Answer:
One inlet stream to the mixer flows at 100.0 kg/hr and is 35wt% species-A and 65wt% species-B.
Explanation:
The process consists of two units, a mixer and a separator. There are three species: A, B, and C.
The goal is to produce an outlet stream with a high concentration of species-A.
How to solveThere are two inlet streams to the mixer. One is 35wt% species-A and 65wt% species-B, and the other is pure species-C. The flow rate of the second stream is unknown.
There are two outlet streams from the separator. One is 85wt% species-A and 10wt% species-B, and the other is 25wt% species-B and 70wt% species-C.
The degree-of-freedom analysis shows that there are 3 equations and 4 unknowns. The order in which the systems of equations must be solved is:
Solve for the flow rate of the stream of pure species-C.
Solve for the compositions of the outlet streams from the separator.
Solve for the compositions of the inlet streams to the mixer.
The solution for all unknown flow rates and compositions is:
The flow rate of the stream of pure species-C is 50.0 kg/hr.
The composition of the outlet stream from the separator that is rich in species-A is 90wt% species-A, 5wt% species-B, and 5wt% species-C.
The composition of the outlet stream from the separator that is rich in species-B is 10wt% species-A, 5wt% species-B, and 85wt% species-C.
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A 3-phase, 50 Hz, 110 KV overhead line has conductors placed in a horizontal plane 3 m apart. Conductor diameter is 2.5 cm. If the line length is 220 km, determine the charging current per phase assuming complete transposition. (6 Marks)
Answer:
A 3-phase, 50 Hz, 110 KV overhead line has conductors
Explanation:
hope it will helps you
A thermal energy storage unit consists of a large rectangular channel, which is well insulated on its outer surface and encloses alternating layers of the storage material and the flow passage. Each layer of the storage material is an aluminum slab of width W=0.05 m, which is at an initial temperature of 25∘C25 ∘C. Consider conditions for which the storage unit is charged by passing a hot gas through the passages, with the gas temperature and the convection coefficient assumed to have constant values of T[infinity]=600∘CT [infinity]=600 ∘C and h=100W/m2⋅Kh=100W/m 2⋅K throughout the channel. How long will it take to achieve 75% of the maximum possible energy storage? What is the temperature of the aluminum at this time?
Answer:
the temperature of the aluminum at this time is 456.25° C
Explanation:
Given that:
width w of the aluminium slab = 0.05 m
the initial temperature [tex]T_1[/tex] = 25° C
[tex]T{\infty} =600^0C[/tex]
h = 100 W/m²
The properties of Aluminium at temperature of 600° C by considering the conditions for which the storage unit is charged; we have ;
density ρ = 2702 kg/m³
thermal conductivity k = 231 W/m.K
Specific heat c = 1033 J/Kg.K
Let's first find the Biot Number Bi which can be expressed by the equation:
[tex]Bi = \dfrac{hL_c}{k} \\ \\ Bi = \dfrac{h \dfrac{w}{2}}{k}[/tex]
[tex]Bi = \dfrac{hL_c}{k} \\ \\ Bi = \dfrac{100 \times \dfrac{0.05}{2}}{231}[/tex]
[tex]Bi = \dfrac{2.5}{231}[/tex]
Bi = 0.0108
The time constant value [tex]\tau_t[/tex] is :
[tex]\tau_t = \dfrac{pL_cc}{h} \\ \\ \tau_t = \dfrac{p \dfrac{w}{2}c}{h}[/tex]
[tex]\tau_t = \dfrac{2702* \dfrac{0.05}{2}*1033}{100}[/tex]
[tex]\tau_t = \dfrac{2702* 0.025*1033}{100}[/tex]
[tex]\tau_t = 697.79[/tex]
Considering Lumped capacitance analysis since value for Bi is less than 1
Then;
[tex]Q= (pVc)\theta_1 [1-e^{\dfrac {-t}{ \tau_1}}][/tex]
where;
[tex]Q = -\Delta E _{st}[/tex] which correlates with the change in the internal energy of the solid.
So;
[tex]Q= (pVc)\theta_1 [1-e^{\dfrac {-t}{ \tau_1}}]= -\Delta E _{st}[/tex]
The maximum value for the change in the internal energy of the solid is :
[tex](pVc)\theta_1 = -\Delta E _{st}max[/tex]
By equating the two previous equation together ; we have:
[tex]\dfrac{-\Delta E _{st}}{\Delta E _{st}{max}}= \dfrac{ (pVc)\theta_1 [1-e^{\dfrac {-t}{ \tau_1}}]} { (pVc)\theta_1}[/tex]
Similarly; we need to understand that the ratio of the energy storage to the maximum possible energy storage = 0.75
Thus;
[tex]0.75= [1-e^{\dfrac {-t}{ \tau_1}}]}[/tex]
So;
[tex]0.75= [1-e^{\dfrac {-t}{ 697.79}}]}[/tex]
[tex]1-0.75= [e^{\dfrac {-t}{ 697.79}}]}[/tex]
[tex]0.25 = e^{\dfrac {-t}{ 697.79}}[/tex]
[tex]In(0.25) = {\dfrac {-t}{ 697.79}}[/tex]
[tex]-1.386294361= \dfrac{-t}{697.79}[/tex]
t = 1.386294361 × 697.79
t = 967.34 s
Finally; the temperature of Aluminium is determined as follows;
[tex]\dfrac{T - T _{\infty}}{T_1-T_{\infty}}= e ^ {\dfrac{-t}{\tau_t}}[/tex]
[tex]\dfrac{T - 600}{25-600}= e ^ {\dfrac{-967.34}{697.79}[/tex]
[tex]\dfrac{T - 600}{25-600}= 0.25[/tex]
[tex]\dfrac{T - 600}{-575}= 0.25[/tex]
T - 600 = -575 × 0.25
T - 600 = -143.75
T = -143.75 + 600
T = 456.25° C
Hence; the temperature of the aluminum at this time is 456.25° C
Air enters a compressor operating at steady state at 176.4 lbf/in.^2, 260°F with a volumetric flow rate of 424 ft^3/min and exits at 15.4 lbf/in.^2, 80°F. Heat transfer occurs at a rate of 6800 Btu/h from the compressor to its surroundings. Assuming the ideal gas model for air and neglecting kinetic and potential energy effects, determine the power input, in hp
Answer:
[tex]W_s =[/tex] 283.181 hp
Explanation:
Given that:
Air enters a compressor operating at steady state at a pressure [tex]P_1[/tex] = 176.4 lbf/in.^2 and Temperature [tex]T_1[/tex] at 260°F
Volumetric flow rate V = 424 ft^3/min
Air exits at a pressure [tex]P_2[/tex] = 15.4 lbf/in.^2 and Temperature [tex]T_2[/tex] at 80°F.
Heat transfer occurs at a rate of 6800 Btu/h from the compressor to its surroundings; since heat is released to the surrounding; then:
[tex]Q_{cv}[/tex] = -6800 Btu/h = - 1.9924 kW
Using the steady state energy in the process;
[tex]h_2 - h_1 + g(z_2-z_1)+ \dfrac{1}{2}(v^2_2-v_1^2) = \dfrac{Q_{cv}}{m}- \dfrac{W_s}{m}[/tex]
where;
[tex]g(z_2-z_1) =0[/tex] and [tex]\dfrac{1}{2}(v^2_2-v_1^2) = 0[/tex]
Then; we have :
[tex]h_2 - h_1 = \dfrac{Q_{cv}}{m}- \dfrac{W_s}{m}[/tex]
[tex]h_2 - h_1 = \dfrac{Q_{cv} - W_s}{m}[/tex]
[tex]{m}(h_2 - h_1) ={Q_{cv} - W_s}[/tex]
[tex]W_s ={Q_{cv} + {m}(h_2 - h_1)[/tex] ----- (1)
Using the relation of Ideal gas equation;
P₁V₁ = mRT₁
Pressure [tex]P_1[/tex] = 176.4 lbf/in.^2 = ( 176.4 × 6894.76 ) N/m² = 1216235.664 N/m²
Volumetric flow rate V = 424 ft^3/min = (424 × 0.0004719) m³ /sec
= 0.2000856 m³ /sec
Temperature = 260°F = (260°F − 32) × 5/9 + 273.15 = 399.817 K
Gas constant R=287 J/kg K
Then;
1216235.664 N/m² × 0.2000856 m³ /sec = m × 287 J/kg K × 399.817 K
[tex]m = \dfrac { 1216235.664 N/m^2 \times 0.2000856 m^3 /sec } {287 J/kg K \times 399.817 K }[/tex]
m = 2.121 kg/sec
The change in enthalpy:
[tex]m(h_1-h_2) = m * C_p * \Delta T= m* C_p * ( T_1 -T_2)[/tex]
[tex]= 2.121* 1.005* ( 399.817 -299.817)[/tex]
= 213.1605 kW
From (1)
[tex]W_s ={Q_{cv} + {m}(h_2 - h_1)[/tex]
[tex]W_s =[/tex] - 1.9924 kW + 213.1605 kW
[tex]W_s =[/tex] 211.1681 kW
[tex]W_s =[/tex] 283.181 hp
The power input is [tex]W_s =[/tex] 283.181 hp
For the following peak or rms values of some important sine waves, calculate the corresponding other value:
(a) 117 V rms, a household-power voltage in North America
(b) 33.9 V peak, a somewhat common peak voltage in rectifier circuits
(c) 220 V rms, a household-power voltage in parts of Europe
(d) 220 kV rms, a high-voltage transmission-line voltage in North America
Answer:
A) V_peak ≈ 165 V
B) V_rms ≈ 24 V
C) V_peak ≈ 311 V
D) V_peak ≈ 311 KV
Explanation:
Formula for RMS value is given as;
V_rms = V_peak/√2
Formula for peak value is given as;
V_peak = V_rms x √2
A) At RMS value of 117 V, peak value would be;
V_peak = 117 x √2
V_peak = 165.46 V
V_peak ≈ 165 V
B) At peak value of 33.9 V, RMS value would be;
V_rms = 33.9/√2
V_rms = 23.97 V
V_rms ≈ 24 V
C) At RMS value of 220 V, peak value is;
V_peak = 220 × √2
V_peak = 311.13 V
V_peak ≈ 311 V
D) At RMS value of 220 KV, peak value is;
V_peak = 220 × √2
V_peak = 311.13 KV
V_peak ≈ 311 KV
Trade-offs can be necessary at any point in time during the life cycle of a project. It is quite possible, and probable, for the criteria for the trade-offs to change over the life cycle of the project. Please also identifies how the relative importance of constraints of time, cost, and performance can change over the life cycle of the project.
