Answer:
Power consume by compressor=113,726.87 KW
Explanation:
Given:[tex]P_{1}=100KPa ,V_{1}=200 m/s,T_{1}=283 K, A_{1} =2m^2[/tex]
[tex]P_{2}=2000KPa ,T_{2}=513 K,A_{2}=0.5m^2[/tex]
Actually compressor is an open system, so here we will use first law of thermodynamics for open system .
We know that first law of thermodynamics for steady flow
[tex]h_{1}+\frac{V_{1} ^{2} }{2}+Q=h_{2}+\frac{V_{2} ^{2} }{2}+W[/tex]
We know that[tex]C_{p}=1.005\frac{Kj}{KgK}[/tex]and we take the air as ideal gas.
System is in steady state then mass flow rate in =mass flow rate out
Mass flow rate= [tex]density\times area\times velocity[/tex]
So mass flow rate =[tex]\rho _{1}V_{1}A_{1}[/tex] ,[tex]\rho =\frac{P}{RT}[/tex]
=1.23×200×2 Kg/s
=541.17 Kg/s
[tex]\rho _{1}V_{1}A_{1}=\rho _{2}V_{2}A_{2}[/tex]
[tex]\rho _{2}=13.58\frac{Kg}{{m}^3}[/tex] ,[tex]\rho =\frac{P}{RT}[/tex]
[tex]V_{2}[/tex]=80.07 m/s
Enthalpy of ideal gas h=[tex]C_{p}\times T[/tex]
So[tex] h_{1}=1.005\times283=284.41\frac{Kj}{Kg}[/tex]
[tex]h_{2}=1.005\times513=515.56\frac{Kj}{Kg}[/tex]
Now by putting the values
[tex]284.41+\frac{220 ^{2} }{2000}+Q=515.56+\frac{80.07 ^{2} }{2000}+W[/tex]
Here Q=0 because heat transfer is zero here.
W= -210.15 KJ/kg
So power consume by compressor=541.17×210.15
=113,726.87 KW
Air initially at 15 psla and 60 F is compressed to 75 psia and 400 F. The power input to air under steady state condition is 5 hp and heat loss of 4 Btu/lbm occurs during the process. If the change in Potential energy and kinetic energles are neglected, what will be the mass flowrate in lbm/min.?
Answer:[tex]\dot{m}=3.46lbm/min[/tex]
Explanation:
Initial conditions
[tex]P_1=15 psia[/tex]
[tex]T_1=60 F^{\circ}[/tex]
Final conditions
[tex]P_2=75 psia[/tex]
[tex]T_2=400F^{\circ}[/tex]
Steady flow energy equation
[tex]\dot{m}\left [ h_1+\frac{v_1^2}{2}+gz_1\right ]+\dot{Q}=\dot{m}\left [ h_2+[tex]\frac{v_2^2}{2}+gz_2\right ]+\dot{W}[/tex]
[tex]\dot{m}\left [ c_pT_1+\frac{0^2}{2}+g0\right ]+\dot{Q}=\dot{m}\left [ c_pT_2+\frac{0^2}{2}+g0\right ]+\dot{W}[/tex]
[tex]\dot{m}c_p\left [ T_1-T_2\right ]+\left [ -5hp\right ]=\dot{W} -5\times 746\times 3.4121[/tex]
[tex]-4\dot{m}-\dot{m}\times 0.24\times \left [ 400-60\right ][/tex]
[tex]-81.6\dot{m}-4\dot{m}=-4.949 BTU/sec[/tex]
[tex]\dot{m}=0.057821lbm/sec[/tex]
[tex]\dot{m}=3.46lbm/min[/tex]
What is “Hardenability” of steels? How it is measured? Why it is important in applications such as Axil rod of cars.
Answer and Explanation :
HARDENABILITY OF STEEL: Hardenability of steel is related to the its ability to form martensite when it is quenched. Hardenability is the measurement of capacity that how hard would be the steel when it is quenched.
The hadenability of steel can be measured as maximum diameter of rod which will have 50% martensite
its application in axial rods of car because it is very hard
Describe the grain structure of a metal ingot that was produced by slow-cooling the metal in a stationary open mold.
Answer:
Explained
Explanation:
In case case of slow cooling of a metal ingot, the micro structure is more like coarse. At the surface due to high heating rate ( surface will to exposed to higher temperature than the inner part and for longer time) the grains are small as grains will get lesser time to cool. Where as we go inside the grains will be gradually elongated. At the center we will find equiaxed grains.
