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3 years ago

7

A hydrogen-filled balloon to be used in high altitude atmosphere studies will eventually be 100 ft in diameter. At 150,000 ft, t

he pressure is 0.14 lb/in2 and the temperature is - 67°F. Assume the balloon is spherical in shape. What is the diameter of the hydrogen balloon at the ground at 14.7 lb/in2 and 68°F

1 answer:

max2010maxim [7]3 years ago

5 0

Answer:

The answer is 11.7 ft

Explanation:

You can use the combined gas law from Boyle's law, Charles's law, and Gay-Lussac's Law. Only because hydrogen behaves like an ideal gas for this conditions.

$\frac{p_1 V_1}{T_1} = \frac{p_2 V_2}{T_2}$

where the subscripts denote the pressure "p", volume "V" and the temperature "T" (in Kelvin) at two different times. Let's consider $t_1$ as the balloom at 150,000 ft so

$p_1 = 0.14 \ lb/in^2$

$V_1 = \frac{4}{3} \pi R_1^3 = 523598.77 \ ft^3$

and $T_1 = -67^\circ F = 218.15\ K$ .

Then, $t_2$ is the moment when the balloon is on the ground.

$p_2 = 14.7 \ lb/in^2$ and $T_2 = 68^\circ F = 293.15\ K$ .

From the first equation,

$V_2 = \frac{p_1 V_1 T_2}{T_1 p_2}$ , then

$V_2 = 6701.07 ft^3$ and the radius would be

$R_2 = \sqrt[3]{\frac{3 V_2}{4 \pi}} = 11.7 \ ft$ .

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Answer:

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Solution:

Now, maximum heat generation, $Q_{m}$ is given by:

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Now, using eqn (1):

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For excess heat generation, Q:

$Q = (Q_{Gi} - Q_{m})\times volume of sphere, V$

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$Q = (46480 - 24924)\times \frac{4}{3}\pi\0.1^{3} = 21556\times \frac{4}{3}\pi\0.1^{3} W/m^{3}$

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$Q = \frac{90.294}{1.163} = 77.638 kcal/h$

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Answer:

Check the explanation

Explanation:

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Kindly check the attached image below to get the step by step explanation to the question above.

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3 years ago

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Answer:

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Explanation:

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