Answer:
According to the Principles of Project management, the three factors which dominate the lifecycle of any project are:
Time;Cost; andPerformance.The relationship between the three is usually governed by trade-offs.
Explanation:
In simple term, in executing a project, one must deal with the factors mentioned above.
It is always desirous for a project to be finished within a stipulated time. If the time required is reduced inconsiderably, it will most likely incur more cost and even impact performance.
On the other hand, if the project is cost-sensitive and is executed to a very minimalistic budget, performance will be impacted and it may take a protracted amount of time.
In addition to the above, if the principal decides to change the original design of the project, the performance expected is altered. This will attract additional time as well as cost.
It is possible for any of the above factors to be renegotiated and readjusted at any time during the project. It usually is a trade-off.. that is one for the other.
Cheers!
The half-wave rectifier below is operating at a frequency of 60 Hz, and the rms value of the transformer output voltage is 6.3 V. (a) What is the value of the dc output voltage VO if the diode voltage drop is 1 V? (b) What is the minimum value of C required to maintain the ripple voltage to less than 0.25 V if R = 0.5Ω?
Given Information:
Frequency = f = 60 Hz
Transformer output voltage = Vrms = 6.3 V
Diode voltage drop = Vd = 1 V
Ripper voltage = Vr = 0.25 V
Load resistance = R = 0.5 Ω
Required Information:
a) dc output voltage = V₀ = ?
b) Capacitane = C = ?
Answer:
a) dc output voltage = V₀ = 2.52 V
b) Capacitane = C = 0.336 F
Explanation:
a) The average or dc output voltage of a half-wave rectifier is given by
[tex]V_0 = V_p/\pi[/tex]
Where Vp is given by
[tex]V_p = (V_{rms} \times \sqrt{2}) - V_d \\\\V_p = (6.3 \times \sqrt{2}) - 1 \\\\V_p = 8.91 - 1 \\\\V_p = 7.91 \: V \\\\[/tex]
So, the dc output voltage is
[tex]V_0 = 7.91/\pi \\\\V_0 = 2.52 \: V[/tex]
b) The minimum value of C required to maintain the ripple voltage to less than 0.25 V is given by
[tex]$ C = \frac{I}{Vr \cdot f} $[/tex]
Where I is current, Vr is the ripple voltage and f is the frequency
[tex]$ I = \frac{V_0}{R} $[/tex]
[tex]$ I = \frac{2.52}{0.5} $[/tex]
[tex]I = 5.04 \: A[/tex]
[tex]$ C = \frac{5.04}{0.25 \cdot 60} $[/tex]
[tex]C = 0.336 \: F[/tex]
Therefore, 0.336 F is the minimum value of capacitance required to maintain the ripple voltage to less than 0.25 V
At steady state, a refrigerator whose coefficient of performance is 3 removes energy by heat transfer from a freezer compartment at 0 degrees C at the rate of 6000 kJ/hr and discharges energy by heat transfer to the surroundings, which are at 20 degrees C. a) Determine the power input to the refrigerator and compare with the power input required by a reversible refrigeration cycle operating between reservoirs at these two temperatures. b) If electricity costs 8 cents per kW-hr, determine the actual and minimum theoretical operating costs, each in $/day
Answer:
(A)0.122 kW (B) Actual cost = 1.056 $/day, Theoretical cost = 0.234 $/day
Explanation:
Solution
Given that:
The coefficient of performance is =3
Heat transfer = 6000kJ/hr
Temperature = 20°C
Cost of electricity = 8 cents per kW-hr
Now
The next step is to find the power input to the refrigerator and compare with the power input considered by a reversed refrigeration cycle operating between reservoirs at the two temperatures.
Thus
(A)The coefficient of performance is given below:
COP = Heat transfer from freezer/Power input
3 =6000/P
P =6000/3
P= 2000
P = 2000 kJ/hr = 2000/(60*60) kW
= 2000 (3600)kW
= 0.55 kW
Thus
The ideal coefficient of performance = T_low/(T_high - T_low)
= (0+273)/(20-0)
= 13.65
So,
P ideal = 6000/13.65 = 439.6 kJ/hr
= 439.6/(60*60) kW
= 0.122 kW
(B)For the actual cost we have the following:
Actual cost = 0.55 kW* 0.08 $/kW-hr = $ 0.044 per hour
= 0.044*24 $/day
= 1.056 $/day
For the theoretical cost we have the following:
Theoretical cost = 0.122 kW* 0.08 $/kW-hr = $ 0.00976 per hour
= 0.00976*24 $/day
= 0.234 $/day
Tengo un problema con steam y es que sale como que no he comprado un juego pero en realidad si lo he comprado pero por una pagina alterna pero no me lo detecta que hago?
Answer:
Cómo forzar a Steam a reconocer los juegos instalados.
1) Vuelva a instalar los juegos sin descargar.
2) Agregue la carpeta Steam Library manualmente.
3) Reconocer juegos de una nueva unidad
4) Utilizar .acf Cache para forzar el reconocimiento de juegos de Steam .acf Cache para forzar el reconocimiento de juegos de Steam.
Explanation:
1) Vuelva a instalar los juegos sin descargar.
- Inicia Steam y ve a Juegos.
- Seleccione y haga clic en instalar para el juego que Steam no pudo reconocer.
- Steam comenzará a descubrir los archivos existentes para el juego.
2) Agregue la carpeta Steam Library manualmente.
- Lanzar Steam.
- Haga clic en Steam y seleccione Configuración.
- Haz clic en la pestaña Descargas.
- Haga clic en las carpetas de la biblioteca de Steam.
- En la ventana emergente, haz clic en Agregar carpeta de biblioteca y selecciona la ubicación donde se guardan todos los datos de tu juego Steam.
- Haga clic en Seleccionar y cerrar la configuración de Steam.
- Salga de la aplicación Steam y reinicie Steam.
- Steam ahora debería reconocer los juegos instalados nuevamente y enumerarlos en la carpeta de juegos.
3) Reconocer juegos de una nueva unidad
- Inicie la aplicación Steam desde el escritorio.
- Haz clic en Steam y selecciona Configuración.
- Haz clic en la pestaña Descargas.
- Haga clic en la carpeta de la biblioteca de Steam en la sección Bibliotecas de contenido.
- Haga clic en Agregar carpeta de biblioteca y navegue a la ubicación donde se mueven sus juegos (nuevo directorio) que es D: / games / your_subdirectory.
- Haga clic en Seleccionar y cerrar para guardar la carpeta de la biblioteca.
- Salga de Steam y reinícielo.
Steam escaneará la carpeta Biblioteca recientemente seleccionada y mostrará todos los juegos instalados.
4) Utilizar .acf Cache para forzar el reconocimiento de juegos de Steam .acf Cache para forzar el reconocimiento de juegos de Steam.
- Asegúrese de haber reinstalado Steam o tener la instalación existente.
- Mueva los datos del juego a C: >> Archivos de programa (x86) >> Steam >> carpeta Steamapps.
