The speed of sound in air is proportional to the square root of the absolute temperature. If the speed of sound is 349 m/s when the air temperature is 20 °C, what is the temperature of the air when the speed of sound is 340 m/s? Give your answer in °C, K, °F, and R (Rankine).

Explanation:

We know that first T-ds equation

 Tds=[tex]C_v[/tex]dT+Pdv

For constant volume process dv=0

⇒Tds=[tex]C_v[/tex]dT

So [tex]\dfrac{dT}{ds}=\dfrac{T}{C_v}[/tex]          ----(1)

We know that second T-ds equation

 Tds=[tex]C_p[/tex]dT-vdP

For constant pressure process dP=0

⇒Tds=[tex]C_p[/tex]dT

So [tex]\dfrac{dT}{ds}=\dfrac{T}{C_p}[/tex]             ---(2)

We know that [tex]C_p[/tex] is always greater than [tex]C_v[/tex].

So from equation (1) and (2) we can say that slope of constant volume process will be always greater than constant pressure process on T-s diagram.

Answer: True

Explanation: Modified Reynolds Analogy is based on heat transfer process in the Plate Heat Exachanger's(PHE) which is done in the phases that have complex form plates shape by being changed into sheet metal. This analogy helps in the increment of the reliability and operation to be carried out easily in the  industries.Thus the given statement is true.

Answer:

a)f=2.25 Hz

b)Time period T=.144 s

c)tex]V_{max}[/tex]=0.42 m/s

d)Phase angle Ф=87.3°

e) [tex]a_{max}=6.0041 [tex]\frac{m}{s^2}[/tex]

Explanation:

a)

Natural frequency

  [tex]\omega _n=\sqrt {\dfrac{K}{m}}[/tex]

[tex]\omega _n=\sqrt {\dfrac{150}{0.75}}[/tex]

[tex]\omega _n[/tex]=14.14 rad/s

w=2πf

⇒f=2.25 Hz

b) Time period

[tex]=\dfrac{2π}{\omega _n}[/tex]

T=[tex]\frac{1}{f}[/tex]

 Time period T=.144 s

c)Displacement equation

[tex]x=Acos\omega _nt+Bsin\omega _nt[/tex]

Boundary condition

t=o,x=0.03 m

t=0,v=.02m/s   , V=[tex]\frac{dx}{dt}[/tex]

Now by using these above conditions

A=0.03,B=0.0014

x=0.03 cos14.14 t+0.0014 sin14.14 t

⇒x=0.03003sin(14.14t+87.3)

[tex]V_{max}=\omega_n X_{max}[/tex]

[tex]V_{max}=14.14\times 0.03003[/tex]=0.42 m/s

d)

Phase angle Ф=87.3°

e)

Maximum acceleration

[tex]a_{max}=(\omega _n )^2X_{max}[/tex]

[tex]a_{max}=(14.14)^20.03003[/tex]=6.0041 [tex]\frac{m}{s^2}[/tex]

Answer:

A. 2.249 hz

B. 0.45 s

C. 0.424 m/s

D. 66⁰

E. 6 m/s^2

Explanation:

Step 1: identify the given parameters

mass of the block (m)= 0.75kg

stiffness constant (k) = 150N/m

Amplitude (A) = 3cm = 0.03m

upward velocity (v) = 2cm/s

Step 2: calculate the natural frequency (F)by applying relevant formula in S.H.M

[tex]f=\frac{1}{2\pi } \sqrt \frac{k}{m}[/tex]

[tex]f=\frac{1}{2\pi } \sqrt \frac{150}{0.75}[/tex]

f = 2.249 hz

Step 3: calculate the period of the oscillation (T)

[tex]period (T) = \frac{1}{frequency}[/tex]

[tex]T = \frac{1}{2.249} (s)[/tex]

T = 0.45 s

Step 4: calculate the maximum velocity,[tex]V_{max}[/tex]

[tex]V_{max} = A\sqrt{\frac{k}{m} }[/tex]

A is the amplitude of the oscilation

[tex]V_{max} = 0.03\sqrt{\frac{150}{0.75} }[/tex]