The exhaust steam from a power station turbine is condensed in a condenser operating at 0.0738 bar(abs). The surface of the heat transfer surface is held at 20°C. What percentage change does the inclusion of the sensible heat correction term make to the estimated heat transfer condensing film coefficient?
Answer:
Percentage change 5.75 %.
Explanation:Given ;
Given
Pressure of condenser =0.0738 bar
Surface temperature=20°C
Now from steam table
Properties of steam at 0.0738 bar
Saturation temperature corresponding to saturation pressure =40°C
[tex]h_f= 167.5\frac{KJ}{Kg},h_g= 2573.5\frac{KJ}{Kg}[/tex]
So Δh=2573.5-167.5=2406 KJ/kg
Enthalpy of condensation=2406 KJ/kg
So total heat=Sensible heat of liquid+Enthalpy of condensation
[tex]Total\ heat\ =C_p\Delta T+\Delta h[/tex]
Total heat =4.2(40-20)+2406
Total heat=2,544 KJ/kg
Now film coefficient before inclusion of sensible heat
[tex]h_1=\dfrac{\Delta h}{\Delta T}[/tex]
[tex]h_1=\dfrac{2406}{20}[/tex]
[tex]h_1=120.3\frac{KJ}{kg-m^2K}[/tex]
Now film coefficient after inclusion of sensible heat
[tex]h_2=\dfrac{total\ heat}{\Delta T}[/tex]
[tex]h_2=\dfrac{2,544}{20}[/tex]
[tex]h_2=127.2\frac{KJ}{kg-m^2K}[/tex]
[tex]So\ Percentage\ change=\dfrac{h_2-h_1}{h_1}\times 100[/tex]
[tex]=\dfrac{127.2-120.3}{120.3}\times 100[/tex]
=5.75 %
So Percentage change 5.75 %.
A piston-cylinder assembly contains ammonia, initially at a temperature of-20°C and a quality of 70%. The ammonia is slowly heated to a final state where the pressure is 6 bar and the temperature is 180°C. While the ammonia is heated, its pressure varies linearly with specific volume. For the ammonia, determine the work and heat transfer, each in kJ/kg.
Answer:
w = -28.8 kJ/kg
q = 723.13 kJ/kg
Explanation:
Given :
Initial properties of piston cylinder assemblies
Temperature, [tex]T_{1}[/tex] = -20°C
Quality, x = 70%
= 0.7
Final properties of piston cylinder assemblies
Temperature, [tex]T_{2}[/tex] = 180°C
Pressure, [tex]P_{2}[/tex] = 6 bar
From saturated ammonia tables at [tex]T_{1}[/tex] = -20°C we get
[tex]P_{1}[/tex] = [tex]P_{sat}[/tex] = 1.9019 bar
[tex]v_{f}[/tex] = 0.001504 [tex]m^{3}[/tex] / kg
[tex]v_{g}[/tex] = 0.62334 [tex]m^{3}[/tex] / kg
[tex]u_{f}[/tex] = 88.76 kJ/kg
[tex]u_{g}[/tex] = 1299.5 kJ/kg
Therefore, for initial state 1 we can find
[tex]v_{1}[/tex] = [tex]v_{f}[/tex]+x ([tex]v_{g}[/tex]-[tex]v_{f}[/tex]
= 0.001504+0.7(0.62334-0.001504)
= 0.43678 [tex]m^{3}[/tex] / kg
[tex]u_{1}[/tex] = [tex]u_{f}[/tex]+x ([tex]u_{g}[/tex]-[tex]u_{f}[/tex]
= 88.76+0.7(1299.5-88.76)
=936.27 kJ/kg
Now, from super heated ammonia at 180°C, we get,
[tex]v_{2}[/tex] = 0.3639 [tex]m^{3}[/tex] / kg
[tex]u_{2}[/tex] = 1688.22 kJ/kg
Therefore, work done, W = area under the curve
[tex]w = \left (\frac{P_{1}+P_{2}}{2} \right )\left ( v_{2}-v_{1} \right )[/tex]
[tex]w = \left (\frac{1.9019+6\times 10^{5}}{2} \right )\left ( 0.3639-0.43678\right )[/tex]
[tex]w = -28794.52[/tex] J/kg
= -28.8 kJ/kg
Now for heat transfer
[tex]q = (u_{2}-u_{1})+w[/tex]
[tex]q = (1688.2-936.27)-28.8[/tex]
= 723.13 kJ/kg
A(n)______ is a device used to ensure positive position of a valve or damper actuator A. calibrator B. positioner C. actuator D. characteristic cam
Answer: C) actuator
Explanation:
Actuator is the device that used to provides the power and manipulate the motion of the moving parts of the valve and damper is used to control the flow of the fluid. Actuator is the device or the mechanism which are used to control valve automatically and valve is a device which is used to control and regulate the fluid by rotating the flow.