- Lanzar Steam.
- En este punto, Steam puede mostrar algunos juegos que están instalados correctamente.
- Para los juegos que se muestran como no instalados, seleccione y haga clic en el botón Instalar.
- Steam comenzará a descubrir todos los archivos existentes.
- Sin embargo, si Steam no reconoce los archivos existentes, comenzará a descargar los archivos y el progreso leerá 0%.
- Pausa la actualización de los juegos y sal de Steam. Vaya a C: >> Archivos de programa (x86) >> Steam >> Steamapps y encuentre todos los archivos .acf actuales.
¡¡¡Espero que esto ayude!!!
You are given a body with no body forces and told that the stress state is given as: ⎡ ⎣ 3αx 5βx2 + αy γz3 5βx2 + αy βx2 0 γz3 0 5 ⎤ ⎦ psi, where (α, β, γ) are constants with the following values: α = 1 psi/in, β = 1 psi/in2, and γ = 1 psi/in3. Does this represent an equilibrium state of stress? Assume the body occupies the domain Ω = [0, 1] × [0, 1] × [0, 1] (in inches).
Answer:
This doesn't represent an equilibrium state of stress
Explanation:
∝ = 1 , β = 1 , y = 1
x = 0 , y = 0 , z = 0 ( body forces given as 0 )
Attached is the detailed solution is and also the conditions for equilibrium
for a stress state to be equilibrium all three conditions has to meet the equilibrum condition as explained in the attached solution
The rectangular frame is composed of four perimeter two-force members and two cables AC and BD which are incapable of supporting compression. Determine the forces in all members due to the load L in position (a) and then in position (b). Forces are positive if in tension, negative if in compression.
Answer:
Your question is lacking some information attached is the missing part and the solution
A) AB = AD = BD = 0, BC = LC
AC = [tex]\frac{5L}{3}T, CD = \frac{4L}{3} C[/tex]
B) AB = AD = BC = BD = 0
AC = [tex]\frac{5L}{3} T, CD = \frac{4L}{3} C[/tex]
Explanation:
A) Forces in all members due to the load L in position A
assuming that BD goes slack from an inspection of Joint B
AB = 0 and BC = LC from Joint D, AD = 0 and CD = 4L/3 C
B) steps to arrive to the answer is attached below
AB = AD = BC = BD = 0
AC = [tex]\frac{5L}{3} T, CD = \frac{4L}{3}C[/tex]
A 30-g bullet is fired with a horizontal velocity of 450 m/s and becomes embedded in block B which has a mass of 3 kg. After the impact, block B slides on a 30-kg carrier C until it impacts the end of the carrier.Knowing the impact between B and C is perfectly plastic determine (a) velocity of the bullet and B after the first impact, (b) the final velocity of the carrier
(Distance between C and B is 0.5 m)
Answer:
a.) 4.46 m/s
b.) 0.41 m/s
Explanation:
a) Given that the mass M of the bullet = 30g = 30/1000 = 0.03 kg
Velocity V = 450 m/s
From conservative of linear momentum,
Sum of momentum before impact = Sum of momentum after impact
0.03 × 450 = (0.03 + 3 ) × v₂
v₂ = 13.5/3.03 = 4.4554 m/s
Therefore the velocity of the bullet and B after the first impact = 4.46 m/s approximately
(b) To calculate the velocity of the carrier, you will consider the conservation of linear momentum again.
(m₁ + m₂)×v₂ = (m₁ + m₂ + m₃)×v₃
Where:
Mass of the carrier m₃ = 30 kg
Substitute all the parameters into the formula
3.03×4.4554 = (3.03 +30) × v₃
v₃ = 13.5 / 33.03 = 0.40872 m/s
Therefore the velocity of the carrier is 0.41 m/s approximately.
A ball bearing has been selected with the bore size specified in the catalog as 35.000 mm to 35.020 mm. Specify appropriate minimum and maximum shaft diameters to provide a locational interference fit.
Answer:
A ball bearing has been selected with the bore size specified in the catalog as 35.000 mm to 35.020 mm. Specify.... ... has been selected with the bore size specified in the catalog as 35.000 mm to 35.020 mm. Specify appropriate minimum and maximum shaft diameters to provide a locational interference fit.
Explanation:
Water vapor at 10bar, 360°C enters a turbine operatingat steady state with a volumetric flow rate of 0.8m3/s and expandsadiabatically to an exit state of 1 bar, 160°C. Kinetic and potentialenergy effects are negligible. Determine for the turbine (a) the powerdeveloped, in kW, (b) the rate of entropy production, in kW/K, and (c)the isentropic turbine efficiency
Answer:
A) W' = 178.568 KW
B) ΔS = 2.6367 KW/k
C) η = 0.3
Explanation:
We are given;
Temperature at state 1;T1 = 360 °C
Temperature at state 2;T2 = 160 °C
Pressure at state 1;P1 = 10 bar
Pressure at State 2;P2 = 1 bar
Volumetric flow rate;V' = 0.8 m³/s
A) From table A-6 attached and by interpolation at temperature of 360°C and Pressure of 10 bar, we have;
Specific volume;v1 = 0.287322 m³/kg
Mass flow rate of water vapour at turbine is defined by the formula;
m' = V'/v1
So; m' = 0.8/0.287322
m' = 2.784 kg/s
Now, From table A-6 attached and by interpolation at state 1 with temperature of 360°C and Pressure of 10 bar, we have;
Specific enthalpy;h1 = 3179.46 KJ/kg
Now, From table A-6 attached and by interpolation at state 2 with temperature of 160°C and Pressure of 1 bar, we have;
Specific enthalpy;h2 = 3115.32 KJ/kg
Now, since stray heat transfer is neglected at turbine, we have;
-W' = m'[(h2 - h1) + ((V2)² - (V1)²)/2 + g(z2 - z1)]
Potential and kinetic energy can be neglected and so we have;
-W' = m'(h2 - h1)
Plugging in relevant values, the work of the turbine is;
W' = -2.784(3115.32 - 3179.46)
W' = 178.568 KW
B) Still From table A-6 attached and by interpolation at state 1 with temperature of 360°C and Pressure of 10 bar, we have;
Specific entropy: s1 = 7.3357 KJ/Kg.k
Still from table A-6 attached and by interpolation at state 2 with temperature of 160°C and Pressure of 1 bar, we have;
Specific entropy; s2 = 8.2828 KJ/kg.k
The amount of entropy produced is defined by;
ΔS = m'(s2 - s1)
ΔS = 2.784(8.2828 - 7.3357)
ΔS = 2.6367 KW/k
C) Still from table A-6 attached and by interpolation at state 2 with s2 = s2s = 8.2828 KJ/kg.k and Pressure of 1 bar, we have;
h2s = 2966.14 KJ/Kg
Energy equation for turbine at ideal process is defined as;
Q' - W' = m'[(h2 - h1) + ((V2)² - (V1)²)/2 + g(z2 - z1)]
Again, Potential and kinetic energy can be neglected and so we have;
-W' = m'(h2s - h1)
W' = -2.784(2966.14 - 3179.46)
W' = 593.88 KW
the isentropic turbine efficiency is defined as;
η = W_actual/W_ideal
η = 178.568/593.88 = 0.3
A long conducting rod of rectangular cross section (20 mm 30 mm) and thermal conductivity k 20 W/m K experiences uniform heat generation at a rate q . 5 107 W/m3, while its surfaces are maintained at 300 K. Using a finite-difference method with a grid spacing of 5 mm, determine the temperature distribution in the rod.
Answer:
Explanation:
We are assuming that there is
a steady state two dimensional conduction
constant properties
uniform volumetric heat generation
From symmetry, we will be determining 6 unknown temperatures.
See attachment for calculation and and tabulation
With T(s) = 300 K, the set of equations were written directly into the IHT work space and solved for nodal temperatures.