[tex]V_{max} = 0.424(\frac{m}{s})[/tex]

Step 5: calculate the phase angle, by applying equation in S.H.M

[tex]X = Acos(\omega{t} +\phi)[/tex]

where X is the displacement; calculated below

Displacement = upward velocity X period of oscillation

[tex]displacement (X) = vt (cm)[/tex]

X = (2cm/s) X (0.45 s)

X = 0.9 cm = 0.009m

where [tex]\omega[/tex] is omega; calculated below

[tex]\omega=\sqrt{\frac{k}{m} }[/tex]

[tex]\omega=\sqrt{\frac{150}{0.75} }[/tex]

[tex]\omega= 14.142[/tex]

[tex]\phi = phase angle[/tex]

Applying displacement equation in S.H.M

[tex]X = Acos(\omega{t}+\phi)[/tex]

[tex]0.009 = 0.03cos(14.142 X 0.45+\phi)[/tex]

[tex]cos(6.364+\phi) = \frac{0.009}{0.03}[/tex]

[tex]cos(6.364+\phi) = 0.3[/tex]

[tex](6.364+\phi) = cos^{-1}(0.3)[/tex]

[tex](6.364+\phi)= 72.5⁰[/tex]

[tex]6.364+\phi =72.5⁰[/tex]

[tex]\phi[/tex] =72.5 -6.364

[tex]\phi[/tex] =66.1⁰

Phase angle, [tex]\phi[/tex] ≅66⁰

Step 6: calculate the maximum acceleration, [tex]a_{max}[/tex]

[tex]a_{max} = \omega^{2}A[/tex]

[tex]a_{max}[/tex] = 14.142 X 14.142 X 0.03

[tex]a_{max}[/tex] = 5.999 [tex](\frac{m}{s^{2} })[/tex]

[tex]a_{max}[/tex] ≅ 6 [tex](\frac{m}{s^{2} })[/tex]

Answer:0.3166N

Explanation:

Given data

Area [tex]\left ( A\right )=500 cm^2[/tex]

Gap below top plate[tex]\left ( y_1\right )=20 mm[/tex]

Gap above bottom plate[tex]\left ( y_2\right )=30 mm[/tex]

SAE 30 oil viscosity =[tex]0.38 N-s/m^2[/tex]

Velocity of middle plate[tex]\left ( v\right )=1 m/s[/tex]

There will viscous force on middle plate i.e. at above surface and below surface

Viscous force[tex]\left ( F\right )=\mu \frac{Av}{y}[/tex]

[tex]Net Force on plate F =\mu Av\left [\frac{1}{y_1} +\frac{1}{y_2}\right ][/tex]

[tex]F=0.38\times 500\times 10^{-4}\left [\frac{1}{20\times 10^{-3}} +\frac{1}{30\times 10^{-3}}\right ][/tex]

[tex]F=31.66\times 10^{-2}=0.3166 N[/tex]

this is force by oil on plate thus we need to apply atleast 0.3166N to move plate

Answer:

Damping ratio  [tex]\zeta =0.0342[/tex]

Explanation:

Given that

m=4.2 kg,K=85.9 N/m,C=1.3 N.s/m

We need to find damping ratio

We know that critical damping co-efficient

 [tex]C_c=2\sqrt {mk}[/tex]

 [tex]C_c=2\sqrt {4.2\times 85.9}[/tex]

 [tex]C_c=37.98[/tex] N.s/m

Damping ratio([tex]\zeta[/tex]) is the ratio of damping co-efficient to the critical damping co-efficient

So [tex]\zeta =\dfrac{C}{C_c}[/tex]

[tex]\zeta =\dfrac{1.3}{37.98}[/tex]

[tex]\zeta =0.0342[/tex]

So damping ratio  [tex]\zeta =0.0342[/tex]

Answer:[tex]\tau _\left ( max\right )[/tex]=11.468MPa

Explanation:

Given data

[tex]power[/tex][tex]\left ( P\right )[/tex]=7 hp=5220 W

N=3200rpm

[tex]\omega [/tex]=[tex]\frac{2\pi\times N}{60}[/tex]=335.14 rad/s

diameter[tex]\left ( d\right )[/tex]=0.75in=19.05mm

we know

P=[tex]Torque\left ( T\right )\omega [/tex]

5220=[tex]T\times 335.14[/tex]

T=15.57 N-m

And

[tex]\tau _\left ( max\right )[/tex]=[tex]\frac{T}{Polar\ modulus}[/tex]

[tex]\tau _\left ( max\right )[/tex]=[tex]\frac{T}{Z_{P}}[/tex]

[tex]\tau _\left ( max\right )[/tex]=[tex]\frac{16\times T}{\pi d^{3}}[/tex]

[tex]\tau _\left ( max\right )[/tex]=11.468MPa

Answer:

Maximum shear stress is 11.47 MPa.

Explanation:

Given:

D=.75 in⇒D=19.05 mm

P=7 hp⇒ P=5219.9 W

N=3200 rpm

We know that

    P=T[tex]\times \omega[/tex]

Where T is the torque and [tex]\omega[/tex] is the speed of shaft.

   P=[tex]\frac{2\pi N\times T}{60}[/tex]

So    5219.9=[tex]\frac{2\pi \times 3200\times T}{60}[/tex]

 T=15.57 N-m

We know that maximum shear stress in shaft

[tex]\tau _{max}=\dfrac{16T}{\pi \times D^3}[/tex]

[tex]\tau _{max}=\dfrac{16\times 15.57}{\pi \times 0.01905^3}[/tex]

[tex]\tau _{max}[/tex]=11.47 MPa

So maximum shear stress is 11.47 MPa.

Answer:

V = 0.5 m/s

Explanation:

given data:

width of channel =  4 m

depth of channel = 2 m

mass flow rate = 4000 kg/s = 4 m3/s

we know that mass flow rate is given as

[tex]\dot{m}=\rho AV[/tex]

Putting all the value to get the velocity of the flow

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

[tex]V = \frac{4000}{1000*4*2}[/tex]

V = 0.5 m/s

The maximum efficiency must be greater than the actual efficiency of the heat engine. These measurements are not reasonable.

The efficiency is defined as the work done by the engine divided by the heat supplied.

So, maximum efficiency η = 1 - T₁/T₂ = W/Qs

Where T₁ is the lower temperature and T₂ is the higher temperature.

An experimentalist claims that, based on his measurements, a heat engine receives 300 Btu of heat from a source of 900 R, converts 160 Btu of it to work, and rejects the rest as waste heat to a sink at 540 R

Substitute the value into the expression , we get

η = 1 -(540 / 900) x 100 %

η = 40 %

η  = Work done/heat supplied

Substitute the value into the expression , we get

η = 160/300 x 100 %

η = 53 %

So, the maximum efficiency must be greater than the actual efficiency of the heat engine.

Thus, these measurements are not reasonable.

Learn more about efficiency.

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The experimentalist's claim that the heat engine converts more work than allowed by the Carnot efficiency is not reasonable because it exceeds the theoretical maximum efficiency, violating the second law of thermodynamics.

The experimentalist's claim that a heat engine receives 300 Btu of heat from a source at 900 R, converts 160 Btu of it to work, and rejects the rest as waste heat to a sink at 540 R can be assessed using the second law of thermodynamics and the concept of Carnot efficiency. The Carnot efficiency can be calculated using the temperatures of the hot and cold reservoirs (Th and Tc):

Carnot Efficiency (EffC) = 1 - (Tc/Th)

Assuming the temperatures are in degrees Rankine (where 0 R is absolute zero), we can calculate the maximum theoretical efficiency for a perfectly reversible engine:

EffC = 1 - (540 R / 900 R) = 1 - 0.6 = 0.4 (or 40%)

However, the experimentalist's claim states that 160 Btu is converted to work out of the 300 Btu received, which represents an actual efficiency (EffA) of:

EffA = Work done / Heat received = 160 Btu / 300 Btu = 0.5333 (or 53.33%)

Since the actual efficiency (53.33%) claimed by the experimentalist exceeds the maximum possible Carnot efficiency (40%) for a heat engine operating between these temperatures, the experimentalist's measurements are not reasonable and violate the second law of thermodynamics.

