Okay bro let’s go man yes yes
Answer:do me ti
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Why not me
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Explanation:
mitski
A turbine of a fossil fuel burning installation delivers 1,500 hp of mechanical energy to a generator. The generator then converts 80.0% of the mechanical energy into electrical energy. If the terminal potential difference of the generator is 1790 V, what current does it deliver (in A)
Answer:
The generator delivers current of 500.11 A
Explanation:
Given the data in the question;
mechanical energy delivered to the generator = 1500 hp
efficiency η = 80.0 %
terminal potential difference of the generator = 1790 V
we know that;
1 hp = 746 W
so
the mechanical energy delivered to the generator will be
Generator Input = ( 1500 × 746 )W = 1119000 W
So the generator output will be;
Generator Output = Generator Input × η
we substitute
Generator Output = 1119000 W × 80.0 %
Generator Output = 1119000 W × 0.8
Generator Output = 895200 W
So the Current will be;
[tex]I[/tex] = Generator Output / terminal potential difference of the generator
we substitute
[tex]I[/tex] = 895200 W / 1790 V
[tex]I[/tex] = 500.11 A
Therefore, The generator delivers current of 500.11 A
In particular, a system may or may not be (1) Memoryless, (2) Time invariant, (3)Linear, (4) Casual, (5) Stable.
a. True
b. False
Answer:
a. True
Explanation:
A system may be sometimes casual, time invariant, memoryless, stable and linear in particular.
Thus the answer is true.
A system is casual when the output of the system at any time depends on the input only at the present time and in the past.
A system is said to be memoryless when the output for each of the independent variable at some given time is fully dependent on the input only at that particular time.
A system is linear when it satisfies the additivity and the homogeneity properties.
A system is called time invariant when the time shift in the output signal will result in the identical time shift of the output signal.
Thus a system can be time invariant, memoryless, linear, casual and stable.
An air conditioner removes heat steadily from a house at a rate of 750 kJ/min while drawing electric power at a rate of 5.25 kW. Determine (a) the COP of this air conditioner and (b) the rate of heat transfer to the outside air.Answers:(a) 2.38, (b) 1065 kJ/min
Answer:
a) the COP of this air conditioner is 2.38
b) the rate of heat transfer to the outside air is 1065 kJ/min
Explanation:
Given the data in the question;
[ Outdoor ] ← Q[tex]_H[/tex] [ W[tex]_{net, in[/tex] ] Q[tex]_L[/tex] ← [ House ]
Rate of heat removed from the house; Q[tex]_L[/tex] = 750 kJ/min = ( 750 kJ/min × ( 1 kW / 60 kJ/min ) ) = 12.5 kW
Net-work input; W[tex]_{net, in[/tex] = 5.25 kW
a) The coefficient of performance of the air conditioner; COP.
COP = Q[tex]_L[/tex] / W[tex]_{net, in[/tex]
we substitute
COP = 12.5 kW / 5.25 kW
COP = 2.38
Therefore, the COP of this air conditioner is 2.38
b) the rate of heat transfer to the outside air.
Q[tex]_H[/tex] = Q[tex]_L[/tex] + W[tex]_{net[/tex]
we substitute
Q[tex]_H[/tex] = 12.5 kW + 5.25 kW
Q[tex]_H[/tex] = 17.75 kW
Q[tex]_H[/tex] = ( 17.75 × 60 ) kJ/min
Q[tex]_H[/tex] = 1065 kJ/min
Therefore, the rate of heat transfer to the outside air is 1065 kJ/min
A brittle material is subjected to a tensile stress of 1.65 MPa. If the specific surface energy and modulus of elasticity for this material are 0.60 J/m2 and 2.0 GPa, respectively. What is the maximum length of a surface flaw that is possible without fracture
Answer:
The maximum length of a surface flaw that is possible without fracture is
[tex]2.806 \times 10^{-4} m[/tex]
Explanation:
The given values are,
σ=1.65 MPa
γs=0.60 J/m2
E= 2.0 GPa
The maximum possible length is calculated as:
[tex]\begin{gathered}a=\frac{2 E \gamma_{s}}{\pi \sigma^{2}}=\frac{(2)\left(2 \times 10^{9} \mathrm{~N} / \mathrm{m}^{2}\right)(0.60 \mathrm{~N} / \mathrm{m})}{\pi\left(1.65\times 10^{6} \mathrm{~N} / \mathrm{m}^{2}\right)^{2}} \\=2.806 \times 10^{-4} \mathrm{~m}\end{gathered}[/tex]
The maximum length of a surface flaw that is possible without fracture is
[tex]2.806 \times 10^{-4} m[/tex]
An intelligence signal is amplified by a 65% efficient amplifier before being combined with a 250W carrier to generate an AM signal. If it is desired to operate at 50% modulation, what must be the dc input power to the final intelligence signal amplifier
Answer:
"192.3 watt" is the right answer.