Which of the following is not a fuel? a)- RP-1 b)- Nitrogen Tetroxide c)- Liquid Hydrogen d)- Methane
Answer: B- Nitrogen Tetroxide
Explanation: Except for the nitrogen tetroxide , other given all options are fuel .Nitrogen Tetroxide is a chemical compound having brownish-red color which is in liquid form having a unpleasant smell, therefore it does not belong to the category of fuel because it cannot be used as a substance for production of heat or power .
0.50 kg of air is heated at constant pressure from 25°C to 100°C. The source of the heat is at 200°C. What is the entropy generation for the process?
Solution:
Given:
mass of air, m = 0.50 Kg
[tex]T_{1}[/tex] = 25°C = 273+25 = 298 K
[tex]T_{2}[/tex] = 100°C = 273+100 = 373 K
[tex]T_{o}[/tex] = 200°C = 273+100 = 473 K
Solution:
Formulae used:
ΔQ = mCΔT (1)
ΔS = [tex]\frac{\Delta Q}{T_{o}}[/tex] (2)
where,
ΔQ = change in heat transfer
ΔS = chane in entropy
C = specific heat
ΔT = change in system temperature
Using eqn (1)
ΔQ = [tex]0.50\times 1.005\times (373-298)[/tex] = 36.687 kJ
Now, for entropy generation, using eqn (2)
ΔS = [tex]\frac{37.687}{473}[/tex] = 0.0796 kJ
A piston-cylinder assembly has initially a volume of 0.3 m3 of air at 25 °C. Mass of the air is 1 kg. Weights are put on the piston until the air reaches to 0.1 m3 and 1,000 °C, in which the air undergoes a polytropic process (PV" const). Assume that heat loss from the cylinder, friction of piston, kinetic and potential effects are negligible. 1) Determine the polytropic constant n. 2) Determine the work transfer in ki for this process, and diseuss its direction. 3) sketch the process in T-V (temperature-volume) diagram.
Answer:
n=2.32
w= -213.9 KW
Explanation:
[tex]V_1=0.3m^3,T_1=298 K[/tex]
[tex]V_2=0.1m^3,T_1=1273 K[/tex]
Mass of air=1 kg
For polytropic process [tex]pv^n=C[/tex] ,n is the polytropic constant.
[tex]Tv^{n-1}=C[/tex]
[tex]T_1v^{n-1}_1=T_2v^{n-1}_2[/tex]
[tex]298\times .3^{n-1}_1=1273\times .1^{n-1}_2[/tex]
n=2.32
Work in polytropic process given as
w=[tex]\dfrac{P_1V_1-P_2V_2}{n-1}[/tex]
w=[tex]mR\dfrac{T_1-T_2}{n-1}[/tex]
Now by putting the values
w=[tex]1\times 0.287\dfrac{289-1273}{2.32-1}[/tex]
w= -213.9 KW
Negative sign indicates that work is given to the system or work is done on the system.
For T_V diagram
We can easily observe that when piston cylinder reach on new position then volume reduces and temperature increases,so we can say that this is compression process.
An air conditioner unit uses an electrical power input of 100W to drive the system and rejects 440W of heat to the kitchen air. Calculate the air conditioner's cooling rate and its coefficient of performance β.??
Answer:
Cooling Rate=340 W
Coefficient of Performance β=3.4
Explanation:
[tex]Desired\ effect= Cooling\ Rate=Q_L= 440-100=340\ W\\ W_{net,in}=Work\ in=100\ W[/tex]
[tex]Coefficient\ of\ performance (\beta) =\frac {Desired\ Out}{Required\ In}=\frac {Cooling\ {Effect}}{Work\ In}=\frac {Q_L}{W_{net,in}}[/tex]
[tex]Coefficient\ of\ performance (\beta) =\frac {440-100}{100}=3.4[/tex]
Name one aluminium alloy used in low pressure die casting and one in high pressure die casting? Explain the major reasons why one is different to the other?
Answer:
Explanation:
Low pressure die casting -
Also called the cold chamber die casting .
Example is -
A380 - having the composition , Al ( > 80% ) , Cu( 3 - 4% ) , Si ( 7.5 - 9.5% )
High Pressure die casting -
Also called hot chamber die casting .
