⭐ 𝑬𝒍𝒆𝒄𝒕𝒓𝒊𝒄 𝒄𝒂𝒓 𝑫𝒖𝒃𝒊𝒏𝒂 𝑬𝒗𝒐 ⭐

Task 8.2

During the movement, the car wheel rose by 10 cm from the normal position. At what angle did the drive shaft rise if its length is 70 cm?

In an isosceles triangle ABC with base AC=10 cm and sides AB=BC=70 cm, angle B (∠ABC) can be found using the cosine theorem:

cos(∠B) = AB^2+BC^2-AC^2/(2·AB·BC)

AC = 10 cm.

AB = BC = 70 cm.

cos(∠B) = 9700/9800 = 0,9898

∠B = arccos(0,9898) ≈ 8°

Task 9.1

The steering knuckle levers are subject to a force of 230 N from the steering gear. What is the moment of force turning the car wheels if the length of the steering knuckle lever is 13 cm?

M = F·r

M – moment of force, N·m;

F – force, N;

r – lever length, m

M = 230 · 0,13 ≈ 30 N·m

Task 9.2

The knuckle of the car suspension is made of aluminum alloy with a density of 2,7 g/cm^3. What is the mass of the knuckle (kg) if its volume is 740 cm^3?

m = V · p

m – mass, kg;

V – volume, m^3;

p – density, kg/m^3

m = 2,7 · 740 = 1998 g ≈ 2 kg

Task 10.1

The outer radius of the wheel hub is 9 cm. What is the centripetal acceleration at the outer radius of the hub if the rotation speed at this point is 20 km/h?

a = V^2 / r

a - centripetal acceleration, m/s^2;

V – speed, m/s;

r – radius, m

V = 20 km/h = 20 · 1000 / 3600 = 5,56 m/s

a = (5,56)^2 / 0,09 = 343,44 m/s^2

Task 10.2

A rolling bearing is installed in the wheel hub. The contact area of ​​the bearing with the hub is 25 cm^2. The force of gravity on the wheel is 3500 N. Determine the pressure at the contact of the bearing with the hub.

P = F / S

P – pressure, Pa;

F – force, N;

S – area, m^2

P = 3500 / 0,0025 = 1400000 N = 1,4 MPa

Task 11.1

Determine the relative elongation of a carbon fiber tube with a diameter of 3 cm and a wall thickness of 4 mm. The tensile force is 5000 N. The Young's modulus of carbon fiber is 2 10^11 Pa.

ε = F / E·S

S = π(D1/2)^2 – π(D2/2)^2

ε – relative elongation

F – tensile force, N

E – Young's modulus for carbon fiber, Pa

S – cross-sectional area, m^2;

D1 – outer diameter, m^2

D2 – inner diameter, m^2

S = 3,14 · (0,015^2 – 0,011^2) = 0,0003267 м^2

ε = 5000 / 2·10^11 · 0,0003267 = 7,65 · 10^-5

Task 11.2

The central tube of the anti-roll bar is made of carbon fiber with a shear modulus of 25 GPa. The length of the tube is 1,6 m, the outer diameter of the tube is 3 cm, the wall thickness of the tube is 4 mm. Determine the torque at the ends of the tube if the twist angle is 3°.

T = α · J · G / L

J = π/2 · (r1^4 – r2^4)

α = 3°· π / 180

T – torque, N·m;

α – twist angle, rad;

J – polar moment of inertia of the tube cross-section, m^4;

G – shear modulus of carbon fiber, Pa;

L – tube length, m;

r1 – tube outer diameter, m;

r2 – tube outer diameter, m

α = 3°· 3,14 / 180 = 0,05236 rad

J = 3,14/2 · (0,015^4 – 0,011^4) = 5,6549·10^-8 m^4

T = 0,05236 · 5,6549·10^-8 · 25 · 10^9  / 1,6 = 74,02 / 1,6 = 46,3 N·m

Task 12.1

The microcomputer in the control unit consumes a current of 2,5 A at a voltage of 5 V. Determine the amount of electrical energy (kJ) consumed by the microcomputer during 6 hours of operation.

W = P · t

P = U · I

W – energy, J;

P – power, W;

t – operating time, sec;

U – voltage, V;

I – current, A

P = 5 · 2,5 = 12,5 W

W = 12,5 · 21600 = 270000 J = 270 kJ

Task 12.2

A car weighing 1200 kg drives under the control of an autopilot at a speed of 80 km/h, it brakes to a speed of 40 km/h in 3 seconds. Determine the force (kN) spent on braking.

