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svetoff [14.1K]

4 years ago

9

A power of 100 kW (105 W) is delivered to the other side of a city by a pair of power lines, between which the voltage is 12,000

V. (a) Use the formula P 5 IV to show that the current in the lines is 8.3 A. (b) If each of the two lines has a resistance of 10 Ω, show that there is a 83-V change of voltage along each line.

1 answer:

vivado [14]4 years ago

3 0

Answer:

I = 8.3 Amp

potential drop = 83 V

Explanation:

Power = 100 KW

V = 12,000 V

R = 10 ohms

a)

Calculate current I in each wire:

P = I*V

I = P / V

I = 100 / 12 = 8.333 A

b)

Calculate potential drop in each wire:

V = I*R

V = (8.3) * (10)

V = 83 V

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5 0

2 years ago

Which line from "On Becoming an Inventor" supports the idea that Dean's time at Worcester Polytechnic Institute was very useful

Stolb23 [73]

Answer:

Explanation:

The line from "On becoming an inventor" that says:

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3 0

3 years ago

Read 2 more answers

A step-up transformer has 20 primary turns and 400 secondary turns. If the primary current is 30 A, what is the secondary curren

-BARSIC- [3]

150
A
Explanation:
V
s
V
p
=
N
s
N
p

(
1
)
N
refers to the number of turns
V
is voltage
s
and
p
refer to the secondary and primary coil.
From the conservation of energy we get:
V
p
I
p
=
V
s
I
s

(
2
)
From
(
1
)
:
V
s
V
p
=
900
00
3
00
=
300
∴
V
s
=
300
V
p
Substituting for
V
s
into
(
2
)
⇒
V
p
I
p
=
300
V
p
×
0.5
∴
I
p
=
150
A
Seems a big current.

3 0

3 years ago

An ideal reheat Rankine cycle with water as the working fluid operates the boiler at 15,000 kPa, the reheater at 2000 kPa, and t

solniwko [45]

Answer:

See the explanation below.

Explanation:

First find the enthalpies h₁, h₂, h₃, h₄, h₅, and h₆.

Find h₁:

Using Saturated Water Table and given pressure p₁ = 100 kPa

h₁ = 417.5 kJ/kg

Find h₂:

In order to find h₂, add the $w_{p}$ to h₁, where $w_{p}$ is the work done by pump and h₁ is the enthalpy computed above h₁ = 417.5 kJ/kg.

But first we need to compute $w_{p}$ To computer

Pressures:

p₁ = 100 kPa

p₂ = 15,000 kPa

and

Using saturated water pressure table, the volume of water $v_{f}$ = 1.0432

Dividing 1.0432/1000 gives us:

Volume of water = v₁ = 0.001043 m³/kg

Compute the value of h₂:

h₂ = h₁ + v₁ (p₂ - p₁)

= 417.5 kJ/kg + 0.001043 m³/kg ( 15,000 kPa - 100 kPa)

= 417.5 + 0.001043 (14900)

= 417.5 + 15.5407

= 433.04 kJ/kg

Find h₃

Using steam table:

At pressure p₃ = 15000 kPa

and Temperature = T₃ = 450°C

Then h₃ = 3159 kJ/kg

The entropy s₃ = 6.14 kJ/ kg K

Find h₄

Since entropy s₃ is equal to s₄ So

s₄ = 6.14 kJ/kgK

To compute h₄

s₄ = $s_{f}$ + $x_{4} s_{fg}$

$x_{4} = s_{4} -s_{f} /s_{fg}$

$x_{4}$ = 6.14 - 2.45 / 3.89

$x_{4}$ = 0.9497

The enthalpy h₄:

h₄ = $h_{f} +x_{4} h_{fg}$

= 908.4 + 0.9497(1889.8)

= 908.4 + 1794.7430

= 2703 kJ/kg

This can simply be computed using the software for steam tables online. Just use the entropy s₃ = 6.14 kJ/ kg K and pressure p₄ = 2000 kPa