Answer:

C) Injection

Explanation:

Injection is a molding process, not a heat transfer mechanism.

Answer:

True.

Explanation:

Yes, zeroes indicates the precision  after the decimal point.

For smallest of the calculation in to get the precise value is very important for that the calculation can have very minute changes in decimal point as more accurate the calculation  is more zeroes will be in the decimal value .

Most of the instrument calculating weights is said to precise by how much decimal they can calculate.

Answer:

In refrigeration cycle heat transfer from inside refrigeration

In heat pump cycle heat transfer from environment

Explanation:

heat cycle is mechanical process use for cool the temperature but

In refrigeration heat transfer from inside of refrigeration that decrease temperature of refrigerator and in heat pump it decrease temperature negligible as compare to refrigerator

Given:

kinetic energy of free particle, KE = 35ev

1eV = [tex]1.6\times 10^{-19}[/tex] J

mass of the particle, m = [tex]2\times 10^{-26}[/tex] Kg

accuracy in velocity= 0.2%= 0.002

Solution:

a) We know that

KE = [tex]\frac{1}{2}mv^{2}[/tex]

v = [tex]\sqrt{\frac{2KE}{m}}[/tex]

⇒ v = [tex]\sqrt{\frac{2\times 35\times1.6\times 10^{-19} }{2\times 10^{-26}}}[/tex]

v = [tex]2.36\times10^{4}[/tex] m/s

b) From Heisenberg's uncertainity principle:

[tex]\Delta x\Delta p = \frac{h}{4\pi}[/tex]

[tex]\Delta x.(mv) = \frac{h}{4\pi}[/tex]

[tex]\Delta x = \frac{h}{4\pi\times 2\times 10^{-26}\times2.36 \times 10^{4}\times 0.002}[/tex]

[tex]\Delta x = 0.56nm[/tex]

Answer:

False

Explanation:

Pascal's law is not for open container it is for enclosed fluid

Pascal's law (also known as the principle of transmission of fluid pressue)

states that when pressure is applied at any part of an enclosed fluid it is transmitted undiminished throughout the fluid as well as to the walls of the container containing the fluid i.e., change in pressure at any point in an enclosed incompressible fluid is transmitted such that the same change occurs everywhere.

Answer:

[tex]p=12\hat{i}+10.58\hat{j}[/tex]

Explanation:

distance of point from the origin is 16 inch

x-axis distance =12 inch

[tex]p=\sqrt{x^2+y^2[/tex]

[tex]y=\sqrt{p^2-x^2} \\y=\sqrt{16^2-12^2}\\ y=\sqrt{112}[/tex]

y=10.58 inch

[tex]p=12\hat{i}+10.58\hat{j}[/tex]

Answer:

a)-True

Explanation:

The drive force for diffusion is 7 Fick's first law can be used to solve the non-steady state diffusion.

This statement is true.

Answer:

A two stroke engine produces twice the power compared to a four stroke engine of same weight and size.

Explanation:

               In a  two stroke engine, all the four processes namely, intake stroke, compression stroke, power stroke and exhaust stroke takes place in one revolution of crankshaft or two strokes of the piston. While in a four stroke engine, all the four processes namely intake stroke, compression stroke, power stroke and exhaust stroke take place in two revolution of crankshaft or four strokes of the piston.

            Therefore, there is one power stroke in one revolutions of the crankshaft in case of a two stroke engine as compared to the four stoke engine where there is one power stroke for two revolutions of the crank shaft.

             So the power developed in a two stroke engine is more ( nearly twice ) as compared to a four stroke engine of the same capacity. When power produced is more, the heat dissipation is also more in case of a two stroke engine. So greater cooling is required to dissipate heat from a two stroke engine as compared to a four stroke engine.

