⭐ 𝑬𝒍𝒆𝒄𝒕𝒓𝒊𝒄 𝒄𝒂𝒓 𝑫𝒖𝒃𝒊𝒏𝒂 𝑬𝒗𝒐 ⭐

Task 1.1

The monocoque is made of carbon prepreg. The length of one prepreg roll is 50 m. The roll width is 1,25 m. The density of carbon prepreg is 422 g/m^2. What is the mass of one prepreg roll (kg)?

The mass of the roll is determined by the formula

m = S · p

m = 422 · (50 · 1,25) = 26375 (g)

m = 26375 / 1000 = 26,4 (kg)

Task 1.2

What is the specific strength of carbon fiber if its ultimate strength is 1300 MPa, density is 1,6 kg/m^3

R = G / p

R = 1300 / 1,6 = 812,5 MPa / kg

Task 2.1

The car frame is made of a round pipe. The outer diameter of the pipe is 4 cm. The inner diameter of the pipe is 3 cm. The length of the pipe is 20 m. What is the volume of the pipe?

The volume of the pipe is determined by the formula

V = π · h · (R^2 – r^2)

V - volume of the pipe;

π - mathematical constant approximately 3,14;

h - length of the pipe, m;

R - outer radius of the pipe, m;

r - inner radius of the pipe, m.

V = 3,14 · 20 · (0,02^2 – 0,015^2) = 3,14 · 20 · 0,000175 = 0,011 м^3

Task 3.1

The battery consists of 16 modules connected in series. One module consists of 60 battery groups connected in parallel. Each group has 6 batteries connected in series. The voltage of one battery is 4.17 V. What is the voltage of the battery?

Ub = 16 · Um

Ug = 4,17 · 6 = 25,02

Ub = 16 · 25,02 = 400,32 ≈ 400 V

Task 3.2

Two point charges Q1 = 4 nC and Q2 = -2 nC are located at a distance of 60 cm from each other. Determine the field strength (V/m) at a point lying in the middle between the charges.

E = E1 + E2

E = 1/4πε · 4/l^2 · (|Q1|+|Q2|)

E – field strength, V/m;

Q1 – charge, 4 · 10^-9 C;

Q2 – charge, -2 · 10^-9 C;

l – distance between charges, m;

1/4πε – electric constant, 9 · 10^9 m/F.

E = 9 · 10^9 · 4/0,6^2 · (|4| + |-2|) = 0,6 kV/m

Task 3.3

A car battery has a nominal voltage of 400 V and a capacity of 85 kWh. How long (hours) can the battery supply the engine with a current of 100 A?

Ca = Cv · 1000 / V

t = Ca / I

Cv – battery capacity, kW·h;

V – battery voltage, V;

I – current, A;

t – time, h.

Ca = 85 · 1000 / 400 = 212 A·h

t = 212 / 100 = 2,12 h

Task 3.4

A lithium-ion battery has a nominal voltage of 3.7 V. The internal resistance of the battery is 0.1 Ohm. The battery is connected to a load with a resistance of 1 Ohm. Determine the power loss inside the battery caused by its internal resistance.

P = I^2 · r

I = U / Rg

Rg = R + r

P – power loss, W;

R – load resistance, Ohm;

r – internal resistance of the battery, Ohm;

U – battery voltage, V.

Rg = 1 + 0,1 = 1,1 Ohm

I = 3,7 / 1,1 = 3,36 A

P = (3,36)^2 · 0,1 = 1,13 W

Task 4.1

What is the moment of force (N·m) relative to the central axis of a 310 mm diameter steering wheel? If the force applied to the rim is 5 N.

The moment of force is found using the formula

M = F · r

М = 5 · 0,115 = 0,775 N·m

Task 4.2

The steering gear ratio is 10. What is the torque on the output shaft of the steering gear if the torque on the steering axis is 0,8 N·m?

M = Ms · i

M - torque on the output shaft, N·m;

Ms - torque on the steering axis, N·m;

i - gear ratio, i = 10.

M = 0,8 · 10 = 8 N·m

Task 5.1

The electric motor has a power of 75 kW and operates at a voltage of 400 V. The internal resistance of the stator winding is 0.05 Ohm. Determine the power loss in the winding due to its resistance (kW).

Pl = I^2 ⋅ Rw

I = P / U

Pl - lost power in the winding, W;

P - motor power, W;

I - current, A;

Rw - internal resistance of the stator winding, Ohm.

U - supply voltage, V;

I = 75000 / 400 = 187,5 A

Pl = (187,5)^2 ⋅ 0,05 = 3515,625 ≈ 3,5 kW

Task 5.2

What amount of heat (J) is generated in the stator winding during 5 seconds if the winding resistance is 0.05 Ohm and the current is 50 A?

Q = I^2 ⋅ R ⋅ t

Q - amount of heat, J;

I - current, A;

R - winding resistance, Ohm;

t - time, s.

