Resistance zero meaning superconductor, so True.
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.
With increases in magnification, which of the following occur? a. The field of view decreases. b. The ambient illumination decreases. c. The larger parts can be measured. d. The eyepiece must be raised.
By increasing magnification you decrease the field of view.
The answer is A.
Hope this helps.
r3t40
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
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 .
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.
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 %.
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
The melting point of W (tungsten) is 3380°C, is the processing at 1100 C hot working or cold working?
Answer:
cold working
Explanation:
Given data in question
melting point tungsten (W) = 3380°C = 3653 K
processing temperature = 1100°C = 1373 K
To find out
process is hot or cold working
solution
we know hot working and cold working process is depend upon the Recrystallization and Recrystallization is a process in which deformed grains of crystal structures are replace with stress-free grains that nucleate and grow till actual grains have been consumed fully
and we know that ratio of processing temperature and melting point tungsten is greater than 60% than the process is start of Recrystallization so we check ratio
ratio = processing temperature / melting point tungsten
ratio = 1373 / 3653
ratio = 0.3758 = 37.58 %
we can see this is less than 60 % so our process is cold working
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
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]
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]
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
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 direct contact heat exchanger (where the fluid mixes completely) has three inlets and one outlet. The mass flow rates of the inlets are 1kg/s, 1.5kg/s and 2 kg/s. The enthalpy of those inlets are the 100kJ/kg, 120kJ/kg, and 500kJ/kg, respectively. What is the enthalpy at the outlet?
Answer:
Enthalpy at outlet=284.44 KJ
Explanation:
[tex]m_1=1 Kg/s,m_2=1.5 Kg/s,m_3=22 Kg/s[/tex]
[tex]h_1=100 KJ/Kg,h_2=120 KJ/Kg,h_3=500 KJ/Kg[/tex]
We need to Find enthalpy of outlet.
Lets take the outlet mass m and outlet enthalpy h.
So from mass conservation
[tex]m_1+m_2+m_3=m[/tex]
m=1+1.5+2 Kg/s
m=4.5 Kg/s
Now from energy conservation
[tex]m_1h_1+m_2h_2+m_3h_3=mh[/tex]
By putting the values
[tex]1\times 100+1.5\times 120+2\times 500=4.5\times h[/tex]
So h=284.44 KJ
The interactions between a closed system and its surroundings include energy transfer by heat, boundary work and flow work. a)True b) False
Answer:
b) False
Explanation:
In close system only energy transfer take place and mass transfer is zero.But on the other hand in open system energy as well mass transfer take place.
Energy transfer means work as well as heat transfer.But we know that in close system there is no any flow of mass so there will not be any flow work,only boundary work will associated with close system.But in open system flow work take place.
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.
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.
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
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 piston-cylinder device contains 1.329 kg of nitrogen gas at 120 kPa and 27 degree C. The gas is now compressed slowly in a polytropic process during which PV^1.49 = constant. The process ends when the volume is reduced by one-half. Determine the entropy change of nitrogen during this process.
Answer:-0.4199 J/k
Explanation:
Given data
mass of nitrogen(m)=1.329 Kg
Initial pressure[tex]\left ( P_1\right )[/tex]=120KPa
Initial temperature[tex]\left ( T_1\right )=27\degree \approx[/tex] 300k
Final volume is half of initial
R=particular gas constant
Therefore initial volume of gas is given by
PV=mRT
V=0.986\times 10^{-3}
Using [tex]PV^{1.49}[/tex]=constant
[tex]P_{1}V^{1.49}[/tex]=[tex]P_2\left (\frac{V}{2}\right )[/tex]
[tex]P_2[/tex]=337.066KPa
[tex]V_2[/tex]=[tex]0.493\times 10^{-3} m^{3}[/tex]
and entropy is given by
[tex]\Delta s[/tex]=[tex]C_v \ln \left (\frac{P_2}{P_1}\right )[/tex]+[tex]C_p \ln \left (\frac{V_2}{V_1}\right )[/tex]
Where, [tex]C_v[/tex]=[tex]\frac{R}{\gamma-1}[/tex]=0.6059
[tex]C_p[/tex]=[tex]\frac{\gamma R}{\gamma -1}[/tex]=0.9027
Substituting values we get
[tex]\Delta s[/tex]=[tex]0.6059\times\ln \left (\frac{337.066}{120}\right )[/tex]+[tex]0.9027 \ln \left (\frac{1}{2}\right )[/tex]
[tex]\Delta s[/tex]=-0.4199 J/k
In thermodynamicsedependent properties means?
Answer:
Explanation:
Thermodynamics properties are the properties which defined the state of any system.
some of the thermodynamics properties are pressure, temperature etc
thermodynamics are broadly divided into two type
1)intensive and
2)extensive properties
Dependent properties are the properties that are dependent on other properties. Extensive property are those which are dependent on the extent of system. Example volume. if size of the system increase or decrease then volume also have same effect according to the changes
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
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]
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]
A dielectric is an insulating material or a very poor conductor of electric current. (True , False )
True.
Dielectric is a material with a low electrical conductivity (σ << 1); that is, an insulator, which has the property of forming electric dipoles inside it under the action of an electric field. Thus, all dielectric materials are insulators but not all insulating materials are dielectric.
Some examples of this type of materials are glass, ceramics, rubber, mica, wax, paper, dry wood, porcelain, some fats for industrial and electronic use and bakelite. As for the gases, the air, nitrogen and sulfur hexafluoride are used as dielectrics.
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.
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.
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.
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
A container filled with a sample of an ideal gas at the pressure of 150 Kpa. The gas is compressed isothermally to one-third of its original volume. What is the new pressure of the gas a)-900 kpa b)- 300 kpa c)- 450 kpa d)- 600 kpa
Answer: c) 450 kPa
Explanation:
Boyle's Law: This law states that pressure is inversely proportional to the volume of the gas at constant temperature and number of moles.
[tex]P\propto \frac{1}{V}[/tex] (At constant temperature and number of moles)
[tex]P_1V_1=P_2V_2[/tex]
where,
[tex]P_1[/tex] = initial pressure of gas = 150 kPa
[tex]P_2[/tex] = final pressure of gas = ?
[tex]V_1[/tex] = initial volume of gas = v L
[tex]V_2[/tex] = final volume of gas = [tex]\frac{v}{3}L[/tex]
[tex]150\times v=P_2\times \frac{v}{3}[/tex]
[tex]P_2=450kPa[/tex]
Therefore, the new pressure of the gas will be 450 kPa.