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sdas [7]
3 years ago
6

Many radios can be operated either by plugging them into the wall or by using batteries. How can a radio use either source of cu

rrent?
Engineering
1 answer:
ch4aika [34]3 years ago
5 0

Answer:

The electric current from the batteries installed in a radio supplies direct current (DC) electricity to the radio components directly as an alternative source to the Alternating Current (AC) converted to DC by the power unit located at the radio end of the cable plugged into the wall outlet.

Explanation:

Part of the power unit in a radio includes an AC to DC converter, which is an electrical circuit that is able to convert the alternating current power input from the wall outlet into a direct current output to the radio with which the radio can work

The alternative source of electric current from the batteries installed in a radio bypasses the AC to DC converter and supplies power directly to the radio so it can also work.

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An inventor claims to have developed a refrigerator that at steady state requires a net power input of 1.1 horsepower to remove
Lynna [10]

Answer:

The inventor's claim is false in the sense that no thermal machine can violate the first thermodynamic law.

Explanation:

The inventor's claim could not be possible as no thermal machine can transfer more heat than the input work consumed. If we expose the thermal efficiency:

n=Q out / W in

Where Q and W both must be in the same power unit, so we will convert the remove heat from BTU/hr to hp:

12000 BTU/hr = 4.72 hp

Therefore by comparing, we notice that the removing heat of 4.75 hp is large than the delivered work of 1.11 hp. By evaluating the efficiency:

[tex]n=4.75 hp / 1.1 hp  = 4.3 > 1[/tex]

6 0
3 years ago
De acuerdo con el tabulador de puestos de una compañía, los salarios mensuales que obtienen los trabajadores son los que se mues
inessss [21]
Sorry don’t understand
8 0
3 years ago
6, Large, centralized power-generating stations are often located near ___. A) areas without natural energy sources B) abundant
Leviafan [203]

Answer: abundant energy sources

Explanation:

A power generating station which is sometimes referred to as the power plant is simply an industrial facility that is used for the generation of power.

Large, centralized power-generating stations are often located near abundant energy sources. It should be noted that most power stations burn fossil fuels like oil, coal, natural gas, in order to be able to generate power and hence, they have to be located where there's abundance of the energy sources.

8 0
3 years ago
An ocean thermal energy conversion system is being proposed for electric power generation. Such a system is based on the standar
Dennis_Churaev [7]

Answer:

a) the heat exchanger area required for the evaporator is 11178.236 m²

b) the required flow rate is 1993630.38 kg/s

Explanation:

Given the data in the question;

Water temperature near the surface = 300 K

temperature at reasonable depths ( cold ) = 280 K

power plant output W' = 2 MW

efficiency η = 3% = 0.03

we know that; efficiency η = W'_{power-out / Q_{supplied

we substitute

0.03 = 2 / Q_{supplied

Q_{supplied = 2 / 0.03

Q_{supplied = 66.667 MW = 66.667 × 10⁶ Watt

Th_{in = 300 K       Th_{out = 292 K

Tc_{in = 290 K       Tc_{out = 290 K    

Now, Heat transfer in evaporator;

Q = UA( LMTD )

so

LMTD = (ΔT₁ - ΔT₂) / ln( ΔT₁ / ΔT₂ )

first we get ΔT₁ and ΔT₂

ΔT₁ = Th_{in - Tc_{out  = 300 - 290 = 10 K

ΔT₂ = Th_{out - Tc_{in  = 292 - 290 = 2 K

so we substitute into our equation;

LMTD = (10 - 2) / ln( 10 / 2 )

LMTD = 8 / ln( 5 )

LMTD = 8 / 1.6094379

LMTD = 4.97

a) Heat transfer Area will be;

Q_H = UA( LMTD )

we substitute

66.667 × 10⁶ = 1200 × A × 4.97

66.667 × 10⁶  = 5964 × A

A = (66.667 × 10⁶) / 5964

A = 11178.236 m²

Therefore, the heat exchanger area required for the evaporator is 11178.236 m²

b) Flow rate  

we know that;

Q_H = m'C_P( T_{in - T_{out )  

specific heat capacity of water Cp = 4.18 (kJ/kg∙°C)

we substitute

66.667 × 10⁶ = m' × 4.18 × ( 300 - 292 )

66.667 × 10⁶ = m' × 33.44

m' = ( 66.667 × 10⁶ ) / 33.44

m' = 1993630.38 kg/s

Therefore, the required flow rate is 1993630.38 kg/s

7 0
3 years ago
Suppose that you are working in a home that has a 15kw oven that operates on 240 V, The oven is on a branch circuit by itself. W
krok68 [10]

Explanation:

please see it my work solution

7 0
3 years ago
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