Q = 50^2 ⋅ 0,05 ⋅ 5 = 2500 ⋅ 0,05 ⋅ 5 = 625 J

Task 5.3

The voltage in the stator winding of an electric motor is 400 V. The current in the stator winding is 150 A. The winding resistance is 0.05 Ohm. What is the current frequency, Hz?

f = U / I ⋅ R

f – current frequency, Hz;

U – voltage, V;

I – current, A;

R – winding resistance, Ohm

f = 400 / 150 ⋅ 0,05 = 53,33 Hz ≈ 53 Hz

Task 5.4

What is the synchronous speed of an electric motor (rpm) with four pole pairs if the frequency of the current in the motor winding is 400 Hz?

The basic formula for the synchronous speed of the motor rotor

ns = f ⋅ 60 / p

ns - synchronous speed (rpm);

f - frequency of the supply current (Hz);

p - number of pole pairs of the motor, p = 4.

ns = 400 ⋅ 60 / 4 = 6000 rpm

Task 5.5

An asynchronous brushless electric motor with neodymium magnets on the rotor has a torque of 450 N⋅m. The magnetic induction in the gap is 1,5 T. The current in the stator winding is 150 A. The length of the active conductor in the magnetic field of the stator windings is 40 m. The winding resistance is 0,05 Ohm. What is the coefficient of the electric motor that characterizes how efficiently the motor converts electrical energy into mechanical energy?

k = M / B ⋅ I ⋅ L ⋅ R

k – coefficient of the electric motor;

M - torque, N⋅m;

B - magnetic induction in the gap, T;

I - current through the stator windings, A;

L - length of the active conductor in the magnetic field, m;

R — rotor radius, m.

k = 450 / 1,5 ⋅ 150 ⋅ 40 ⋅ 0,05 = 1

Task 5.6

An ideal electric motor with a power of 75 kW rotates at a frequency of 1500 rpm. Determine the torque on the motor shaft.

M = P / ω

ω = 2 π n / 60

M - torque, N·m;

P - motor power, W;

ω - rotor angular velocity, rad/s.

ω = 2π ⋅ 1500 / 60 = 157 rad/s.

M = 75000 / 157 ≈ 478 N·m.

Task 5.7

An ideal electric motor with a power of 75 kW has a torque of 478 N·m at a rotation speed of 1500 rpm. Determine the torque of the electric motor at a rotation speed of 3000 rpm?

M2 = M1 · n1 / n2

M1, M2 - torque at a speed of n1, n2, N·m;

n1, n2 - rotation speed, rpm;

M2 = 478 · 1500 / 3000 = 239 N·m.

Task 6.1

In an isosceles triangle, sides AB = AC = 45 cm. Angle ∠A = 45°. What is side BC?

BC = ((AB^2 + AC^2 – 2 ⋅ AB ⋅ AC ⋅ cos (∠A))^1/2

BC = (45^2 + 45^2 – 2 ⋅ 45 ⋅ 45 ⋅ cos (45°))^1/2 = ((2025 + 2025 – 2 · 2025 ⋅ cos (45°))^1/2 = ((2025 + 2025 – 2 · 2025 ⋅ (2^1/2)/2)^1/2 = (4050 − 2862,9)^1/2 = 34,47 cm

Task 6.2

Circles with radii of 45 cm and 40 cm have equal lengths of arcs formed between the radii. The angle between the radii of the first circle is 15°. What is the angle formed between the radii of the second circle if the lengths of the arcs are proportional to their radii and the angles between the corresponding radii?

l1 = l2

l1 = r1 ⋅ θ1

l2 = r2 ⋅ θ2

r1 ⋅ θ1 = r2 ⋅ θ2

θ2 = r1 ⋅ θ1 / r2

l1, l2 – arc lengths, cm;

r1, r2 – circle radii, cm;

θ1, θ2 – angles, rad.

θ2 = (45 ⋅ π / 12) / 40 = 45 π / 480) = 3 π / 32 rad.

Conversion to degrees

θ2 = 3 π ⋅ 180 / 32 ⋅ π = ​​540 / 32 = 16,875°

Task 6.3

In an isosceles trapezoid, the side is 45 cm. The length of the longer base is 43 cm. The length of the shorter base is 20 cm. What is the height of the trapezoid?

h = (c^2 – ¼ (b-a)^2)^1/2

h - height of the trapezoid, cm;

c - side, cm;

b - length of the longer base, cm;

a - length of the shorter base, cm.

h = (45^2 – 1 / 4 · (43 - 20)^2)^1/2 = (2025 – 132,25)^1/2 = 43,5 cm.

Task 6.4

In a right triangle the legs are 40 cm and 30 cm. What is the hypotenuse of the triangle?

c = (a^2 + b^2)^1/2

c - length of the hypotenuse, cm;

a, b - lengths of the legs, cm;

c = (40^2 + 30^2)^1/2 = (1600 + 900)^1/2 = 2500^1/2 = 50 cm.

Task 6.5

In a right trapezoid, base AD = 45, BC = 40. The shorter side AB = 20. What is the side CD?

