Answer:
a. [tex]\eta _{th}[/tex] = 77.65%
b. bwr = 6.5%
c. 3538.986 kW
d. -163.169 kJ
Explanation:
a. The given property are;
P₂/P₁ = 10, P₂ = 10 * 100 kPa = 1000 kPa
p₄/p₁ = 10
P₂/P₁ = p₄/p₃ = √10
p₂ = 100·√10
[tex]T_{2s}[/tex] = T₁×(√10)^(0.4/1.4) = 300 × (√10)^(0.4/1.4) = 416.85 K
T₂ = T₁ + ([tex]T_{2s}[/tex] - T₁)/[tex]\eta _c[/tex] = 300 + (416.85 - 300)/0.8 = 446.0625 K
p₄ = 10×p₁ = 10×100 = 1000 kPa
p₄/p₃ = √10 =
p₃ = 100·√10
T₃ = 300 K
T₃/[tex]T_{4s}[/tex] = (P₂/P₁)^((k - 1)/k) = (√10)^(0.4/1.4)
[tex]T_{4s}[/tex] = T₃/((√10)^(0.4/1.4) ) = 300/((√10)^(0.4/1.4)) = 215.905 K
T₄ = T₃ + ([tex]T_{4s}[/tex] - T₃)/[tex]\eta _c[/tex] = 300 + (215.905- 300)/0.8 = 194.881 K
The efficiency = 1 - (T₄ - T₁)/(T₃ - T₂) = 1 - (194.881 -300)/(300 -446.0625 ) = 0.28
T₄ = 446.0625 K
T₆ = 1400 K
[tex]T_{7s}[/tex]/T₆ = (1/√10)^(0.4/1.4)
[tex]T_{7s}[/tex] = 1400×(1/√10)^(0.4/1.4) = 1007.6 K
T₇ = T₆ - [tex]\eta _t[/tex](T₆ - [tex]T_{7s}[/tex]) = 1400 - 0.8*(1400 - 1007.6) = 1086.08 K
T₈ = 1400 K
T₉ = 1086.08 K
T₅ = T₄ + [tex]\epsilon _{regen}[/tex](T₉ - T₄) = 446.0625 +0.8*(1086.08 - 446.0625) = 958.0765 K
[tex]\eta _{th}[/tex] =(((T₆ - T₇) + (T₈ - T₉)) -((T₂ - T₁) + (T₄ - T₃)))/((T₆ - T₅) + (T₈ - T₇))
(((1400 - 1086.08) + (1400 -1086.08 ))-((446.0625 - 300)+(194.881 - 300)))/((1400 -958.0765 ) + (1400 -1086.08 )) = 0.7765
[tex]\eta _{th}[/tex] = 77.65%
b. Back work ratio, bwr = [tex]bwr = \dfrac{w_{c,in}}{w_{t,out}}[/tex]
((446.0625 - 300)+(194.881 - 300))/((1400 - 1086.08) + (1400 -1086.08 ))
40.9435/627.84 = 6.5%
c. [tex]w_{net, out} = c_p[(T_6 -T_7) + (T_8 - T_9)] - [(T_2 - T_1) + (T_4 -T_3)][/tex]
Power developed is given by the relation;
[tex]\dot m \cdot w_{net, out}[/tex]
[tex]\dot m \cdot w_{net, out}[/tex]= 6*1.005*(((1400 - 1086.08) + (1400 -1086.08 ))-((446.0625 - 300)+(194.881 - 300))) = 3538.986 kW
d. Exergy destruction = 6*(1.005*(300-446.0625 ) - 300*1.005*(-0.3966766)
-163.169 kJ
Air enters the compressor of a cold air-standard Brayton cycle with regeneration and reheat at 100 kPa, 300 K, with a mass flow rate of 6 kg/s. The compressor pressure ratio is 10, and the inlet temperature for each turbine stage is 1400 K. The pressure ratios across each turbine stage are equal. The turbine stages and compressor each have isentropic efficiencies of 80% and the regenerator effectiveness is 80%. For k= 1.4.
Calculate:
a. the thermal efficiency of the cycle.
b. the back work ratio.
c. the net power developed, in kW.
