Answer:
C) Injection
Explanation:
Injection is a molding process, not a heat transfer mechanism.
A free particle has kinetic energy equal to 35eV a)- What is the velocity of the particle? b)- If this velocity is known to within 0.2% accuracy, what is r of the position that particle? Assume that the mass of the particle is 2 x 10^-26 kg. Use h and give the answer in nm.
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]
Cylinder cushions at the ends of the cylinder speeds the piston up at the end of the stroke. a)- True b) False
What is the difference between a refrigeration cycle and a heat pump cycle?
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
What is an isentropic process?
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.
A Carnot refrigeration cycle absorbs heat at -12 °C and rejects it at 40 °C. a)-Calculate the coefficient of performance of this refrigeration cycle b)-If the cycle is absorbing 15 kW at the -12C temperature, how much power is required? c)-If a Carnot heat pump operates between the same temperatures as the above refrigeration cycle, determine the COP of heat pump? d)-What is the rate of heat rejection at the 40°C temperature if the heat pump absorbs 15 kw at the -12 C temperature?
Answer:
a)COP=5.01
b)[tex]W_{in}=2.998[/tex] KW
c)COP=6.01
d)[tex]Q_R=17.99 KW[/tex]
Explanation:
Given
[tex]T_L[/tex]= -12°C,[tex]T_H[/tex]=40°C
For refrigeration
We know that Carnot cycle is an ideal cycle that have all reversible process.
So COP of refrigeration is given as follows
[tex]COP=\dfrac{T_L}{T_H-T_L}[/tex] ,T in Kelvin.
[tex] COP=\dfrac{261}{313-261}[/tex]
a)COP=5.01
Given that refrigeration effect= 15 KW
We know that [tex]COP=\dfrac{RE}{W_{in}}[/tex]
RE is the refrigeration effect
So
5.01=[tex]\dfrac{15}{W_{in}}[/tex]
b)[tex]W_{in}=2.998[/tex] KW
For heat pump
So COP of heat pump is given as follows
[tex]COP=\dfrac{T_h}{T_H-T_L}[/tex] ,T in Kelvin.
[tex] COP=\dfrac{313}{313-261}[/tex]
c)COP=6.01
In heat pump
Heat rejection at high temperature=heat absorb at low temperature+work in put
[tex]Q_R=Q_A+W_{in}[/tex]
Given that [tex]Q_A=15[/tex]KW
We know that [tex]COP=\dfrac{Q_R}{W_{in}}[/tex]
[tex]COP=\dfrac{Q_R}{Q_R-Q_A}[/tex]
[tex]6.01=\dfrac{Q_R}{Q_R-15}[/tex]
d)[tex]Q_R=17.99 KW[/tex]
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. Are these measurements reasonable? Why?
The maximum efficiency must be greater than the actual efficiency of the heat engine. These measurements are not reasonable.
What is Efficiency?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.
Explanation: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.
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 cubic transmission casing whose side length is 25cm receives an input from the engine at a rate of 350 hp. If the vehicle's velocity is such that the average heat transfer coefficient is 230 W/m'K and the efficiency of the transmission is n=0.95 (i.e. the remainder is heat), calculate the surface temperature if the ambient temperature is 15°C. Assume that heat transfer only occurs on one face of the cube. NOTE: 1 hp=745 W
Answer:
The surface temperature is 921.95°C .
Explanation:
Given:
a=25 cm ,P=350 hp⇒P=260750 W
Power transmitted [tex]0.95\times 260750[/tex]W and remaining will lost in the form of heat.This heat transmitted to air by the convection.
h=230[tex]\frac{W}{m^2-K},\eta =0.95[/tex]
Actually heat will be transmit by the convection.
In convection Q=hA[tex]\Delta T[/tex]
So [tex]P=\Delta T\times Q[/tex]
[tex]0.05\times 260750=230\times0.25^2\(T-15)[/tex]
T=921.95°C
So the surface temperature is 921.95°C .
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.
The equation of motion is not valid without the assumption of an inertial frame. a) True b)- false
Answer:
True
Explanation:
when we write equation of motion we have to assume frame of reference as because frame of reference is the property that in frame of reference the body is not accelerated and net force acting on the body is zero.
when body is assumed to be any frame of reference we can assume the body is at rest and moving with constant speed.
Answer:
true
Explanation:
hope this helped , God bless
A fluid should be changed a. when its viscosity changes b. when it becomes contaminated c. at operating temperature d. when its acidity increases e. all of the above
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 .
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! :)
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
List fabrication methods of composite Materials.
Answer:
Fabrication method of composite materials varies for one product of material to other product. It is basically developed to meet the product requirement.
