Determining the power of an engine The power of each engine (mechanical) is written on its plate and is given in Hp or KW (1 hp = 0.736KW).
If it is selected less than required the engine will overload and burn out. Although much larger then elected the efficiency and the motor power factor is small, we have large starting current with larger ducts and insurance and of course increased installation costs
We give below uncomplicated calculation of engine power (eg cranes, hoists , pumps, machine tools, etc.)
Linear motion
According to this the motor load moves in a straight line
P (hp) = F * v / 75 * n and P (KW) = F * v / 102 * n
where:
F: the maximum lifting load in Kg
v: speed of the load in m / sec
n: total degree of efficiency of the driving device
When calculating the face lifts the weight of each person is calculated at 75kg
The maximum lifting load F = F΄ + QG where:
F΄: cabin weightQ: weight of peopleG : counterweight
The weight of the counterweight should be equal to G = F΄ + 1 / 2Q
In order for the engine not to work at full load and to make noise or heat we choose an engine with power 25% higher than the required
Example
Calculate the power 4-person elevator motor with chamber weight F΄ = 300Kg, moving at a speed of 0.6m / sec. The efficiency of the engine is n (kin) = 0.6 and the efficiency of the gearbox n (μ) = 0.5
The weight of the atoms Q = 4 people * 75Κg = 300KgB counterweight G = F΄ + 1 / 2Q = 300 + 1/2 * 300 = 450Kg
The maximum lifting load F = F΄ + QG = 300 + 300-450 = 150Kg
O total efficiency of the drive n = n (kin) * n (μ) = 0,6 * 0,5 = 0,3
So the required motor power will be:
Ρ (ΚW) = F * v / 102 * n = (150 * 0.6) / (102 * 0.3) = 2.94KW We
choose an engine with power 25% higher than required, ie 2.94+ (2.94 * 0.25) = 3.67KW or 5 HP
Pump applications
The required motor power will be: P = (ε * h * Q) / (n * 3600) where:
ε: the specific gravity of the fluid (N / m3) (9810N / m3 for water and 8000 for oil).
h: the geometric height hγ (suction height + discharge height) + friction height hτ in m (you can see the calculation of the friction heighthere ) Q: pump flow in m3 / hn: the total efficiency of the electric motor / pump system
If the specific gravity of the fluid is given in Kp / dm3) (0.98 Kp / dm3 for water and 0.8 Kp / dm3 for oil) and the flowQ in m3 / sec, then the formula will be: P = (9,8 * ε * h * Q) / n
Example
Water pumping unit (ε = 9810N / m3) operates with flow Q = 72 m3 / h. The geometric height of the installation (suction height and discharge height) is 9m and the frictions are equivalent to 1m. What is the engine power if the engine and pump efficiency degrees are respectively n (kin) = 0.8 and n (pump) = 0.7
The efficiency of the pump-motor system will be n = n (kin) * n (pump) = 0.8 * 0.7 = 0.56
P = (ε * h * Q) / (n * 3600) = ( 9810 * 10 * 72) / (0.56 * 3600) = 3503 W or 3.5 KW
Fan or air compressor
drive The motors for these applications are calculated based on the equation:
P = (27.7777 * Q * dp) / n where:
Q: the air supply in m3 / h.dp: the pressure difference in bar = 105 Pa (N / m2)
h: the total efficiency of the electric motor / fan assembly (or air compressor)
Rotary motion
During this useful work occurs in rotational energy (ie expressed with a torque M)
P (W) = (T * v) / (9.55 * n)
T: torque in Nm
v: number of revolutions per minute rpm / min
or: degree system performance
Example An
asynchronous three-phase short-circuit motor has a power factor of coefficient = 0.8 and a degree of efficiency n (kin) = 0.85. It rotates a machine tool that exerts at its output torque T = 450Nm with speed v = 30 rpm. The efficiency of the machine tool is n (erg) = 0.7. What should be the required engine power.
The required motor power will be P (W) = (T * v) / (9.55 * n)
The efficiency of the motor-machine tool system will be: n = n (kin) * n (erg) = 0, 85 * 0,7 = 0,595
So P (W) = (450 * 30) / (9,55 * 0,595) = 2375,8W
Calculation of normal operating current of different loads
When studying an installation and for the selection of line cross-sections and locking devices we must calculate the normal operating currents of the loads.
The loads can be single-phase or three-phase and can also be ohmic, inductive or capacitive.
A single-phase load is one that needs a phase and a neutral to operate (also all loads need to have a protection conductor which is not involved in the operation of the loads but only in our protection against electric shock).
A three-phase load is one that needs three phases or three phases and a neutral to operate.