Answer:
a. 47.48%
b. 35.58%
c. 2957.715 KW
Explanation:
[tex]T_2 =T_1 + \dfrac{T_{2s} - T_1}{\eta _c}[/tex]
T₁ = 300 K
[tex]\dfrac{T_{2s}}{T_1} = \left( \dfrac{P_{2}}{P_1} \right)^{\dfrac{k-1}{k} }[/tex]
[tex]T_{2s} = 300 \times (10) ^{\dfrac{0.4}{1.4} }[/tex]
[tex]T_{2s}[/tex] = 579.21 K
T₂ = 300+ (579.21 - 300)/0.8 = 649.01 K
T₃ = T₂ + [tex]\epsilon _{regen}[/tex](T₅ - T₂)
T₄ = 1400 K
Given that the pressure ratios across each turbine stage are equal, we have;
[tex]\dfrac{T_{5s}}{T_4} = \left( \dfrac{P_{5}}{P_4} \right)^{\dfrac{k-1}{k} }[/tex]
[tex]T_{5s}[/tex] = 1400×[tex]\left( 1/\sqrt{10} \right)^{\dfrac{0.4}{1.4} }[/tex] = 1007.6 K
T₅ = T₄ + ([tex]T_{5s}[/tex] - T₄)/[tex]\eta _t[/tex] = 1400 + (1007.6- 1400)/0.8 = 909.5 K
T₃ = T₂ + [tex]\epsilon _{regen}[/tex](T₅ - T₂)
T₃ = 649.01 + 0.8*(909.5 - 649.01 ) = 857.402 K
T₆ = 1400 K
[tex]\dfrac{T_{7s}}{T_6} = \left( \dfrac{P_{7}}{P_6} \right)^{\dfrac{k-1}{k} }[/tex]
[tex]T_{7s}[/tex] = 1400×[tex]\left( 1/\sqrt{10} \right)^{\dfrac{0.4}{1.4} }[/tex] = 1007.6 K
T₇ = T₆ + ([tex]T_{7s}[/tex] - T₆)/[tex]\eta _t[/tex] = 1400 + (1007.6 - 1400)/0.8 = 909.5 K
a. [tex]W_{net \ out}[/tex] = cp(T₆ -T₇) = 1.005 * (1400 - 909.5) = 492.9525 KJ/kg
Heat supplied is given by the relation
cp(T₄ - T₃) + cp(T₆ - T₅) = 1.005*((1400 - 857.402) + (1400 - 909.5)) = 1038.26349 kJ/kg
Thermal efficiency of the cycle = (Net work output)/(Heat supplied)
Thermal efficiency of the cycle = (492.9525 )/(1038.26349 ) =0.4748 = 47.48%
b. [tex]bwr = \dfrac{W_{c,in}}{W_{t,out}}[/tex]
bwr = (T₂ -T₁)/[(T₄ - T₅) +(T₆ -T₇)] = (649.01 - 300)/((1400 - 909.5) + (1400 - 909.5)) = 35.58%
c. Power = 6 kg *492.9525 KJ/kg = 2957.715 KW
Anytime scaffolds are assembled or __________, a competent person must oversee the operation.
a. Drawn
b. Disassembled
c. Thought
d. Made
When scaffolds are now being construct or deconstruct, a competent person must supervise the work and train everybody who'll be assisting, and the further discussion can be defined as follows:
The competent person is also responsible for proposing whether fall protection is required for each scaffold erected. In constructing a scaffold, there are specific criteria for the ground the scaffold is constructed. On the products and components used to build the scaffold, its height in relation to the foundation. It's platform's design, and whether or not high efficiency is needed to supervise the installation.Therefore, the final answer is "Option B".
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An example of a transient analysis involving the 1st law of thermodynamics and conservation of mass is the filling of a compressed air tank. Assume that an air tank is being filled using a compressor to a pressure of 5 atm, and that it is being fed with air at a temperature of 25°C and 1 atm pressure. The compression process is adiabatic. Will the temperature of the air in the tank when it is done being filled i.e. once the pressure in the tank reaches 5 atm), be greater than, equal to, or less that the temperature of the 25°C air feeding the compressor?
A. Greater than 25°C
B. Unable to determine
C. Same as 25°C
D. Less than 25°C
Answer:
The temperature will be greater than 25°C
Explanation:
In an adiabatic process, heat is not transferred to or from the boundary of the system. The gain or loss of internal heat energy is solely from the work done on the system, or work done by the system. The work done on the system by the environment adds heat to the system, and work done by the system on its environment takes away heat from the system.
mathematically
Change in the internal energy of a system ΔU = ΔQ + ΔW
in an adiabatic process, ΔQ = 0
therefore
ΔU = ΔW
where ΔQ is the change in heat into the system
ΔW is the work done by or done on the system
when work is done on the system, it is conventionally negative, and vice versa.
also W = pΔv
where p is the pressure, and
Δv = change in volume of the system.
In this case, work is done on the gas by compressing it from an initial volume to the new volume of the cylinder. The result is that the temperature of the gas will rise above the initial temperature of 25°C
Select True/False for each of the following statements regarding aluminum / aluminum alloys: (a) Aluminum alloys are generally not viable as lightweight structural materials in humid environments because they are highly susceptible to corrosion by water vapor. (b) Aluminum alloys are generally superior to pure aluminum, in terms of yield strength, because their microstructures often contain precipitate phases that strain the lattice, thereby hardening the alloy relative to pure aluminum. (c) Aluminum is not very workable at high temperatures in air, in terms of extrusion and rolling, because a non-protective oxide grows and consumes the metal, converting it to a hard and brittle ceramic. (d) Compared to most other metals, like steel, pure aluminum is very resistant to creep deformation. (e) The relatively low melting point of aluminum is often considered a significant limitation for high-temperature structural applications.
Explanation:
(a) Aluminum alloys are generally not viable as lightweight structural materials in humid environments because they are highly susceptible to corrosion by water vapor.
False, aluminium is not susceptible to any corrosion by the presence of water vapor.
