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AysviL [449]
2 years ago
10

A heating system must maintain the interior of a building at TH = 20 °C when the outside temperature is TC = 2 °C. If the rate o

f heat transfer from the building through its walls and roof is 16.4 kW, determine the electrical power required, in kW, to heat the building using (a) electrical-resistance heating, (b) a heat pump whose coefficient of performance is 3.0, (c) a reversible heat pump operating between hot and cold reservoirs at 20 °C and 2 °C, respectively.
Engineering
1 answer:
nasty-shy [4]2 years ago
5 0

Answer:

a)Q_H=16.4kW

b) w=5.467

c)  w=1.2345

Explanation:

From the question we are told that:

Interior temperature TH = 20 °C

outside temperature is TC = 2 °C

Rate H=16.4kW

a)

Generally electric resistance heating is the heat transfer from interior building

Therefore

Q_H=R

Q_H=16.4kW

b)

COP =3

Generally the equation for coefficient of performance is mathematically given by

COP=\frac{Q_H}{w}

Therefore

w=\frac{Q_H}{COP}

w=16.4/3

w=5.467

c)

Generally the equation for coefficient of performance is mathematically given by

COP_r=\frac{TH}{TH-TC}

COP_r=\frac{20+273}{(20+273)-(2+273)}

COP_r=16.2

Therefore

w=1.2345

w=20/16.2

w=1.2345

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NeTakaya

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3 years ago
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7 0
2 years ago
Air at 80 kPa and 10°C enters an adiabatic diffuser steadily with a velocity of 150 m/s and leaves with a low velocity at a pre
il63 [147K]

Answer:

The exit temperature is 293.74 K.

Explanation:

Given that

At inlet condition(1)

P =80 KPa

V=150 m/s

T=10 C

Exit area is 5 times the inlet area

Now

A_2=5A_1

If consider that density of air is not changing from inlet to exit then by using continuity equation

A_1V_1=A_2V_2

So   A_1\times 150=5A_1V_2

V_2=30m/s

Now from first law for open system

h_1+\dfrac{V_1^2}{2}+Q=h_2+\dfrac{V_2^2}{2}+w

Here Q=0 and w=0

h_1+\dfrac{V_1^2}{2}=h_2+\dfrac{V_2^2}{2}

When air is treating as ideal gas  

h=C_pT

Noe by putting the values

h_1+\dfrac{V_1^2}{2}=h_2+\dfrac{V_2^2}{2}

1.005\times 283+\dfrac{150^2}{2000}=1.005\times T_2+\dfrac{30^2}{2000}

T_2=293.74K

So the exit temperature is 293.74 K.

7 0
3 years ago
What is the voltage output (in V) of a transformer used for rechargeable flashlight batteries, if its primary has 515 turns, its
kow [346]
<h2>Answer:</h2>

7532V

<h2>Explanation:</h2>

For a given transformer, the ratio of the number of turns in its primary coil (N_{p}) to the number of turns in its secondary coil (N_{s}) is equal to the ratio of the input voltage (V_{p}) to the output voltage (V_{s}) of the transformer. i.e

\frac{N_p}{N_s} = \frac{V_p}{V_s}            ----------------(i)

<em>From the question;</em>

N_{p} = number of turns in the primary coil = 8 turns

N_{s} = number of turns in the secondary coil = 515 turns

V_{p} = input voltage = 117V

<em>Substitute these values into equation (i) as follows;</em>

\frac{8}{515} = \frac{117}{V_s}

<em>Solve for </em>V_{s}<em>;</em>

V_{s} = 117 x 515 / 8

V_{s} = 7532V

Therefore, the output voltage (in V) of the transformer is 7532

6 0
3 years ago
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