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Iteru [2.4K]
3 years ago
14

(a) Calculate the rate of heat conduction through house walls that are 13.0 cm thick and that have an average thermal conductivi

ty twice that of glass wool. Assume there are no windows or doors. The surface area of the walls is 120 m2 and their inside surface is at 18.0 oC, while their outside surface is at 5.0 oC. (b) How many 1-kW room heaters would be needed to balance the heat transfer due to conduction
Physics
1 answer:
Mkey [24]3 years ago
6 0

Answer:

a. 960 W b. One 1 kW room heater

Explanation:

a. The rate of heat conduction P = kA(T₂ - T₁)/d where k = 2 × 0.040 W/m-K =  0.080 W/m-K since the thermal conductivity of glass wool is 0.040 W/m-K and that of the material is twice the thermal conductivity of glass wool, A = area of walls = 120 m², T₁ = outside surface temperature = 5.0 °C, T₂ = inside surface temperature = 18.0 °C and d = thickness of wall = 13.0 cm = 0.13 m

P = kA(T₂ - T₁)/d

= 0.080 W/m-K × 120 m²(18.0 °C - 5.0 °C)/0.13 m

= 9.6 Wm/K × 13 K/0.13 m

= 124.8 Wm/0.13 m

= 960 W

b. The number of 1 kW room heater required will be

n = rate of heat conduction/power of one room heater = 960 W/ 1 kW = 960 W/1000 W = 0.96 ≅ 1

So we need only one 1 kW room heater.

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to 10 Hz. Superimposed on this signal is 60-Hz noise with an amplitude of 0.1 V. It is desired to attenuate the 60-Hz signal to
givi [52]

Answer:

G \sqrt{1 +(\frac{f}{f_c})^{2n}} = 1

If we square both sides we got:

G^2 (1+\frac{f}{f_c})^{2n}= 1

We divide both sides by G^2 and we got:

(1+\frac{f}{f_c})^{2n} = \frac{1}{G^2}

Now we can apply log on both sides and we got:

2n ln(1+\frac{f}{f_c}) = ln (\frac{1}{G^2})

And solving for n we got:

n = \frac{ ln (\frac{1}{G^2})}{2ln(1+\frac{f}{f_c})}

And replacing we got:

n = \frac{ln (\frac{1}{0.1^2})}{2ln(1+\frac{60}{10})}

n = \frac{4.60517}{3.8918}=1.18

And since n needs to be an integer the correct answer would be n=2 for the filter order.

Explanation:

For this case we can use the formula for the Butterworth filter gain given by:

[tec] G = \frac{1}{\sqrt{1 +(\frac{f}{f_c})^{2n}}}[/tex]

Where:

G represent the transfer function and we want that G =0.1 since the desired signal is less than 10% of it's value

f_c = 10 Hz represent the corner frequency

f= 60 Hz represent the original frequency

n represent the filter order and that's the variable that we need to find

G \sqrt{1 +(\frac{f}{f_c})^{2n}} = 1

If we square both sides we got:

G^2 (1+\frac{f}{f_c})^{2n}= 1

We divide both sides by G^2 and we got:

(1+\frac{f}{f_c})^{2n} = \frac{1}{G^2}

Now we can apply log on both sides and we got:

2n ln(1+\frac{f}{f_c}) = ln (\frac{1}{G^2})

And solving for n we got:

n = \frac{ ln (\frac{1}{G^2})}{2ln(1+\frac{f}{f_c})}

And replacing we got:

n = \frac{ln (\frac{1}{0.1^2})}{2ln(1+\frac{60}{10})}

n = \frac{4.60517}{3.8918}=1.18

And since n needs to be an integer the correct answer would be n=2 for the filter order.

7 0
3 years ago
An object is initially moving at an unknown velocity. It accelerates at a rate of 2.0 m/s2 to a new velocity of 45 m/s in 10s. W
katrin2010 [14]

Answer:

Initial velocity=25m/s

Explanation:

acceleration(a)=2m/s^2

Final velocity(v)=45m/s

Time(t)=10seconds

Initial velocity(u)=?

V=u+axt

45=u+2x10

45=u+20

u=45-20

u=25m/s

6 0
3 years ago
Cierto volumen de gas se encuentra a 60°c de temperatura y 5atm de presión, es calentado hasta 140°c, estado en el cual ocupa un
earnstyle [38]

Answer:

V1 = 2221.33 L

Explanation:

The system is about a ideal gas. Then you can use the equation for ideal gases for a volume V1, temperature T1 and pressure P1:

P_1V_1=nRT_1   (1)

And also for the situation in which the variables T, V and P has changed:

P_2V_2=nRT_2   (1)

R: constant of ideal gases = 0.082 L.atm/mol.K

For both cases (1) and (2) the number of moles are the same. Next, you solve for n in (1) and (2):

n=\frac{P_1V_1}{RT_1}\\\\n=\frac{P_2V_2}{RT_2}

Next, you equal these equations an solve for T2:

\frac{P_1V_1}{RT_1}=\frac{P_2V_2}{RT_2}\\\\V_1=\frac{P_2V_2T_1}{P_1T_2}

Finally you replace the values of P2, V2, T1 and T2:

V_1=\frac{(7atm)(680L)(140\°C)}{(60\°C)(5atm)}=2221.33\ L

Hence, the initial volume of the gas is 2221.33 L

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