(b) Aluminum alloys are generally superior to pure aluminum, in terms of yield strength, because their micro structures often contain precipitate phases that strain the lattice, thereby hardening the alloy relative to pure aluminum.
True.
(c) Aluminum is not very workable at high temperatures in air, in terms of extrusion and rolling, because a non-protective oxide grows and consumes the metal, converting it to a hard and brittle ceramic.
False, aluminium is stable at high temperatures and does not oxidizes.
(d) Compared to most other metals, like steel, pure aluminum is very resistant to creep deformation.
False,pure aluminium is not resistant to the creep deformation.
(e) The relatively low melting point of aluminum is often considered a significant limitation for high-temperature structural applications.
False.
In this exercise, we have to analyze the statements that deal with aluminum and its properties, thus classifying it as true or false:
A) False
B) True
C) False
D) False
E) True
Analyzing the statements we can classify them as:
(a) For this statement we can say that it is False, aluminium is not susceptible to any corrosion by the presence of water vapor.
(b) For this statement we can say that it is True.
(c) For this statement we can say that it is False, aluminium is stable at high temperatures and does not oxidizes.
(d) For this statement we can say that it is False, pure aluminium is not resistant to the creep deformation.
(e) For this statement we can say that it is True.
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A furnace wall composed of 200 mm, of fire brick. 120 mm common brick 50mm 80% magnesia and 3mm of steel plate on the outside. If the inside surface temperature is 1450 °C and outer surface temperature is 90°C, estimate the temperature between layers and calculate the heat loss in KJ/h-m2. Assume k for fire brick 4 KJ/m-h°C, k for common brick= 2.8 KJ/m-h°C, k for 85% magnesia = 0.25 KJ/m-h°C and k for steel 240 KJ/m-h°C, k
Answer:
fire brick / common brick : 1218 °Ccommon brick / magnesia : 1019 °Cmagnesia / steel : 90.06 °Cheat loss: 4644 kJ/m^2/hExplanation:
The thermal resistance (R) of a layer of thickness d given in °C·m²·h/kJ is ...
R = d/k
so the thermal resistances of the layers of furnace wall are ...
R₁ = 0.200/4 = 0.05 °C·m²·h/kJ
R₂ = 0.120 2.8 = 3/70 °C·m²·h/kJ
R₃ = 0.05/0.25 = 0.2 °C·m²·h/kJ
R₄ = 0.003/240 = 1.25×10⁻⁵ °C·m²·h/kJ
So, the total thermal resistance is ...
R₁ +R₂ +R₃ +R₄ = R ≈ 0.29286 °C·m²·h/kJ
__
The rate of heat loss is ΔT/R = (1450 -90)/0.29286 = 4643.70 kJ/(m²·h)
__
The temperature drops across the various layers will be found by multiplying this heat rate by the thermal resistance for the layer:
fire brick: (4543.79 kJ/(m²·h))(0.05 °C·m²·h/kJ) = 232 °C
so, the fire brick interface temperature at the common brick is ...
1450 -232 = 1218 °C
For the next layers, the interface temperatures are ...
common brick to magnesia = 1218 °C - (3/70)(4643.7) = 1019 °C
magnesia to steel = 1019 °C -0.2(4643.7) = 90.06 °C
_____
Comment on temperatures
Most temperatures are rounded to the nearest degree. We wanted to show the small temperature drop across the steel plate, so we showed the inside boundary temperature to enough digits to give the idea of the magnitude of that.
Describe experimental factors that could be modified, and unalterable properties of materials used.
Answer:
a. mechanical properties
b. thermal properties
c. chemical properties
d. electical properties
e. magnetic properties
Explanation:
a. The mechanical properties of a material are those properties that involve a reaction to an applied load.The most common properties considered are strength, ductility, hardness, impact resistance, and fracture toughness, elasticity, malleability, youngs' modulus etc.
b. Thermal properties such as boiling point , coefficient of thermal expansion , critical temperature , flammability , heat of vaporization , melting point ,thermal conductivity , thermal expansion ,triple point , specific heat capacity
c. Chemical properties such as corrosion resistance , hygroscopy , pH , reactivity , specific internal surface area , surface energy , surface tension
d. electrical properties such as capacitance , dielectric constant , dielectric strength , electrical resistivity and conductivity , electric susceptibility , nernst coefficient (thermoelectric effect) , permittivity etc.
e. magnetic properties such as diamagnetism, hysteresis, magnetostriction , magnetocaloric coefficient , magnetoresistance , permeability , piezomagnetism , pyromagnetic coefficient
If the contact surface between the 20-kg block and the ground is smooth, determine the power of force F when t = 4 s. Initially, the block is at rest
Answer:
The power of force F is 115.2 W
Explanation:
Use following formula
Power = F x V
[tex]F_{H}[/tex] = F cos0
[tex]F_{H}[/tex] = (30) x 4/5
[tex]F_{H}[/tex] = 24N
Now Calculate V using following formula
V = [tex]V_{0}[/tex] + at
[tex]V_{0}[/tex] = 0
a = [tex]F_{H}[/tex] / m
a = 24N / 20 kg
a = 1.2m / [tex]S^{2}[/tex]
no place value in the formula of V
V = 0 + (1.2)(4)
V = 4.8 m/s
So,
Power = [tex]F_{H}[/tex] x V
Power = 24 x 4.8
Power = 115.2 W
Compute the volume percent of graphite, VGr, in a 3.2 wt% C cast iron, assuming that all the carbon exists as the graphite phase. Assume densities of 7.9 and 2.3 g/cm3 for ferrite and graphite, respectively.