C1D = AD – BC

CD = (AB^2 + C1D^2)^1/2

C1D – leg of right triangle C1CD;

C1D = 45 - 40 = 5

CD = (20^2 + 5^2)^1/2 = 425^1/2 = 20,62

Task 6.6

In a right triangle, legs AC = 20, BC = 40. How many degrees is ∠ B?

tan(∠B) = AC/BC

tan(∠B) = 20/40 = 0,5

∠B = arctan(0,5) ≈ 27°

Task 6.7

In an isosceles trapezoid, the length of the larger base is 43, the length of the smaller base is 20. What is the length of the segment connecting the midpoints of the sides?

m = a + b / 2

m - length of the midline;

a - length of the upper base;

b - length of the lower base.

m = 43 + 20 / 2 = 31,5

Task 7.1

A weight of 400 kg presses on a steel shock absorber spring. The spring is compressed by 6 cm under the weight. What is the spring's elasticity coefficient?

k = F / Δx

k - elasticity coefficient, N/m;

F - force applied to the spring, N;

Δx - change in the length of the spring, m.

F = 400 kg ⋅ 9,8 m/s^2 = 3920 N

Δx = 6 cm = 0,06 m

k = 3920 / 0,06 ≈ 65333 N/m.

Task 7.2

In a shock absorber, nitrogen gas is in a 0.2 liter tube at a temperature of 300 K and a pressure of 7 MPa. How many grams of nitrogen gas are in the shock absorber?

m = ρ ⋅ V

ρ = P ⋅ M / R · T

m - mass of the gas, kg;

p - density of the gas, kg/m^3,

V - volume of the gas, m^3.

P - gas pressure, Pa;

M - molar mass of nitrogen, M = 28 ⋅ 10^−3 kg / mol;

R - universal gas constant, R = 8,31 J / K ⋅ mol;

T — gas temperature, K.

ρ = 7 ⋅ 10^6 ⋅ 28 ⋅ 10^−3 / 8,31 · 300 = 78,62 kg/m^3

m = 78,62 ⋅ 0,0002 = 0,01572 kg.

m = 15,72 g.

Task 7.3

In a shock absorber, nitrogen gas is in a tube with a volume of 0.2 liters at a temperature of 300 K and a pressure of 7 MPa. After the shock absorber is compressed, the gas volume decreases to 0,1 liters. What is the gas pressure if the gas temperature does not change?

P1 ⋅ V1 = P2 ⋅ V2

P2 = P1 ⋅ V1 / V2

P1 - initial pressure, Pa;

P2 - final pressure, Pa;

V1 - initial volume, m^3;

V2 - final volume, m^3

P2 = 7 ⋅ 10^6 ⋅ 0,0002 / 0,0001 = 14 MPa

Task 7.4

A shock absorber tube contains 16 grams of nitrogen gas at a temperature of 300 K and a pressure of 3 MPa. How many liters does the gas occupy?

P ⋅ V = n ⋅ R ⋅ T

V = n ⋅ R ⋅ T / P

n = m / M

P - gas pressure, Pa;

V - gas volume, m^3;

n - amount of gas substance, mol;

R - universal gas constant, R = 8,31 J/(K ⋅ mol)

T - temperature, K;

m - mass of gas, kg;

M – molar mass of nitrogen, kg/mol

m = 16 g = 0,016 kg

M = 28 g/mol = 0,028 kg/mol

n = 0,016 / 0,028 ≈ 0,571 mol

V = 0,571 ⋅ 8,31 ⋅ 300 / 3 ⋅ 10^6 = 1424,49 / 3 ⋅ 10^6 ≈ 0,0004748 m^3

V = 0,475 liters

Task 7.5

In a gas shock absorber, the spring wire diameter is 12 mm. The average spring diameter is 80 mm. The number of working spring turns is 6. The shear modulus of spring steel is 78500 MPa. What is the spring stiffness of the shock absorber (N/mm)?

c = G · d^4 / 8 · D^3 · n

с – spring stiffness, N/mm;

G – shear modulus, MPa;

d – rod diameter, mm;

D – average spring diameter, mm;

n – number of working spring turns

c = 78500 · 12^4 / 8 · 80^3 · 6 = 1628090 / 24576 = 66,25 N/mm

Task 7.6

The shock absorber spring has a stiffness of 65 N/mm. The spring diameter is 80 mm. The spring wire diameter is 12 mm. The shear modulus of spring steel is 78500 MPa. What is the number of working turns of the spring?

n = G · d^4 / 8 · D^3 · c

n - number of working turns of the spring

G - shear modulus, MPa;

d - rod diameter, mm;

D - average spring diameter, mm;

c - spring stiffness, N/mm;

c = 78500 · 12^4 / 8 · 80^3 · 65 = 6

Task 8.1

What is the moment of inertia of a solid cylinder relative to its axis of rotation (kg ·m^2) if the mass of the cylinder is 2 kg and its radius is 2 cm?

I = 1/2m·r^2

I - moment of inertia, kg·m^2,

m - mass of the cylinder, kg,

r - radius of the cylinder, m.

I = 1/2·2·0,02^2 = 0,0004 kg·m^2