A triangular plate with a base 5 ft and altitude 3 ft is submerged vertically in water. If the base is in the surface of water, find the force against onr side of the plate. Express the hydrostatic force against one side of the plate as an integral and evaluate it. (Recall that the weight density of water is 62.5 lb/ft3.)

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:

- 46.5171kW

Explanation:

FIrst, the value given:

P1 = 1.05 bar (Initial pressure)

P2 = 12 bar (final pressure)

Heat transfer, Q = - 3.5 kW (It is negative because the compressor losses heat to the surroundings)

Mgaseous nitrogen = Mair = 28.0134 Kg/mol (constant)

Universal gas constant, Ru = 8.3143 Kj/Kgmolk

Specific gas constant, R = 0.28699 Kj/KgK

Initial temperature, T1 = 300 K

Final temperature, T2 = 400 K

Finding the volume:

P1V1 = RT1

V1 = RT1 ÷ P1

= (0.28699 Kj/KgK X 300k) ÷ 105

Note convert bar to Kj/Nm by multiply it by 100

V1 =  0.81997 m3/Kg

To get the mass flow rate:

m = volumetric flow rate / V1

= (21 m3/min x 1/60seconds) ÷ 0.81997 m3/Kg

= 0.4268Kg/s

Using tables for the enthalpy,

hT1 = 300.19 KJ/Kg

hT2 = 400.98 KJ/Kg

The enthalpy change = hT2 - hT1

= 100.79 KJ/Kg

Power, P = Q - (m X enthalpy change)

= - 3.5 - (0.4268 X 100.79)

= - 46.5171kW

Answer: As to the more sophisticated way of detecting "smoke" from an object a human may use in hotel rooms, this sensor called a Fresh Air Sensor does not just detect and, and but alerts the management about a smoking incident in a hotel room

Answer:

that not even a question

Answer:

Printing press , lenses etc,

Explanation:

Almost all of the Renaissance innovations that influenced the world, printing press was one of them. The printing press is widely regarded as the Renaissance's most significant innovation. The press was invented by Johannes Gutenberg, and Germany's culture was altered forever.

Astrophysics, humanist psychology, the printed word, vernacular vocabulary of poetry, drawing, and sculpture practice are only a few of the Renaissance's main inventions.

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.

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.