Answer:
a) ∝ and β
The phase compositions are :
C[tex]_{\alpha }[/tex] = 5wt% Sn - 95 wt% Pb
C[tex]_{\beta }[/tex] = 98 wt% Sn - 2wt% Pb
b)
The phase is; ∝
The phase compositions is; 82 wt% Sn - 91.8 wt% Pb
Explanation:
a) 15 wt% Sn - 85 wt% Pb at 100⁰C.
The phases are ; ∝ and β
The phase compositions are :
C[tex]_{\alpha }[/tex] = 5wt% Sn - 95 wt% Pb
C[tex]_{\beta }[/tex] = 98 wt% Sn - 2wt% Pb
b) 1.25 kg of Sn and 14 kg Pb at 200⁰C
The phase is ; ∝
The phase compositions is; 82 wt% Sn - 91.8 wt% Pb
Csn = 1.25 * 100 / 1.25 + 14 = 8.2 wt%
Cpb = 14 * 100 / 1.25 + 14 = 91.8 wt%
Các đặc điểm chính của đường dây dài siêu cao áp .
Answer:
Đường dây siêu cao áp 500kV: Những chuyện giờ mới kể ... Ngày 27/5/1994, hệ thống đường dây điện siêu cao áp 500kV Bắc - Nam chính thức đưa ... Tại thời điểm đó, các nước như Pháp, Úc, Mỹ khi xây dựng đường dây dài nhất ... và chế ra các máy kéo dây theo đặc thù công việc của từng đơn vị.
Explanation:
A masonry chimney should be braced with horizontal metal straps every few feet against the structure and into reinforced points such as wall studs to stabilize the chimney from the shaking force of an earthquake.
a. True
b. False
Answer: True
Explanation:
The statement that "a masonry chimney should be braced with horizontal metal straps every few feet against the structure and into reinforced points such as wall studs to stabilize the chimney from the shaking force of an earthquake" is true.
In a scenario whereby the chimney isn't braced with the horizontal metal straps every few feet, this can lead to its collapse in case of an earthquake. Therefore, the correct option is "true".
a. Name the major strengthening mechanisms in metals and explain the working principle under each mechanism.Give the relevant equations corresponding to the mechanisms.
b. Briefly explain why small-angle grain boundaries are not as effective in interfering with the slip process as are high-angle grain boundaries.
Answer:
a) Solid solution strengthening and alloying, Precipitation hardening, work hardening
b) Absence of enough crystallographic misalignment in the grain boundary region for a small-angle
Explanation:
A) strengthening mechanism
i) Solid solution strengthening and alloying:
In solid solution strengthening and alloying mechanism there is an addition of one atom of solute to another during this process, there might be substitution of interstitial point defect in crystal
also the shear stress required can be represented as: Δz = Gb√Ce^3/2
where : C = solute concentration , e = strain on material
ii) Precipitation hardening:
During precipitation hardening the alloying above the concentrate will lead to the formation of a second phase also under precipitation hardening a second phase can also be created via thermal treatments
particle bowing cab be written as : Δz = Gb / L-2x
iii) work hardening :
Dislocation caused by stress fields been generated hardens metals under the work hardening mechanism
dislocation can be represented as ; Gb √ p
where : G = shear modulus , b = Burgess vector, p = dislocation density
B) The small angle grain boundaries are not effective enough because there is less crystallographic misalignment in the grain boundary region for a small-angle
The worst case signal-to-noise ratio at the output of an FM detector occurs when: ________
a. the desired signal is 90 degrees out of phase with the intelligence signal.
b. the desired signal is 90 degrees out of phase with the noise signal.
c. the noise signal is 90 degrees out of phase with the resultant signal of adding the signal to the noise.
d. the desired signal is 90 degrees out of phase with the resultant signal of adding the signal to the noise.
Answer:
a. the desired signal is 90 degrees out of phase with the intelligence signal.
Explanation:
The signal to noise ratio of FM detector is defined as function of modulation index for SSB FM signal plus narrow band Gaussian noise at input. The ratio is usually higher than 1:1 which indicates more signals than noise.