Example is -
ZAMAK 2 - having composition , Al ( 3.5 - 4.3% ) , Cu ( 2.5 - 3.5% ) , Zn( > 90% )
Low pressure die casting -
This type of die casting is perfect for the metals with high melting point , for example aluminium . during this process , the metal is liquefied by very high temperature in the furnace and then loaded in to the cold chamber to be injected to the die.
High Pressure die casting -
The metal is melted in a container and then a piston injects the liquid metal under high pressure into the die . low melting point metals that don not chemically attack are ideal for this die casting , example Zinc.
The following yield criteria are dependent on hydrostatic stress (a) Maximum distortion energy and maximum normal stress (b) Tresca and Mohr-Coulomb (c) Tresca and von-Mises (d) Maximum normal stress and Mohr-Coulomb
Answer:
c). Tresca and von-Mises
Explanation:
Tresca yield criteria states that when maximum shear stress becomes greater than the yield strength, the materials starts to yield.
Von -Mises is also known as Distortion energy theory. This theory states that failure occurs when a body is acted upon to a bi axial stresses or tri axial stresses when at any point the strain energy of distortion by unit volume of the body equal to the specimen of the strain energy of distortion by unit volume when yielding starts in tension test.
Thus most successful and commonly used yield criteria are the Von-Mises criteria and Tresca criteria.
If 65 gallons of hydraulic oil weighs 350lb, what is the specific weight of the oil in lb/ft^3?
Answer:
55.655 lb/ft³
Explanation:
Given data in question
oil weight i.e. w = 350 lb
oil volume i.e. v = 65 gallons = 6.68403 ft³
To find out
the specific weight of the oil
Solution
We know the specific weight formula is weight / volume
we have given both value so we will put weight and volume value in
specific weight formula i.e.
specific weight = weight / volume
specific weight = 372 / 6.68403 = 55.6550
specific weight = 55.655 lb/ft³
A heat pump with refrigerant-134a as the working fluid is used to keep aspace at 25°C by absorbing heat from geothermal water that enters the evaporator at 60°C at a rate of 0.065 kg/s and leaves at 40°C. Refrigerant enters the evaporator at 12°C with a quality of 15 percent and leaves at the same pressure as saturated vapor.If the compressor consumes 1.6 kW of power, determine (a) the mass flow rate of the refrigerant, (b) the rate of heat supply, (c) the COP.
Answer:
(a) [tex]m_{R-134a}=0.0338kg/s[/tex]
(b) [tex]Q_H=7.03kW[/tex]
(c) [tex]COP=4.39[/tex]
Explanation:
Hello,
(a) In this part, we must know that the energy provided by the water equals the energy gained by the refrigerant-134a, thus:
[tex]m_{R-134a}(h_2-h1)=m_{H_2O}Cp_{H_2O}*\Delta T_{H_2O}[/tex]
Now, water's heat capacity is about 4.18kJ/kg°C and the enthalpies at both the first and second state for the refrigerant-134a are computed as shown below, considering the first state as a vapor-liquid mixture (VLM) at 12°C and the second state as a saturated vapor (ST) at the same conditions:
[tex]h_{1/12^0C/VLM}=68.18kJ/kg+0.15*189.09kJ/kg=96.54kJ/kg\\h_{2,SV}=257.27kJ/kg[/tex]
Next, solving the mass of water one obtains:
[tex]m_{R-134a}=\frac{m_{H_2O}Cp_{H_2O}*\Delta T_{H_2O}}{(h_2-h1)}=\frac{0.065kg/s*4.18kJ/kg^0C*(60^0C-40^0C)}{257.27kJ/kg-96.54kJ/kg} \\m_{R-134a}=0.0338kg/s[/tex]
(b) Now, the energy balance allows us to compute the heat supply:
[tex]Q_L+W_{in}=Q_H\\Q_H=0.0338kg/s*(257.27kJ/kg-96.54kJ/kg)+1.6kW\\Q_H=7.03kW[/tex]
(c) Finally, the COP (coefficient of performance) is computed via:
[tex]COP=\frac{Q_H}{W_{in}}=\frac{7.03kW}{1.6kW}\\COP=4.39[/tex]
Best regards.
How does a 2.5 MW wind turbine costing $ 4 million compare to a 5-kw wind turbine $3 /W? a) Same $/w b) Smaller $/w c) Larger $/w
A workpiece of 2000 mm length and 300 mm width was machined by a planning operation with the feed set at 0.3 mm/stroke. If the machine tool executes 10 double strokes/min, the planning time for a single pass will be?
Answer:
The planning time of the planner machine is 100 minute
Explanation:
Planning machine
A planning machine is a metal working machine that gives a flat surface to the work piece. Here the work-piece reciprocates and the feed is given to the tool. A planning machine is used for heavy duty work and are often large in size.