The result is seen in the second attachment
A photograph of the NASA Apollo 16 Lunar Module (abbreviated by NASA as the LM is shown on the surface of the Moon. Such spacecraft made six Moon landings during 1,969 - 72. A simplified model for one of the four landing gear assemblies of the LM is shown. If the LM has 13,500 kg mass, and rests on the surface of the Moon where acceleration due to gravity is 1.82 m/s^2, determine the force supported by members AB, AC, and AD. Assume the weight of the LM is uniformly supported by all four landing gear assemblies, and neglect friction between the landing gear and the surface of the Moon. TAB =N TAC = TAD =N A ( 2.6, 2.6, -2.2 ) m B(1.5, 1.5, 0)m C(2,1,-1.2)m D(1,2,-1.2)m
Answer:
[tex]\mathbf{F_{AB} = 13785.06 N }[/tex]
[tex]\mathbf{F_{AC} = -5062.38 N }[/tex]
[tex]\mathbf{F_{AD} = -5062.38 N }[/tex]
Explanation:
From the given information:
Let calculate the position vector of AB, AC, and AD
To start with AB; in order to calculate the position vector of AB ; we have:
[tex]r_{AB}^{\to} = r _{OA}^{\to} - r_{OB}^{\to} \\ \\ r_{AB}^{\to} = (2.6 \ \hat i + 2.6 \ \hat j - 2.2 \ \hat k ) - ( 1.5 \ \hat i + \ 1. 5 \hat j ) \\ \\ r_{AB}^{\to} = ( 2.6 \ \hat i - 1.5 \ \hat i + 2.6 \ \hat j - 1.5 \ \hat j - 2.2 \ \hat k) \\ \\ r_{AB}^{\to} = (1.1 \ \hat i + 1.1 \ \hat j - 2.2 \ \hat k ) m[/tex]
To calculate the position vector of AC; we have:
[tex]r_{AC}^{\to} = r _{OA}^{\to} - r_{OC}^{\to} \\ \\ r_{AC}^{\to} = (2.6 \ \hat i + 2.6 \ \hat j - 2.2 \ \hat k ) - ( 2\ \hat i + \ \hat j - 1.2 \ \hat k) \\ \\ r_{AC}^{\to} = ( 2.6 \ \hat i - 2\ \hat i + 2.6 \ \hat j - \ \hat j - 2.2 \ \hat k + 1.2 \ \hat k) \\ \\ r_{AC}^{\to} = (0.6 \ \hat i + 1.6 \ \hat j - \ \hat k ) m[/tex]
To calculate the position vector of AD ; we have:
[tex]r_{AD}^{\to} = r _{OA}^{\to} - r_{OD}^{\to} \\ \\ r_{AC}^{\to} = (2.6 \ \hat i + 2.6 \ \hat j - 2.2 \ \hat k ) - ( \hat i + \ 2 \hat j - 1.2 \ \hat k) \\ \\ r_{AD}^{\to} = ( 2.6 \ \hat i - \hat i + 2.6 \ \hat j - 2 \ \hat j - 2.2 \ \hat k + 1.2 \ \hat k) \\ \\ r_{AD}^{\to} = (1.6 \ \hat i + 0.6 \ \hat j - \ \hat k ) m[/tex]
However; let's calculate the force in AB, AC and AD in their respective unit vector form;
To start with unit vector AB by using the following expression; we have:
[tex]F_{AB}^{\to} = F_{AB} \dfrac{ r _{AB}^{\to} }{|r_{AB}^{\to}} \\ \\ \\ F_{AB}^{\to} = F_{AB} \dfrac{(1.1 \ \hat i + 1.1 \ \hat j - 2.2 \ \hat k ) }{\sqrt{ (1.1)^2 + (1.1)^2 + (-2.2 )^2 }} \\ \\ \\ F_{AB}^{\to} = F_{AB} \dfrac{(1.1 \ \hat i + 1.1 \ \hat j - 2.2 \ \hat k ) }{ \sqrt{7.26}} \\ \\ \\ F_{AB}^{\to} = F_{AB} \dfrac{(1.1 \ \hat i + 1.1 \ \hat j - 2.2 \ \hat k ) }{ 2.6944} \\ \\ \\ F_{AB}^{\to} = F_{AB} (0.408 \ \hat i+ 0.408 \ \hat j - 0.8165 \ \hat k ) N\\[/tex]
The force AC in unit vector form is ;
[tex]F_{AC}^{\to} = F_{AC} \dfrac{ r _{AC}^{\to} }{|r_{AC}^{\to}} \\ \\ \\ F_{AC}^{\to} = F_{AC} \dfrac{(0.6 \ \hat i + 1.6 \ \hat j - \ \hat k ) }{\sqrt{ (0.6)^2 + (1.6)^2 + (-1 )^2 }} \\ \\ \\ F_{AC}^{\to} = F_{AC} \dfrac{(0.6 \ \hat i + 1.6 \ \hat j - \ \hat k ) }{ \sqrt{3.92}} \\ \\ \\ F_{AC}^{\to} = F_{AC} \dfrac{(0.6 \ \hat i + 1.6 \ \hat j - \ \hat k ) }{1.9798} \\ \\ \\ F_{AC}^{\to} = F_{AC} (0.303 \ \hat i+ 0.808 \ \hat j - 0.505 \ \hat k ) N\\[/tex]
The force AD in unit vector form is ;
[tex]F_{AD}^{\to} = F_{AD} \dfrac{ r _{AD}^{\to} }{|r_{AD}^{\to}|} \\ \\ \\ F_{AD}^{\to} = F_{AD} \dfrac{(1.6 \ \hat i + 0.6 \ \hat j - \ \hat k ) }{\sqrt{ (1.6)^2 + (0.6)^2 + (-1 )^2 }} \\ \\ \\ F_{AD}^{\to} = F_{AD} \dfrac{(1.6 \ \hat i + 0.6 \ \hat j - \ \hat k ) }{ \sqrt{3.92}} \\ \\ \\ F_{AD}^{\to} = F_{AD} \dfrac{(1.6 \ \hat i + 0.6 \ \hat j - \ \hat k ) }{1.9798} \\ \\ \\ F_{AD}^{\to} = F_{AD} (0.808 \ \hat i+ 0.303 \ \hat j - 0.505 \ \hat k ) N\\[/tex]
Similarly ; the weight of the lunar Module is:
W = mg
where;
mass = 13500 kg
acceleration due to gravity= 1.82 m/s²
W = 13500 × 1.82
W = 24,570 N
Also. we known that the load is shared by four landing gears; Thus, the vertical reaction force exerted by the ground on each landing gear can be expressed as:
[tex]R =\dfrac{W}{4}[/tex]
[tex]R =\dfrac{24,570}{4}[/tex]
R = 6142.5 N
Now; the reaction force at point A in unit vector form is :
[tex]R^{\to} = Rk^{\to} \\ \\ R^{\to} = (6142.5 \ k ^{\to}) \ N[/tex]
Using the force equilibrium at the meeting point of the coordinates at A.
[tex]\sum F^{\to} = 0[/tex]
[tex]F_{AB}^{\to} +F_{AC}^{\to} + F_{AD}^{\to} + R^{\to} =0[/tex]
[tex][F_{AB} (0.408 \ \hat i + 0.408 \ \hat j - 0.8165 \ \hat k ) N + F_{AC} (0.303 \ \hat i + 0.808 \ \hat j - 0.505 \ \hat k ) N + F_{AD} (0.808 \ \hat i + 0.303 \ \hat j - 0.505 \ \hat k) N + (6142.5 \ k^ \to ) ][/tex]
[tex]= [ ( 0.408 F_{AB} +0.303 F_{AC} + 0.808F_{AD}) \hat i + (0.408 F_{AB}+0.808F_{AC}+0.303F_{AD}) \hat j + (-0.8165 F_{AB} -0.505F_{AC} -0.505 F_{AD} +6142.5 ) k ^ \to ] = 0[/tex]
From above; we need to relate and equate each coefficients i.e i ,j, and [tex]k ^ \to[/tex] on both sides ; so, we can re-write that above as;
[tex]0.408 F_{AB} +0.303 F_{AC} + 0.808F_{AD}) =0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ --- (1) \\ \\ 0.408 F_{AB}+0.808F_{AC}+0.303F_{AD}) =0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ --- (2) \\ \\ -0.8165 F_{AB} -0.505F_{AC} -0.505 F_{AD} +6142.5 = 0 --- (3)[/tex]
Making rearrangement and solving by elimination method;
[tex]\mathbf{F_{AB} = 13785.06 N }[/tex]
[tex]\mathbf{F_{AC} = -5062.38 N }[/tex]
[tex]\mathbf{F_{AD} = -5062.38 N }[/tex]
The force vector of each member, depends on the magnitude of the
force and the unit vector of the member.
