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3 years ago

8

An engineer is investigating energy loss through windows. The windowpane of interest is 0.650 cm thick, has dimensions of 1.19 m

× 2.25 m, and has a thermal conductivity of 0.8 w/(m·°C). On a given cold day, the outside temperature is 0°C and the temperature of the interior surface of the glass is 24.00. (a) Determine the rate (in W) at which heat energy is transferred through the window (b) Determine the amount of energy (in J) transferred through the window in one day, assuming the temperature on the surfaces remains constant.

1 answer:

krok68 [10]3 years ago

8 0

Answer:

7908.92307 W

683330953.248 J

Explanation:

k = Heat conduction coefficient = 0.8 W/(m·°C)

A = Area = $1.19\times 2.25\ m^2$

l = Thickness = 0.65 cm

$T_2$ = 24°C

$T_1$ = 0°C

Rate of heat transfer is given by

$Q=\frac{kA(T_2-T_1)}{l}\\\Rightarrow Q=\frac{0.8\times 1.19\times 2.25(24-0)}{0.65\times 10^{-2}}\\\Rightarrow Q=7908.92307\ W$

The rate of heat transfer is 7908.92307 W

Amount of energy is given by

$E=Qt\\\Rightarrow E=7908.92307\times 24\times 3600\\\Rightarrow E=683330953.248\ J$

The energy transferred through the window in one day is 683330953.248 J

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Explanation:

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Where:

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$T$ - Tension in the cable connecting the jet and glider, measured in newtons.

$m_{G}$ , $m_{J}$ - Masses of the glider and the jet, measured in kilograms.

$a$ - Acceleration of the glider-jet system, measured in meters per square second.

If we know that $m_{G} = 260\,kg$ , $m_{J} = 1,940\,kg$ and $a = 2.20\,\frac{m}{s^{2}}$ , then the solution of this system of equations:

By (2):

$T = (260\,kg)\cdot \left(2.20\,\frac{m}{s^{2}} \right)$

$T = 572\,N$

By (1):

$F = T+m_{J}\cdot a$

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Substitute equation (2) into equation (3)

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Answer:

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Given that,

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Frequency f' = 650 Hz

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Using formula of organ pipe

$f=\dfrac{nv}{2L}$

For 550 Hz,

$550=\dfrac{n\times340}{2L}$ ...(I)

For 650 Hz,

$650=\dfrac{(n+1)\times340}{2L}$ ...(II)

From equation (I) and (II)

$550-650=\dfrac{340}{2L}-\dfrac{340}{L}$

$L=\dfrac{340}{2\times100}$

$L=1.7\ m$

(a). We need to calculate the fundamental frequency for n = 1

Using formula of fundamental frequency

$=f=\dfrac{n\lambda}{2L}$

put the value of L

$f=\dfrac{1\times340}{2\times1.7}$

$f=100\ Hz$

Hence, The fundamental frequency and length of the pipe are 100 Hz and 1.7 m.

4 0

3 years ago