Explanation:
Given:
Efficient amplifier,
= 65%
or,
= 0.65
Power,
[tex]P_c=250 \ watt[/tex]
As we know,
⇒ [tex]P_t=P_c(1+\frac{\mu^2}{2} )[/tex]
By putting the values, we get
[tex]=P_c(1+\frac{1}{2} )[/tex]
[tex]=1.5 \ P_c[/tex]
Now,
⇒ [tex]P_i=(P_t-P_c)[/tex]
[tex]=1.5 \ P_c-P_c[/tex]
[tex]=\frac{P_c}{2}[/tex]
DC input (0.65) will be equal to "[tex](\frac{P_c}{2} )[/tex]".
hence,
The DC input power will be:
= [tex]\frac{250}{2}\times \frac{1}{0.65}[/tex]
= [tex]\frac{125}{0.65}[/tex]
= [tex]192.3 \ watt[/tex]
The number of pulses per second from IGBTs is referred to as
4. An aluminum alloy fin of 12 mm thick, 10 mm width and 50 mm long protrudes from a wall, which is maintained at 120C. The ambient air temperature is 22C. The heat transfer coefficient and conductivity of the fin material are 140 W/m2K and 55 W/mk respectively. Determine a. Temperature at the end of the fin b. Temperature at the middle of the fin. c. Calculate the heat dissipation energy of the fin
Answer:
a) 84.034°C
b) 92.56°C
c) ≈ 88 watts
Explanation:
Thickness of aluminum alloy fin = 12 mm
width = 10 mm
length = 50 mm
Ambient air temperature = 22°C
Temperature of aluminum alloy is maintained at 120°C
a) Determine temperature at end of fin
m = √ hp/Ka
= √( 140*2 ) / ( 12 * 10^-3 * 55 )
= √ 280 / 0.66 = 20.60
Attached below is the remaining answers
Can you use isentropic efficiency for a non-adiabatic compressor?
Can you use isothermal efficiency for an adiabatic compressor?
Why does the ceramic made from Thorium and Oxygen have the chemical ratio of 2 oxygen atoms to every thorium atom (ThO2)
Propane is to be compressed from 0.4 MPa and 360 K to 4 MPa using a two-stage compressor. An interstage cooler returns the temperature of the propane to 360 K before it enters the second compressor. The intermediate pressure is 1.2 MPa. Both adiabatic compressors have a compressor efficiency of 80%.(a) What is the work required in the first compressor per kg of propane?(b) What is the temperature at the exit of the first compressor?(c) What is the cooling requirement in the interstage cooler per kg of propane?
Answer:
a. 81 kj/kg
b. 420.625K
c. 101.24kj/kg
Explanation:
[tex]\frac{t2}{t1} =[\frac{p2}{p1} ]^{\frac{y-1}{y} }[/tex]
t1 = 360
p1 = 0.4mpa
p2 = 1.20
y = 1.13
substitute these values into the equation
[tex]\frac{t2}{360} =[\frac{1.20}{0.4} ]^{\frac{1.13-1}{1.13} }[/tex]
[tex]\frac{t2}{360} =[\frac{1.2}{0.4} ]^{0.1150}\\\frac{t2}{360} =1.1347[/tex]
when we cross multiply
t2 = 360 * 1.1347
= 408.5
a. the work required in the firs compressor
w=c(t2-t1)
c=1.67x10³
t1 = 360
t2 = 408.5
w = 1670(408.5-360)
= 1670*48.5
= 80995 J
= 81KJ/kg
b. [tex]n=\frac{t2-t1}{t'2-t1}[/tex]
n = 80%
t2 = 408.5
t1 = 360
0.80 = 408.5-360 ÷ t'2-360
[tex]0.80 =\frac{48.5}{t'2-360}[/tex]
cross multiply to get the value of t'2
0.80(t'2-360) = 48.5
0.80t'2 - 288 = 48.5
0.8t'2 = 48.5+288
0.8t'2 = 336.5
t'2 = 336.5/0.8
= 420.625
this is the temperature at the exit of the first compressor
c. cooling requirement
w = c(t2-t1)
= 1.67x10³(420.625-360)
= 1670*60.625
= 101243.75
= 101.24kj/kg
Can some people answer these questions so i can get to know the age group i an making my target market for DT-GSCE thankyou if you do my deadline is tomorrow :D
Answer:
I think you might have forgotten to post the problems
Water flows through a converging pipe at a mass flow rate of 25 kg/s. If the inside diameter of the pipes sections are 7.0 cm and 5.0 cm, find the volume flow rate and the average velocity in each pipe section.