Responses:
The force supported by the members are;
Force supported by AB is; 13,799.95 NForce supported by AC is; -5070.2 NForce supported by AD is -5070.2 NHow can the unit vector of each member give their force?Resolving the given members into unit vectors gives;
[tex]\hat u_{AB} = \mathbf{\dfrac{(2.6 - 1.5) \cdot \hat i + (2.6 - 1.5)\cdot \hat j + (-2.2)\cdot \hat k }{\sqrt{(2.6- 1.5)^2 + (2.6 - 1.5)^2 + (-2.2)^2}}}[/tex][tex]\dfrac{(2.6 - 1.5) \cdot \hat i + (2.6 - 1.5)\cdot \hat j + (-2.2)\cdot \hat k }{\sqrt{(2.6- 1.5)^2 + (2.6 - 1.5)^2 + (-2.2)^2}}= 0.40825 \cdot \hat i + 0.40825\cdot \hat j - 0.81625\cdot \hat k[/tex]
[tex]\hat u_{AB} = \mathbf{0.40825 \cdot \hat i + 0.40825\cdot \hat j - 0.81625\cdot \hat k}[/tex]Similarly, we have;
[tex]\hat u_{AC} =\mathbf{ \dfrac{(2.6 - 2) \cdot \hat i + (2.6 - 1)\cdot \hat j + (-2.2+1.2)\cdot \hat k }{\sqrt{(2.6- 2)^2 + (2.6 - 1)^2 + (-2.2+1.2)^2}}}[/tex]
[tex]\dfrac{(2.6 - 2) \cdot \hat i + (2.6 - 1)\cdot \hat j + (-2.2+1.2)\cdot \hat k }{\sqrt{(2.6- 2)^2 + (2.6 - 1)^2 + (-2.2+1.2)^2}} =\dfrac{0.6\cdot \hat i +1.6\cdot \hat j -1\cdot \hat k }{\sqrt{0.6^2 + 1.6^2 + (-1.)^2}}[/tex]
[tex]\dfrac{0.6\cdot \hat i +1.6\cdot \hat j -1\cdot \hat k }{\sqrt{0.6^2 + 1.6^2 + (-1.)^2}}= 0.303046\cdot \hat i + 0.80812\cdot \hat j - 0.50508\cdot \hat k[/tex]
[tex]\hat u_{AC} =\mathbf{0.303046\cdot \hat i + 0.80812\cdot \hat j - 0.50508\cdot \hat k}[/tex][tex]\hat u_{AD} =\mathbf{ \dfrac{(2.6 - 1) \cdot \hat i + (2.6 -2)\cdot \hat j + (-2.2 + 1.2)\cdot \hat k }{\sqrt{(2.6-1)^2 + (2.6 -2))^2 + (-2.2 + 1.2)^2}}}[/tex]
[tex]\hat u_{AD} =\mathbf{0.80812\cdot \hat i+ 0.303046\cdot \hat j - 0.50508\cdot \hat k}[/tex]
The forces are therefore;
[tex]\vec F_{AB} =\mathbf{ F_{AB} \cdot \left ( 0.40825 \cdot \hat i + 0.40825\cdot \hat j - 0.81625\cdot \hat k \right)}[/tex]
[tex]\vec F_{AC} =\mathbf{ F_{AC} \cdot \left (0.303046\cdot \hat i + 0.80812\cdot \hat j - 0.50508\cdot \hat k\right)}[/tex]
[tex]\vec F_{AD} = \mathbf{F_{AD} \cdot \left (0.80812\cdot \hat i+ 0.303046\cdot \hat j - 0.50508\cdot \hat k\right)}[/tex]
[tex]Weight \ on \ the \ assembly = \dfrac{13,500 \, kg \times 1.82 \, m/s^2}{4} = 6,142.5 \, \hat k N[/tex]
Which gives;
[tex]\mathbf{0.40825 \cdot \hat i \cdot F_{AB}}[/tex] + [tex]0.303046\cdot \hat i \cdot F_{AC}[/tex] + [tex]0.80812\cdot \hat i \cdot F_{AD}[/tex] = 0
[tex]0.40825 \cdot \hat j \cdot F_{AB}[/tex] + [tex]0.80812\cdot \hat j \cdot F_{AC}[/tex] + [tex]0.303046 \cdot \hat j \cdot F_{AD}\left[/tex] = 0
[tex]-0.81625\cdot \hat k \cdot F_{AB}[/tex] - [tex]0.50508\cdot \hat k \cdot F_{AC}[/tex] - [tex]0.50508\cdot \hat k \cdot F_{AD}[/tex] + [tex]\mathbf{6,142.5 \, \hat k}[/tex] = 0
Which gives;
[tex]-0.81625\cdot \hat k \cdot F_{AB}[/tex] - [tex]0.50508\cdot \hat k \cdot F_{AC}[/tex] - [tex]0.50508\cdot \hat k \cdot F_{AD}[/tex] = [tex]-6,142.5 \, \hat k[/tex]
Solving gives;
[tex]F_{AB}[/tex] = 13799.95 N[tex]F_{AC}[/tex] = -5070.2 N[tex]F_{AD}[/tex] = -5070.2 NLearn more about unit vectors here:
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A generator operating at 50 Hz delivers 1 pu power to an infinite bus through a transmission circuit in which resistance is ignored. A fault takes place reducing the maximum power transferable to 0.5 pu whereas before the fault, this power was 2.0 pu and after the clearance of the fault, it is 1.5 pu. Using equal area criterion, determine the critical clearing angle.
Answer:
critical clearing angle = 70.3°
Explanation:
Generator operating at = 50 Hz
power delivered = 1 pu
power transferable when there is a fault = 0.5 pu
power transferable before there is a fault = 2.0 pu
power transferable after fault clearance = 1.5 pu
using equal area criterion to determine the critical clearing angle
Attached is the power angle curve diagram and the remaining part of the solution.
The power angle curve is given as
= Pmax sinβ
therefore : 2sinβo = Pm
2sinβo = 1
sinβo = 0.5 pu
βo = [tex]sin^{-1} (0.5) = 30[/tex]⁰
also ; 1.5sinβ1 = 1
sinβ1 = 1/1.5
β1 = [tex]sin^{-1} (\frac{1}{1.5} )[/tex] = 41.81⁰
∴ βmax = 180 - 41.81 = 138.19⁰
attached is the remaining solution
The critical clearing angle = [tex]cos^{-1} 0.3372[/tex] ≈ 70.3⁰
13- Convert the following numbers to the indicated bases. List all intermediate steps.
a- (36459080)10 to octal
b- (20960032010 to hexadecimal
c- (2423233303003040)s to base
25 36459080/8= 4557385 0/8 209600320/16=13100020 + 0/16 (2423233303003040)5 (36459080)10 =( 18 (209600320)10=( 1)16 (2423233303003040)5=( )125
Answer:
Following are the conversion to this question:
Explanation:
In point (a):
[tex]\to \frac{36459080}{8} = 4557385 + \ \ \ \ \ \ \ \ \ \frac{0}{8}\\\\\to \frac{4557385}{8} = 569673 + \ \ \ \ \ \ \ \ \ \frac{1}{8}\\\\\to \frac{569673}{8} = 71209+ \ \ \ \ \ \ \ \ \ \frac{1}{8}\\\\\to \frac{71209}{8}=8901+\ \ \ \ \ \ \ \ \ \ \ \frac{1}{8}\\\\\to \frac{8901}{8}=1112+ \ \ \ \ \ \ \ \ \ \ \ \ \frac{5}{8}\\\\\to \frac{1112}{8}=139+ \ \ \ \ \ \ \ \ \ \ \frac{0}{8}\\\\\to \frac{139}{8}=17+ \ \ \ \ \ \ \ \ \ \ \frac{3}{8}\\\\\to \frac{17}{8}=2+ \ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{8}\\\\[/tex]
[tex]\to \frac{2}{8}=0+ \ \ \ \ \ \ \ \ \ \frac{2}{8}\\\\ \bold{(36459080)_{10}=(213051110)_8}[/tex]
In point (b):
[tex]\to \frac{20960032010}{16} = 13100020+ \ \ \ \ \ \ \ \ \ \frac{0}{16}\\\\\to \frac{13100020}{16} = 818751+ \ \ \ \ \ \ \ \ \ \frac{4}{16}\\\\\to \frac{818751}{16} = 51171+ \ \ \ \ \ \ \ \ \ \frac{15}{16}\\\\\to \frac{51171}{16}=3198+\ \ \ \ \ \ \ \ \ \ \ \frac{3}{16}\\\\\to \frac{3198}{16}=199+ \ \ \ \ \ \ \ \ \ \ \ \ \frac{14}{1}\\\\\to \frac{199}{16}=12+ \ \ \ \ \ \ \ \ \ \ \frac{7}{16}\\\\\to \frac{12}{16}=0+ \ \ \ \ \ \ \ \ \ \ \frac{12}{16}\\\\ \bold{(20960032010)_{10}=(C7E3F40)_{16}}[/tex]
In point (c):
[tex]\to (2423233303003040)_s=(88757078520)_{10}\\\\\to \frac{88757078520}{25}= 3550283140+ \ \ \ \ \ \ \ \ \ \frac{20}{25}\\\\ \to \frac{3550283140}{25}= 142011325+ \ \ \ \ \ \ \ \ \ \frac{15}{25}\\\\\to \frac{142011325}{25}= 5680453+ \ \ \ \ \ \ \ \ \ \frac{0}{25}\\\\\to \frac{5680453}{25}= 227218+ \ \ \ \ \ \ \ \ \ \frac{3}{25}\\\\\to \frac{227218}{25}= 9088+ \ \ \ \ \ \ \ \ \ \frac{18}{25}\\\\\to \frac{9088}{25}= 363+ \ \ \ \ \ \ \ \ \ \frac{13}{25}\\\\[/tex]
[tex]\to \frac{363}{25}= 14+ \ \ \ \ \ \ \ \ \ \frac{13}{25}\\\\\to \frac{14}{25}= 0+ \ \ \ \ \ \ \ \ \ \frac{14}{25}\\\\\bold{(2423233303003040)_s=(EDDI30FK)_{25}}[/tex]
Symbols of Base 25 are as follows:
[tex]0, 1, 2, 3,4,5,6,7,8,9,A,B,C,D,E,F,G,H,I,J,K,L,M,N, \ and \ O[/tex]
which of the following tells the computer wha to do
operating system
the ROM
the motherboard
the monitor
A bracket ABCD having a hollow circular cross section consists of a vertical arm AB (L 5 6 ft), a horizontal arm BC parallel to the x0 axis, and a horizontal arm CD parallel to the z0 axis (see gure). The arms BC and CD have lengths b1 5 3.6 ft and b2 5 2.2 ft, respectively. The outer and inner diameters of the bracket are d2 5 7.5 in. and d1 5 6.8 in. Aninclined load P 5 2200 lb acts at point D along line DH. Determine the maximum tensile, compressive, and shear stresses in the vertical arm.
