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-Dominant- [34]
4 years ago
6

A three-phase source delivers 4.8 kVA to a wye-connected load with a phase voltage of 214 V and a power factor of 0.9 lagging. C

alculate the source line current and the source line voltage.

Engineering
1 answer:
BlackZzzverrR [31]4 years ago
5 0

Answer:

a) 7.5A

b) 370V

Explanation:

Note that the Line current is the current through any one line between a three-phase source and load.

From the given question, The source line current =7.5A(approximately)

While the Source line voltage = 370V.

Kindly go through the attached file for a step by step detail of how I arrived at these answers.

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A gas is compressed from an initial volume of 0.42 m3 to a final volume of 0.12 m3. During the quasi-equilibrium process, the pr
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W=-52 800\ \text{J}=-52.8\ \text{kJ}

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First I sketched the compression of the gas with the help of the given pressure change process relation. That is your pressure change due to change in volume.

To find the area underneath the curve (the same as saying to find the work done) you should integrate the given relation for pressure change:

W=\int_{0.42}^{0.12}-1200V+500dV=-52.8\ \text{kJ}

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An overhead 25m long, uninsulated industrial steam pipe of 100mm diameter is routed through a building whose walls and air are a
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Answer:

a) he rate of heat loss from the steam line is 18.413588 kW

b) the annual cost of heat loss from line is $12904.25

Explanation:

a)

first we find the area

A = πdL

d is the diameter (0.1m) and L is the length (25m)

so

A = π ×  0.1 × 25

A = 7.85 m²

Now rate of heat loss through convection

qconv = hA(Ts -Ta)

h is the convective heat transfer coefficient (10 W/m²K), Ts is surface temperature (150°), Ta is temperature of air (25°)

so we substitute

qconv = 10 W/m²K × 7.85 m² × ( 150° - 25°)

qconv = 9817.477 J/s

Now heat lost through radiation

qrad = ∈Aα ( Ts⁴ - Ta⁴)

∈ is the emissivity (0.8), α is the boltzmann constant ( 5.67×10⁻⁸m⁻²K⁻⁴ ),

first we shall covert our temperatures from Celsius to kelvin scale

Ts is surface temperature (150 + 273K ), Ta is temperature of air (25 + 273K)

so we substitute

qrad = 0.8 × 7.854 × 5.67×10⁻⁸ × ( (423)⁴ - (298)⁴ )

qrad = 3.5625×10⁻⁷ × 2.413×10¹⁰

qrad = 8596.112 J/s

Now to get the total rate of heat loss through convection and radiation, we say

q = qconv + qrad

q = 9817.477 + 8596.112

q = 18413.588 J/s ≈ 18.413588 kW

Therefore the rate of heat loss from the steam line is 18.413588 kW

b)

annual cost of heat lost rate

A = C × q/n × ( 3600 × 24 × 365 )

C is the cost of heat per MJ( $0.02/10⁶) n is broiler efficiency ( 0.9)

so we substitute

A = 0.02/10⁶  × 18413.588/0.9 × ( 3600 × 24 × 365 )

A = $12904.25

Therefore the annual cost of heat loss from line is $12904.25

4 0
3 years ago
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