An ohmic load is what basically consists of resistors, which, leaking from the electric current, heat up and produce useful work (they light our house, they cook our food, they heat us) and in which the current with the voltage is confocal (e.g. x electric water heater or a halogen lamp). That is, the voltage and the intensity form an angle 0ο and the power factor is equal to συνφ = συν0ο = 1
Inductive load is that which contains windings (coils). Such a charge can be a motor, a transformer, and generally a device that uses magnetism (electromagnetism more correctly) in order to move e.g. the washing machine bucket, or to turn the mains voltage to another lower one, so that the garden lighting installation is safer (from 230 volts to 42 volts), or to turn on a fluorescent lamp (start chuck) . At purely inductive load the voltage precedes the current by 90o. That is, the voltage and the intensity form an angle 90ο and the power factor is equal to συνφ = συν90ο = 0
Capacitive charge is that which consists mainly of capacitors (in which electric charge can be stored) and in which the voltage is delayed by a current of 90o. That is, the voltage and the intensity form an angle of 90ο and the power factor is equal to συνφ = συν90ο = 0
Usually the real consumers are a combination of two or all of the above three elements with one of the three "behaviors" being quantitatively superior to the others and so to give its own characteristics to the whole electrical device of the consumer.
Actual loads have all three of these "properties", and as a rule some of them are prevalent, e.g. In an electric stove (pine cone) the ohmic behavior prevails "overwhelmingly", but nevertheless there is some induction due to the fact that there are resistance wires in the form of a spiral and even wrapped around the ceramic of the "pine cone".
On the contrary, in a washing machine the inductive behavior prevails due to the bucket motor (and this only when the resistance that heats the water does not work). But beyond that, there is the ohmic resistance of the conductors that make up the windings of its engine.
In them the voltage and the intensity form an angle 0ο <φ <90ο with a power factor 0 <συνφ <1
So let's see how we calculate the operating current of the above loads: Usually in an electrical installation we have single-phase and three-phase ohms (household appliances and lamps) and inductive electrical loads (motors and transformers).
1. SINGLE-PHASE LOAD
1a) Omic single-phase load
I = P / V where
I: operating current
P: load power
V: operating voltage of the load
Eg a 4 KW electric water heater will leak current of current I = P / V = 4000/230 = 17,39 Α
The above formula also applies to direct current
1β) Inductive single-phase load
Ι = Ρ / (V * συνφ) where
I: operating current
P: load power
V: operating voltage of the load
coefficient: power factor
Eg a device 800 W and power factor 0.95 will leak current intensity I = P / (V * coefficient) = 1000 / (230 * 0.95) = 4.57A
We must pay attention to case of a single-phase induction motor where we are given in its label the mechanical power Pmix (the power that comes out on its axis) and not the electrical power Rill that absorbs from the network. In this case the above formula becomes:
Ι = Ρμηχ / (V * συνφ * n) where
Ι: operating current
Ρμχ: the mechanical power of the motor
V: operating voltage of the motor
συνφ: power factor
n: degree of efficiency of the motor
For example, a single-phase induction motor with a power of 2 KW (mechanical), a power factor of 0.85 and an efficiency of 0.9 will have a normal operating current I = Pmx / (V * synf * n) = 2000 / (230 * 0.85 * 0.9) = 11.36A
2. THREE-PHASE LOAD
2a) Omic three-phase load
I = P / (1.73 * V) where
I: operating current
P: load power
V: operating voltage of the load
Eg a three-phase 10 KW power cooker will leak from current I = P / (1.73 * V) = 10000 / (1.73 * 400) = 14.45 A (in each phase)
2b) Inductive three-phase load
I = P / ( 1.73 * V * sync) where
I: operating current
P: load power
V: operating voltage of the load
sync: power factor
eg a device 10000 W and power factor 0.95 will leak current intensity I = P / (1.73 * V * conf) = 10000 / (1.73 * 400 * 0.95) = 15.2A
Will we must be careful in the case of a three-phase induction motor where we are given in its label the mechanical power Pmix (the power that comes out on its axis) and not the electrical power Rill that absorbs from the network. In this case the above formula becomes:
Ι = Ρμηχ / (1,73 * V * συνφ * n) where
Ι: operating current
Ρμχ: the mechanical power of the motor
V: operating voltage of the motor
συνφ: power factor
n: degree of efficiency of the engine
For example, a three-phase induction motor with a power of 10 KW (mechanical), a power factor of 0.85 and an efficiency of 0.9 will have a normal operating current I = Pmax / (1.73 * V * co * n) = 10000 / (1 , 73 * 400 * 0.85 * 0.9) = 18.89A
For three-phase motors there is an approximate formula to find the normal operating current which applies more accurately to the medium power.
