Answer:690.21 mm of Hg
Explanation:
Given data
depth of lake[tex]\left (h\right )[/tex]=237.3m
density of lake water[tex]\left ( \rho\right )[/tex]=[tex]989kg/m^3[/tex]
acceleration due to gravity[tex]\left ( g\right )[/tex]=[tex]9.806m/s^2[/tex]
air above water has a barometric pressure of 29.83in of Hg=711.2 mm of Hg
Altitude correct for pressure=1.87in of Hg=47.5 mm of Hg
Absolute pressure above water =711.2-47.5mm of Hg=663.7 mm of Hg
Absolute pressure at bottom of Lake=Pressure at surface level +[tex]\left ( \rho \times g\times h\right )[/tex]
Absolute pressure at bottom of Lake=663.7+[tex]\frac{989\times 9.81\273.3}{10^{5}}[/tex]
Absolute pressure at bottom of Lake=690.21 mm of Hg
What does STP and NTP stands for in temperature measurement?
STP stands for standard temperature pressure and NTP stands for normal temperature pressure
What is the output of a system having the transfer function G = 2/[(s + 3) x(s + 4)] and subject to a unit impulse?
Answer:
output=[tex]\frac{2}{(s+3)(s+4)}[/tex]
Explanation:
output =transfer function ×input
here transfer function G=[tex]\frac{2}{(S+3)(S+4)} {}[/tex]
input = unit impulse
in S domain unit impulse =1
so output =[tex]\frac{2}{(S+3)(S+4)} {}[/tex]
=[tex]\frac{2}{(s+3)(s+4)}[/tex]
The Poisson effect does not apply to shear strains. a)True b)- False
Answer:
true
Explanation:
Shear strains and direct strains are independent components in a strain tensor at a point. Also the poisson ratio is defined as
μ = - [tex]\frac{lateral strain}{longitudinal strain}[/tex]
Which is independent of shear strains Thus this affect does not apply on shear strains
The turbine blade tip speed v for r=1m and 5 rev/sec is a) 5 m/s b) 0.25 m/s c) 0.20 m/s
Answer:
(a) 5 m/sec
Explanation:
we have given r=1 meter
angular velocity ω =5 revolution/sec
we have to find the velocity of turbine blade tip
the velocity of turbine blade tip is given by v =rω
where v = velocity of turbine blade tip
r = radius
ω = angular velocity
so v =5×1= 5 meter/sec
so the option (a) will be the correct option as in option (a) velocity is given as 5 meter/sec
What is the Thermodynamic (Absolute) temperature scale?
Answer:
0 K
Explanation:
The thermodynamic absolute temperature is that temperature at which there is an infinite cooling and so there is no movement of any molecules or particles it is given by kelvin. 0 K is that temperature at which there is no movement of molecule so 0 K or -273°C in degree Celsius is the absolute temperature in thermodynamic temperature scale.
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 .
An alternating current E(t) =120 sin(12t) has been running through a simple circuit for a long time. The circuit has an inductance of L = 0.2 henrys, a resistor of R = 5 ohms and a capacitor of capcitance C = 0.043 farads. What is the amplitude of the current I?
Answer:
Explanation:
we have given E(t)=120 sin(12t)
R=5 ohm
L=0.2 H
ω=12 ( from expression of E)
[tex]X_L=0.2\times 12=2.4[/tex] ohm
[tex]X_C=\frac{1}{\omega \times C}=\frac{1}{12\times 0.043}=1.9379\ ohm[/tex]
[tex]Z=\sqrt{R^2+\left ( \omega L-\frac{1}{\omega C} \right )^2}[/tex]
[tex]Z=\sqrt{5^2+\left ( \2.4-1.9379 )^2}[/tex]
=5.021 ohm
so amplitude of current = [tex]\frac{v}{z}=\frac{120}{5.021}=23.89[/tex]
A mass of 7 kg undergoes a process during which there is heat transler frorn the mass at a rate of 2 kJ per kg, an elevation decrease of 40 m, and an increase in velocity from 13 m/s to 23 m/s. The specific internal energy decreases by 4 kJ/kg and the acceleration of gravity is constant at 9.7 m/s2. Determine the work for the process, in k.J
Answer:44.61 KJ
Explanation:
Let h_1,V_1,Z_1 be the initial specific enthalpy,velocity&elevation of the system
and h_2,V_2,Z_2 be the Final specific enthalpy,velocity&elevation of the system
mass(m)=7kg
Applying Steady Flow Energy Equation
[tex]m\left [ h_1+\frac{v_1^2}{2g}+gZ_1\right ][/tex]+Q=[tex]\left [ h_2+\frac{v_2^2}{2g}+gZ_2\right ][/tex]+W
[tex]h_1-h_2=4 KJ/kg[/tex]
[tex]V_1=13m/s[/tex]
[tex]V_2=23m/s[/tex]
[tex]Z_1-Z_2=40m[/tex]
substituting values
[tex]7\left [ h_1+\frac{13^{2}}{2g}+gZ_1\right ]+7\times2[/tex] = [tex]\left [ h_2+\frac{23^2}{2g}+gZ_2\right ][/tex]+W
W=[tex]7\left [h_1-h_2+\frac{V_1^2-V_2^2}{2000g}+g\frac{\left (Z_1-Z_2 \right )}{1000}\right ][/tex]+Q
W=[tex]7\left [4+\frac{13^2-23^2}{2000g}+g\frac{\left (40 \right )}{1000}[\right ][/tex]+[tex]2\times 7[/tex]
W=44.61KJ
The melting point of Pb (lead) is 327°C, is the processing at 20°C hot working or cold working?
