The assembly consists of two sections of galvanized steel pipe connected together using a reducing coupling at B. The smaller pipe has an outer diameter of 0.75 in. and an inner diameter of 0.68 in., whereas the larger pipe has an outer diameter of 1 in. and an inner diameter of 0.86 in. If the pipe is tightly secured to the wall at C, determine the maximum shear stress in each section of the pipe when the couple is applied to the handles of the wrench.

Answer:

2901 vehicles

Explanation:

We are given;

Percentage of large trucks & buses; p_t = 7% = 0.07

Percentage of recreational vehicles; p_r = 3% = 0.03

PHF = 0.90

Driver population adjustment; f_p = 0.92

First of all, let's Calculate the heavy vehicle factor from the formula;

f_hv = 1/[1 + p_t(e_t - 1) + p_r(e_r - 1)]

Where;

e_t = passenger car equivalents for trucks and buses

e_r = passenger car equivalents for recreational vehicles

From the table attached, for a mountainous terrain, e_t = 6 and e_r = 4. Thus;

f_hv = 1/[1 + 0.07(6 - 1) + 0.03(4 - 1)]

f_hv = 1.44

Let's now calculate the initial hourly volume from the formula;

v_p = V1/(PHF × N × f_hv × f_p)

Where;

v_p = 15-minute passenger-car equivalent flow rate

V1 = hourly volume

N = number of lanes in each direction

From online tables of LOS criteria for multilane freeway segments, v_p = 1300 pc/hr/ln

Thus;

1300 = V1/(0.9 × 3 × 1.44 × 0.92)

V1 = 1300 × (0.9 × 3 × 1.44 × 0.92)

V1 = 4650 veh/hr

Now, let's Calculate the final hourly volume;

From online sources, the maximum capacity of a 6 lane highway with free-flow speed of 50 mi/h is 2000 pc/hr/ln.

We are told the online peak-hour factor increases to 0.95 and so PHF = 0.95.

Thus;

2000 = V2/(0.95 × 3 × 1.44 × 0.92)

V2 = 2000(0.95 × 3 × 1.44 × 0.92)

V2 = 7551 veh/hr

Number of vehicles added to the highway = V2 - V1 = 7551 - 4650 = 2901 vehicles

Answer:

(2) ( (x+y)⁴)³(3)+(3)x(4x+y)

Explanation:

correct me if I'm wrong^_^

Answer:

hahahaha

Explanation:

Answer:

C = $0.0032 per day

Explanation:

We are given;

Dimension of cell phone; 50 mm × 45 mm × 20 mm

Temperature of charger; T1 = 33°C = 306K

Emissivity; ε = 0.92

convection heat transfer coefficient; h = 4.5 W/m².K

Room air temperature; T∞ = 22°C = 295K

Wall temperature; T2 = 20°C = 293 K

Cost of electricity; C = $0.18/kW.h

Chargers are usually in the form of a cuboid, and thus, surface Area is;

A = (50 × 45) + 2(50 × 20) + 2(45 × 20)

A = 6050 mm²

A = 6.05 × 10^(-3) m²

Formula for total heat transfer rate is;

E_t = hA(T1 - T∞) + εσA((T1)⁴ - (T2)⁴)

Where σ is Stefan Boltzmann constant with a value of; σ = 5.67 × 10^(-8) W/m².K⁴

Thus;

E_t = 4.5 × 6.05 × 10^(-3) (306 - 295) + (0.92 × 6.05 × 10^(-3) × 5.67 × 10^(-8)(306^(4) - 293^(4)))

E_t = 0.7406 W = 0.7406 × 10^(-3) KW

Now, we know C = $0.18/kW.h

Thus daily cost which has 24 hours gives;

C = 0.18 × 0.7406 × 10^(-3) × 24

C = $0.0032 per day

Answer:

The rate at which entropy is produced within the turbine is 22.762 kilojoules per kilogram-Kelvin.

Explanation:

By either the Principle of Mass Conservation and First and Second Laws of Thermodynamics, we model the steam turbine by the two equations described below:

Principle of Mass Conservation

[tex]\dot m_{in} - \dot m_{out} = 0[/tex] (1)

First Law of Thermodynamics

[tex]-\dot Q_{out} + \dot m \cdot \left[h_{in}-h_{out}+ \frac{1}{2}\cdot (v_{in}^{2}-v_{out}^{2}) - w_{out} \right] = 0[/tex] (2)

Second Law of Thermodynamics

[tex]-\frac{\dot Q_{out}}{T_{out}} + \dot m\cdot (s_{in}-s_{out}) + \dot S_{gen} = 0[/tex] (3)

By dividing each each expression by [tex]\dot m[/tex], we have the following system of equations:

[tex]-q_{out} + h_{in}-h_{out} + \frac{1}{2}\cdot (v_{in}^{2}-v_{out}^{2}) - w_{out} = 0[/tex] (2b)

[tex]-\frac{q_{out}}{T_{out}} + s_{in}-s_{out} + s_{gen} = 0[/tex] (3b)

Where:

[tex]\dot Q_{out}[/tex] - Heat transfer rate between the turbine and its surroundings, in kilowatts.

[tex]q_{out}[/tex] - Specific heat transfer between the turbine and its surroundings, in kilojoules per kilogram.