I (A) = 1.5 * P (HP) or I (A) = 2 * P (KW)
If it is selected less than required the engine will overload and burn out. Although much larger then elected the efficiency and the motor power factor is small, we have large starting current with larger ducts and insurance and of course increased installation costs
We give below uncomplicated calculation of engine power (eg cranes, hoists , pumps, machine tools, etc.)
Linear motion
According to this the motor load moves in a straight line
P (hp) = F * v / 75 * n and P (KW) = F * v / 102 * n
where:
F: the maximum lifting load in Kg
v: speed of the load in m / sec
n: total degree of efficiency of the driving device
When calculating the face lifts the weight of each person is calculated at 75kg
The maximum lifting load F = F΄ + QG where:
F΄: cabin weightQ: weight of peopleG : counterweight
The weight of the counterweight should be equal to G = F΄ + 1 / 2Q
In order for the engine not to work at full load and to make noise or heat we choose an engine with power 25% higher than the required
Example
Calculate the power 4-person elevator motor with chamber weight F΄ = 300Kg, moving at a speed of 0.6m / sec. The efficiency of the engine is n (kin) = 0.6 and the efficiency of the gearbox n (μ) = 0.5
The maximum lifting load F = F΄ + QG = 300 + 300-450 = 150Kg
O total efficiency of the drive n = n (kin) * n (μ) = 0,6 * 0,5 = 0,3
So the required motor power will be:
Ρ (ΚW) = F * v / 102 * n = (150 * 0.6) / (102 * 0.3) = 2.94KW We
choose an engine with power 25% higher than required, ie 2.94+ (2.94 * 0.25) = 3.67KW or 5 HP
Pump applications
The required motor power will be: P = (ε * h * Q) / (n * 3600) where:
ε: the specific gravity of the fluid (N / m3) (9810N / m3 for water and 8000 for oil).
h: the geometric height hγ (suction height + discharge height) + friction height hτ in m (you can see the calculation of the friction heighthere ) Q: pump flow in m3 / hn: the total efficiency of the electric motor / pump system
If the specific gravity of the fluid is given in Kp / dm3) (0.98 Kp / dm3 for water and 0.8 Kp / dm3 for oil) and the flowQ in m3 / sec, then the formula will be: P = (9,8 * ε * h * Q) / n
Example
Water pumping unit (ε = 9810N / m3) operates with flow Q = 72 m3 / h. The geometric height of the installation (suction height and discharge height) is 9m and the frictions are equivalent to 1m. What is the engine power if the engine and pump efficiency degrees are respectively n (kin) = 0.8 and n (pump) = 0.7
The efficiency of the pump-motor system will be n = n (kin) * n (pump) = 0.8 * 0.7 = 0.56
P = (ε * h * Q) / (n * 3600) = ( 9810 * 10 * 72) / (0.56 * 3600) = 3503 W or 3.5 KW
Fan or air compressor
drive The motors for these applications are calculated based on the equation:
P = (27.7777 * Q * dp) / n where:
Q: the air supply in m3 / h.dp: the pressure difference in bar = 105 Pa (N / m2)
h: the total efficiency of the electric motor / fan assembly (or air compressor)
Rotary motion
During this useful work occurs in rotational energy (ie expressed with a torque M)
P (W) = (T * v) / (9.55 * n)
T: torque in Nm
v: number of revolutions per minute rpm / min
or: degree system performance
asynchronous three-phase short-circuit motor has a power factor of coefficient = 0.8 and a degree of efficiency n (kin) = 0.85. It rotates a machine tool that exerts at its output torque T = 450Nm with speed v = 30 rpm. The efficiency of the machine tool is n (erg) = 0.7. What should be the required engine power.
The required motor power will be P (W) = (T * v) / (9.55 * n)
The efficiency of the motor-machine tool system will be: n = n (kin) * n (erg) = 0, 85 * 0,7 = 0,595
So P (W) = (450 * 30) / (9,55 * 0,595) = 2375,8W
Calculation of normal operating current of different loads
When studying an installation and for the selection of line cross-sections and locking devices we must calculate the normal operating currents of the loads.
The loads can be single-phase or three-phase and can also be ohmic, inductive or capacitive.
A single-phase load is one that needs a phase and a neutral to operate (also all loads need to have a protection conductor which is not involved in the operation of the loads but only in our protection against electric shock).
A three-phase load is one that needs three phases or three phases and a neutral to operate.