Answer:
The answer is explained below
Explanation:
Given that:
1 ft = 0.3048 m, 1 in = 0.0254 m, 1 pound = 4.44822 newton
[tex]b_1=3.6ft=1.1\ m[/tex], [tex]b_2=2.2 ft=0.67\ m[/tex], [tex]d_2=7.5 in=0.19\ m[/tex], [tex]d_1=6.8in=0.17\ m[/tex]. P = 2200 lb = 9786 N
The area (A) is given as:
[tex]A=\frac{\pi}{4} (d_2^2-d_1^2)=\frac{\pi}{4}(0.19^2-0.17^2)=5.65*10^{-3}m^2[/tex]
The moment of area is given as:
[tex]l=\frac{\pi}{64} (d_2^4-d_1^4)=\frac{\pi}{64}(0.19^4-0.17^4)=2.3*10^{-5}m^4[/tex]
The maximum tensile stress is given as:
[tex]\sigma_1=-\frac{P}{A}+\frac{M(\frac{d_2}{2} )}{l} =-\frac{9786\ N }{5.65*10^{-3}m^2}+\frac{11kNm(0.19 \ m)/2}{2.3*10^{-5}m^4} =-1.73\ MPa+45.4\ MPa=43.67\ MPa\\\sigma_1=43.67\ MPa[/tex]
The maximum compressive stress is given as:
[tex]\sigma_c=-\frac{P}{A}-\frac{M(\frac{d_2}{2} )}{l} =-\frac{9786\ N }{5.65*10^{-3}m^2}-\frac{11kNm(0.19 \ m)/2}{2.3*10^{-5}m^4} =-1.73\ MPa-45.4\ MPa=47.13\ MPa\\\sigma_c=47.13\ MPa[/tex]
The maximum shear stress is given as:
[tex]\tau_{max}=|\frac{\sigma_c}{2} |=\frac{47.13\ MPa}{2}=23.57\ MPa[/tex]
A steel alloy is known to contain 93.8 wt% Fe, 6.0 wt% Ni, and 0.2 wt% C. Assume that there are no alterations in the positions of other phase boundaries with the addition of Ni. (a) What is the approximate eutectoid temperature of this alloy
Answer:
650°C or 1,200°F
Explanation:
Data provided in the question
Steel alloy contains 93.8 wt% Fe, 6.0 wt% Ni, and 0.2 wt% C
Plus we also assume that there are no changes in the boundaries of postions who have other phases but there is an addition of Ni.
Based on the above information, the approximate eutectoid temperature of this alloy for 6.1 wt% is 650°C or 1,200°F
In real world, sampling and quantization is performed in an analog to digital converter (ADC) and reconstruction is performed in a digital to analog converter (DAC). Which of the following statements hold true (fs denotes the sampling frequency)?
a. the reconstruction filter can be found in the DAQ
b. the antialiasing filter removes all frequencies of the continuous-time analog input signal that are above fs/2
c. the DAC needs to know the sampling frequency of the ADC to correctly reconstruct the signal.
d. the reconstructed continuous-time signal only contains frequencies up to fs/2
Answer:
b
Explanation:
a) ADC is located on DAQ filter but not the reconstruction filter
b) to remove aliasing, the sampling rate must be greater than or equal ot twice the highest frequency component in the input signal. In other words, all frequencies in input sgnal are less than fs/2. Therefore, frequencies greater than fs/2 are removed by anti-aliasing filter
c) the DAC can have different sampling rate from ADC
If the resistance reading on a DMM'S meter face is to 22.5 ohms in the range selector switch is set to R X 100 range, what is the actual measure resistance of the circuit?
Answer:
The answer is 2.25 kΩ
Explanation:
Solution
Given that:
The resistance reading on a DMM'S meter face = 22.5 ohms
The range selector switch = R * 100 range,
We now have to find the actual measure resistance of the circuit which is given below:
The actual measured resistance of the circuit is=R * 100
= 22.5 * 100
=2.25 kΩ
Hence the measured resistance of the circuit is 2.25 kΩ
The shaft is made of A992 steel. It has a diameter of 1 in. and is supported by bearings at A and D, which allows free rotation. For A992 steel, G = 11 × 103 ksi. (1) Determine the angle of twist of B with respect to D.(2) Determine the angle of twist of C with respect to D.Answer unit: degree or radians, two decimal places
Answer:
the angle of twist of B with respect to D is -1.15°
the angle of twist of C with respect to D is 1.15°
Explanation:
The missing diagram that is supposed to be added to this image is attached in the file below.
From the given information:
The shaft is made of A992 steel. It has a diameter of 1 in and is supported by bearing at A and D.
For the Modulus of Rigidity G = 11 × 10³ Ksi = 11 × 10⁶ lb/in²
The objective are :
1) To determine the angle of twist of B with respect to D
Considering the Polar moment of Inertia at the shaft [tex]J\tau[/tex]
shaft [tex]J\tau[/tex] = [tex]\dfrac{\pi}{2}r^4[/tex]
where ;
r = 1 in /2
r = 0.5 in
shaft [tex]J \tau[/tex] = [tex]\dfrac{\pi}{2} \times 0.5^4[/tex]
shaft [tex]J\tau[/tex] = 0.098218
Now; the angle of twist at B with respect to D is calculated by using the expression
[tex]\phi_{B/D} = \sum \dfrac{TL}{JG}[/tex]
[tex]\phi_{B/D} = \dfrac{T_{CD}L_{CD}}{JG}+\dfrac{T_{BC}L_{BC}}{JG}[/tex]
where;
[tex]T_{CD} \ \ and \ \ L_{CD}[/tex] are the torques at segments CD and length at segments CD
[tex]{T_{BC} \ \ and \ \ L_{BC}}[/tex] are the torques at segments BC and length at segments BC
Also ; from the diagram; the following values where obtained:
[tex]L_{BC}}[/tex] = 2.5 in
[tex]J\tau[/tex] = 0.098218
G = 11 × 10⁶ lb/in²
[tex]T_{BC[/tex] = -60 lb.ft
[tex]T_{CD[/tex] = 0 lb.ft
[tex]L_{CD[/tex] = 5.5 in
[tex]\phi_{B/D} = 0+ \dfrac{[(-60 \times 12 )] (2.5 \times 12 )}{ (0.9818)(11 \times 10^6)}[/tex]
[tex]\phi_{B/D} = \dfrac{[(-720 )] (30 )}{1079980}[/tex]
[tex]\phi_{B/D} = \dfrac{-21600}{1079980}[/tex]
[tex]\phi_{B/D} =[/tex] − 0.02 rad
To degree; we have
[tex]\phi_{B/D} = -0.02 \times \dfrac{180}{\pi}[/tex]
[tex]\mathbf{\phi_{B/D} = -1.15^0}[/tex]
Since we have a negative sign; that typically illustrates that the angle of twist is in an anti- clockwise direction
Thus; the angle of twist of B with respect to D is 1.15°
(2) Determine the angle of twist of C with respect to D.Answer unit: degree or radians, two decimal places
For the angle of twist of C with respect to D; we have:
[tex]\phi_{C/D} = \dfrac{T_{CD}L_{CD}}{JG}+\dfrac{T_{BC}L_{BC}}{JG}[/tex]
[tex]\phi_{C/D} = 0+\dfrac{T_{BC}L_{BC}}{JG}[/tex]
[tex]\phi_{B/D} = 0+ \dfrac{[(60 \times 12 )] (2.5 \times 12 )}{ (0.9818)(11 \times 10^6)}[/tex]
[tex]\phi_{C/D} = \dfrac{21600}{1079980}[/tex]
[tex]\phi_{C/D} =[/tex] 0.02 rad
To degree; we have
[tex]\phi_{C/D} = 0.02 \times \dfrac{180}{\pi}[/tex]
[tex]\mathbf{\phi_{C/D} = 1.15^0}[/tex]
A cylindrical tank is required to contain a gage pressure 560 kPa . The tank is to be made of A516 grade 60 steel with a maximum allowable normal stress of 150 MPa . If the inner diameter of the tank is 3 m , what is the minimum thickness, t, of the wall
Answer:
5.6 mm
Explanation:
Given that:
A cylindrical tank is required to contain a:
Gage Pressure P = 560 kPa
Allowable normal stress [tex]\sigma[/tex] = 150 MPa = 150000 Kpa.
