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2 years ago

What is a combination circuit? A combination circuit:

Engineering

1 answer:

Anon25 [30]2 years ago

7 0

Answer:

Combination circuit; The basic strategy for the analysis of combination circuits involves using the meaning of equivalent resistance for parallel branches to transform the combination circuit into a series circuit.

Example:

The use of both series and parallel connections within the same circuit. In this case, light bulbs A and B are connected by parallel connections and light bulbs C and D are connected by series connections. This is an example of a combination circuit.

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What is the activation energy (Q) for a vacancy formation if 10 moles of a metal have 2.3 X 10^13 vacancies at 425°C?

Yakvenalex [24]

Answer:

$Activation\ Energy=2.5\times 10^{-19}\ J$

Explanation:

Using the expression shown below as:

$N_v=N\times e^{-\frac {Q_v}{k\times T}$

Where,

$N_v$ is the number of vacancies

N is the number of defective sites

k is Boltzmann's constant = $1.38\times 10^{-23}\ J/K$

${Q_v}$ is the activation energy

T is the temperature

Given that:

$N_v=2.3\times 10^{13}$

N = 10 moles

1 mole = $6.023\times 10^{23}$

So,

N = $10\times 6.023\times 10^{23}=6.023\times 10^{24}$

Temperature = 425°C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (425 + 273.15) K = 698.15 K

T = 698.15 K

Applying the values as:

$2.3\times 10^{13}=6.023\times 10^{24}\times e^{-\frac {Q_v}{1.38\times 10^{-23}\times 698.15}$

$ln[\frac {2.3}{6.023}\times 10^{-11}]=-\frac {Q_v}{1.38\times 10^{-23}\times 698.15}$

$Q_v=2.5\times 10^{-19}\ J$

4 0

2 years ago

Two streams of air enter a control volume: stream 1 enters at a rate of 0.05 kg / s at 300 kPa and 380 K, while stream 2 enters

alex41 [277]

Answer:

0.08kg/s

Explanation:

For this problem you must use 2 equations, the first is the continuity equation that indicates that all the mass flows that enter is equal to those that leave the system, there you have the first equation.

The second equation is obtained using the first law of thermodynamics that indicates that all the energies that enter a system are the same that come out, you must take into account the heat flows, work and mass flows of each state, as well as their enthalpies found with the temperature.

finally you use the two previous equations to make a system and find the mass flows

I attached procedure

5 0

3 years ago

What is polarized electrical receptacle used for

lidiya [134]

Answer:

they are used for electrical currents so that they can flow along the appropriate wires in the circuit

Explanation:

4 0

3 years ago

11–17 A long, thin-walled double-pipe heat exchanger with tube and shell diameters of 1.0 cm and 2.5 cm, respectively, is used t

lana [24]

Answer:

the overall heat transfer coefficient of this heat exchanger is 1855.8923 W/m²°C

Explanation:

Given:

d₁ = diameter of the tube = 1 cm = 0.01 m

d₂ = diameter of the shell = 2.5 cm = 0.025 m

Refrigerant-134a

20°C is the temperature of water

h₁ = convection heat transfer coefficient = 4100 W/m² K

Water flows at a rate of 0.3 kg/s

Question: Determine the overall heat transfer coefficient of this heat exchanger, Q = ?

First at all, you need to get the properties of water at 20°C in tables:

k = 0.598 W/m°C

v = 1.004x10⁻⁶m²/s

Pr = 7.01

ρ = 998 kg/m³

Now, you need to calculate the velocity of the water that flows through the shell:

$v_{w} =\frac{m}{\rho \pi (\frac{d_{2}^{2}-d_{1}^{2} }{4} )} =\frac{0.3}{998*\pi (\frac{0.025^{2}-0.01^{2} }{4}) } =0.729m/s$

It is necessary to get the Reynold's number:

$Re=\frac{v_{w}(d_{2}-d_{1}) }{v} =\frac{0.729*(0.025-0.01)}{1.004x10^{-6} } =10891.4343$

Like the Reynold's number is greater than 10000, the regime is turbulent. Now, the Nusselt's number:

$Nu=0.023Re^{0.8} Pr^{0.4} =0.023*(10891.4343)^{0.8} *(7.01)^{0.4} =85.0517$

The overall heat transfer coefficient:

$Q=\frac{1}{\frac{1}{h_{1} }+\frac{1}{h_{2} } }$

Here

$h_{2} =\frac{kNu}{d_{2}-d_{1}} =\frac{0.598*85.0517}{0.025-0.01} =3390.7278W/m^{2}C$

Substituting values:

$Q=\frac{1}{\frac{1}{4100}+\frac{1}{3390.7278} } =1855.8923W/m^{2} C$

5 0

3 years ago

What are the four categories of engineering materials used in manufacturing?

alexgriva [62]

Answer:

metals, composite, ceramics and polymers.

Explanation:

The four categories of engineering materials used in manufacturing are metals, composite, ceramics and polymers.

i) Metals: Metals are solids made up of atoms held by matrix of electrons. They are good conductors of heat and electricity, ductile and strong.

ii) Composite: This is a combination of two or more materials. They have high strength to weight ratio, stiff, low conductivity. E.g are wood, concrete.

iii) Ceramics: They are inorganic, non-metallic crystalline compounds with high hardness and strength as well as poor conductors of electricity and heat.

iv) Polymers: They have low weight and are poor conductors of electricity and heat

8 0

2 years ago