The fabrication of composite parts are depends upon different factors that are:
Depend on the Characteristics of strengthening and matrices. The details of the product and the shape and size also.Application or end uses.Types of fabrication methodologies are :
Press moldingCompression moldingContact moldOpen molding Tube rolling
The use of zeroes after a decimal point are an indicator of accuracy. a)True b)- False
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.
The area under the moment diagram is shear force. a)-True b)-False
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
A student checks her car's tyre air pressure at a petrol station and finds it is 31 psi. Re-express this as an absolute pressure in kpa. Note any assumptions you make.
Answer:
Absolute Pressure=315.06256 kPa
Explanation:
Gauge pressure= 31 psi
Atmospheric Pressure at Sea level= 1 atm=101.325 kPa
[tex]1\ psi=6.89476\ kPa\\\Rightarrow 31\ psi=31\times 6.89476\\\Rightarrow 31\ psi=213.73756\ kPa=Gauge\ Pressure\\Absolute\ Pressure=Gauge\ Pressure+Atmospheric\ Pressure\\\Rightarrow Absolute\ Pressure=213.73756+101.325\\\therefore Absolute\ Pressure=315.06256\ kPa[/tex]
Modified Reynolds Analogy is often used in the Heat Exchanger industry. a) True b) False
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.
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 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]
As the car gets older, will its engine's compression ratio change, if yes how.
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.
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]
Define the terms (a) thermal conductivity, (b) heat capacity and (c) thermal diffusivity
Explanation:
(a)
The measure of material's ability to conduct thermal energy (heat) is known as thermal conductivity. For examples, metals have high thermal conductivity, it means that they are very efficient at conducting heat. The SI unit of heat capacity is W/m.K.
The expression for thermal conductivity is:
[tex]q=-\kappa \bigtriangledown T[/tex]
Where,
q is the heat flux
[tex]\kappa[/tex] is the thermal conductivity
[tex] \bigtriangledown T[/tex] is the temperature gradient.
(b)
Heat capacity for a substance is defined as the ratio of the amount of energy required to change the temperature of the substance and the magnitude of temperature change. The SI unit of heat capacity is J/K.
The expression for Heat capacity is:
[tex]C=\frac{E}{\Delta T}[/tex]
Where,
C is the Heat capacity
E is the energy absorbed/released
[tex]\Delta T[/tex] is the change in temperature
(c)
Thermal diffusivity is defined as the thermal conductivity divided by specific heat capacity at constant pressure and its density. The Si unit of thermal diffusivity is m²/s.
The expression for thermal diffusivity is:
[tex]\alpha=\frac{\kappa}{C_p \times \rho}[/tex]
Where,
[tex]\alpha[/tex] is thermal diffusivity
[tex]\kappa[/tex] is the thermal conductivity
[tex]C_p[/tex] is specific heat capacity at constant pressure
[tex]\rho[/tex] is density
Nanocomposite coatings: a)-Are mostly composed of metals b)-They comprise at least two immiscible phases c)-They are used to improve appearance d)-They are used to improve finish
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.
What is the function of the following: 1- Oil rings 2- Flywheel 3- Timing gears
Answer:
Explanation:
The functions of the following are as follows:
1). Oil Rings:
Regulates oil within cylinder walls.Helps to keep cylinder walls lubricatedPrevent Heat transferReduce friction between piston and cylinder2). Fly Wheel:
It is used to store rotational energyIt resist any change in rotational speed due to its moment of inertiaIt helps to control the orientation of the mechanical system3). Timing gears:
It provides synchronization in the rotation of crankshaft and the camshaft so as to provide proper valve opening and closing time during each cylinder's stroke(intake and exhaust).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
What does STP and NTP stands for in temperature measurement?
STP stands for standard temperature pressure and NTP stands for normal temperature pressure
If 100 J of heat is added to a system so that the final temperature of the system is 400 K, what is the change in entropy of the system? a)- 0.25 J/K b)- 2.5 J/K c)- 1 J/K d)- 4 J/K
Answer:
0.25 J/K
Explanation:
Given data in given question
heat (Q) = 100 J
temperature (T) = 400 K
to find out
the change in entropy of the given system
Solution
we use the entropy change equation here i.e
ΔS = ΔQ / T ...................a
Now we put the value of heat (Q) and Temperature (T) in equation a
ΔS is the entropy change, Q is heat and T is the temperature,
so that
ΔS = 100/400 J/K
ΔS = 0.25 J/K
If a point is positioned 16 inches from the origin on the x-y plane, and it is 12 inches above the x axis, what is its position in terms of the unit vectors i and j?
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]
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.