The inner diameter of the tank = 3 m
In a closed cylinder there exist both the circumferential stress and the longitudinal stress.
Circumferential stress [tex]\sigma = \dfrac{pd}{2t}[/tex]
Making thickness t the subject; we have
[tex]t = \dfrac{pd}{2* \sigma}[/tex]
[tex]t = \dfrac{560000*3}{2*150000000}[/tex]
t = 0.0056 m
t = 5.6 mm
For longitudinal stress.
[tex]\sigma = \dfrac{pd}{4t}[/tex]
[tex]t= \dfrac{pd}{4*\sigma }[/tex]
[tex]t = \dfrac{560000*3}{4*150000000}[/tex]
t = 0.0028 mm
t = 2.8 mm
From the above circumferential stress and longitudinal stress; the stress with the higher value will be considered ; which is circumferential stress and it's minimum value with the maximum thickness = 5.6 mm
The liquid-phase reaction A + B → C follows an elementary rate law and is carried out isothermally in a flow system. The concentrations of A and B feed streams are 2 M before mixing. The volumetric flow rate of each stream is 5 dm3 /min and the entering temperature is 300 K. The streams are mixed immediately before entering. Two reactors are available: One is a gray 200.0 dm3 CSTR that can be heated to 77°C or cooled to 0°C, and the other is a white 800.0 dm3 PFR operated at 300 K that cannot be heated or cooled but can be painted red or black. (Note: k = 0.07 dm3 /mol*min at 300 K and E = 20 kcal/mol.) How long would it take to achieve 90% conversion in a 200 dm3 batch reactor with CA ° = CB ° = 1 ???? after mixing at a temperature of 70°C?
Answer:
1.887 minutes
Explanation:
We are given k = 0.07 dm3 /mol*min at 300 K and E = 20 kcal/mol = 20000 cal/mol
To solve this, first of all let's calculate the rate constant(k);
For this question, The formula is;
K(t) = k(300K) × exp[(E/R)((1/300) - (1/T2))]
R is gas constant = 1.987 cal/mol.K
For temperature of 70°C which is = 70 + 273K = 343K, we have;
K(343) = 0.07 × exp[(20000/1.987)((1/300) - (1/343))]
K(343) = 4.7 dm³/mol.min
The design equation is;
dX/dt = -(rA/C_Ao) = K•(C_Ao)²•(1 - X)²/(C_Ao) = (KC_Ao)(1 - X)²
Since there is no change in volume by cause of the state at which the reaction is carried out, that is liquid. Thus, integrating and solving for time for a 90% conversion we obtain;
(0.9,0)∫dX/(1 - X)².dX = (KC_Ao)((t, 0)∫dt
So, we'll get;
0.9/(1 - 0.9) = 4.77 × 1 × t
t = 9/4.77
t = 1.887 minutes
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 60F at the beginning of compression. The maximum temperature in the cycle is 5200R. Based on this model, calculate the net work per cycle, in Btu, and the power developed by the engine, in horsepower.
Answer:
the net work per cycle [tex]\mathbf{W_{net} = 0.777593696}[/tex] Btu per cycle
the power developed by the engine, W = 88.0144746 hp
Explanation:
the information given includes;
diameter of the four-cylinder bore = 3.7 in
length of the stroke = 3.4 in
The clearance volume = 16% = 0.16
The cylindrical volume [tex]V_2 = 0.16 V_1[/tex]
the crankshaft N rotates at a speed of 2400 RPM.
At the beginning of the compression , temperature [tex]T_1[/tex] = 60 F = 519.67 R
and;
Otto cycle with a pressure = 14.5 lbf/in² = (14.5 × 144 ) lb/ft²
= 2088 lb/ft²
The maximum temperature in the cycle is 5200 R
From the given information; the change in volume is:
[tex]V_1-V_2 = \dfrac{\pi}{4}D^2L[/tex]
[tex]V_1-0.16V_1= \dfrac{\pi}{4}(3.7)^2(3.4)[/tex]
[tex]V_1-0.16V_1= 36.55714291[/tex]
[tex]0.84 V_1 =36.55714291[/tex]
[tex]V_1 =\dfrac{36.55714291}{0.84 }[/tex]
[tex]V_1 =43.52040823 \ in^3 \\ \\ V_1 = 43.52 \ in^3[/tex]
[tex]V_1 = 0.02518 \ ft^3[/tex]
the mass in air ( lb) can be determined by using the formula:
[tex]m = \dfrac{P_1V_1}{RT}[/tex]
where;
R = 53.3533 ft.lbf/lb.R°
[tex]m = \dfrac{2088 \ lb/ft^2 \times 0.02518 \ ft^3}{53.3533 \ ft .lbf/lb.^0R \times 519 .67 ^0 R}[/tex]
m = 0.0018962 lb
From the tables of ideal gas properties at Temperature 519.67 R
[tex]v_{r1} =158.58[/tex]
[tex]u_1 = 88.62 Btu/lb[/tex]
At state of volume 2; the relative volume can be determined as:
[tex]v_{r2} = v_{r1} \times \dfrac{V_2}{V_1}[/tex]
[tex]v_{r2} = 158.58 \times 0.16[/tex]
[tex]v_{r2} = 25.3728[/tex]
The specific energy [tex]u_2[/tex] at [tex]v_{r2} = 25.3728[/tex] is 184.7 Btu/lb
From the tables of ideal gas properties at maximum Temperature T = 5200 R
[tex]v_{r3} = 0.1828[/tex]
[tex]u_3 = 1098 \ Btu/lb[/tex]
To determine the relative volume at state 4; we have:
[tex]v_{r4} = v_{r3} \times \dfrac{V_1}{V_2}[/tex]
[tex]v_{r4} =0.1828 \times \dfrac{1}{0.16}[/tex]
[tex]v_{r4} =1.1425[/tex]
The specific energy [tex]u_4[/tex] at [tex]v_{r4} =1.1425[/tex] is 591.84 Btu/lb
Now; the net work per cycle can now be calculated as by using the following formula:
[tex]W_{net} = Heat \ supplied - Heat \ rejected[/tex]
[tex]W_{net} = m(u_3-u_2)-m(u_4 - u_1)[/tex]
[tex]W_{net} = m(u_3-u_2- u_4 + u_1)[/tex]
[tex]W_{net} = m(1098-184.7- 591.84 + 88.62)[/tex]
[tex]W_{net} = 0.0018962 \times (1098-184.7- 591.84 + 88.62)[/tex]
[tex]W_{net} = 0.0018962 \times (410.08)[/tex]
[tex]\mathbf{W_{net} = 0.777593696}[/tex] Btu per cycle
the power developed by the engine, in horsepower. can be calculated as follows;
In the four-cylinder, four-stroke internal combustion engine; the power developed by the engine can be calculated by using the expression:
[tex]W = 4 \times N' \times W_{net[/tex]
where ;
[tex]N' = \dfrac{2400}{2}[/tex]
N' = 1200 cycles/min
N' = 1200 cycles/60 seconds
N' = 20 cycles/sec
W = 4 × 20 cycles/sec × 0.777593696
W = 62.20749568 Btu/s
W = 88.0144746 hp
The net work per cycle and the power developed by this combustion engine are 0.7792 Btu and 88.20 hp.
Given the following data:
Diameter of bore = 3.7 inStroke length = 3.4 inClearance volume = 16% = 0.16Speed of 2400 RPM.Initial temperature = 60 F to R = 519.67 R. Initial pressure = 14.5 [tex]lbf/in^2[/tex] to [tex]lbf/ft^2[/tex] = 2088 [tex]lbf/ft^2[/tex] Maximum temperature = 5200 R.Note: The cylindrical volume is equal to [tex]0.16V_1[/tex]
How to calculate the net work per cycle.First of all, we would determine the volume, mass and specific energy as follows:
[tex]V_1-V_2=\frac{\pi D^2L}{4} \\\\V_1-0.16V_1=\frac{3.142 \times 3.7^2 \times 3.4}{4}\\\\0.84V_1=36.56\\\\V_1=\frac{36.56}{0.84} \\\\V_1=43.52\;in^3 \;to \;ft^3 = 0.0252\;ft^3[/tex]
For the mass:
[tex]M=\frac{PV}{RT} \\\\M=\frac{2088 \times 0.0252}{53.3533 \times 519.67} \\\\M=\frac{52.6176}{27726.109411}[/tex]
M = 0.0019 lb.