The machining time or the planning time of a planning machine is given by,
[tex]t_{m}[/tex] = [tex]\frac{L_{w}}{N_{s}\times f}[/tex]
where, [tex]L_{w}[/tex] is the total length of travel of job
= width of the job
= 300 mm
[tex]N_{s}[/tex] is number of strokes per min
= 10 double strokes per min
f is feed of the tool, mm per stroke
= 0.3 mm per stroke
Therefore, [tex]t_{m}[/tex] = [tex]\frac{L_{w}}{N_{s}\times f}[/tex]
= [tex]\frac{300}{10\times 0.3}[/tex]
= 100 min
Therefore the planning time is 100 minute.
In determining liquid propellant performance a combination of A. High chemical energy & high molecular weight B. High chemical energy & low molecular weight C. Low chemical energy & low molecular weight D. Great taste & less filling
Answer: B) High chemical energy and low molecular weight
Explanation: Liquid propellant is a a single chemical compound or mix of other chemicals as well. It is used in the rocket that uses them as a major part for the fuel. A liquid propellant is supposed to have high chemical energy and low molecular weight so that is can be ignited with ease. High chemical energy can release good amount of heat in a chemical reaction and thus is good igniting compound for the liquid propellant rocket.
_____The coefficients, i.e. a and b of van der Waals equation can be determined by (A) critical condition of the gas, (B) curve fit of p-v-t experimental data points, (C) statistical analysis
Answer:
(B) Curve fit of p-v-t experimental data points
Explanation:
The constants a and b have positive values and are characteristic of the individual gas. The van der Waals state equation approximates the ideal gas law PV = nRT as the value of these constants approaches zero. The constant a provides a correction for intermolecular forces. The constant b is a correction for finite molecular size and its value is the volume of one mole of atoms or molecules.
A cubic shaped box has a side length of 1.0 ft and a mass of 10 lbm is sliding on a frictionless horizontal surface towards a 30 upward incline. The horizontal velocity of the box is 20 ft/s. Determine how far up the incline the box will travel (report center of mass distance along the inclined surface, not vertical distance)
Explanation & answer:
Assuming a smooth transition so that there is no abrupt change in slopes to avoid frictional loss nor toppling, we can use energy considerations.
Initially, the cube has a kinetic energy of
KE = mv^2/2 = 10 lbm * 20^2 ft^2/s^2 / 2 = 2000 lbm-ft^2 / s^2
At the highest point when the block stops, the gain in potential energy is
PE = mgh = 10 lbm * 32.2 ft/s^2 * h ft = 322 lbm ft^2/s^2
By assumption, there was no loss in energies, we equate PE = KE
322h lbm ft^2/s^2 = 2000 lbm ft^2/s^2
=>
h = 2000 /322 = 6.211 (ft)
distance up incline = h / sin(30) = 12.4 ft
What do you understand by the term redundant work?
Answer:
Redundant work refers to the work done during the process of deformation due to friction. It happens during the wire drawing. Redundant work per unit volume increases when the radial position becomes higher. The redundant work factor is defined as increased strain of the deformation to the stress. It is basically related to the deformation area geometry.
For tool A, Taylor's tool life exponent (n) is 0.45 and constant (K) is 90. Similarly for tool B, n = 0.3 and K = 60. The cutting speed (in m/min) above which tool A will have a higher tool life than tool B is (a) 26.7 (b) 42.5 (c) 80.7 (d) 142.9
Answer:
26.667
Explanation:
Given Data
For Tool A
Life exponent [tex]{\ n_1}[/tex]=0.45
Constant [tex]{C_1}[/tex]=90
For tool B
Life exponent [tex]{n_2}[/tex]=0.3
Constant [tex]{C_2}[/tex]=60
and tool life equation is
[tex]VT^{n}=c[/tex]
[tex]VT_{A}^{0.45}=90[/tex]
[tex]T_{A}^{0.45}=\frac{90}{V}[/tex]
[tex]T_{A}=\frac{90}{V}^{\frac{1}{0.45}}[/tex]
[tex]For Tool B[/tex]
[tex]VT_{A}^{0.3}=60[/tex]
[tex]T_{B}^{0.3}=\frac{60}{V}[/tex]
[tex]T_{B}=\frac{60}{V}^{\frac{1}{0.3}}[/tex]
[tex]T_{A}>T_{B}[/tex]
[tex]\frac{90}{V}^{\frac{1}{0.45}}>\frac{60}{V}^{\frac{1}{0.3}}[/tex]
[tex]V>26.667[/tex]
The most advantage of fuel cells is that it can produce electrical energy directly (___)
Answer:The most advantage of fuel cells is that can produce electrical energy directly from chemical energy of hydrogen or other fuel.
