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AysviL [449]
3 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]3 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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Answer:

a) 0.26

b) 1077 MPa

Explanation:

a) The following equation can be used to determine the volume fraction:

\frac{F_f}{F_m} =\frac{E_fV_f}{E_m(1-V_f)}

\frac{0.97}{1-0.97} =\frac{260V_f}{2.8(1-V_f)}

32.3 = \frac{260V_f}{2.8-2.8V_f}

V_f = 0.26

b) Tensile strength can be found by using the following equation:

\sigma_{cl} = \sigma_m(1-V_f)+\sigma_fV_f = 50*(1-0.26)+4000*0.26 = 1077 MPa

5 0
4 years ago
If a master cylinder piston applies pressure to a brak piston that has twice as much square area what will be the force from the
sp2606 [1]

Explanation:

Pressure = force/area

Pressure stays the same.

If the area is doubled the force is doubled.

5 0
3 years ago
I need help on what’s and input and output to make the flowchart
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Input: what is put in, taken in, or operated on by any process or system.

Output: the amount of something produced by a person, machine, or industry.

3 0
3 years ago
A horizontal curve is being designed for a new two-lane highway (12-ft lanes). The PI is at station 250 + 50, the design speed i
kogti [31]

Answer

given,

Speed of vehicle = 65 mi/hr

                            = 65 x 1.4667 = 95.33 ft/s

e = 0.07 ft/ft

f is the lateral friction, f = 0.11

central angle,Δ = 38°

The PI station is

PI = 250 + 50

   = 25050 ft

using super elevation formula

e + f = \dfrac{v^2}{rg}

0.07 + 0.11 =\dfrac{95.33^2}{r\times 32.2}

r = \dfrac{95.33^2}{32.2\times 0.18}

  r = 1568 ft

As the road is two lane with width 12 ft

R = 1568 + 12/2

R = 1574 ft

Length of the curve

L = \dfrac{\piR\Delta}{180}

L = \dfrac{\pi\times 1574\times 38}{180}

L = 1044 ft

Tangent of the curve calculation

  T = R tan(\dfrac{\Delta}{2})

  T = 1574 tan(\dfrac{38}{2})

      T = 542 ft

The station PC and PT are

 PC = PI - T

 PC = 25050 - 542

       = 24508 ft

       = 245 + 8 ft

PT = PC + L

     = 24508 + 1044

     =25552

     = 255 + 52 ft

the middle ordinate calculation

MO = R(1-cos\dfrac{\Delta}{2})

MO = 1574\times (1-cos\dfrac{38}{2})

     MO = 85.75 ft

degree of the curvature

D = \dfrac{5729.578}{R}

D = \dfrac{5729.578}{1574}

D = 3.64°

8 0
4 years ago
A car hits a tree at an estimated speed of 10 mi/hr on a 2% downgrade. If skid marks of 100 ft. are observed on dry pavement (F=
Dafna11 [192]

Answer:

v_{o} = 22.703\,\frac{m}{s} \left(50.795\,\frac{m}{s}\right)

Explanation:

The deceleration of the car on the dry pavement is found by the Newton's Law:

\Sigma F = -\mu_{k,1}\cdot m\cdot g \cdot \cos \theta + m\cdot g \cdot \sin \theta = m\cdot a_{1}

Where:

a_{1} = (-\mu_{k,1}\cdot \cos \theta + \sin \theta)\cdot g

a_{1} = (-0.33\cdot \cos 1.146^{\textdegree}+\sin 1.146^{\textdegree})\cdot \left(9.807\,\frac{m}{s^{2}} \right)

a_{1} = -3.040\,\frac{m}{s^{2}}

Likewise, the deceleration of the car on the unpaved shoulder is:

a_{2} = (-\mu_{k,2}\cdot \cos \theta + \sin \theta)\cdot g

a_{2} = (-0.28\cdot \cos 1.146^{\textdegree}+\sin 1.146^{\textdegree})\cdot \left(9.807\,\frac{m}{s^{2}} \right)

a_{2} = -2.549\,\frac{m}{s^{2}}

The speed just before the car entered the unpaved shoulder is:

v_{o} = \sqrt{\left(4.469\,\frac{m}{s} \right)^{2}-2\cdot \left(-2.549\,\frac{m}{s^{2}} \right)\cdot (60.88\,m)}

v_{o} = 18.175\,\frac{m}{s}

And, the speed just before the pavement skid was begun is:

v_{o} = \sqrt{\left(18.175\,\frac{m}{s} \right)^{2}-2\cdot \left(-3.040\,\frac{m}{s^{2}} \right)\cdot (30.44\,m)}

v_{o} = 22.703\,\frac{m}{s} \left(50.795\,\frac{m}{s}\right)

8 0
3 years ago
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