F = m · a

a = (Vi – Vf) / t

F – force, N;

m – vehicle mass, kg;

a – acceleration, m/s^2;

Vi – initial speed, m/s;

Vf – final speed, m/s;

t – time, sec

a = (22,22 – 11,11) / 3 = 3,7 m/s^2

F = 1200 · 3,7 = 4440 Н = 4,44 kN

Task 13.1

An electric motor with 4 pole pairs is controlled by a speed controller. What current frequency should the speed controller produce for the motor to rotate at 1000 rpm?

f = n·p/60

f – current frequency, Hz;

n – rotor speed, rpm;

p – number of pole pairs

f = 1000·4 / 60 = 66,67 Hz

Task 13.2

The speed controller of an electric motor supplies a current of 150 A to a copper stator winding 40 meters long with a specific resistance of 0,01724 Ohm·mm^2/m. By how many degrees will the temperature of the stator winding change if the current is supplied for 5 seconds? The specific heat capacity of copper is 385 J/ kg·°C. The density of copper is 8,96 kg/m^3;

ΔT = Q/m⋅c

Q = I^2 ⋅ R ⋅ t

m = pm⋅V = pm⋅A⋅L​

R = p⋅L/A

ΔT – temperature change, °C;

Q – amount of heat released, J;

m – mass of copper winding, kg;

c – specific heat capacity of copper, 385 J/ kg⋅°C;

I – current, A;

R – winding resistance, Ohm;

t – current supply time, sec;

pm – copper density, 8,96 kg/m^3;

V – conductor volume, m^3;

A – cross-sectional area, mm^2;

L – winding length, m;

p – specific resistance of copper, 0,01724 Ohm⋅mm^2/m

R = 0,01724 ⋅ 40/35 = 0,0197 Ohm

Q = 150^2 ⋅ 0,0197 ⋅ 5 = 22500 ⋅ 0,0197 ⋅ 5 = 2212,125 J

m = 9,960 ⋅ 35 ⋅ 10^-6 ⋅ 40 = 12,54 kg

ΔT = 2212,125/12,54 ⋅ 385 = 0,458 °C

Task 13.3

The speed controller supplies a current of 50 A to the stator winding. The stator length is 25 cm. The number of winding turns is 800. Determine the magnetic induction.

B = µ ⋅ N ⋅ I / l

B - magnetic induction, T;

µ - magnetic constant, µ = 4π⋅10^-7 H/m;

N - number of turns;

I - current, A;

l - stator length, m

B = 4 ⋅ 3,14 ⋅ 10^-7 ⋅ 800 ⋅ 50 / 0,25 ≈ 0,2 T

Task 13.4

To smooth out pulsations and maintain a stable voltage of 400 V in the inverter circuit, a power capacitor with a capacity of 100 μF is used. Determine the energy stored in the capacitor.

W = C ⋅ U^2 / 2

W – energy, J

C – capacitance, F

U – voltage, V

W = 100 ⋅ 10^-6 ⋅ 400^2 / 2 = 8 J

Task 13.5

The regulator uses a MOSFET with a resistance of 0,005 Ohm. What is the heat loss if a current of 100 A passes through it?

P = I^2 ⋅ R

P - heat loss, W;

I - current, A;

R - resistance, Ohm

P = 100^2 ⋅ 0,005 = 50 W

Task 13.6

A capacitor with a capacity of 100 μF is connected to a circuit with a voltage of 400 V. What is the current flowing through the capacitor if the voltage changes at a constant rate of 200 V/s

I = C ⋅ dU/dt

I - current, A;

C - capacitance of the capacitor, F;

dU/dt - rate of change of voltage, V/s

I = 100 · 10^-6 ⋅ 200 = 0,02 A

Task 14.1

A slider potentiometer with a total resistance of 10 kOhm is connected to a voltage of 12 V. After moving the slider by 5 cm, the resistance of the potentiometer is 5 kOhm. Determine the voltage after moving the slider.

Vx = Vo · Rx / Ro

Vx - voltage after moving the slider, V;

Vo - total voltage, V;

Rx - resistance of the potentiometer section, Ohm;

Ro - total resistance, Ohm

Vx = 12 · 5 / 10 = 6 V

Task 14.2

A slider potentiometer with a total resistance of 10 kOhm is connected to a voltage of 12 V. Determine the resistance of the potentiometer at a voltage of 8 V.

Rx = Vx · Ro / Vo

Rx - resistance of the potentiometer section, Ohm;

Ro - total resistance, Ohm

Vx - voltage after the slider is shifted, V;

Vo - total voltage, V;

Vx = 8 · 10 / 12 ≈ 6,7 V

Task 15.1

An air conditioner cools the air in a car from 30 °C to 20 °C. The air flow rate is 0.5 kg/s. Find the power consumed by the air conditioning system if the specific heat capacity of air is c = 1005 J/(kg\K)

Q = m ⋅ c ⋅ ΔT

Q - power of the air conditioner, W;

m - mass flow rate of air, kg/s;

c - specific heat capacity of air, J/kg⋅K;

ΔT - change in temperature, K

Q = 0.5 ⋅ 1005 ⋅ (30 − 20) = 5025 ≈ 5.025 kW

Task 15.2

In a closed car with a volume of 3 m^3. At a pressure of 100 kPa there are 2.7 kg of nitrogen and 0.8 kg of oxygen. Assuming the gases are ideal, determine the temperature (°C) in the car.