Answer:
The volume percentage of graphite is 10.197 per cent.
Explanation:
The volume percent of graphite is the ratio of the volume occupied by the graphite phase to the volume occupied by the graphite and ferrite phases. The weight percent in the cast iron is 3.2 wt% (graphite) and 96.8 wt% (ferrite). The volume percentage of graphite is:
[tex]\%V_{gr} = \frac{V_{gr}}{V_{gr}+V_{fe}} \times 100\,\%[/tex]
Where:
[tex]V_{gr}[/tex] - Volume occupied by the graphite phase, measured in cubic centimeters.
[tex]V_{fe}[/tex] - Volume occupied by the graphite phase, measured in cubic centimeters.
The expression is expanded by using the definition of density and subsequently simplified:
[tex]\%V_{gr} = \frac{\frac{m_{gr}}{\rho_{gr}} }{\frac{m_{gr}}{\rho_{gr}}+\frac{m_{fe}}{\rho_{fe}}}\times 100\,\%[/tex]
Where:
[tex]m_{fe}[/tex], [tex]m_{gr}[/tex] - Masses of the ferrite and graphite phases, measured in grams.
[tex]\rho_{fe}, \rho_{gr}[/tex] - Densities of the ferrite and graphite phases, measured in grams per cubic centimeter.
[tex]\%V_{gr} = \frac{1}{1+\frac{\frac{m_{fe}}{\rho_{fe}} }{\frac{m_{gr}}{\rho_{gr}} } }\times 100\,\%[/tex]
[tex]\%V_{gr} = \frac{1}{1 + \left(\frac{\rho_{gr}}{\rho_{fe}} \right)\cdot\left(\frac{m_{fe}}{m_{gr}} \right)} \times 100\,\%[/tex]
If [tex]\rho_{gr} = 2.3\,\frac{g}{cm^{3}}[/tex], [tex]\rho_{fe} = 7.9\,\frac{g}{cm^{3}}[/tex], [tex]m_{gr} = 3.2\,g[/tex] and [tex]m_{fe} = 96.8\,g[/tex], the volume percentage of graphite is:
[tex]\%V_{gr} = \frac{1}{1+\left(\frac{2.3\,\frac{g}{cm^{3}} }{7.9\,\frac{g}{cm^{3}} } \right)\cdot \left(\frac{96.8\,g}{3.2\,g} \right)} \times 100\,\%[/tex]
[tex]\%V_{gr} = 10.197\,\%V[/tex]
The volume percentage of graphite is 10.197 per cent.
Following are the solution to the given points:
[tex]\to C_{Gr} = 100\\\\ \to C_{\alpha}= 0[/tex]From [tex]Fe-F_{\frac{e}{3}} c[/tex] diagram.
[tex]\to W_{\alpha} =\frac{C_{Gr}-C_{o}}{C_{Gr}-C_{\alpha}}[/tex]
[tex]= \frac{100-3.6}{100-0} \\\\= \frac{100-3.6}{100} \\\\= \frac{96.4}{100} \\\\=0.964[/tex]
Calculating the weight fraction of graphite:
[tex]\to W_{Gr}=\frac{C_0 - c_d}{C_{Gr} -c_d}[/tex]
[tex]= \frac{3.6-0}{100-0} \\\\ = \frac{3.6}{100} \\\\= 0.036[/tex]
Calculating the volume percent of graphite:
[tex]\to V_{Gr}=\frac{\frac{W_{Gr}}{P_{Gr}}}{\frac{w_{\alpha}}{P_{\alpha}}+ \frac{W_{Gr}}{P_{Gr}}}[/tex]
[tex]=\frac{\frac{0.036}{2.3}}{\frac{0.964}{7.9}+\frac{0.036}{2.3}}\\\\=0.11368 \times 100\%\\\\=11.368\%[/tex]
Therefore, the final answer is "0.964, 0.036, and 11.368%"
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A closed, 5-m-tall tank is filled with water to a depth of 4 m. The top portion of the tank is filled with air which, as indicated by a pressure gage at the top of the tank, is at a pressure of 20 kPa. Determine the pressure that the water exerts on the bottom of the tank.
Answer:
The pressure that the water exerts on the bottom of the tank is 59.2 kPa
Explanation:
Given;
height of tank, h = 5m
height of water in the tank, [tex]h_w[/tex] = 4m
pressure at the top of the tank, [tex]P_{top}[/tex] = 20 kPa
The pressure exerted by water at the bottom of the tank is the sum of pressure on water surface and pressure due to water column.
