Answer:
A 3-phase, 50 Hz, 110 KV overhead line has conductors
Explanation:
hope it will helps you
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]
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
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
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:
which of the following tells the computer wha to do
operating system
the ROM
the motherboard
the monitor
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 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
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
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