Find h₅

Using steam table:

At pressure p₅ = 2000 kPa

and Temperature = T₅ = 450°C

Then h₅ = 3358 kJ/kg

Find h₆:

Since the entropy s₅ = 7.286 kJ/kgK is equal s₆ to So

s₆ = 7.286 kJ/kgK = 7.29 kJ/kgK

To compute h₆

s₆ = $s_{f}$ + $x_{6} s_{fg}$

$x_{6} = s_{6} -s_{f} /s_{fg}$

$x_{6}$ = 7.29 - 1.3028 / 6.0562

$x_{6}$ = 0.988

The enthalpy h₆:

h₆ = $h_{f} +x_{6} h_{fg}$

= 417.51 + 0.988 (2257.5)

= 417.51 + 2230.41

h₆ = 2648 kJ/kg

This can simply be computed using the software for steam tables online. Just use the entropy s₅ = 7.286 kJ/kgK and pressure p₅ = 2000 kPa

Compute power used by pump:

$P_{p}$ is found by using:

mass flow rate = m = 1.74 kg/s

Volume of water = v₁ = 0.001043 m³/kg

p₁ = 100 kPa

p₂ = 15,000 kPa

$P_{p}$ = ( m ) ( v₁ ) ( p₂ - p₁ )

= (1.74 kg/s) (0.001043 m³/kg) (15,000 kPa - 100 kPa)

= (1.74 kg/s) (0.001043 m³/kg) (14900)

= 27.04

$P_{p}$ = 27 kW

Compute heat added $q_{a}$ and heat rejected $q_{r}$ from boiler using computed enthalpies:

$q_{a}$ = ( h₃ - h₂ ) + ( h₅ - h₄ )

= ( 3159 kJ/kg - 433.04 kJ/kg ) + ( 3358 kJ/kg - 2703 kJ/kg )

= 2726 + 655

= 3381 kJ/kg

$q_{r}$ = h₆ - h₁

= 2648 kJ/kg - 417.5 kJ/kg

= 2232 kJ/kg

Compute net work

W $_{net}$ = $q_{a}$ - $q_{r}$

= 3381 kJ/kg - 2232 kJ/kg

= 1150 kJ/kg

Compute power produced by the cycle

mass flow rate = m = 1.74 kg/s

W $_{net}$ = 1150 kJ/kg

P = m * W $_{net}$

= 1.74 kg/s * 1150 kJ/kg

= 2001 kW

Compute rate of heat transfer in the reheater

Q = m * ( h₅ - h₄ )

= 1.74 kg/s * 655

= 1140 kW

Compute Thermal efficiency of this system

μ $_{t}$ = 1 - $q_{r}$ / $q_{a}$

= 1 - 2232 kJ/kg / 3381 kJ/kg

= 1 - 0.6601

= 0.34

= 34%

7 0

3 years ago

A tank with a volume of 8 m3 containing 4 m3 of 20% (by volume) NaOH solution is to be purged by adding pure water at a rate of

lawyer [7]

Answer:

The time necessary to purge 95% of the NaOH is 0.38 h

Explanation:

Given:

vfpure water(i) = 3 m³/h

vNaOH = 4 m³

xNaOH = 0.2

vfpure water(f) = 2 m³/h

pwater = 1000 kg/m³

pNaOH = 1220 kg/m³

The mass flow rate of the water is = 3 * 1000 = 3000 kg/h

The mass of NaOH in the solution is = 0.2 * 4 * 1220 = 976 kg

When the 95% of the NaOH is purged, thus the NaOH in outlet is = 0.95 * 976 = 927.2 kg

The volume of NaOH in outlet after time is = 927.2/1220 = 0.76 m³

The time required to purge the 95% of the NaOH is = 0.76/2 = 0.38 h

4 0

4 years ago