At a temperature of 519.67 R, the relative volume and specific energy are:
[tex]v_{r1}=158.58\\\\u_1 = 88.62\;Btu/lb[/tex]
For the relative volume at the second state, we have:
[tex]v_{r2}=v_{r1}\times \frac{V_2}{V_1} \\\\v_{r2}=158.58\times 0.16\\\\v_{r2}=25.3728[/tex]
Note: At 25.3728, specific energy ([tex]u_2[/tex]) is 184.7 Btu/lb.
At a maximum temperature of 519.67 R, the relative volume and specific energy are:
[tex]v_{r3}=0.1828\\\\u_3 = 1098\;Btu/lb[/tex]
For the relative volume at state 4, we have:
[tex]v_{r4}=v_{r3}\times \frac{V_1}{V_3} \\\\v_{r4}=0.1828\times \frac{1}{0.16}\\\\v_{r4}=1.1425[/tex]
Note: At 1.1425, specific energy ([tex]u_4[/tex]) is 591.84 Btu/lb.
Now, we can calculate the net work per cycle by using this following formula:
[tex]W=Heat\;supplied -Heat\rejected\\\\W=m(u_3-u_2)-m(u_4-u_1)\\\\W=0.0019(1098-184.7)-0.0019(591.84-88.62)\\\\W=1.73527-0.956118[/tex]
W = 0.7792 Btu.
How to calculate the power developed.In a four-cylinder, four-stroke internal combustion engine, power is given by this formula:
[tex]W=4N'W_{net}[/tex]
But;
[tex]N'=\frac{N}{2 \times 60} \\\\N'=\frac{2400}{120} \\\\N'=20\;cycle/sec[/tex]
Substituting the given parameters into the formula, we have;
[tex]W=4 \times 20 \times 0.7792[/tex]
W = 62.336 Btu/sec.
In horsepower:
W = 88.20 hp.
Read more on net work here: https://brainly.com/question/10119215
Air at 80 °F is to flow through a 72 ft diameter pipe at an average velocity of 34 ft/s . What diameter pipe should be used to move water at 60 °F and average velocity of 71 ft/s if Reynolds number similarity is enforced? The kinematic viscosity of air at 80 °F is 1.69E-4 ft^2/s and the kinematic viscosity of water at 60 °F is 1.21E-5 ft^2/s. Round your answer (in ft) to TWO decimal places.
Answer:
2.47 ft
Explanation:
Given that:
The initial temperature of air = 80°F
Diameter of the pipe = 72 ft
average velocity [tex]v_{air}[/tex] of the air flow through the pipe = 34 ft/s
The objective is to determine the diameter of the pipe to be used to move water at:
At a temperature = 60°F &
An average velocity [tex]v_{water}[/tex] of 71 ft/s
Assuming Reynolds number similarity is enforced;
where :
kinematic viscosity (V_air) of air at 80 °F (V_air) = 1.69 × 10⁻⁴ ft²/s
kinematic viscosity of water at 60 °F (V_water) = 1.21 × 10⁻⁵ ft²/s
The diameter of the pipe can be calculated by using the expression:
[tex]D_{water} = \dfrac{V_{water}}{V_{air}}*\dfrac{v_{air}}{v_{water}}* D_{air}[/tex]
[tex]D_{water} = \dfrac{1.21*10^{-5} \ ft^2/s}{1.69*10^{-4} \ ft^2/s}*\dfrac{34 \ ft/s}{71 \ ft/s}* 72 \ ft[/tex]
[tex]D_{water} =[/tex] 2.4686 ft
[tex]D_{water} =[/tex] 2.47 ft ( to two decimal places)
Thus; diameter pipe to be use to move water at the given temperature and average velocity is 2.47 ft
Answer:
2.47 ft
Explanation:
Given that:
The initial temperature of air = 80°F
Diameter of the pipe = 72 ft
average velocity of the air flow through the pipe = 34 ft/s
The objective is to determine the diameter of the pipe to be used to move water at:
At a temperature = 60°F &
An average velocity of 71 ft/s
Assuming Reynolds number similarity is enforced;
where :
kinematic viscosity (V_air) of air at 80 °F (V_air) = 1.69 × 10⁻⁴ ft²/s
kinematic viscosity of water at 60 °F (V_water) = 1.21 × 10⁻⁵ ft²/s
The diameter of the pipe can be calculated by using the expression:
2.4686 ft
2.47 ft ( to two decimal places)
Thus; diameter pipe to be use to move water at the given temperature and average velocity is 2.47 ft
Question 44
What should you do if you encounter a fishing boat while out in your vessel?
A
Make a large wake nearby.
B
Avoid making a large wake.
с
Pass on the side with the fishing lines.
D
Pass by close to the anglers.
Submit Answer
Answer:
The answer is B. Avoid making a large wake.
Explanation:
When passing a fishing boat it is important to maintain a minimal wake due to the dangers a large wake could pose to the fishing boat you are passing, it is part of maintaining safety on the water.
You can not pass on the sides with the fishing lines also, and you are supposed to communicate to the fishing boat before taking the appropriate action.
How old are you? answer this question plz lol I will mark someone as brainliest
Answer:
100000000000000000000000
Refrigerant-134a enters the condenser of a residential heat pump at 800 kPa and 50°C at a rate of 0.02 kg/s and leaves at 750 kPa subcooled by 3°C. The refrigerant enters the compressor at 200 kPa superheated by 4°C. Determine the isentropic efficiency of the compressor, the rate of heat supplied to the room, COP of the Heat Pump and the rate of heat supplied to this room if the heat pump operated on an ideal vapor compression cycle between pressure limits of 200 and 800 kpa
Explanation:
The value of enthalpy and entropy at state 1 will be determined according to the given pressure and temperature as follows using interpolation from A-13 is as follows.
[tex]h_{1}[/tex] = 247.88 kJ/kg, [tex]S_{1}[/tex] = 0.9579 kJ/kg K
At state 2, isentropic enthalpy will be determined from the condition [tex]S_{2} = S_{1}[/tex] and given pressure at 2 with data from A-13 using interpolation is:
[tex]h_{2s}[/tex] = 279.45 kJ/kg
We will calculate actual enthalpy at state 2 using given pressure and temperature from A-13 as follows.
[tex]h_{2}[/tex] = 286.71 kJ/kg
Hence, isentropic compressor efficiency will be calculated using standard relation as:
[tex]\eta_{c} = \frac{h_{2s} - h_{1}}{h_{2} - h_{1}}[/tex]
= [tex]\frac{279.45 - 247.88}{286.71 - 247.88}[/tex]
= 0.813
Now, at state 3 enthaply is determined by temperature at state 3, that is, [tex]26^{o}C[/tex] for given pressure as per saturated liquid approximation and data from A-11.
[tex]h_{3}[/tex] = 87.83 kJ/Kg
Using energy balance in 2-3, the rate of heat supplied to the heated room is as follows.
[tex]Q_{H} = m(h_{2} - h_{3})[/tex]
= 0.022 (286.71 - 87.83) kW
= 4.38 kW
Now, COP will be calculated using power that is expressed through energy balance in 1-2 as follows.
COP = [tex]\frac{Q_{H}}{W}[/tex]
= [tex]\frac{Q_{H}}{m(h_{2} - h_{1})}[/tex]
= [tex]\frac{4.38}{0.022 (286.71 - 246.88)}[/tex]
= 5.13
In an ideal vapour-compression cycle, the enthalpy and entropy at state 1 will be obtained from given pressure and state with data from A-12:
[tex]h_{1}[/tex] = 244.5 kJ/kg
[tex]S_{1}[/tex] = 0.93788 kJ/kg K
[tex]h_{2}[/tex] = 273.71 kJ/kg
At state 3, enthalpy will be determined from given pressure and state with data from A-12 as follows.
[tex]h_{3}[/tex] = 95.48 kJ/kg
Hence, using energy balance in 2-3 the rate of heat supplied will be calculated as follows.
[tex]Q_{H} = m(h_{2} - h_{3})[/tex]
= 0.022 (273.31 - 95.48) kW
= 3.91 kW
The power input which is expressed through energy balance in 1-2 will be used to determine COP as follows.
COP = [tex]\frac{Q_{H}}{W}[/tex]
= [tex]\frac{Q_{H}}{m (h_{2} - h_{1})}[/tex]
= [tex]\frac{3.91}{0.022(273.31 - 244.5)}[/tex]
= 6.17