Answer:
volumetric flow rate = [tex]0.0251 m^3/s[/tex]
Velocity in pipe section 1 = [tex]6.513m/s[/tex]
velocity in pipe section 2 = 12.79 m/s
Explanation:
We can obtain the volume flow rate from the mass flow rate by utilizing the fact that the fluid has the same density when measuring the mass flow rate and the volumetric flow rates.
The density of water is = 997 kg/m³
density = mass/ volume
since we are given the mass, therefore, the volume will be mass/density
25/997 = [tex]0.0251 m^3/s[/tex]
volumetric flow rate = [tex]0.0251 m^3/s[/tex]
Average velocity calculations:
Pipe section A:
cross-sectional area =
[tex]\pi \times d^2\\=\pi \times 0.07^2 = 3.85\times10^{-3}m^2[/tex]
mass flow rate = density X cross-sectional area X velocity
velocity = mass flow rate /(density X cross-sectional area)
[tex]velocity = 25/(997 \times 3.85\times10^{-3}) = 6.513m/s[/tex]
Pipe section B:
cross-sectional area =
[tex]\pi \times d^2\\=\pi \times 0.05^2= 1.96\times10^{-3}m^2[/tex]
mass flow rate = density X cross-sectional area X velocity
velocity = mass flow rate /(density X cross-sectional area)
[tex]velocity = 25/(997 \times 1.96\times10^{-3}) = 12.79m/s[/tex]
Safety Issues for Operators of Oil and Gas Exploiting Equipment when working off rigs
Answer:
Safety First, Safety Always. Safety stands out as a core value for the oil and natural gas industry, embedded in every process and decision for operations. The oil and natural gas industry and the federal government are working together to continuously improve the safety of offshore operations. ...
A ceramic specimen with an elastic modulus of 300 GPa is under a tensile stress of 800 MPa. Will it fracture if its most severe flaw is an internal crack of 0.30 mm long with a tip radius of curvature in the amount of 0.0015 mm? Please justify your conclusion. (Hint: Compare the largest stress in the specimen around the crack to the theoretical strength which is roughly E/10).
Answer:
16Gpa < 30 Gpa
there would be no fracture
Explanation:
fracture can occur if the maximum strength at the top of the biggest flaw is more than the theoretical fracture
to get the theoretical strength =
e/10 = 300/10
= 30 Gpa
we get the magnitude at the buggest flaw
= 2σ√a/ρt
σ = 800
ρτ = 0.0015
a= 0.3/2
[tex]=2*800\sqrt{\frac{\frac{0.3}{2} }{0.0015} }[/tex]
= [tex]=2*800*\sqrt{100} \\=2*800*10\\=16000MPa[/tex]
= 16Gpa < 30 Gpa
the fracture is not going to happen given that the maximum strenght is smaller than the theoretical fracture strength.
A cylindrical space capsule lands in the ocean. This capsule is 2.44 m long, 1.10 m in diameter, and weighted at one end so that it floats with its long central axis vertical and 0.820 m of its length above the water surface. The mass density of sea water is 1025 kg/m3.What is the magnitude of the buoyant force exerted on the capsule?
Answer:
The correct answer is "15456.8 N".
