What kind of wire should be used to install a 15kw motor Is it OK to use 6 square meters or more?

What kind of wire should be used to install a 15kw motor Is it OK to use 6 square meters or more?

15kw triangle connection current is:
15*1000/（380*1.732*0.85*0.85）=31.5A
Wire with 4-6 square copper wire can be
An electric iron with a resistance value of 220 Ω is connected to a 220 V power supply. When it is powered on for 1 hour, how many joules does the current work, and the power consumption is kilowatt hours
How many hours can a 220 V, 0.2 a lamp shine normally with 0.88 kilowatt hours of electric energy
If the voltage at both ends of the conductor resistance is 6V and the current passing through the conductor is 0.3A, then the conductor resistance is several Ω. If the power on time is 1min, how much Joule is the current passing through the conductor
1、 An electric iron with a resistance value of 220 Ω is connected to a 220 V power supply. When it is powered on for 1 hour, how many joules does the current work and how much energy is consumed____ Kilowatt hour
Electric irons are pure resistance appliances
∴W=Pt=U²·t/R=(220V)²×1h/220Ω=0.22kwh=7.92×10^5J
A: the work done by current is 7.92 × 10 ^ 5J, and the power consumption is 0.22kwh
2、 How many hours can a 220 V, 0.2 a electric lamp shine normally with 0.88 kwh of electric energy?
∵W=Pt
∴t=W/P=W/UI=0.88kwh/220V×0.2A=20h
Answer: the lamp can shine normally for 20 hours
3、 If the voltage at both ends of the conductor resistance is 6V and the current passing through the conductor is 0.3A, how many Ω is the conductor resistance? If the power on time is 1min, how much Joule is the current passing through the conductor?
The conductor is pure resistance
∴R=U/I=6V/0.3A=20Ω
∵1min=60s
∴W=UIt=6V×0.3A×60s=108J
A: the resistance of the conductor is 20 Ω; the work done by the current is 108j
The above is the standard process, if you have any questions, please ask
I hope my answer will help you_ ∩)O
The circuit is connected in series with a constant value resistor. Is the voltage at both ends of the constant value resistor equal to the power supply voltage
If it is a question, do not consider the internal resistance of the power supply. If it is considered, then the voltage at both ends of the constant resistance is less than the power supply voltage; if it is not considered, it is an ideal situation, and the voltage at both ends of the constant resistance is equal to the power supply voltage
There must be internal resistance in life
U setting = u electricity * r setting / (r setting + R inner)
If the power supply has no internal resistance, that is, r = 0, then u = u, that is, the voltage at both ends of the constant resistance is equal to the power supply voltage.
In addition, u set
How many kilowatts of electricity equals one watt?
Watt is the unit of power, degree is the unit of electric energy. They can't be directly converted. A watt is equal to 1000 hours of electricity consumption
A 200V '300W electric iron is connected to a 220V AC power supply and used for 30 minutes. Try to calculate: (1) how many joules does the electric iron consume? (2)
When I was studying, I didn't make this joke seriously. Please forgive me! I'd better use the formula and the result! & nbsp;
Time: T = 30 minutes = 1800 seconds, energy: w = Pt = 300W * 1800 seconds = 540000 joules, please click to adopt. If it is really 200V, the process is as follows: (at this time, the actual voltage is higher than the rated voltage
A copper wire is 0.56mm in diameter and 3000mm in length?
It's urgent
R=ρ*L/S^2,
Here, ρ = 0.017 Ω (length 1 meter, cross section 1 square millimeter),
L = 3M, s = π * 0.28 * 0.28 = 0.246300864041439789547124124911m2,
Therefore, r = 0.20706382902261893174012938602036 Ω
R=pl/S
p=1.7 × 10-8Ω m
s=3.14*0.56*0.56/4/1000/1000
R=1.7*0.00000001*3000/1000/(3.14*0.56*0.56/4/1000/1000)=0.000000051/0.000000246176=0.2072Ω
R = ρ * L / S ^ 2, ρ = 0.017 Ω, 1 meter long, section
R = 0.02856 Ω
How much electricity is one watt?
I'm very interested in this problem
Watt is the unit of power, while degree is the unit of electric work. They cannot be converted to each other. If the power is multiplied by another time, the result is work. The units of work are Joule and kilowatt hour. Kilowatt hour is usually called "degree". The relationship between them is as follows: 1 joule = 1 watt · s, 1 kilowatt hour = 1 kilowatt · H
Can't convert to each other ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~!!!!!!!!!!!!
1 / 1000 kWh
1 degree = 1000 W / h
1000 W / h = 1 degree
1000 W / h = 1 degree
Watt is the unit of power, degree is the unit of electric energy, one thousand watt hours = one degree of electricity
1 W = 1 / 1000 degree
Big brother, their units are different, Watt is power unit, degree is electric work unit!!
1000 W / h = 1 degree
Known a 220 V 300 W electric iron, connected to the 220 V power supply to work for 30 minutes, try to calculate: what is the electric energy of the electric iron
First of all, we should understand the working principle of the electric iron: generally, there is a constant temperature switch. When the temperature does not reach the set value, the heater will work at a power consumption of 0.3 kwh / h. When the temperature reaches the set value, the heater will stop using electricity. When the temperature drops a little, it will start heating again. So, once the temperature is cut off, the power consumption of your 300W iron will not exceed 0.15 kwh
The maximum power consumption P = w x H = 300 x 0.5 = 150 watt hour, that is 0.15 degree electricity. Because the constant temperature insulation starts after reaching the preset temperature, the heat dissipation power consumption at this time is difficult to calculate, so the actual power consumption can only be said to be less than 0.15 degree.
ha-ha... Will ask questions in Baidu, but will not do this simple exercise..
W = Pt = 0.3kw * 0.5h = 1.5kwh, i.e. 1.5 kwh.
300W by 30 by 60s
A ton of copper wire, its diameter is 1 mm, find the length of copper wire
The teacher will check it tomorrow
The diameter is 1 mm = 0.001 M,
Radius: r = 0.0005m
Therefore, the cross-sectional area s = 3.14 × R ^ 2 = 7.85 × 10 ^ - 7m2
Mass m = 1t = 1000kg
ρ＝m/v,
v=m/ρ,ρ＝8900kg/m3,v=Sh
So: SH = m / ρ
That is: H = m / ρ s = 1000 / (8900 × 7.85 × 10 ^ - 7m2)
h=143133.2m
Let me tell you the calculation steps. Specifically, you can calculate by yourself. First, check the density of copper, and then calculate the volume through the formula: ρ = m / v. Then, the height h is calculated by the volume formula v = π R ^ 2 × h. The cross-sectional area s = π R ^ 2, r = 1 / 2 diameter, π is about 3.1415
m=ρV
V=SL=π(d/2)^2*L
The results show that: l = 4m / (π ρ d ^ 2) = 4 * 1000 / (3.14 * 8.9 * 10 ^ 3 * 0.001 ^ 2) = 1.43 * 10 ^ 5m
If you know the density,
H = mass / (density * cross-sectional area) the density of copper is 8.9 × 10 ~ (- 179); kg / (- 179); but the cross-sectional area is Can you be more detailed? 1000 kg / (8900 * Pai * 0.000001) is about 35765.16m, isn't it 143133.2m? You're a bit off the mark. The diameter is used as the radius... Well, anyway, that answer is wrong. It's 143
If you know the density,
H = mass / (density * cross-sectional area): the density of copper is 8.9 × 10 ᦉ 179; kg / ᦉ 179; but the cross-sectional area is Can you be more detailed?
1000 watts of electrical appliances, a few hours of electricity
One hour per kilowatt hour