Explanation: Fuel cell utilizes the chemical energy from the hydrogen or any other fuel and then converts it to the electrical energy. A fuel like hydrogen is supplied to the anode part and air is supplied to the cathode part . For hydrogen fuel cell there is a catalyst at anode side which divides hydrogen molecules in protons and electrons, which split and take go in different direction to cathode side. Thus the fuel cell works and generate the electrical energy
Which of the following is not an example of heat generation? a)- Exothermic chemical reaction in a solid b)- Endothermic Chemical Reactions in a solid c)- Nuclear reaction in nuclear fuel rods d)- Electric resistance heater
Answer:
b) Endothermic Chemical Reactions in a solid
Explanation:
Endothermic reactions consume energy, which will result in a cooler solid when the reaction finishes.
At winter design conditions, a house is projected to lose heat at a rate of 60,000 Btu/h. The internal heat gairn from people, lights, and appliances is estimated to be 6000 Btuh Ifthis house is to be heated by electric resistance heaters, determine the required rated power of these heaters in kW to maintain the house at constant temperature.
Answer:
15.8529 kW
Explanation:
Rate of heat loss = 60000 Btu/h
Internal heat gain = 6000 Btu/h
Rate of heat required to be supplied
[tex]P_{Sup}=\text{Rate of heat loss}-\text{Internal heat gain}\\\Rightarrow P_{Sup}=60000-6000\\\Rightarrow P_{Sup}=54000\ Btu/h[/tex]
Converting 54000 Btu/h to kW (kJ/s)
1 Btu = 1.05506 kJ
1 h = 3600 s
[tex]P_{Sup}=54000\times \frac{1.05506}{3600}\\\Rightarrow P_{Sup}=15.8529\ kW[/tex]
∴ Required rated power of these heaters is 15.8529 kW
Answer:
Q = 15.8 kW
Explanation:
Given data:
Heat loss rate is 60,000 Btu/h
Heat gain is 6000 Btu/h
Rate of heat required is computed as
Q = (60000 - 6000) Btu/h
Q = 54000 Btu/h
change Btu/h to Kilo Watts
[tex]Q = 54000 Btu/h (\frac{1W}{3.412142\ Btu/h})[/tex]
[tex]Q = 15825.8 W(\frac{1 kW}{1000 W})[/tex]
Q = 15.8 kW
A bronze statue weighing 4 tonnes with a base of area 0.8 m2 is placed on a granite museum floor. The yield strength of the bronze is 240 MPa. What is the true area of contact, between the base and the floor?
Answer:
true area of contact is 1.7 * [tex]10^{-4}[/tex] m²
Explanation:
Given data
mass (m) = 4000 tonnes
yield strength = 240 MPa i.e. = 240 * [tex]10^{6}[/tex]
base area = 0.8 m²
To find out
the true area of contact
Solution
we have given yield strength and weight
so with we can find contact area directly we know that
area is equal to weight / yield strength
so we will put weight and yield strength value in this formula
and weight = mass * 9.81 = 4 * 9.81 = 39.24 tonnes = 39240 N
area = weight / yield strength
area = 39240 / 240 * [tex]10^{6}[/tex]
true area of contact = 1.7 * [tex]10^{-4}[/tex] m²
Air with a mass flow rate of 2.3 kg/s enters a horizontal nozzle operating at steady state at 420 K, 350 kPa, and velocity of 3 m/s. At the exit, the temperature is 300 K and the velocity is 460 m/s. Using the ideal gas model for air with constant ep=1.011 k/kg. K, determine: (a) the area at the inlet, in m2 (b) the heat transfer to the nozzle from its surroundings, in kW.