[tex]P_{bottom} = \gamma h + P_{top}\\\\P_{bottom} = (9.8*10^3*4 \ \ + \ 20*10^3)Pa\\\\P_{bottom} = 59200 \ Pa\\\\P_{bottom} = 59.2 \ kPa[/tex]
Therefore, the pressure that the water exerts on the bottom of the tank is 59.2 kPa
For a bolted assembly with six bolts, the stiffness of each bolt is kb=Mlbf/in and the stiffness of the members is km=12Mlbf/in. An external load of 80 kips is applied to the entire joint. Assume the load is equally distributed to all the bolts. It has been determined to use 1/2 in- 13 UNC grade 8 bolts with rolled threads. Assume the bolts are preloaded to 75% of the proof load. Clearly state any assumptions.
(a) Determine the yielding factor of safety,
(b) Determine the overload factor of safety,
(c) Determine the factor of safety baserd on joint seperation.
Answer:
nP ≈ 4.9 nL = 1.50Explanation:
GIVEN DATA
external load applied (p) = 85 kips
bolt stiffness ( Kb ) = 3(10^6) Ibf / in
Member stiffness (Km) = 12(10^6) Ibf / in
Diameter of bolts ( d ) = 1/2 in - 13 UNC grade 8
Number of bolts = 6
assumptions
for unified screw threads UNC and UNF
tensile stress area ( A ) = 0.1419 in^2
SAE specifications for steel bolts for grade 8
we have
Minimum proff strength ( Sp) = 120 kpsi
Minimum tensile strength (St) = 150 Kpsi
Load Bolt (p) = external load / number of bolts = 85 / 6 = 14.17 kips
Given the following values
Fi = 75%* Sp*At = (0.75*120*0.1419 ) = 12.771 kip
Preload stress
αi = 0.75Sp = 0.75 * 120 = 90 kpsi
stiffness constant
C = [tex]\frac{Kb}{Kb + Km}[/tex] = [tex]\frac{3}{3+2}[/tex] = 0.2
A) yielding factor of safety
nP = [tex]\frac{sPAt}{Cp + Fi}[/tex] = [tex]\frac{120* 0.1419}{0.2*14.17 + 12.771}[/tex]
nP = 77.028 / 15.605 = 4.94 ≈ 4.9
B) Determine the overload factor safety
[tex]nL = \frac{SpAt - Fi}{CP}[/tex] = ( 120 * 0.1419) - 12.771 / 0.2 * 14.17
= 17.028 - 12.771 / 2.834
= 1.50
Identify the correct statements in the context of friction factors of laminar and turbulent flows
a) In turbulent flow, the tubes with rough surfaces have much higher friction factors than the tubes with smooth surfaces
b) In turbulent flow, the tubes with rough surfaces have much lower friction factors than the tubes with smooth surfaces.
c) In laminar flow, the friction factor is dependent on the surface roughness
d) In laminar flow, the friction factor is independent of the surface roughness.
Answer:
a) In turbulent flow, the tubes with rough surfaces have much higher friction factors than the tubes with smooth surfaces.
Explanation:
Turbulent flow is a type of fluid flow in which fluid will undergo irregular fluctuations. The tubes with rough surfaces have higher friction factors than the tubes with smooth surfaces. In laminar flow the effect of effect of surface roughness is negligible on friction factors.
For a fluid flowing through a pipe assuming that pressure drop per unit length of pipe (P/L) depends on the diameter of the pipe , the velocity of fluid, the density of fluid and the viscosity of the fluid. Show that = ∅ ൬ ൰
Answer:
Explanation:
La vaca
El pato
As steam is slowly injected into a turbine, the angular acceleration of the rotor is observed to increase linearly with the time t. Know that the rotor starts from rest at t = 0 and that after 10 s the rotor has completed 20 revolutions.Choose the correct equations of motion for the rotor. (You must provide an answer before moving on to the next part.)
a) a = 2kt, w = 3krº, and 8 = 4kr
b) a = {kt, w = ke?, and 0 = }ke?
c) a = kr?, w = jke', and 0 = krº
d) a = kt, w = jke?, and 0 kr
Answer:
α = kt
ω = [tex]\frac{kt^2}{2}[/tex]
θ = [tex]\frac{kt^3}{6}[/tex]
Explanation:
given data
Initial velocity ω = 0
time t = 10 s
Number of revolutions = 20
solution
we will take here first α = kt .....................1
and as α = [tex]\frac{d\omega}{dt}[/tex]
so that
[tex]\frac{d\omega}{dt}[/tex] = kt ..................2
now we will integrate it then we will get
∫dω = [tex]\int_{0}^{t} kt\ dt[/tex] .......................3
so
ω = [tex]\frac{kt^2}{2}[/tex]
and
ω = [tex]\frac{d\theta}{dt}[/tex] ..............4
so that
[tex]\frac{d\theta}{dt}[/tex] = [tex]\frac{kt^2}{2}[/tex]
now we will integrate it then we will get
∫dθ = [tex]\int_{0}^{t}\frac{kt^2}{2} \ dt[/tex] ...............5
solve it and we get
θ = [tex]\frac{kt^3}{6}[/tex]
A long corridor has a single light bulb and two doors with light switch at each door. design logic circuit for the light; assume that the light is off when both switches are in the same position.
