out} = 21.456 \frac{kJ}{h} [/tex]

Answer:

25.4 kW

Explanation:

There are 15 people doing exercises, each will dissipate 740 W of heat, so they will disspiate a total of p = 15*740 = 11100 W = 11.1 kW

There are 7 treadmills, each has a 2.5 hp motor (1.86 kW) running at a load factor of 0.7 with an efficiency of 0.77. So their total power would be p = 7*1.86*0.77/0.7 =14.3 kW

So the total heat dissipated would be 11.1 + 14.3 = 25.4 kW.

The rate of heat gain from people and equipment at peak load is approximately 14.955 kW, including treadmill and people's heat dissipation.

To calculate the rate of heat gain of the exercise room from people and equipment at peak load conditions, we need to consider the heat dissipated by both the exercising equipment and the people in the room.

1. Heat dissipation from exercising equipment:

  - For the treadmills: Each treadmill has a 2.5-hp motor operating at 0.7 load factor and 0.77 efficiency. So, the power consumed by each treadmill is [tex]\( P_{\text{treadmill}} = 2.5 \, \text{hp} \times 0.7 \times 0.77 = 1.925 \, \text{kW} \).[/tex]

  - For the weight-lifting machines: Since they have no motors, they do not contribute to heat dissipation.

2. Heat dissipation from people:

  - There are two people doing light exercises. Assuming each person dissipates heat at a rate of 740 W, the total heat dissipated by people is [tex]\( P_{\text{people}} = 2 \times 740 \, \text{W} = 1480 \, \text{W} \).[/tex]

Now, we can calculate the total rate of heat gain of the exercise room:

[tex]\[ \text{Total heat gain} = P_{\text{treadmill}} \times \text{number of treadmills} + P_{\text{people}} \][/tex]

Given there are 7 treadmills, we have:

[tex]\[ \text{Total heat gain} = 1.925 \, \text{kW} \times 7 + 1480 \, \text{W} \]\[ \text{Total heat gain} = 13.475 \, \text{kW} + 1480 \, \text{W} \]\[ \text{Total heat gain} \approx 13.475 \, \text{kW} + 1.48 \, \text{kW} \]\[ \text{Total heat gain} \approx 14.955 \, \text{kW} \][/tex]

So, the rate of heat gain of the exercise room from people and equipment at peak load conditions is approximately 14.955 kW.

Answer:

(a) Total mass = 14.931 kg

(b) Weight % of platinum = 17.29 %

Explanation:

(a) We have given moles of platinum = 0.0633 moles

Molar mass of platinum = 195.084 unit

Now mass of platinum = number of moles × molar mass = 0.0633×195.084 = 12.3488 gram

We have given moles of nickle = 0.044

Molar mass of nickle = 58.6934 unit

So mass of nickle = 0.044×59.6394 =2.5825 gram

So total mass = 12.3488+2.5825 = [tex]14.931[/tex]

(b) Weight % of platinum = [tex]\frac{weight\ of\ platinum}{total\ weight}=\frac{2.5825}{14.93}=0.1729[/tex] = 17.29 %

Answer:

The tube diameter is 2.71 mm.

Explanation:

Given:

Open glass tube is inserted into a pan of fresh water at 20°C.

Height of capillary raise is four times tube diameter.

h = 4d

Assumption:

Take water as pure water as the water is fresh enough. So, the angle of contact is 0 degree.

Take surface tension of water at 20°C as [tex]72.53\times 10^{-3}[/tex] N/m.

Take density of water as 100 kg/m3.

Calculation:

Step1

Expression for height of capillary rise is gives as follows:

[tex]h=\frac{4\sigma\cos\theta}{dg\rho}[/tex]

Step2

Substitute the value of height h, surface tension, density of water, acceleration due to gravity and contact angle in the above equation as follows:

[tex]4d=\frac{4\times72.53\times10^{-3}\cos0^{\circ}}{d\times9.81\times1000}[/tex]

[tex]d^{2}=7.39\times10^{-6}[/tex]

[tex]d=2.719\times10^{-3}[/tex] m.