Explanation:
According to the question,
The inside volume will be:
= [tex]3.14\times (\frac{1.1}{2} )^2\times (2.44-0.82)[/tex]
= [tex]3.14\times \frac{1.21}{4}\times 1.62[/tex]
= [tex]3.14\times 03025\times 1.62[/tex]
= [tex]1.538757 \ m^3[/tex]
hence,
The buoyant force will be:
= [tex]V\times Pw\times g[/tex]
= [tex]1.538757\times 1025\times 9.8[/tex]
= [tex]15456.8 \ N[/tex]
A particle which moves in two-dimensional curvilinear motion has coordinates in millimeters which vary with time t in seconds according to X=2t^2 +3t–1 and y = 5t - 2. Determine the coordinates of the center of curvature C at time t = 1s.
Answer:
The answer is "22.501,-22.899"
Explanation:
Just as in the previous problems find the angle the velocity makes with the x-axis and radius of curvature.
[tex]x= 2t^2 + 3t — 1\\\\y=5t-2\\\\x=4t+3\\\\y=5\\\\\tan \alpha (t = 1) =\frac{y}{x}=\frac{5}{4+3}=\frac{5}{7} \to alpha=35.54^{\circ}\\\\[/tex]
For the radius of curvature, we can use the expression from the last two problems, but first express the position and derivatives as y(x).
[tex]y(x)=2(\frac{y+2}{5})^2+3(\frac{y+2}{5})-1=\frac{1}{25}(2y^2+23y+13)\\\\y'(x)=\frac{1}{25}(4y+23)\\\\y''(x)=\frac{4}{25}\\\\\rho(t=1)=\frac{[1+(\frac{dy}{dx})^2]^{\frac{3}{2}}}{\frac{d^2y}{dx^2}}=\frac{(1+(\frac{35}{25})^2)^{\frac{3}{2}}}{4}25=31.828[/tex]
The position for the center of the radius of curvature [tex]\vec{r}[/tex], (finding this expression is easy and is left as an exercise for the reader.)
[tex]\to \vec{r} = \hat{x}(x + \rho \sin \alpha) + \hat{y}(y- \rho \cos \alpha)\\\\= (4 + 18.501, 3-25.899)\\\\=(22.501, -22.899)[/tex]
A 2.0-in-thick slab is 10.0 in wide and 12.0 ft long. Thickness is to be reduced in three steps in a hot rolling operation. Each step will reduce the slab to 75% of its previous thickness. It is expected that for this metal and reduction, the slab will widen by 3% in each step. If the entry speed of the slab in the first step is 40 ft/min, and roll speed is the same for the three steps, determine: (a) length and (b) exit velocity of the slab after the final reduction
Answer:
26.02 ft
86.7690 ft/min
Explanation:
After 3 steps
0.75³(2.0 thickness)
T = 0.84375
W = (1+0.03)³10
= 10.92727 inches
A To get length
2.0 x 10 x 12 x12 = 0.84375 x 10.92727x lf
= 2880 = 9.21988Lf
Lf = 2880/9.21988
= 312.368 inches
Convert to feet
322.368 x 0.0833
= 26.02 ft
B.
= 2 x 10 x 40 = 0.84375 x 10.92727 x vf
800 = 9.21988vf
Vf = 800/9.21988
Vf = 86.7690 ft/min
Even though the content of many alcohol blends doesn’t affect engine drive ability using gasoline with alcohol in warm weather may cause
are there engineering students here?
Technician A says white smoke in the exhaust of a diesel engine can be the result of a cylinder misfire in a warm engine. Technician B says blue smoke in the exhaust of a diesel engine can be caused by scored cylinder walls. Who is correct?
Answer:
Both
Explanation:
Because of water, fuel does not burn completely. This brings about water fumes that are white in color and looks like white smoke. If engine is cold and water is heating, it leads to steam formation like water vapor. The white times are because of not firing properly in the heated engine. Technician A is right.
Blue fine is caused by this scoring. It is also caused by dirty oil. Technician b is right too
The net potential energy between two adjacent ions, EN, may be represented by Where A, B, and n are constants whose values depend on the particular ionic system. Calculate the bonding energy E0 in terms of the parameters A, B, and n using the following procedure:
a. Differentiate EN with respect to r, and then set the resulting expression equal to zero, since the curve of EN versus r is a minimum at Eo.
b. Solve for r in terms of A, B, and n, which yields r0, the equilibrium interionic spacing.
c. Determine the expression for E0 by substitution of r0 into the above equation for EN.