Answer and Explanation:
Let A denote its switch first after that we will assume B which denotes the next switch and then we will assume C stand for both the bulb. we assume 0 mean turn off while 1 mean turn on, too. The light is off, as both switches are in the same place. This may be illustrated with the below table of truth:
A B C (output)
0 0 0
0 1 1
1 0 1
1 1 0
The logic circuit is shown below
C = A'B + AB'
If the switches are in multiple places the bulb outcome will be on on the other hand if another switches are all in the same place, the result of the bulb will be off. This gate is XOR. The gate is shown in the diagram adjoining below.
A four-cylinder, four-stroke internal combustion engine has a bore of 3.7 in. and a stroke of 3.4 in. The clearance volume is 16% of the cylinder volume at bottom dead center and the crankshaft rotates at 2400 RPM. The processes within each cylinder are modeled as an air-standard Otto cycle with a pressure of 14.5 lbf/in. 2 and a temperature of 60 8 F at the beginning of compression. The maximum temperature in the cycle is 5200 8 R.
Based on this model,
1- Write possible Assumptions no less than three assumptions
2- Draw clear schematic for this problem
3- Determine possible Assumptions no less than three assumptions
4- Draw clear schematic for this problem.
5- calculate the net work per cycle, in Btu, and the power developed by the engine, in horsepower.
Answer:
1) The three possible assumptions are
a) All processes are reversible internally
b) Air, which is the working fluid circulates continuously in a closed loop
cycle
c) The process of combustion is depicted as a heat addition process
2) The diagrams are attached
5) The net work per cycle is 845.88 kJ/kg
The power developed in horsepower ≈ 45374 hP
Explanation:
1) The three possible assumptions are
a) All processes are reversible internally
b) Air, which is the working fluid circulates continuously in a closed loop
cycle
c) The process of combustion is depicted as a heat addition process
2) The diagrams are attached
5) The dimension of the cylinder bore diameter = 3.7 in. = 0.09398 m
Stroke length = 3.4 in. = 0.08636 m.
The volume of the cylinder v₁= 0.08636 ×(0.09398²)/4 = 5.99×10⁻⁴ m³
The clearance volume = 16% of cylinder volume = 0.16×5.99×10⁻⁴ m³
The clearance volume, v₂ = 9.59 × 10⁻⁵ m³
p₁ = 14.5 lbf/in.² = 99973.981 Pa
T₁ = 60 F = 288.706 K
[tex]\dfrac{T_{2}}{T_{1}} = \left (\dfrac{v_{1}}{v_{2}} \right )^{K-1}[/tex]
Otto cycle T-S diagram
T₂ = 288.706*[tex]6.25^{0.393}[/tex] = 592.984 K
The maximum temperature = T₃ = 5200 R = 2888.89 K
[tex]\dfrac{T_{3}}{T_{4}} = \left (\dfrac{v_{4}}{v_{3}} \right )^{K-1}[/tex]
T₄ = 2888.89 / [tex]6.25^{0.393}[/tex] = 1406.5 K
Work done, W = [tex]c_v[/tex]×(T₃ - T₂) - [tex]c_v[/tex]×(T₄ - T₁)
0.718×(2888.89 - 592.984) - 0.718×(1406.5 - 288.706) = 845.88 kJ/kg
The power developed in an Otto cycle = W×Cycle per second
= 845.88 × 2400 / 60 = 33,835.377 kW = 45373.99 ≈ 45374 hP.
Consider a double-pipe counter-flow heat exchanger. In order to enhance its heat transfer, the length of the heat exchanger is doubled. Will the effectiveness of the exchanger double?
Answer:
effectiveness of the heat exchanger will not be double when the length of the heat exchanger is doubled.
Because effectiveness depends on NTU and not necessarily the length of the heat exchanger
The effectiveness of a heat exchanger is defined as the ratio of the maximum possible heat transfer rate to the actual heat transfer rate.
a. True
b. False
Answer:
False
Explanation:
Because
The effectiveness (ϵ) of a heat exchanger is defined as the ratio of the actual heat transfer to the maximum possible heat transfer.
A ramp from an expressway with a design speed of 30 mi/h connects with a local road, forming a T intersection. An additional lane is provided on the local road to allow vehicles from the ramp to turn right onto the local road without stopping. The turning roadway has stabilized shoulders on both sides and will provide for a onelane, one-way operation with no provision for passing a stalled vehicle. Determine the width of the turning roadway if the design vehicle is a single-unit truck. Use 0.08 for superelevation.
Answer:
the width of the turning roadway = 15 ft
Explanation:
Given that:
A ramp from an expressway with a design speed(u) = 30 mi/h connects with a local road
Using 0.08 for superelevation(e)
The minimum radius of the curve on the road can be determined by using the expression:
[tex]R = \dfrac{u^2}{15(e+f_s)}[/tex]
where;
R= radius
[tex]f_s[/tex] = coefficient of friction
From the tables of coefficient of friction for a design speed at 30 mi/h ;
[tex]f_s[/tex] = 0.20
So;
[tex]R = \dfrac{30^2}{15(0.08+0.20)}[/tex]
[tex]R = \dfrac{900}{15(0.28)}[/tex]
[tex]R = \dfrac{900}{4.2}[/tex]
R = 214.29 ft
R ≅ 215 ft
However; given that :
The turning roadway has stabilized shoulders on both sides and will provide for a onelane, one-way operation with no provision for passing a stalled vehicle.