T = PV/nR

n = n(N2) + n(O2)

T – temperature, °C;

P – pressure, Pa;

V – volume, m^3;

n – amount of substance, mol;

R – universal gas constant, 8.31 J/mol⋅K;

M(N2) = 0.028 kg/mol

M(O2) = 0.032 kg/mol

n(N2) = 2.7/0.028 ≈ 96.43 mol

n(O2) = 0.8/0.032 ≈ 25 mol

n = 96.43 + 25 = 121.43 mol

T = 100000 · 3/121.43 · 8.31 ≈ 297 K = 24 °C

Task 15.3

A passenger compartment heating radiator is made of aluminum. The heat transfer coefficient between the radiator and the air is 15 W/m^2 K. The surface area of ​​the radiator is 0.4 m^2. The temperature of the radiator wall is 90 °C. The air temperature in the passenger compartment is 20 °C. Determine the thermal power of the radiator.

Q = k · A · (Tr – Ta)

Q – thermal power, W;

k – heat transfer coefficient, W/m^2 K;

A – heat exchange area of ​​the radiator, m^2;

Tr – radiator temperature, °C;

Ta – temperature in the passenger compartment, °C

Q = 15 · 0.4 · (90 – 20) = 420 W

Task 16.1

What is the volume of a rectangular parallelepiped if its height is 0.2 m and the sides of the base are 1.5 m and 0.3 m?

V = a · b · h

V = 1.5 · 0.3 · 0.2 = 0.09 m^3

Task 16.2

An airbag located in the dashboard of a car deploys to a distance of 0.5 m with an acceleration of 300 m/s^2. How long does it take for the airbag to deploy?

t = (2S/a)^1/2

t – deployment time, s;

S – distance, m;

a – acceleration, m/s^2

t = (2 · 0.5 / 300)^1/2 = 0.058 s

Task 16.3

The car dashboard has a display with a brightness of 300 cd/m², which operates on a current of 1 A. The display consumes 12 W of power. Determine the voltage of the current?

V = P / I

V – voltage, V;

P – power, W;

I – current, A

V = 12 / 1 = 12 V

Task 17.1

The diameter of a homogeneous disk is 310 mm, the mass is 1.5 kg. Determine the moment of inertia of the disk relative to the axis passing through its center of mass.

I = (m · R^2) / 2

I – moment of inertia, kg·m^2;

m – mass, kg;

R – radius, m

I = (1.5·0.155^2) / 2 ≈ 0.018 kg·m^2

Task 17.2

A steering wheel with a diameter of 300 mm forms an arc of 15 cm when turning. At what angle did the steering wheel turn?

Ɵ = l / R

Ɵ – steering wheel angle, rad;

l – arc length, m;

R – radius, m

Ɵ = 0.15 / 0.15 = 1 rad ≈ 57.3°

Task 17.3

When steering, the driver applies a force of 10 N to a steering wheel with a diameter of 30 cm. Determine the moment of force on the steering wheel.

M = F · r

M = 10 · 0.15 = 1.5 N·m

Task 18.1

What is the current power in a three-phase network with a voltage of 400 V and a current of 30 A?

P = 3^1/2 · U · I

P = 3^1/2 · 400 · 30 ≈ 21 kW

Task 18.2

The resistivity of copper at 20 °C is 0.017 Ohm·mm^2/m. What is the resistivity of copper at 100 °C if the temperature coefficient of resistance of copper is 0.0039 °C^-1

pt = p · (1 + α · (T – 20))

pt - resistivity of copper at 100 °C

p - resistivity of copper at 20 °C

α - temperature coefficient of resistance of copper

T - temperature of copper

pt = 0.017 · (1 + 0.0039 · (100 – 20)) = 0.0223 Ohm·mm^2/m

Task 18.3

A copper wire with a cross-sectional area of ​​4 mm^2 has a resistance of 0.021 Ohm. The resistivity of copper at 20°C is 1.7⋅10⁻⁸ Ohm⋅m. What is the length of the wire?

l = RA/ρ

l - length of the wire, m;

R - resistance of the wire, Ohm;

A – cross-sectional area, m^2

ρ – resistivity of copper at 20°C, Ohm⋅m

l = (0.021 Ohm) × (4 × 10⁻⁶ m²) / (1.7 × 10⁻⁸ Ohm⋅m)

l = 84 × 10⁻⁶ m² / (1.7 × 10⁻⁸ Ohm⋅m)

l ≈ 4.94 m

Task 18.4

A wire with a cross-sectional area of ​​8 mm^2 is heated to 180 °C by the current passing through it. What should the cross-sectional area of ​​the wire be for the heating temperature to be 20 °C?

T2 = T1 · (S1/S2)^2

S2 = S1 · (T1/T2)^1/2

S2 = 8 · (180/20)^1/2 = 24 mm^2