Answer:
0.1 J/K
Explanation:
entropy change equation is as followed:
[tex]\Delta S=\frac{\Delta Q}{T}[/tex]
where ΔS=entropy change
Q=Heat transfer
T= temperature
[tex]\Delta S=\frac{\Delta Q}{T}[/tex]
[tex]\Delta S=\frac{30}{300}[/tex]
[tex]\Delta S=0.1 J/K[/tex]
hence the change in entropy of system which is [tex]\Delta S[/tex]is equal to 0.1 J/K
A thin flat plate of surface area, 500 cm², is located between two fixed plates such that the gap below the top plate is 20 mm and the gap above the bottom plate is 30 mm. The gap between the fixed plates contain SAE 30 oil at 15.6°C. Determine the force needed to push the middle plate at a velocity of 1.0 m/s.
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
A heat engine operates between a hot reservoir at 2000°C and the atmosphere (cold reservoir) at 25°C. it produces 50 MW of power while rejecting 40 MW of waste heat. Determine the maximum possible thermal efficiency of the engine in percent.
Answer:
55.56%
Explanation:
Given data
Temprature of hot reservior =2000°c=2273k
Temprature of Cold reservior=25°c=298k
Power produced by engine=50MW
Heat rejected =40MW
we know that Effeciency(η) of heat engine=[tex]\frac{Work produced}{heat supplied}[/tex]
Also we know that
heat supplied[tex]\left ({Q_s} \right )=work produced{W}+Heat rejected{Q_r}[/tex]
Q_s=50+40=90MW
η=[tex]\frac{W}{Q_s}[/tex]
η=[tex]\frac{50}{90}[/tex]
η=55.55%
Define Viscosity. What are the main differences between viscous and inviscid flows?
1. Define Viscosity
In physics, Viscosity refers to the level of resistance of a fluid to flow due to internal friction, in other words, viscosity is the result of the magnitude of internal friction in a fluid, as measured by the force per unit area resisting uniform flow. For example, the honey is a fluid with high viscosity while the water has low viscosity.
What are the main differences between viscous and inviscid flows?
Viscous flows are flows that has a thick, sticky consistency between solid and liquid, contain and conduct heat, does not have a rest frame mass density and whose motion at a fixed point always remains constant. Inviscid flows, on the other hand, are flows characterized for having zero viscosity (it does not have a thick, sticky consistency), for not containing or conducting heat, for the lack of steady flow and for having a rest frame mass density
Furthermore, viscous flows are much more common than inviscid flows, while this latter is often considered an idealized model since helium is the only fluid that can become inviscid.
A solid 0.75 in diameter steel shaft transmits 7 hp at 3,200 rpm. Determine the maximum shear stress magnitude produced in the shaft. Hint: Use P=Tω and convert hp to ft-lbf/s. Find τ by using Tc/J. Recall max shear stress will be on the outer most surface.
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.
Air is heated from 50 F to 200 F in a rigid container with a heat transfer of 500 Btu. Assume that the air behaves as an ideal gas. Determine the volume of air [ft3] if the initial pressure is 2 atm. Also show the process on a P-v state diagram. Use the following temperature conversion: T[R] = T[F] + 460.
Answer:
[tex]V=68.86ft^3[/tex]
Explanation:
[tex]T_1[/tex] =10°C,[tex]T_2[/tex] =93.33°C
Q=500 btu=527.58 KJ
[tex]P_1= 2atm[/tex]
If we assume that air is ideal gas PV=mRT, ΔU=[tex]mC_v(T_2-T_1)[/tex]
Actually this is closed system so work will be zero.
Now fro first law
Q=ΔU=[tex]mC_v(T_2-T_1)[/tex]+W
⇒Q=[tex]mC_v(T_2-T_1)[/tex]
527.58 =[tex]m\times 0.71(200-50)[/tex]
m=4.9kg
PV=mRT
[tex]200V=4.9\times 0.287\times (10+273)[/tex]
[tex]V=1.95m^3[/tex] ([tex]V=1m^3=35.31ft^3[/tex])
[tex]V=68.86ft^3[/tex]
The two stroke engines has greater cooling and lubrication requirements than four stroke engine. Explain why?
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.
Water is flowing at a rate of 0.15 ft3/s in a 6 inch diameter pipe. The water then goes through a sudden contraction to a 2 inch diameter pipe. What is the head loss through this contraction
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
The drilling pipe on an oil rig is made from steel pipe which having thickness of 5-mm and an outside diameter of 90-mm. Calculate the maximum shear stress occur in the pipe if the pipe is turning at 650 rev/min while being powered using 12kW motor.
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]
A shaft made of stainless steel has an outside diameter of 42 mm and a wall thickness of 4 mm. Determine the maximum torque T that may be applied to the shaft if the allowable shear stress is equal to 100 MPa.
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
Pascal's law tells us that, pressure is transmitted undiminished throughout an open container. a)- True b) False
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.