Consider a venturi with a throat-to-inlet area ratio of 0.75, mounted on the side of an airplane fuselage. The airplane is in flight at standard sea level. If the static pressure at the throat is 2050 lb/ft2 , calculate the velocity of the airplane.
This question is incomplete, the complete question is;
Consider a venturi with a throat-to-inlet area ratio of 0.75, mounted on the side of an airplane fuselage. The airplane is in flight at standard sea level. If the static pressure at the throat is 2050 lb/ft2 , calculate the velocity of the airplane.
Note that standard sea level density and pressure are 1.23 kg/m3 (0.002377 slug/ft3) and 1.01 x 105 N/m2 (2116lb/ft3), respectively.
Answer:
the velocity of the airplane is 267.2 ft/s
Explanation:
Given the data in the question;
throat-to-inlet area ratio A₂/A₁ = 0.75
density of air ρ = 0.002377 slug/ft³
the pressure at inlet p₁ = 2116 lb/ft³
the pressure at the throat p₂ = 2050 lb/ft³
Now, for a venturi duct, the velocity of the airplane V is given as;
V = √[ (2( p₁ - p₂ )) / (ρ( [A₁/A₂]² - 1 )) ]
so we substitute in our values
V = √[ (2( 2116 - 2050 )) / (0.002377 ( [1/0.75]² - 1 )) ]
V = √[ ( 2 × 66 ) / (0.002377 ( 1.7778 - 1 )) ]
V = √[ ( 2 × 66 ) / (0.002377 × 0.7778 ) ]
V = √[ 132 / 0.0018488 ]
V = √[ 71397.663349 ]
V = 267.2 ft/s
Therefore, the velocity of the airplane is 267.2 ft/s
If the two 304-stainless-steel carriage bolts of the clamp each have a diameter of 10 mmmm, and they hold the cylinder snug with negligible force against the rigid jaws, determine the temperature at which the average normal stress in either the magnesium or the steel first becomes 12.0 MPaMPa .
Answer: hello some data related to your question is missing attached below is the missing data
answer:
T2 = 265°C
Explanation:
First step : calculate sum of vertical forces
∑ y = 0
Fmg - 2(0.5 Fst ) = 0
∴Fmg = ( 12 * 10^6 ) ( 2 * π/4 (0.01)^2 )
= 1884.96 N
Also determine the Compatibility equation in order to determine the change in Temperature
ΔT = 250°C
therefore Temperature at which average normal stress becomes 12.0 MPa
ΔT = T2 - T1
250°C = T2 - 15°C
T2 = 250 + 15 = 265°C
attached below is the detailed solution
Please choose a specific type of stability or control surface (e.g., a canard) and explain how it is used, what it is used for, and the pros and cons of the device or system.
Answer:
small forewing
pro : Can be used in place of tail plane configuration
con : Can be very complex and difficult to use
Explanation:
A canard is generally used to provide some form of stability to an unstable or semi stable system.
An example of a Canard is a small forewing placed in an aircraft that will help stabilize the aircraft when in motion( in air ). because an airplane is generally an unstable system on its own
pro : Can be used in place of tail plane configuration
con : Can be very complex and difficult to use
A cold air standard gas turbine engine with a thermal efficiency of 56.9 % has a minimum pressure of 100 kP
a. Calculate the pressure at the inlet to the expansion process in the turbine. Enter your answer in kPa to one decimal place.
b. Calculate the thermal efficiency of a cold air standard Otto cycle engine with a compression ratio of 10.9. Enter your answer in percent with one decimal place.