From the tables of "Design widths of pavement for turning roads"
For a One-way operation with no provision for passing a stalled vehicle; this criteria falls under Case 1 operation
Similarly; we are told that the design vehicle is a single-unit truck; so therefore , it falls under traffic condition B.
As such in Case 1 operation that falls under traffic condition B in accordance with the Design widths of pavement for turning roads;
If the radius = 215 ft; the value for the width of the turning roadway for this conditions = 15ft
Hence; the width of the turning roadway = 15 ft
The value of an SMT capacitor is signified by a
Answer:
Working volttage
Explanation:
SMT electrolytic capacitors are marked with working voltage. The value of these capacitors is measured in micro farads. It is a surface mount capacitor which is used for high volume manufacturers. They are small lead less and are widely used. They are placed on modern circuit boards.
A shaft consisting of a steel tube of 50-mm outer diameter is to transmit 100 kW of power while rotating at a frequency of 34 Hz. Determine the tube thickness that should be used if the shearing stress is not to exceed 60 MPa.
Answer:
25 - [tex]\sqrt[4]{26.66*10^{-8} }[/tex] mm
Explanation:
Given data
steel tube : outer diameter = 50-mm
power transmitted = 100 KW
frequency(f) = 34 Hz
shearing stress ≤ 60 MPa
Determine tube thickness
firstly we calculate the ; power, angular velocity and torque of the tube
power = T(torque) * w (angular velocity)
angular velocity ( w ) = 2[tex]\pi[/tex]f = 2 * [tex]\pi[/tex] * 34 = 213.71
Torque (T) = power / angular velocity = 100000 / 213.71 = 467.92 N.m/s
next we calculate the inner diameter using the relation
[tex]\frac{J}{c_{2} } = \frac{T}{t_{max} }[/tex] = 467.92 / (60 * 10^6) = 7.8 * 10^-6 m^3
also
c2 = (50/2) = 25 mm
[tex]\frac{J}{c_{2} }[/tex] = [tex]\frac{\pi }{2c_{2} } ( c^{4} _{2} - c^{4} _{1} )[/tex] = [tex]\frac{\pi }{0.050} [ ( 0.025^{4} - c^{4} _{1} ) ][/tex]
therefore; 0.025^4 - [tex]c^{4} _{1}[/tex] = 0.050 / [tex]\pi[/tex] (7.8 *10^-6)
[tex]c^{4} _{1}[/tex] = 39.06 * 10 ^-8 - ( 1.59*10^-2 * 7.8*10^-6)
39.06 * 10^-8 - 12.402 * 10^-8 =26.66 *10^-8
[tex]c_{1} = \sqrt[4]{26.66 * 10^{-8} }[/tex] =
THE TUBE THICKNESS
[tex]c_{2} - c_{1}[/tex] = 25 - [tex]\sqrt[4]{26.66*10^{-8} }[/tex] mm
Commutation is the process of converting the ac voltages and currents in the rotor of a dc machine to dc voltages and currents at its terminals. True False
Answer:
false
Explanation:
the changing of a prisoner sentence or another penalty to another less severe
A steam turbine receives 8 kg/s of steam at 9 MPa, 650 C and 60 m/s (pressure, temperature and velocity). It discharges liquid-vapor mixture with a quality of 0.94 at a pressure of 325 kPa and a velocity of 15 m/s. In addition, there is heat transfer from the turbine to the surroundings for 560 kW. Find the power produced by the turbine and express it in kW?
Answer:
The power produced by the turbine is 23309.1856 kW
Explanation:
h₁ = 3755.39
s₁ = 7.0955
s₂ = sf + x₂sfg =
Interpolating fot the pressure at 3.25 bar gives;
570.935 +(3.25 - 3.2)/(3.3 - 3.2)*(575.500 - 570.935) = 573.2175
2156.92 +(3.25 - 3.2)/(3.3 - 3.2)*(2153.77- 2156.92) = 2155.345
h₂ = 573.2175 + 0.94*2155.345 = 2599.2418 kJ/kg
Power output of the turbine formula =
[tex]Q - \dot{W } = \dot{m}\left [ \left (h_{2}-h_{1} \right )+\dfrac{v_{2}^{2}- v_{1}^{2}}{2} + g(z_{2}-z_{1})\right ][/tex]
Which gives;
[tex]560 - \dot{W } = 8\left [ \left (2599.2418-3755.39 \right )+\dfrac{15^{2}- 60^{2}}{2} \right ][/tex]
= -8*((2599.2418 - 3755.39)+(15^2 - 60^2)/2 ) = -22749.1856
[tex]- \dot{W }[/tex] = -22749.1856 - 560 = -23309.1856 kJ
[tex]\dot{W }[/tex] = 23309.1856 kJ
Power produced by the turbine = Work done per second = 23309.1856 kW.