Internal flow is one in which the flow is not bounded. a) True b) False
False it is External
The specific gravity of a fluid with a weight density of 31.2 lb/ft is a. 2.00 b. 0.969 c. 0.500 d. 1.03
Answer:
Answer is c 0.500
Explanation:
[tex]SpecificGravity=\frac{\rho _{fluid}*g}{\rho _{water}*g}[/tex]
We know that [tex]\rho_{water}=62.42lb/ft^{3}[/tex]
Applying values we get
[tex]SpecificGravity=\frac{31.2}{62.4}=0.5[/tex]
Fluid power is a. The technology that deals with the generation, control, and transmission of power-using pressurized fluids b. muscle that moves industry. c. used to push, pull, regulate, or drive virtually all the machine of modern industry d. probably as old as civilization itself e. all of the above
Answer: a) The technology that deals with the generation, control and transmission of power using pressurized fluids
Explanation: Fluid power is defined as the fluids which are under pressure and then are used for generation,control and transmit the power. Fluid power systems produces high forces as well as power in small amount . These systems usually tend to have better life if maintained properly. The force that are applied on this system can be monitored by gauges as well as meter.
A block of mass 0.75 kg is suspended from a spring having a stiffness of 150 N/m. The block is displaced downwards from its equilibrium position through a distance of 3 cm with an upward velocity of 2 cm/sec. Determine: a)- The natural frequency b)-The period of oscillation c)-The maximum velocity d)-The phase angle e)-The maximum acceleration
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]
In an air standard diesel cycle compression starts at 100kpa and 300k. the compression ratio is 16 to 1. The maximum cycle temperature is 2031K. Determine the thermal efficiency.
Answer:
[tex]\eta[/tex]=0.60
Explanation:
Given :Take [tex]\gamma[/tex]=1.4 for air
[tex]P_1=100 KPa ,T_1=300K[/tex]
[tex]\frac{V_1}{V_2}[/tex]=r ⇒ r=16
As we know that
[tex]T_2=T_1(r^{\gamma-1})[/tex]
So [tex]T_2=300\times 16^{\gamma-1}[/tex]
[tex]T_2[/tex]=909.42K
Now find the cut off ration [tex]\rho[/tex]
[tex]\rho=\frac{V_3}{V_2}[/tex]
[tex]\frac{V_3}{V_2}=\frac{T_3}{T_2}[/tex]
[tex]\rho=\frac{2031}{909.42}[/tex]
[tex]\rho=2.23[/tex]
So efficiency of diesel engine
[tex]\eta =1-\dfrac{\rho^\gamma-1}{\gamma\times r^{\gamma-1}(\rho-1)}[/tex]
Now by putting the all values
[tex]\eta =1-\dfrac{2.23^{1.4}-1}{1.4\times 16^{1.4-1}(2.23-1)}[/tex]
So [tex]\eta[/tex]=0.60
So the efficiency of diesel engine=0.60
Prove that the slope of a constant-volume line is steeper than that for a constant-pressure line for a given state (point) on the T-s diagram.
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.
A particle moves along a circular path of radius 300 mm. If its angular velocity is θ = (2t) rad/s, where t is in seconds, determine the magnitude of the particle's acceleration when t= 2 s.
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]
The efficiency of a transformer is mainly dependent on: a)- Core losses b)- Copper losses c)- Stray losses d)- Dielectric losses
The efficiency of a transformer is mainly dependent on a)- Core losses
Hope this helps! :)
The drive force for diffusion is 7 Fick's first law can be used to solve the non-steady state diffusion. a)-True b)-False
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.
Which of the following is not a method of heat transfer? A. Conduction B. Convection C. Injection D. Radiation is desirable.
Answer:
C) Injection
Explanation:
Injection is a molding process, not a heat transfer mechanism.
An aluminum electrical cable is 20 mm in diameter is covered by a plastic insulation (k = 1 W/m-k) of critical thickness. This wire is placed in a room with an air flow heat transfer coefficient of 50 W/m^2-K. Compared to the bare aluminum wire, the heat loss from this insulated wire will be a) LESS b) MORE c) SAME d) ZERO
Answer:
the heat loss from this insulated wire is less
Explanation:
Given data in question
diameter of cable (d) = 20 mm
( K ) = 1 W/m-k
heat transfer coefficient (h) = 50 W/m²-K
To find out
the heat loss from this insulated wire
solution
we will find out thickness of wire
heat loss is depend on wire thickness also
we have given dia 20 mm
so radius will be d/2 = 20/ 2 = 10 mm
Now we find the critical thickness i.e.
critical thickness = K / heat transfer coefficient
critical thickness = 1 / 50 = 0.02 m i.e. 20 mm
now we can see that critical thickness is greater than radius 10 mm
so our rate of heat loss will be decreasing
so we can say our correct option is (a) less
Consider an open loop 1-degree-of-freedom mass-spring damper system. The system has mass 4.2 kg, and spring stiffness of 85.9 N/m, and damping coefficient of 1.3 N.s/m. What is the non-dimensional damping ratio of the system? Use at least 4 significant digits after the decimal point.