           Also in a two stroke engine, the lubricating oil is used with the oil whereas a four stroke engine has a separate tank for lubricating oil. So the lubricating oil gets burnt quickly in a two stroke engine.

Thus, to dissipate more heat, a two stroke engines has greater cooling and lubrication requirements than a four stroke engines as power produce in a two stroke engine is more than a four stoke engine with same weight or size.

The efficiency of a transformer is mainly dependent on a)- Core losses

Hope this helps! :)

Answer:

Explanation:

Using equation of pure torsion

[tex]\frac{T}{I_{polar} }=\frac{t}{r}[/tex]

where

T is the applied Torque

[tex]I_{polar}[/tex] is polar moment of inertia of the shaft

t is the shear stress at a distance r from the center

r is distance from center

For a shaft with

[tex]D_{0} =[/tex] Outer Diameter

[tex]D_{i} =[/tex] Inner Diameter

[tex]I_{polar}=\frac{\pi (D_{o} ^{4}-D_{in} ^{4}) }{32}[/tex]

Applying values in the above equation we get

[tex]I_{polar} =\frac{\pi(0.042^{4}-(0.042-.008)^{4})}{32}\\

I_{polar}= 1.74[/tex] x [tex]10^{-7} m^{4}[/tex]

Thus from the equation of torsion we get

[tex]T=\frac{I_{polar} t}{r}[/tex]

Applying values we get

[tex]T=\frac{1.74X10^{-7}X100X10^{6}  }{.021}[/tex]

T =829.97Nm

Answer:

x3=0.100167

Explanation:

Let's find the answer.

Because we are going to find the solution for sin(Ф)=0.1 then:

f(x)=sin(Ф)-0.1 and:

f'(x)=cos(Ф)

Because 0<Ф<π/2 let's start with an initial guess of 0.001 (x0), so:

x1=x0-f(x0)/f'(0)

x1=0.001-(sin(0.001)-0.1)/cos(0.001)

x1= 0.100000

x2=0.100000-(sin(0.100000)-0.1)/cos(0.100000)

x2=0.100167

x3=0.100167

Answer:

Head loss=0.00366 ft

Explanation:

Given :Water flow rate Q=0.15 [tex]\frac{ft^{3}}{sec}[/tex]

         [tex]D_{1}[/tex]= 6 inch=0.5 ft

        [tex]D_{2}[/tex]=2 inch=0.1667 ft

As we know that Q=AV

[tex]A_{1}\times V_{1}=A_{2}\times V_{2}[/tex]

So [tex]V_{2}=\frac{Q}{A_2}[/tex]

     [tex]V_{2}=\dfrac{.015}{\frac{3.14}{4}\times 0.1667^{2}}[/tex]

     [tex]V_{2[/tex]=0.687 ft/sec

We know that Head loss due to sudden contraction

           [tex]h_{l}=K\frac{V_{2}^2}{2g}[/tex]

If nothing is given then take K=0.5

So head loss[tex]h_{l}=(0.5)\frac{{0.687}^2}{2\times 32.18}[/tex]

                                    =0.00366 ft

So head loss=0.00366 ft

Answer:

yes

Explanation:

yes, it is correct as car gets older  the engine's compression ratio changes .

compression ratio is the volume in maximum compression chamber to the volume in the piston at full compression .

More and more engine will be used if the ring wears compression goes down

or there can be one condition if carbon build up on the piston then this can alter the compression ratio.

Answer: b) They comprise at least two immiscible phases

Explanation: Nano composite coating are the coating that are the mixture of one or more phases that are irregular or immiscible in nature. They are in the nano- material form which helps in the improvement of chemical as well as physical properties. The phase that are usually present in nano composite coating are of nano crystalline phase or amorphous phase or two nano crystalline phases  which are different from each other.

Answer:

Explanation:

The functions of the following are as follows:

1). Oil Rings:

2). Fly Wheel:

3). Timing gears:

Answer: Isentropic process is the process in fluids which have a constant entropy.

Explanation: The  isentropic process is considered as the ideal thermodynamical  process and has both adiabatic as well as reversible processes in internal form.This process supports no transfer of heat and  no transformation of matter .The entropy of the provided mass also remains unchanged or consistent.These processes are usually carried out on material on  the efficient device.