Answer:
(a)[tex]A_1=0.26 m^2[/tex]
(b)Q= -35.69 KW
Explanation:
Given:
[tex]P_1=350 KPa,T_1=420 K,V_1=3 m/s,T_2=300 K,V_2=460 m/s[/tex]
We know that foe air [tex]C_p=1.011\frac{KJ}{kg-k}[/tex]
Mass flow rate for air =2.3 kg/s
(a)
By mass balancing [tex]\dot{m}=\dot{m_1}\dot{m_2}[/tex]
[tex]\dot{m}=\rho AV[/tex]
[tex]\rho_1A_1V_1=\rho_2A_2V_2[/tex]
[tex]\rho_1 =\dfrac {P_1}{RT_1},R=0.287\frac{KJ}{kg-K}[/tex]
[tex]\rho_1 =\dfrac {350}{0.287\times 420}[/tex]
[tex]\rho_1=2.9\frac{kg}{m^3}[/tex]
[tex]\dot{m}=\rho_1 A_1V_1[/tex]
[tex]2.3=2.9\times A_1\times 3[/tex]
[tex]A_1=0.26 m^2[/tex]
(b)
Now from first law for open(nozzle) system
[tex]h_1+\dfrac{V_1^2}{2}+Q=h_2+\dfrac{V_2^2}{2}[/tex]
Δh=[tex]C_p(T_2-T_1)\frac{KJ}{kg}[/tex]
[tex]1.011\times 420+\dfrac{3^2}{2000}+Q=1.011\times 300+\dfrac{460^2}{2000}[/tex]
Q=-15.52 KJ/s
⇒[tex]Q= -15.52\times 2.3[/tex] KW
Q= -35.69 KW
If heat will loss from the system then we will take negative and if heat will incoming to the system we will take as positive.
Answer:
A) A1 ==0.2829 m^2
B) [tex]\frac{dQ}{dt} = -105.5 kW[/tex]
Explanation:
A) we know from continuity equation
[tex]\frac{dm}{dt} = \frac{A_1 v_1}{V_1}[/tex]
solving for A1
[tex]A_1 = \frac{\frac{dm}{dt} V_1}{v_1}[/tex]
we know V = \frac{RT}{P} as per ideal gas equation, so we have
[tex]A_1 = = \frac{\frac{dm}{dt} \frac{RT_1}{P_1}}{v_1}[/tex]
[tex]= \frac{2.3 \frac{0.287 \times 450}{350}}{3}[/tex]
=0.2829 m^2
b) the energy balanced equation is
[tex]\frac{dQ}{dt} = \frac{dm}{dt} ( Cp(T_2 -T_1) + \frac{V_2^2 - V_1^2}{2})[/tex]
[tex]= 2.3 ( 1.011(300 - 450) + [\frac{460^2+3^2}{2}])[/tex]
[tex]\frac{dQ}{dt} = -105.5 kW[/tex]
Radioactive wastes are temporarily stored in a spherical container, the center of which is buried a distance of 10 m below the earth's air-soil surface. The outside diameter of the container is 2.0 m, and 500 W of heat are released as a result of radloactive decay. If the soll surface temperature is 25*C, what is the outslde surface temperature of the contalner?
Answer:
Outside temperature =88.03°C
Explanation:
Conductivity of air-soil from standard table
K=0.60 W/m-k
To find temperature we need to balance energy
Heat generation=Heat dissipation
Now find the value
We know that for sphere
[tex]q=\dfrac{2\pi DK}{1-\dfrac{D}{4H}}(T_1-T_2)[/tex]
Given that q=500 W
so
[tex]500=\dfrac{2\pi 2\times .6}{1-\dfrac{2}{4\times 10}}(T_1-25)[/tex]
By solving that equation we get
[tex]T_2[/tex]=88.03°C
So outside temperature =88.03°C
Amorphous material is characterized by by a) organized crystalline structure; b) high hardness and ductility c)the chaotic arrangement of atoms or high hardness; d) excellent magnetic, electrical properties, atomic chaotic layout, high hardness.
Answer:
C.The chaotic arrangement of atoms or high hardness
Explanation:
We know that atomic arrangement in Solids are of two types
1)Crystalline
2)Amorphous
Crystalline arrangement have periodic arrangement where as Amorphous arrangement have random arrangement.
Generally all metal have Crystalline arrangement and material like wood ,glass have random arrangement ,that is why wood and glass is called Amorphous.We know that wood act as a insulator for conductivity and glass is a brittle and hard material.
So from above we can say that Amorphous material have chaotic arrangement of atoms or have high harness,so our option c is right.
Define the difference between elastic and plastic deformation in terms of the effect on the crystal lattice structure.
Elastic deformation is a temporary, reversible change in a material's crystal lattice under stress, following Hooke's law, and is depicted as a linear response on a stress-strain graph. Plastic deformation results in permanent, irreversible changes in the crystal structure, typically involving dislocations and is characterized by the yield point and yield stress. Factors such as temperature and rate of stress application influence a rock's response to stress.
