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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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A fatigue test was conducted in which the mean stress was 90 MPa (13050 psi), and the stress amplitude was 190 MPa (27560 psi).
Gwar [14]

Answer:

a) 280MPa

b) -100MPa

c) -0.35

d) 380 MPa

Explanation:

GIVEN DATA:

mean stress \sigma_m = 90MPa

stress amplitude \sigma_a = 190MPa

a) \sigma_m =\frac{\sigma_max+\sigma_min}{2}

    90 =\frac{\sigma_{max}+\sigma_{min}}{2} --------------1

\sigma_a =\frac{\sigma_{max}-\sigma_{min}}{2}

   190 = \frac{\sigma_{max}-\sigma_{min}}{2} -----------2

solving 1 and 2 equation we get

\sigma_{max} = 280MPa

b) \sigma_{min} = - 100MPa

c)

stress ratio=\frac{\sigma_{min}}{\sigma_{max}}

=\frac{-100}{280} = -0.35

d)magnitude of stress range

                      =(\sigma_{max} -\sigma_{min})

                       = 280 -(-100) = 380 MPa

3 0
3 years ago
Identify parts of the E-Cig that constitute voltage, current, and resistance. Discuss the role each plays in the E-Cig and typic
Leno4ka [110]

Answer: c

Explanation:

7 0
2 years ago
A completely reversible heat pump produces heat at a rate of 300 kW to warm a house maintained at 24 °C. The exterior air, which
trapecia [35]

Answer:

No.

Explanation:

The Coefficient of Performance of the reversible heat pump is determined by the Carnot's cycle:

COP_{HP} = \frac{T_{H}}{T_{H}-T_{L}}

COP_{HP} = \frac{297.15\,K}{297.15\,K-280.15\,K}

COP_{HP} = 3.339

The power required to make the heat pump working is:

\dot W = \frac{300\,kW}{3.339}

\dot W = 89.847\,kW

The heat absorbed from the exterior air is:

\dot Q_{L} = 300\,kW - 89.847\,kW

\dot Q_{L} = 210.153\,kW

According to the Second Law of Thermodynamics, the entropy generation rate in a reversible cycle must be zero. The formula for the heat pump is:

\frac{\dot Q_{L}}{T_{L}} - \frac{\dot Q_{H}}{T_{H}} + \dot S_{gen} = 0

\dot S_{gen} = \frac{\dot Q_{H}}{T_{H}} - \frac{\dot Q_{L}}{T_{L}}

\dot S_{gen} = \frac{300\,kW}{297.15\,K}-\frac{210.153\,kW}{280.15\,K}

\dot S_{gen} = 0.259\,\frac{kW}{K}

Which contradicts the reversibility criterion according to the Second Law of Thermodynamics.

5 0
2 years ago
External crack of length of 3.0 mm was detected on the surface of the shaft of wind turbine made from 4340 steel. The diameter o
sveticcg [70]

Answer:

The correct answer is "K_c=6.0369 \ MPa\sqrt{m}".

Explanation:

Given:

Maximum load,

P = 50,000 N

Crack length,

a = 3mm

or,

  = 3×10⁻³ m

Diameter,

d = 32 mm

As we know,

⇒  Maximum stress, \sigma=\frac{P}{A}

                                      =\frac{50000}{(\frac{\pi}{4}\times 32^2)}

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Now,

⇒  Fracture tougness, K_c=Y \sigma\sqrt{\pi a}

On substituting the values, we get

                                           =1\times 62.20\times \sqrt{3.14\times 3\times 10^{-3}}

                                           =6.0369 \ MPa\sqrt{m}

4 0
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Answer:

-8.34\times 10^{11}\ \text{J}

Explanation:

m = Mass of satellite = 15000 kg

h = Distance above Earth = 800 km

R = Radius of Earth = 6371 km

M = Mass of Earth = 5.972\times 10^{24}\ \text{kg}

G = Gravitational constant = 6.674\times 10^{-11}\ \text{Nm}^2/\text{kg}^2

Potential energy is given by

U=-\dfrac{GMm}{R+h}\\\Rightarrow U=-\dfrac{6.674\times 10^{-11}\times 5.972\times 10^{24}\times 15000}{(6371+800)\times 10^3}\\\Rightarrow U=-8.34\times 10^{11}\ \text{J}

The potential energy of the satellite is -8.34\times 10^{11}\ \text{J}.

6 0
2 years ago
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