Answer:
a) 5.2 kPa
b) 49.3%
Explanation:
Given data:
Thermal efficiency ( л ) = 56.9% = 0.569
minimum pressure ( P1 ) = 100 kpa
a) Determine the pressure at inlet to expansion process
P2 = ?
r = 1.4
efficiency = 1 - [ 1 / (rp)[tex]\frac{r-1}{r}[/tex] ]
0.569 = 1 - [ 1 / (rp)^0.4/1.4
1 - 0.569 = 1 / (rp)^0.285
∴ (rp)^0.285 = 0.431
rp = 0.0522
note : rp = P2 / P1
therefore P2 = rp * P1 = 0.0522 * 100 kpa
= 5.2 kPa
b) Thermal efficiency
Л = 1 - [ 1 / ( 10.9 )^0.285 ]
= 0.493 = 49.3%
khái niệm về môi trường nhiệt nóng và môi trường nhiện lạnh ?
ảnh hưởng của môi trường và môi trường nhiệt lạnh đến con người như thế nào ?
Theo em môi trường nào gây nguy hiểm hơn đối với con người ? Vì sao ?
Explanation:
उह्ह्नमजज्ल्ह्ह्बनुतनकुहक्जो
What method is most likely to be used to measure the
perature of a liquid contained in an open vessel?
Answer:
Many techniques have been developed for the measurement of pressure and vacuum. Instruments used to measure and display pressure in an integral unit are called pressure meters or pressure gauges or vacuum gauges.
R-134a is throttled in a line flowing at 25oC, 750 kPa with negligible kinetic energy to a pressure of 165 kPa. Find the exit temperature and the ratio of the exit pipe diameter to that of the inlet pipe (Dex/Din) so that the velocity stays constant.
Solution :
For R-134a, we are given :
[tex]$T_i = 25^\circ C$[/tex]
[tex]$P_i=750 \ kPa$[/tex]
[tex]$P_e=165 \ kPa$[/tex]
Now we have one inlet and one exit flow, no work and no heat transfer. The energy equation is :
[tex]$h_e+\frac{1}{2}.v_e^2= h_i+\frac{1}{2}.v_i^2 $[/tex]
We also know that the gas is throttled and there is no change in the kinetic energy.
So, [tex]$v_e=v_i$[/tex]
Now from the energy equation above, we can see that the inlet and the exit enthalpies are also the same. Therefore,
[tex]$h_i=h_e$[/tex]
From the saturated R-134a table, corresponding to [tex]P_e = 165 \ kPa[/tex], we can find the exit saturation temperature.
[tex]$T_e=-15^\circ C$[/tex]
From the saturated R-134a table, corresponding to [tex]P_e = 165 \ kPa[/tex], we can find the specific enthalpies :
[tex]$h_f = 180.19 \ kJ/kg$[/tex]
[tex]$h_{fg} = 209 \ kJ/kg$[/tex]
Calculating the exit flow quality factor,
[tex]$x_e=\frac{h_e-h_f}{h_{fg}}$[/tex]
[tex]$=\frac{234.59-180.19}{209}$[/tex]
= 0.26
From the saturated R-134a table, corresponding to [tex]P_e = 165 \ kPa[/tex], we can find the specific volumes :
[tex]$v_f = 0.00746 \ m^3/kg$[/tex]
[tex]$v_{fg} = 0.11932 \ m^3/kg$[/tex]
Calculating the exit specific volume :
[tex]$v_e=v_f+x_e(v_{fg})$[/tex]
= 0.000746 + 0.26 (0.11932)
= 0.0318 [tex]m^3/kg[/tex]
The mass flow is equal to :
[tex]$\dot{m} = A_i . \frac{v}{v_i}$[/tex]
[tex]$=A_e . \frac{v}{v_e}$[/tex]
So, [tex]$\frac{A_e}{A_i}=\frac{v_e}{v_i}$[/tex]
Therefore, the ratio of the exit pipe and the inlet pipe diameter is equal to
[tex]$\frac{D_e}{D_i}=\sqrt{\frac{A_e}{A_i}}$[/tex]
[tex]$\frac{D_e}{D_i}=\sqrt{\frac{v_e}{v_i}}$[/tex]
[tex]$\frac{D_e}{D_i}=\sqrt{\frac{0.0318}{0.000829}}$[/tex]
[tex]$\frac{D_e}{D_i}=6.19$[/tex]