An ohmic load is what basically consists of resistors, which, leaking from the electric current, heat up and produce useful work (they light our house, they cook our food, they heat us) and in which the current with the voltage is confocal (e.g. x electric water heater or a halogen lamp). That is, the voltage and the intensity form an angle 0ο and the power factor is equal to συνφ = συν0ο = 1
Inductive load is that which contains windings (coils). Such a charge can be a motor, a transformer, and generally a device that uses magnetism (electromagnetism more correctly) in order to move e.g. the washing machine bucket, or to turn the mains voltage to another lower one, so that the garden lighting installation is safer (from 230 volts to 42 volts), or to turn on a fluorescent lamp (start chuck) . At purely inductive load the voltage precedes the current by 90o. That is, the voltage and the intensity form an angle 90ο and the power factor is equal to συνφ = συν90ο = 0
Capacitive charge is that which consists mainly of capacitors (in which electric charge can be stored) and in which the voltage is delayed by a current of 90o. That is, the voltage and the intensity form an angle of 90ο and the power factor is equal to συνφ = συν90ο = 0
Usually the real consumers are a combination of two or all of the above three elements with one of the three "behaviors" being quantitatively superior to the others and so to give its own characteristics to the whole electrical device of the consumer.
Actual loads have all three of these "properties", and as a rule some of them are prevalent, e.g. In an electric stove (pine cone) the ohmic behavior prevails "overwhelmingly", but nevertheless there is some induction due to the fact that there are resistance wires in the form of a spiral and even wrapped around the ceramic of the "pine cone".
On the contrary, in a washing machine the inductive behavior prevails due to the bucket motor (and this only when the resistance that heats the water does not work). But beyond that, there is the ohmic resistance of the conductors that make up the windings of its engine.
In them the voltage and the intensity form an angle 0ο <φ <90ο with a power factor 0 <συνφ <1
So let's see how we calculate the operating current of the above loads: Usually in an electrical installation we have single-phase and three-phase ohms (household appliances and lamps) and inductive electrical loads (motors and transformers).
1. SINGLE-PHASE LOAD
1a) Omic single-phase load
I = P / V where
I: operating current
P: load power
V: operating voltage of the load
Eg a 4 KW electric water heater will leak current of current I = P / V = 4000/230 = 17,39 Α
The above formula also applies to direct current
1β) Inductive single-phase load
Ι = Ρ / (V * συνφ) where
I: operating current
P: load power
V: operating voltage of the load
coefficient: power factor
Eg a device 800 W and power factor 0.95 will leak current intensity I = P / (V * coefficient) = 1000 / (230 * 0.95) = 4.57A
We must pay attention to case of a single-phase induction motor where we are given in its label the mechanical power Pmix (the power that comes out on its axis) and not the electrical power Rill that absorbs from the network. In this case the above formula becomes:
Ι = Ρμηχ / (V * συνφ * n) where
Ι: operating current
Ρμχ: the mechanical power of the motor
V: operating voltage of the motor
συνφ: power factor
n: degree of efficiency of the motor
For example, a single-phase induction motor with a power of 2 KW (mechanical), a power factor of 0.85 and an efficiency of 0.9 will have a normal operating current I = Pmx / (V * synf * n) = 2000 / (230 * 0.85 * 0.9) = 11.36A
2. THREE-PHASE LOAD
2a) Omic three-phase load
I = P / (1.73 * V) where
I: operating current
P: load power
V: operating voltage of the load
Eg a three-phase 10 KW power cooker will leak from current I = P / (1.73 * V) = 10000 / (1.73 * 400) = 14.45 A (in each phase)
2b) Inductive three-phase load
I = P / ( 1.73 * V * sync) where
I: operating current
P: load power
V: operating voltage of the load
sync: power factor
eg a device 10000 W and power factor 0.95 will leak current intensity I = P / (1.73 * V * conf) = 10000 / (1.73 * 400 * 0.95) = 15.2A
Will we must be careful in the case of a three-phase induction motor where we are given in its label the mechanical power Pmix (the power that comes out on its axis) and not the electrical power Rill that absorbs from the network. In this case the above formula becomes:
Ι = Ρμηχ / (1,73 * V * συνφ * n) where
Ι: operating current
Ρμχ: the mechanical power of the motor
V: operating voltage of the motor
συνφ: power factor
n: degree of efficiency of the engine
For example, a three-phase induction motor with a power of 10 KW (mechanical), a power factor of 0.85 and an efficiency of 0.9 will have a normal operating current I = Pmax / (1.73 * V * co * n) = 10000 / (1 , 73 * 400 * 0.85 * 0.9) = 18.89A
For three-phase motors there is an approximate formula to find the normal operating current which applies more accurately to the medium power.
I (A) = 1.5 * P (HP) or I (A) = 2 * P (KW)
