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2 years ago

10

A heat recovery system (HRS) is used to conserve heat from the surroundings and supply it to the Mars Rover. The HRS fluid loop

s use Freon as the working fluid. The instrumentation must be kept as a temperature greater than negative 65 degrees Fahrenheit [°F] to avoid damage. The temperature in the area of Mars where the rover is exploring is 192 kelvins [K]. If the system must remove 46.9 British Thermal Units [BTU] of energy, what volume of Freon is needed in units of liters [L]?

1 answer:

Annette [7]2 years ago

7 0

Answer: Volume of Freon = 12.23. $10^{3}$ L

Explanation: First, the temperatures have to be in the same unit, so all the tmeperatures is transformed into Kelvin:

T₁ = 192K

T₂ = $\frac{5}{9}$ . (F - 32)+273.15

T₂ = $\frac{5}{9}$ . (- 65 - 32)+273.15

T₂ = 219.26K.

Second, BTU is an unit of energy. It can be transformed into Joules by the relation 1BTU = 1055.06J.

Q = 1055.06 . 46.9 = 49482.314J

The letter Q represents Heat and is calculated as Q = m.Cp.ΔT, where:

m is mass of the element;

Cp is the heat capacity specific for each element;

ΔT is the difference between the final and initial temperature;

Third, it will be needed the properties of the Freon:

Freon (CF2Cl2) = 120.91g/mol; Cp = 74J/mol.K; ρ = 1.49kg/m³

Fourth, we calculate the mass of Freon necessary for the remove of 46.9BTU of energy from the system:

Q = m.Cp.ΔT

m= $\frac{Q}{c(T-T_{0} )}$

m = $\frac{49482.314}{74(219.26-192)}$

m = 18228.214g or m=18.228Kg

Fifth, using the density of Freon, Volume can be found

ρ = $\frac{m}{V}$

V = m / ρ

V = 18.228 / 1.49

V = 12.23m³

As SI for density is Kg/m³, the volume found is in m³.

Cubic meters is related to Liters as the following: 1m³=1000l.

So, volume of Freon = 12.23. $10^{3}$ L

The volume of Freon needed is 12.23. $10^{3}$ L.

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Un tubo de acero de 40000 kilómetros forma un anillo que se ajusta bien a la circunferencia de la tierra. Imagine que las person

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Answer:

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Explanation:

In order to find the distance of the steel ring to the ground, when its temperature has raised by 1°C, you first calculate the radius of the steel tube before its temperature increases.

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you solve for r in the equation (1):

$r=\frac{C}{2\pi}=\frac{4*10^7m}{2\pi}=6,366,197.724m$

Next, you use the following formula to calculate the change in the length of the tube, when its temperature increases by 1°C:

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L: final length of the tube = ?

Lo: initial length of the tube = 4*10^7m

ΔT = change in the temperature of the steel tube = 1°C

α: thermal coefficient expansion of steel = 13*10^-6 /°C

You replace the values of the parameters in the equation (2):

$L=(4*10^7m)(1+(13*10^{-6}/ \°C)(1\°C))=40,000,520m$

With the new length of the tube, you can calculate the radius of a ring formed with the tube. You again solve the equation (1) for r:

$r'=\frac{C}{2\pi}=\frac{40,000,520m}{2\pi}=6,366,280.484m$

Finally, you compare both r and r' radius:

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Hence, the distance to the ring from the ground is 82.76m

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1 year ago

if m represents mass in kg, v represents speed in m/s, and r represents radius in m, show that the force F in the equation F=mv^

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2 years ago

Sunlight strikes a piece of crown glass at an angle of incidence of 38.0°. Calculate the difference in the angle of refraction b

zhuklara [117]

Answer:

Difference in the angle of refraction = 0.3°

41.04° is the minimum angle of incidence.

Explanation:

Angle of incidence = 38.0°

For yellow light :

Using Snell's law as:

$\frac {sin\theta_2}{sin\theta_1}=\frac {n_1}{n_2}$

Where,

Θ₁ is the angle of incidence

Θ₂ is the angle of refraction

n₁ is the refractive index for yellow light which is 1.523

n₂ is the refractive index of air which is 1

So,

$\frac {sin\theta_2}{sin{38.0}^0}=\frac {1.523}{1}$

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Angle of refraction for yellow light = sin⁻¹ 0.9377 = 69.67°.

For green light :

Using Snell's law as:

$\frac {sin\theta_2}{sin\theta_1}=\frac {n_1}{n_2}$

Where,

Θ₁ is the angle of incidence

Θ₂ is the angle of refraction

n₁ is the refractive index for green light which is 1.526

n₂ is the refractive index of air which is 1

So,

$\frac {sin\theta_2}{sin{38.0}^0}=\frac {1.526}{1}$

${sin\theta_2}=0.9395$

Angle of refraction for green light = sin⁻¹ 0.9395 = 69.97°.

The difference in the angle of refraction = 69.97° - 69.67° = 0.3°

Calculation of the critical angle for the yellow light for the total internal reflection to occur :

The formula for the critical angle is:

${sin\theta_{critical}}=\frac {n_r}{n_i}$

Where,

${\theta_{critical}}$ is the critical angle

$n_r$ is the refractive index of the refractive medium.

$n_i$ is the refractive index of the incident medium.

n₁ is the refractive index for yellow light which is 1.523 (incident medium)

n₂ is the refractive index of air which is 1 (refractive medium)

Applying in the formula as:

${sin\theta_{critical}}=\frac {1}{1.523}$

The critical angle is = sin⁻¹ 0.6566 = 41.04°

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2 years ago