Answer:
Explained
Explanation:
Cold working: It is plastic deformation of material at temperature below recrystallization temperature. whereas hot working is deforming material above the recrystallization temperature.
Given melting point temp of lead is 327° C and lead recrystallizes at about
0.3 to 0.5 times melting temperature which will be higher that 20°C. Hence we can conclude that at 20°C lead will under go cold working only.
Which of the following are not related to a materials structure? (Mark all that apply) a)- Atomic bonding b)- Crystal structure c)- Atomic number d)- Microstructure
The answer is c) atomic number
A spherical steel container 3 feet in diameter is buried in a land fill. The container is filled with a chemical that keeps the outer surface of the container at 100°F, whereas the earth's surface is at 50°F. Determine the heat transfer from the container if it is buried under 3 feet of earth.
Answer:
Q = 378.247 Bt/hr
Explanation:
given data:
diameter of container = 3 m
so r = 1.5 m
T1 = 50°C
T2 = 100°C
depth y = 3 ft
Heat transfer is given as Q
[tex]Q = SK\Delta T[/tex]
Where
S = Shape factor for the object
[tex]S = \frac{4\pi r}{1-\frac{r}{2y}}[/tex]
[tex]S = \frac{4\pi *1.5}{1-\frac{1.5}{2*3}}[/tex]
S = 25.132 ft
[tex]Q = SK\Delta T[/tex]
Q = 25.132*0.301 *(100-50)
Q = 378.247 Bt/hr
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]
A gear pump has a 4.25in outside diameter, a 3.25in inside diameter, and a 2in width. If the actual pump flow rate is 1800rpm and rated pressure is 29gpm, what is the volumetric efficiency?
Answer:
volumetric efficeincy = 0.315%
Explanation:
Given data:
outside diameter = 4.25 inch
inside diameter = 3.25 inch
flow rate V = 1800 rpm
actual flow rate Qa = 29gpm = 6699 inch3/min
volume of pump can be determine by using below formula
[tex]VOLUME = \frac{\pi}{4}*(D_{0}^{2}-D_{1}^{2})L[/tex]
[tex]=\frac{\pi}{4}*(4.25^{2}-3.25^{2})2[/tex]
[tex]VOLUME = 11.78 inc^{3}[/tex]
theoretical flow rate is given as Qt
[tex]Q_{T} = V.N[/tex]
= 11.78*1800
=[tex]21204 inch ^{3}/ min[/tex]
[tex]volumetric efficeincy = \frac{Q_{A}}{Q_{T}}[/tex]
[tex]=\frac{6699}{21204}[/tex]
volumetric efficeincy = 0.315%
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]
Air is compressed in an isentropic process from an initial pressure and temperature of P1 = 90 kPa and T1=22°C to a final pressure of P2=900 kPa. Determine: a)- The final temperature of the air. b)-The work done per kg of air during the process.
Answer:
a) [tex]T_2=569.35 K[/tex]
b)Work done per kg of air=196.84 KJ/Kg
Explanation:
Given: [tex]\gamma =1.4[/tex] for air.