Or

[tex]d=(2.719\times10^{-3}m)(\frac{1000mm}{1m})[/tex]

d=2.719 mm

Thus, the tube diameter is 2.719 mm.

Answer:

Factor of safety is defines as the ratio between the strength of a material and the maximum stress that is developed in the material as a result of any given loading.

Mathematically we can write

[tex]F.O.S=\frac{Strength}{Stress_{working}}[/tex]

Consider any member of a machine that is under state of stress due to a general external loading, now we know that the material has a definite value of strength meaning that there is a definite value of force/stress at which the material will fail by fracture or by yielding as an example say a rod of steel will fail if we apply a load of 1000 Newton's on it.

Since as a basic principle of design we do not want any chance of failure in the machine at any cost we limit the maximum value of the stress that is developed in the machine part by designing the part in such a way that during the application of maximum load in the machine part the maximum stresses that are developed in the machine are well below the strength of the material for safety considerations. The factor by which the working stresses are less than the strength is termed as factor of safety.

Another significance of factor of safety is that if we need to estimate the loads on the machine that may act on it and we are not absolutely sure about the magnitude of the loads then we tend to increase the loads by a factor so that we design the structure to withstand greater loads hence if in case in the lifetime of the structure the loads exceed the normal value our structure will still remain structurally safe.  

Answer:

SI unit of system is  superior.

Explanation:

The metric system is used internationally for measurement of unit.This metric system of unit is easier to understand .

Metric system of unit also known as international system of unit.The international system of unit is also represent in short form like SI unit.This unit system is internationally accepted.

So SI unit of system is  superior.

Answer:

froce by 6 gallon liquid on moon surface is  2.86 lbf

Explanation:

given data:

at earth surface

volume of an incompressible liquid = Ve = 25 gallons

force by liquid = 70 lbf

on moon

volume of  liquid = Vm = 6 gallons

gravitational acceleration on moon is am = 5.51 ft/s2

Due to incompressibility , the density remain constant.

mass of liquid on surface of earth[tex]= \frac{ force}{ acceleration}[/tex]

[tex]mass = \frac{70lbf}{32.2 ft/s2}[/tex]

mass = 2.173 pound

[tex]density \rho = \frac{mass}{volume}[/tex]

                  [tex]= \frac{2.173}{25} = 0.0869 pound/ gallon[/tex]

froce by 6 gallon liquid on moon surface is

Fm = mass * acceleration

      = density* volume * am

      = 0.0869 *6* 5.51

      = 2.86 lbf

Answer:

0.948 Btu

Explanation:

1 Btu = 1055 J so [tex]\frac{1000}{1055}[/tex] = 0.948 Btu

1 BTU = 1055.06 joules **

(1000 J) x (1 BTU / 1055.06 J) = 0.9478 BTU

** Note:  You'll have to accept my conversion factor, since I went and looked it up and you didn't.

Answer:

[tex]\eta =\dfrac{1}{mL}[/tex]

[tex]R=\sqrt {\dfrac{KA}{hP}}[/tex]

Explanation:

Fin efficiency:

 Fin efficiency is the ratio of actual heat transfer through fin to the maximum heat transfer through fin.

For infinite long fin:

 Actual heat transfer

[tex]q=\sqrt{hPKA}\Delta[/tex]

So the efficiency of fin

[tex]\eta =\dfrac{1}{mL}[/tex]

Where

[tex]m =\sqrt{\dfrac{hP}{KA}}[/tex]

h is heat transfer coefficient,P is the perimeter,K is the thermal conductivity and A is the cross sectional area.

[tex]q=\sqrt{hPKA}\Delta[/tex]\

From equation we can say that fin resistance

[tex]R=\sqrt {\dfrac{KA}{hP}}[/tex]

Thermal resistance offer resistance to flow of heat.