[tex]T_{out}[/tex] - Outer surface temperature of the turbine, in Kelvin.

[tex]\dot m[/tex] - Mass flow rate through the turbine, in kilograms per second.

[tex]h_{in}[/tex], [tex]h_{out}[/tex] - Specific enthalpy of water at inlet and outlet, in kilojoules per kilogram.

[tex]v_{in}[/tex], [tex]v_{out}[/tex] - Speed of water at inlet and outlet, in meters per second.

[tex]w_{out}[/tex] - Specific work of the turbine, in kilojoules per kilogram.

[tex]s_{in}[/tex], [tex]s_{out}[/tex] - Specific entropy of water at inlet and outlet, in kilojoules per kilogram-Kelvin.

[tex]s_{gen}[/tex] - Specific generated entropy, in kilojoules per kilogram-Kelvin.

By property charts for steam, we get the following information:

Inlet

[tex]T = 400\,^{\circ}C[/tex], [tex]p = 3000\,kPa[/tex], [tex]h = 3231.7\,\frac{kJ}{kg}[/tex], [tex]s = 6.9235\,\frac{kJ}{kg\cdot K}[/tex]

Outlet

[tex]T = 100\,^{\circ}C[/tex], [tex]p = 101.42\,kPa[/tex], [tex]h = 2675.6\,\frac{kJ}{kg}[/tex], [tex]s = 7.3542\,\frac{kJ}{kg\cdot K}[/tex]

If we know that [tex]h_{in} = 3231.7\,\frac{kJ}{kg}[/tex], [tex]h_{out} = 2675.6\,\frac{kJ}{kg}[/tex], [tex]v_{in} = 160\,\frac{m}{s}[/tex], [tex]v_{out} = 100\,\frac{m}{s}[/tex], [tex]w_{out} = 540\,\frac{kJ}{kg}[/tex], [tex]T_{out} = 350\,K[/tex], [tex]s_{in} = 6.9235\,\frac{kJ}{kg\cdot K}[/tex] and [tex]s_{out} = 7.3542\,\frac{kJ}{kg\cdot K}[/tex], then the rate at which entropy is produced withing the turbine is:

[tex]q_{out} = h_{in} - h_{out} + \frac{1}{2}\cdot (v_{in}^{2}-v_{out}^{2})-w_{out}[/tex]

[tex]q_{out} = 3231.7\,\frac{kJ}{kg} - 2675.6\,\frac{kJ}{kg} + \frac{1}{2}\cdot \left[\left(160\,\frac{m}{s} \right)^{2}-\left(100\,\frac{m}{s} \right)^{2}\right] - 540\,\frac{kJ}{kg}[/tex]

[tex]q_{out} = 7816.1\,\frac{kJ}{kg}[/tex]

[tex]s_{gen} = \frac{q_{out}}{T_{out}}+s_{out}-s_{in}[/tex]

[tex]s_{gen} = \frac{7816.1\,\frac{kJ}{kg} }{350\,K} + 7.3542\,\frac{kJ}{kg\cdot K} - 6.9235\,\frac{kJ}{kg\cdot K}[/tex]

[tex]s_{gen} = 22.762\,\frac{kJ}{kg\cdot K}[/tex]

The rate at which entropy is produced within the turbine is 22.762 kilojoules per kilogram-Kelvin.

Answer:

Horizontal force = 89.2 N

Explanation:

The frictional force = coefficient of friction * magnitude of the force (weight of the body) * cos theta

Substituting the given values, we get -

Frictional Force = 0.3*300 * cos 25 = 89.2 N

Horizontal force = 89.2 N

Answer: hello some parts of your question is missing attached below is the missing information

The radiator of a car is a type of heat exchanger. Hot fluid coming from the car engine, called the coolant, flows through aluminum radiator tubes of thickness d that release heat to the outside air by conduction. The average temperature gradient between the coolant and the outside air is about 130 K/mm . The term ΔT/d  is called the temperature gradient which is the temperature difference ΔT between coolant inside and the air outside per unit thickness of tube

answer : Total surface area = 3/2 * area of old radiator

Explanation:

we will use this relation

K = [tex]\frac{Qd }{A* change in T }[/tex]

change in T =  ΔT  

therefore New Area  ( A ) = 3/2 * area of old radiator

Given that the thermal conductivity is the same in the new and old radiators

Answer:

L = Henry

C = Farad

Explanation:

The electrical parameter represented as L is the inductance whose unit is Henry(H).

The electrical parameter represented as C is the inductance whose unit is Farad

Resonance frequency occurs when the applied period force is equal to the natural frequency of the system upon which the force acts :

To obtain :

At resonance, Inductive reactance = capacitive reactance

Equate the inductive and capacitive reactance

Inductive reactance(Xl) = 2πFL

Capacitive Reactance(Xc) = 1/2πFC

Inductive reactance(Xl) = Capacitive Reactance(Xc)

2πFL = 1/2πFC

Multiplying both sides by F

F * 2πFL = F * 1/2πFC

2πF²L = 1/2πC

Isolating F²

F² = 1/2πC2πL

F² = 1/4π²LC

Take the square root of both sides to make F the subject

F = √1 / √4π²LC

F = 1 /2π√LC

Hence, the proof.