[tex]P_1=90 KPa ,T_=22^\circ C,P_2=900 KPa[/tex]
We know that
[tex]\dfrac{T_2}{T_1}=\left (\frac{P_2}{P_1}\right )^{\dfrac{{\gamma-1}}{\gamma}}[/tex]
So [tex]\dfrac{T_2}{295}=\left (\frac{900}{90}\right )^{\dfrac{{1.4-1}}{1.4}}[/tex]
[tex]T_2=569.35 K[/tex]
(a) [tex]T_2=569.35 K[/tex]
(b)Work for adiabatic process
W=[tex]\frac{P_1V_1-P_2V_2}{\gamma -1}[/tex]
We know that PV=mRT for ideal gas.
W=[tex]mR\frac{T_1-T_2}{\gamma -1}[/tex]
Now by putting values
work per kg of air=[tex]0.287\times \frac{295-569.35}{1.4 -1}[/tex]
Work w=-196.84 KJ/Kg (Negative sign indicate work given to input.)
So work done per kg of air=196.84 KJ/Kg
What is the advantage to use a multistage compression refrigeration system over a single stage compression system?
Answer:
the advantages of using multistage compression refrigeration system over a single stage compression system are as follows:
Explanation:
It results in increased volumetric efficiency of compressor due to decrease in pressure ratio in each stage.Cost of operation is comparatively lowUniformity in torque is achieved thus reducing the size of the flywheel.Reduced size of condensor as a result of heat removal during condensationLower temperature at the end of compression resulting in effective lubrication and increased compressor life.The shear force diagram is always the slope of the bending moment diagram. a)True b)- False
Answer:
True
Explanation:
Shear force diagram is a diagram which is drawn by calculating the shear force either to the right of the section or to the left of the section .
shear force is also be define as the change of moment w.r.t to distance
V=[tex]\frac{\mathrm{d}M }{\mathrm{d} x}[/tex]
where V is shear force and M is bending moment.
from the equation we can clearly say that shear force diagram is slope of bending moment diagram.
What is the no-slip condition? What causes it?
The no-slip condition is the viscosity of a fluid.This is most likely brought on by stress or stressful situations.
Answer/Explanation:
The no-slip condition for viscous fluids assumes that at a solid boundary, the fluid will have zero velocity relative to the boundary. Along with a flow when adhesion is stronger than cohesion. ... Basically the molecules of the fluid crash into the molecules of the wall and get stopped.
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
An object of mass 521 kg, initially having a velocity of 90 m/s, decelerates to a final velocity of 14 m/s. What is the change in kinetic energy of the object, in kJ?
Answer:2058.992KJ
Explanation:
Given data
Mass of object[tex]\left ( m\right )[/tex]=521kg
initial velocity[tex]\left ( v_0\right )[/tex]=90m/s
Final velocity[tex]\left ( v\right )[/tex]=14m/s
kinetic energy of body is given by=[tex]\frac{1}{2}[/tex][tex]m[/tex][tex]v^{2}[/tex]
change in kinectic energy is given by substracting final kinetic energy from initial kinetic energy of body.
Change in kinetic energy=[tex]\frac{1}{2}\times m[/tex][tex]\left ( V_0^{2}-V^2\right )[/tex]
Change in kinetic energy=[tex]\frac{1}{2}\times521[/tex][tex]\left ( 90^{2}-14^2\right )[/tex]
Change in kinetic energy=2058.992KJ
The "view factor" Fij depends on surface emissivity and surface geometry. a) True b) False
Answer:
(B) FALSE
Explanation:
view factor [tex]F_{ij}[/tex] depends on the surface emissivity and the surface of geometry view factor is the term used in radiative heat transfer. View factor is depends upon the radiation which leave the surface and strike the surface.View factor is also called shape factor configuration factor it is denoted by [tex]F_{ij}[/tex]
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
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
Describe the design experiments method for product and process optimization
Answer:
Design of Experiment also known as DoE is a systematic methodology to understand how any process or parameters of any product affects the response variables like physical properties or performance of any product, etc. It serves the purpose of making the job easier.
It is technique to generate valuable information required with minimum experimentation with the use of these:
Statistical methodologyMathematical analysis to predict the output within the limits of experiments at any point.Experimental extremities
Process Optimization:
It includes the following:
Product prototype should be used in designingOptimization of significant value adding activitiesDetection and minimization of errorStrong built concept for the designTesting and validation of the process for further improvement in efficiencyTime, quality, environmental, manufacting and overall cost, safety and operational constraints should be duly noted.To measure the voltage drop across a resistor, a _______________ must be placed in _______________ with the resistor. Ammeter; Series Ammeter; Parallel Voltmeter; Series Voltmeter; Parallel
Answer:
The given blanks can be filled as given below
Voltmeter must be connected in parallel
Explanation:
A voltmeter is connected in parallel to measure the voltage drop across a resistor this is because in parallel connection, current is divided in each parallel branch and voltage remains same in parallel connections.