How to Cancel prescription
The purpose of the international residential code is to
Answer:
The International Building Code (IBC) is a model code that provides minimum requirements to safeguard the public health, safety and general welfare of the occupants of new and existing buildings and structures.
Explanation:
An engineer is applying dimensional analysis to study the flow of air through this horizontal sudden contraction for the purpose of characterizing the pressure drop. The flow is being modeled as constant density and steady. What is the functional relationship of the variables that characterize this situation
Answer:
The answer is " [tex]\Delta p = f(V1, p, V2, d, D, L)[/tex]"
Explanation:
Please find the complete question in the attached file.
Its change in temperature in pipes depends on rate heads and loss in pipes owing to pipe flow, contractual loss, etc.
The temperature change thus relies on V1 v2 p d D L.
Other than applying a strain to the gauge, what is the primary external/environmental factor that will influence the readings of a strain gauge?
Answer:
Temperature of the environment
Explanation:
Aside the strain applied to the strain gauge there are several other factors that might affect the reading of the strain gauge, and they are either external or internal factors like ; resistivity, length, and material of the strain gauge.
But the primary external factor is the Temperature of the environment when reading is taken
Determine the horsepower required to compress 1 lbm/min of ethylene oxide from 70 oF and 1 atm to 250 psia. The compressor has an efficiency of 75%. The molar heat capacity of ethylene oxide is given by Cp
Complete Question:
Problem 8 Determine the horsepower required to compress 1 lbm/min of ethylene oxide from 70 °F and 1 atm to 250 psia. The compressor has an efficiency of 75%. The molar heat capacity of ethylene oxide is given by
C_p=10.03+0.0184T C_p[=]Btu/lbmole- "F ; T[=] °F C,
Answer:
[tex]P'=0.377hp[/tex]
Explanation:
From the question we are told that:
Initial Temperature T_1=70 F
Final Temperature [tex]T_2=250pisa =114.94F[/tex]
Efficiency [tex]E=75\%=0.75[/tex]
Generally the equation for Work-done is mathematically given by
[tex]W=\int C_pT[/tex]
[tex]W=10.03(114.94-70 )+0.0184((114.94)^2-70^2 )[/tex]
[tex]W=527.21btu/ibmole[/tex]
[tex]W=11.982btu/ibm[/tex]
Generally the equation for Efficiency is mathematically given by
[tex]E=\frac{isotropic Power}{Actual P'}[/tex]
[tex]E=\frac{P}{P'}[/tex]
Since
Isotropic Power
[tex]P=0.0167*11.982btu/ibm[/tex]
[tex]P=0.2btu/s[/tex]
Therefore
[tex]P'=\frac{0.2}{0.75}[/tex]
[tex]P'=0266btu/s[/tex]
Since
[tex]1btu/s=1.4148hp[/tex]
Therefore
[tex]P'=0.377hp[/tex]
Type the correct answer in each box. Spell all words correctly. According to the priority matrix, which tasks should an entrepreneur complete first? According to the priority matrix, entrepreneurs should first complete tasks that are blank and important.
Answer:
Development of creative and develop ideas
Explanation:
First task as an entrepreneur is to be creative and develop ideas. The person must design the product based on which he will develop the business strategy.
The remaining activities such as marketing, fund raising, recruitment etc. comes at a later stage.
to check for ripple voltage from the alternator, connect a digital multimeter and select
Answer:
isn't it summer? sjsushsiansudndd
An AM signal having a carrier frequency of 460 kHz is to be mixed with a local oscillator signal at a frequency of 1135 kHz. What does the output of the IF amplifier consist of
Answer:
the output of the IF amplifier consist of 675 kHz
Explanation:
Given the data in the question;
AM signal carrier frequency [tex]_{RF[/tex] = 460 kHz
Local oscillator frequency[tex]_{lo[/tex] = 1135 kHz
Now, The output of the IF amplifier consists of difference of local oscillator frequency & AM carrier signal frequency;
FREQUECY[tex]_{IF[/tex] = FREQUECY[tex]_{lo[/tex] - FREQUECY[tex]_{RF[/tex]
so we substitute in our given values
FREQUECY[tex]_{IF[/tex] = 1135 kHz - 460 kHz
FREQUECY[tex]_{IF[/tex] = 675 kHz
Therefore, the output of the IF amplifier consist of 675 kHz