Deformation in materials can be categorized into two types: elastic deformation and plastic deformation. Elastic deformation refers to temporary changes in the crystal lattice that are reversible when the applied stress is removed. It follows Hooke's law, where the force is proportional to the displacement, and is shown as a linear region on a stress-strain graph. On the other hand, plastic deformation results in permanent changes to the lattice structure. Dislocations play a significant role in this process, where planes of atoms slip past one another. The moment when deformation transitions from elastic to plastic is known as the yield point, with associated yield stress. Beyond this point, deformation is irrevocable, and with continued stress, the material will eventually fracture.
At a microscopic level, plastic deformation involves the movement of dislocations and the introduction of an extra plane of atoms in the crystal structure, which allows atoms to move more easily under stress. This results in a permanently altered lattice configuration. Elastic deformation, by contrast, can be envisioned as if atoms were connected by springs that return to their original positions after the removal of stress.
The ability of rocks to deform elastically or plastically before breaking depends on several factors including temperature, water content in clay-bearing rocks, the rate at which stress is applied, and the inherent strength of the rock. A fundamental understanding of these principles is essential in geology and materials science.
Elastic deformation is reversible, maintaining lattice structure. Plastic deformation is irreversible, causing permanent lattice rearrangement.
Elastic and plastic deformation are two different responses of materials to applied stress, and they affect the crystal lattice structure differently:
1. Elastic Deformation :
- Elastic deformation occurs when a material is subjected to stress, but it returns to its original shape and size once the stress is removed.
- In elastic deformation, the atomic or molecular bonds within the crystal lattice are stretched or compressed, causing the material to temporarily change shape.
- Within the elastic limit, the crystal lattice structure remains intact, and the atoms or molecules maintain their relative positions.
- The deformation is reversible, meaning the material returns to its original state when the applied stress is released.
2. Plastic Deformation :
- Plastic deformation occurs when a material is subjected to stress beyond its elastic limit, causing permanent changes in shape or size even after the stress is removed.
- In plastic deformation, the atomic or molecular bonds within the crystal lattice undergo significant rearrangement or sliding.
- Plastic deformation leads to the permanent displacement of atoms or molecules within the lattice structure, resulting in the material maintaining a new shape or size.
- The material undergoes irreversible changes in its crystal lattice structure due to dislocation movement, grain boundary sliding, or other mechanisms.
- Plastic deformation is characteristic of materials undergoing permanent deformation, such as metals being shaped or formed through processes like forging, rolling, or extrusion.
In summary, the difference between elastic and plastic deformation lies in the extent of the changes to the crystal lattice structure and whether the deformation is reversible or permanent. Elastic deformation involves temporary changes within the elastic limit, whereas plastic deformation involves permanent changes beyond the elastic limit.
A diesel engine with CR= 20 has inlet at 520R, a maximum pressure of 920 psia and maximum temperature of 3200 R. With cold air properties find the cutoff ratio, the expansion ratio v4/v3, and the exhaust temperature.
Answer:
Cut-off ratio[tex]\dfrac{V_3}{V_2}=6.15[/tex]
Cxpansion ratio[tex]\dfrac{V_4}{V_3}=3.25[/tex]
The exhaust temperature[tex]T_4=1997.5R[/tex]
Explanation:
Compression ratio CR(r)=20
[tex]\dfrac{V_1}{V_2}=20[/tex]
[tex]P_2=P_3=920 psia[/tex]
[tex]T_1=520 R ,T_{max}=T_3,T_3=3200 R[/tex]
We know that for air γ=1.4
If we assume that in diesel engine all process is adiabatic then
[tex]\dfrac{T_2}{T_1}=r^{\gamma -1}[/tex]
[tex]\dfrac{T_2}{520}=20^{1.4 -1}[/tex]
[tex]T_2=1723.28R[/tex]
[tex]\dfrac{V_3}{V_2}=\dfrac{T_3}{T_2}[/tex]
[tex]\dfrac{V_3}{V_2}=\dfrac{3200}{520}[/tex]
So cut-off ratio[tex]\dfrac{V_3}{V_2}=6.15[/tex]
[tex]\dfrac{V_1}{V_2}=\dfrac{V_4}{V_3}\times\dfrac{V_3}{V_2}[/tex]
Now putting the values in above equation
[tex]\dfrac20=\dfrac{V_4}{V_3}\times 6.15[/tex]
[tex]\dfrac{V_4}{V_3}=3.25[/tex]
So expansion ratio[tex]\dfrac{V_4}{V_3}=3.25[/tex].
[tex]\dfrac{T_4}{T_3}=(expansion\ ratio)^{\gamma -1}[/tex]
[tex]\dfrac{T_3}{T_4}=(3.25)^{1.4 -1}[/tex]
[tex]T_4=1997.5R[/tex]
So the exhaust temperature[tex]T_4=1997.5R[/tex]