Therefore, in order to measure the same voltage across the voltmeter as that of the voltage drop across resistor, voltmeter must be connected in parallel.
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.
In a horizontal pipeline a 150 mm diameter pipe is connected to a 250 mm diameter pipe. The flow rate in the pipeline is 0.15 m^3/s. Take the connection as a sudden enlargement and determine: (a) The pressure head loss vhen the vater flovs from the large pipe to the smaller pipe (take Cc=0.64) (b) The pressure head loss when vater flovs from the small pipe to the larger pipe (c) The loss of power in both cases
Answer:
a) [tex]h_L=1.17m[/tex]
b)[tex]h_L=1.52m[/tex]
c)[tex]P_1=1.721 kN[/tex]
[tex]P_2=2.236 kN[/tex]
Explanation:
velocities of the pipe;
velocity of small dia pipe
[tex]v_{small}=\frac{Q}{A_{small}} =\frac{0.15}{\frac{\pi}{4}\times d^2 }= \frac{0.15}{\frac{\pi}{4}\times 0.15^2 }=8.52m/s[/tex]
velocity of larger dia pipe
[tex]v_{large}=\frac{Q}{A_{large}} =\frac{0.15}{\frac{\pi}{4}\times d^2 }= \frac{0.15}{\frac{\pi}{4}\times 0.25^2 }=3.05m/s[/tex]
a) pressure head loss when water is flowing from large to smaller pipe
[tex]h_L=(\frac{1}{C_c} -1)^2 \times \frac{v^2}{2g}[/tex]
[tex]h_L=(\frac{1}{0.64} -1)^2 \times \frac{8.52^2}{2\times 9.81} = 1.17m[/tex]
b) pressure head loss when water flow from small pipe to large pipe
[tex]h_L= \frac{v^2}{2g}(1-\frac{A_{small}}{A_{large}} )^2[/tex]
[tex]h_L=\frac{8.52^2}{2\times9.81}(1-\frac{\frac{\pi}{4}\times 0.15^2}{\frac{\pi}{4}\times 0.25^2}} )^2= 1.52 m[/tex]
c) power loss in both cases are
[tex]P_1= \rho g Q h_{L1}= 1000 \times 9.81\times 0.15\times 1.17= 1.721 kN[/tex]
[tex]P_2= \rho g Q h_{L2}= 1000 \times 9.81\times 0.15\times 1.52= 2.236 kN[/tex]
Describe the basic types of chips produced in metal-cutting operations.
Answer:
Chips are of three types --
1. Continuous chip
2. Discontinuous or segmental chip
3. Continuous chip with built up edge
Explanation:
Conventional machining process always removes some excess part of the metal in the form of Chips. Every machinist should be well aware of the type of chip formed as it gives the knowledge of the machining process. The chips forms give the knowledge of --
1. Dimension of tool
2. feed rate
3. cutting speed
4. nature of tool
5. Friction between tool and work piece
the different types of chips are :
1. Continuous chips :
Continuous chips are long ribbon like coil that are bonded together. The continuous chips undergoes plastic deformation continuously. This is the most desirable form of chip produced. When such chips are formed, the cutting is smooth with good surface finish. Mostly ductile material forms continuous chips.
2. Discontinuous chips :
Discontinuous chips are formed when metals are machined and the material gets deformed easily. Brittle materials forms discontinuous chips. Discontinuous chips are in the form of loose broken chips that are not continuous. Discontinuous chips are formed when depth of cut and feed is large and cutting sped is low.
3. Continuous chips with built up edge :
These chips are similar to the continuous chips where surface finish is not smooth. When ductile materials are machined at low cutting speed, a portion of work material tends tends to stick at the rake face of the tool due to the friction between the tools and the chip. This is known as built up edge.
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
List irreversibilities
Answer:
Some of the irreversibilities are listed below:
Plastic deformation of solidsTransfer of heat over finite difference of temperatureWhen two fluids are mixed together the process is irreversibleCombustion of a gasCurrent flowing through a finite resistor Diffusion and free compression or expansion of gasRelative motion of body with force of frictionProcesses involving chemical reactions(spontaneous)