Answer:

[tex]\tau_{max}= 3.28 \ MPa[/tex]

Explanation:

outside diameter = 90 mm

inside diameter = 90- 2× t=  90- 2× 5 = 80mm

where t is thickness of pipe.

power (P)  = 12 kW

Revolution (N)= 650 rev/min

we

Power = torque × angular velocity

P= T× ω

ω =  [tex]\frac{2 \pi N}{60}[/tex]

[tex]P=T \times \frac{2\pi N}{60}\\12 \times 10^3=T\times \frac{2\pi \times 650}{60}[/tex]

T=  176.3 Nm

for maximum shear stress

[tex]\frac{\tau_{max}}{y_{max}}=\frac{T}{I_p}[/tex]

where ymax is maximum distance from neutral axis.

[tex]y_{max}=\frac{d_0}{2}= \frac{90}{2}[/tex]= 45 mm

[tex]I_p[/tex]= polar moment area

          = [tex]\frac{\pi}{32} (d_o^4-d_i^4)=\frac{\pi}{32} (90^4-80^4)[/tex]

          = 2,420,008 mm⁴

[tex]\dfrac{\tau_{max}}{45}=\dfrac{176.3 \times 10^3}{2,420,008}[/tex]

[tex]\tau_{max}= 3.28 \ MPa[/tex]

Answer:

e.All of the above

Explanation:

A fluid should be change

a.When its acidity increase

b.When its become contaminated

c.At operating temperature

Generally fluid is used in fluid power.Fluid power means transmission of power by using pressurized fluids.

Fluid power works on Pascal's and Bernoulli law.

From the above option we can say that all option is right for fluid  .

Answer:

Drag Coefficient:

It is a dimensionless quantity that is used to quantify air friction or fluid friction i.e., drag and is commonly denoted by [tex]C_{d}[/ tex].It is used in Drag Equation, where a lower drag coefficient indicates that the object will have lower value of air friction or fluid friction(i.e., aerodynamic or hydrodynamic drag). It is always associated with a particular surface area.

Drag Coefficient is given by:

[tex]C_{d} = \frac{2D}{A\rho v^{2}}[/tex]

Skin Friction Coefficient:

It is a dimensionless quantity which represents skin shear stress due to dynamic pressure of a free stream.

Skin Friction Coefficient is given by:

[tex]C_{f} = \frac{2\tau _{w}}{\rho V^{2}}[/tex]

where,

[tex]\tau _{w}[/tex]= skin shear stress

v = free stream speed

Answer:

4.83m/[tex]s^{2}[/tex]

Explanation:

For a particle moving in a circular path the resultant  acceleration at any point is the vector sum of radial and the tangential acceleration

Radial acceleration is given by [tex]a_{radial}=w^{2}[/tex]r

Applying values we get  [tex]a_{radial}=(2t)^{2}[/tex]X0.3m

Thus [tex]a_{radial}=1.2t^{2}[/tex]

At time = 2seconds [tex]a_{radial}= 4.8m/s^{2}[/tex]

The tangential acceleration is given by [tex]a_{tangential} =\frac{dV}{dt}=\frac{d(wr)}{dt}[/tex]

[tex]a_{tangential}=\frac{d(2tr)}{dt}[/tex]

[tex]a_{tangential}= 2r[/tex]

[tex]a_{tangential}=0.6m/s^{2}[/tex]

Thus the resultant acceleration is given by

[tex]a_{res} =\sqrt{a_{rad}^{2}+a_{tangential}^{2}}[/tex]

[tex]a_{res} =\sqrt{4.8^{2}+0.6^{2}  } =4.83m/s^{2}[/tex]

Answer:

False

Explanation:

as we know that [tex]V(x)=\frac{dM}{dx}\\ \\=> M(x) = \int\limits^x_o {V(x)} \, dx \\\\[/tex]

=> Area under shear diagram gives the moment at any point but the reverse cannot be established from the same relation