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3 years ago

An electric kettle is required to heat 0.64 kg of water from 15.4°C to 98.2°C in six

Engineering

1 answer:

skelet666 [1.2K]3 years ago

7 0

Answer:

Almost done

Explanation:

I am just finishing up my work

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Following is attached the solution or the question given.

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Explanation:

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3 years ago

9 Consider fully developed conditions in a circular tube with constant surface temperature Ts Tm. Determine whether a small- or

mojhsa [17]

Answer:

Hello your question is incomplete attached below is the complete question

answer:

Considering Laminar flow

Q ( heat ) will be independent of diameter

Considering Turbulent flow

The heat transfer will increase with decreasing "dia" for the turbulent

heat transfer = f(d^-0.8 )

Explanation:

attached below is the detailed solution

Considering Laminar flow

Q ( heat ) will be independent of diameter

Considering Turbulent flow

The heat transfer will increase with decreasing "dia" for the turbulent

heat transfer = f(d^-0.8 )

6 0

3 years ago

Determine the enthalpy, volume and density of 1.0 kg of steam at a pressure of 0.5 MN/m2 and with a dryness fraction of 0.96

Viktor [21]

Answer:

Enthalpy, hsteam = 2663.7 kJ/kg

Volume, Vsteam = 0.3598613 m^3 / kg

Density = 2.67 kg/ m^3

Explanation:

Mass of steam, m = 1 kg

Pressure of the steam, P = 0.5 MN/m^2

Dryness fraction, x = 0.96

At P = 0.5 MPa:

Tsat = 151.831°C

Vf = 0.00109255 m^3 / kg

Vg = 0.37481 m^3 / kg

hf = 640.09 kJ/kg

hg = 2748.1 kJ/kg

hfg = 2108 kJ/kg

The enthalpy can be given by the formula:

hsteam = hf + x * hfg

hsteam = 640.09 + ( 0.96 * 2108)

hsteam = 2663.7 kJ/kg

The volume of the steam can be given as:

Vsteam = Vf + x(Vg - Vf)

Vsteam = 0.00109255 + 0.96(0.37481 - 640.09)

Vsteam = 0.3598613 m^3 / kg

From the steam table, the density of the steam at a pressure of 0.5 MPa is 2.67 kg/ m^3

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3 years ago

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Marizza181 [45]

Answer:

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7 0

3 years ago

The symmetrical load below is connected to a three-phase network. A line current of 25A has been measured. The load resistors ha

boyakko [2]

Answer:

The line voltage of the three phase network is 346.41 V

Explanation:

Star Connected Load

Resistance, R₁ = R₂ = R₃ = 18 Ω

For a star connected load, the line current = the phase current, that is we have

$I_L = 25 \, A = I_{Ph}$

Whereby the the voltage across each resistance = $V_R$ is given by the relation;

$V_R$ = $I_{Ph}$ × R

Hence;

$V_{Ph}$ = $V_R$ = $I_{Ph}$ × R = 25 × 8 = 200 V

Therefore we have;

The line voltage, $V_{L}$ = √3 × $V_{Ph}$ = √3 × 200 = 346.41 V.

Hence, the line voltage of the three phase network = 346.41 V.

3 0

4 years ago