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]
Sandwich materials typically use a high density core with non-structural cover plates. a)True b)- False
Answer: False
Explanation: Sandwich materials are usually in composite material form which has a fabrication of two thin layers which are stiff in nature and have light weighing and thick core .The construction is based on the ratio that is of stiffness to the weight .Therefore, the density of the material in the core is not high and are only connected with the skin layer through adhesive .So the given statement is false that sandwich materials typically use a high density core with non- structural cover plates.
How to convert a friction to decimal ?
Answer:
To convert a fraction to a decimal, divide the numerator by the denominator.
What is the output of a system with the transfer function s/(s + 3)^2 and subject to a unit step input at time t = 0?
Answer:
0
Explanation:
output =transfer function H(s) ×input U(s)
here H(s)=[tex]\frac{s}{(s+3)^2}[/tex]
U(s)=[tex]\frac{1}{s}[/tex] for unit step function
output =H(s)×U(s)
=[tex]\frac{s}{(s+3)^2}[/tex]×[tex]\frac{1}{s}[/tex]
=[tex]\frac{1}{(s+3)^2}[/tex]
taking inverse laplace of output
output=t×[tex]e^{-3t}[/tex]
at t=0 putting the value of t=0 in output
output =0
Use Newton's method to determine the angle θ, between 0 and π/2 accurate to six decimal places. for which sin(θ) = 0.1. Show your work until you start computing x1, etc. Then just write down what your calculator gives you.
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
The mass flow rate in a 4.0-m wide, 2.0-m deep channel is 4000 kg/s of water. If the velocity distribution in the channel is linear with depth what is the surface velocity of flow in the channel?
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
A horizontal pipe is fitted with a nozzle. The inlet diameter of the nozzle is 40 mm and the outlet diameter is 20 mm. The flow rate in the pipe is 1.2 m3 /min and water density is 1000 kg/m3 . Determine the force exerted by the nozzle on the water.
Answer:
969.68N
Explanation:
d₁=0.04 m A₁=[tex]\frac{\pi d^2_{1} }{4}[/tex]
[tex]A_{1} =\frac{\pi \times .04^2}{4}= 0.00125m^{2} \[/tex]
d₂=0.02 m A₂=[tex]\frac{\pi d^2_{2} }{4}[/tex]
[tex]A_{2} =\frac{\pi \times .02^2}{4}= 0.00031m^{2} \[/tex]
Q=1.2m³/min Q=1.2/60=0.02m³/s
using continuity equation
Q₁=A₁v₁
v₁=Q₁/A₁=0.02/0.00125=16m/s
Q₂=A₂v₂
v₂=Q₂/A₂=0.02/0.00031=64.5m/s
[tex]F_{inlet}=\rho A_{1}v_1^{2}[/tex]
[tex]F_{inlet}=1000\times 0.00125\times16^{2}=320N[/tex]
[tex]F_{outlet}=\rho A_{2}v_2^{2}[/tex]
[tex]F_{outlet}=1000\times 0.00031\times64.5^{2}=1289.68N[/tex]
Force on the nozzle=F_{outlet}-F_{inlet}
= 1289.68-320
=969.68N
What is the difference between the drag coefficient and skin friction coefficient for flow along flat plate.
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
A hollow steel shaft with and outside diameter of (do)-420 mm and an inside diameter of (di) 350 mm is subjected to a torque of 300 KNm, as shown. The modulus of rigidity G for the steel is 80 GPa. Determine: (a) The maximum shearing stress in the shaft. (b) The shearing stress on a traverse cross section at the inside surface of the shaft (c) The magnitude of the angle of twist for a (L) -2.5 m length.
Answer:
a. [tex]\tau=51.55 MPa[/tex]
b.[tex]\tau=42.95MPa[/tex]
c.[tex]\theta=7.67\times 10^{-3}[/tex] rad.
Explanation:
Given: [tex]D_i=350 mm,D_o=420 mm,T=300 KN-m ,G=80 G Pa [/tex]
We know that
[tex]\dfrac{\tau}{J}=\dfrac{T}{r}=\dfrac{G\theta}{L}[/tex]
J for hollow shaft [tex]J=\dfrac{\pi (D_o^4-D_i^4)}{64}[/tex]
(a)
Maximum shear stress [tex]\tau =\dfrac{16T}{\pi Do^3(1-K^4)}[/tex]
[tex]K=\dfrac{D_i}{D_o}[/tex]⇒K=0.83
[tex]\tau =\dfrac{16\times 300\times 1000}{\pi\times 0.42^3(1-.88^4)}[/tex]
[tex]\tau=51.55 MPa[/tex]
(b)
We know that [tex]\tau \alpha r[/tex]
So [tex]\dfrac{\tau_{max}}{\tau}=\dfrac{R_o}{r}[/tex]
[tex]\dfrac{51.55}{\tau}=\dfrac{210}{175}[/tex]
[tex]\tau=42.95MPa[/tex]
(c)
[tex]\dfrac{\tau_{max}}{R_{max}}=\dfrac{G\theta }{L}[/tex]
[tex]\dfrac{51.55}{210}=\dfrac{80\times 10^3\theta }{2500}[/tex]
[tex]\theta=7.67\times 10^{-3}[/tex] rad.