10 kg/s Propane at 10 bar and 20 C is directed to an adiabatic rigid mixer and is mixed with 20 kg/s Propane at 10 bar and 40 C. What is the final volumetric flow rate in (m3/s) of the resulting mixture.
Answer:
The final volumetric flow rate will be "76.4 m³/s".
Explanation:
The given values are:
[tex]\dot{m_{1}}=10 \ kg/s[/tex]
[tex]\dot{m_{2}}=20 \ Kg/s[/tex]
[tex]T_{1}=293 \ K[/tex]
[tex]T_{2}=313 \ K[/tex]
[tex]P_{1}=P_{2}=P_{3}=10 \ bar[/tex]
As we know,
⇒ [tex]E_{in}=E_{out}[/tex]
[tex]\dot{m_{1}}h_{1}+\dot{m_{2}}h_{2}=\dot{m_{3}}h_{3}[/tex]
[tex]e_{1}\dot{v_{1}}h_{1}+e_{2}\dot{v_{2}}h_{2}=e_{3}\dot{v_{3}}h_{3}[/tex]
[tex]\frac{P_{1}}{RP_{1}}\dot{v_{1}} \ C_{p}T_{1}+ \frac{P_{2}}{RP_{2}}\dot{v_{2}} \ C_{p}T_{1}=\frac{P_{3}}{RP_{3}}\dot{v_{3}} \ C_{p}T_{3}[/tex]
⇒ [tex]\dot{v_{3}}=\dot{v_{1}}+\dot{v_{2}}[/tex]
[tex]=\frac{\dot{m_{1}}}{e_{1}}+\frac{\dot{m_{2}}}{e_{2}}[/tex]
On substituting the values, we get
[tex]=\frac{10}{10\times 10^5}\times 8314\times 293+\frac{20\times 8314\times 313}{10\times 10^5}[/tex]
[tex]=76.4 \ m^3/s[/tex]
One kg of an idea gas is contained in one side of a well-insulated vessel at 800 kPa. The other side of the vessel is under vacuum. The two sides are separated by a piston that is initially held in place by the pins. The pins are removed and the gas suddenly expands until it hits the stops. What happens to the internal energy of the gas?
a. internal energy goes up
b. internal energy goes down
c. internal energy stays the same
d. we need to know the volumes to make the calculation
Answer:
Option C = internal energy stays the same.
Explanation:
The internal energy will remain the same or unchanged because this question has to do with a concept in physics or classical chemistry (in thermodynamics) known as Free expansion.
So, the internal energy will be equals to the multiplication of the change in temperature, the heat capacity (keeping volume constant) and the number of moles. And in free expansion the internal energy is ZERO/UNCHANGED.
Where, the internal energy, ∆U = 0 =quantity of heat, q - work,w.
The amount of heat,q = Work,w.
In the concept of free expansion the only thing that changes is the volume.
Identify the advantages of using 6 tube passes instead of just 2 of the same diameter in shell-and-tube heat exchanger.What are the advantages and disadvantages of using 6 tube passes instead of just 2 of the same diameter?
Answer:
Please check explanation for answer
Explanation:
Here, we are concerned with stating the advantages and disadvantages of using a 6 tube passes instead of a 2 tube passes of the same diameter:
Advantages
* By using a 6 tube passes diameter, we are increasing the surface area of the heat transfer surface
* As a result of increasing the heat transfer surface area, the rate of heat transfer automatically increases too
Thus, from the above, we can conclude that the heat transfer rate of a 6 tube passes is higher than that of a 2 tube passes of the same diameter.
Disadvantages
* They are larger in size and in weight when compared to a 2 tube passes of the same diameter and therefore does not find use in applications where space conservation is quite necessary.
* They are more expensive than the 2 tube passes of the same diameter and thus are primarily undesirable in terms of manufacturing costs
. The job of applications engineer for which Maria was applying requires (a) excellent technical skills with respect to mechanical engineering, (b) a commitment to working in the area of pollution control, (c) the ability to deal well and confidently with customers who have engineering problems, (d) a willingness to travel worldwide, and (e) a very intelligent and well-balanced personality. List 10 questions you would ask when interviewing applicants for the job.
Answer:
Tell us about your self Are your confident that you are the right candidate for this positionwhy should i hire youDo you like working under supervisionHow do you like to work ( in a group or individually )What is your ultimate workplace goalwhat are your future plansWhat do you expect from the Organization when given the jobDo you like taking on critical problemsHow long can you work in this positionExplanation:
For a job of applications engineer which require excellent technical skills, commitment to working , ability to deal well and confidently with customers a willingness to travel and very intelligent and well-balanced personality.
The ten questions you should ask Maria to determine if she is qualified for the job are :
Tell us about your self ( functions you have )Are your confident that you are the right candidate for this positionwhy should i hire youDo you like working under supervisionHow do you like to work ( in a group or individually )What is your ultimate workplace goalwhat are your future plansWhat do you expect from the Organization when given the jobDo you like taking on critical problemsHow long can you work in this position