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sladkih [1.3K]
3 years ago
12

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 −67 degrees Fahrenheit​ [°F] to avoid damage.
The temperature in the area of Mars where the rover is exploring is 189 kelvins​ [K]. If the system must remove 47.4
British Thermal Units​ [BTU] of​ energy, what volume of Freon is needed in units of liters​ [L]?

Some facts about​ Freon:
The specific heat of Freon is 74
joules per mole kelvin J/(mol °K)]
The specific gravity of Freon is 1.49
The molecular mass of Freon is 120 grams per mole​ [g/mol]
Engineering
1 answer:
blagie [28]3 years ago
7 0

Answer:

0.304 L of Freon is needed

Explanation:

Q = mCT

Q is quantity of energy that must be removed = 47 BTU = 47×1055.06 = 49587.82 J

C is specific heat of Freon = 74 J/mol.K = 74 J/mol.K × 1 mol/120 g = 0.617 J/g.K

T is temperature in the area of Mars = 189 K

m = Q/CT = 49587.82/(0.617×189) = 452.23 g = 452.24/1000 = 0.45223 kg

Density of Freon = specific gravity of Freon × density of water = 1.49 × 1000 kg/m^3 = 1490 kg/m^3

Volume of Freon = mass/density = 0.45223/1490 = 0.000304 m^3 = 0.000304×1000 = 0.304 L

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Answer:

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An air compressor of mass 120 kg is mounted on an elastic foundation. It has been observed that, when a harmonic force of amplit
kupik [55]

Answer:

equivalent stiffness is 136906.78 N/m

damping constant is 718.96 N.s/m

Explanation:

given data

mass = 120 kg

amplitude = 120 N

frequency = 320 r/min

displacement = 5 mm

to find out

equivalent stiffness and damping

solution

we will apply here frequency formula that is

frequency ω = ω(n) √(1-∈ ²)      ......................1

here  ω(n) is natural frequency i.e = √(k/m)

so from equation 1

320×2π/60 = √(k/120) × √(1-2∈²)

k × ( 1 - 2∈²) = 33.51² ×120

k × ( 1 - 2∈²) = 134752.99    .....................2

and here amplitude ( max ) of displacement is express as

displacement = force / k  ×  (  \frac{1}{2\varepsilon \sqrt{1-\varepsilon ^2}})

put here value

0.005 = 120/k   ×  (  \frac{1}{2\varepsilon \sqrt{1-\varepsilon ^2}})  

k ×∈ × √(1-2∈²) = 1200       ......................3

so by equation 3 and 2

\frac{k\varepsilon \sqrt{1-\varepsilon^2})}{k(1-2\varepsilon^2)} = \frac{12000}{134752.99}

\varepsilon^{2} - \varepsilon^{4}  = 7.929 * 10^{-3} - 0.01585 * \varepsilon^{2}

solve it and we get

∈ = 1.00396

and

∈ = 0.08869

here small value we will consider so

by equation 2 we get

k × ( 1 - 2(0.08869)²) = 134752.99

k  = 136906.78 N/m

so equivalent stiffness is 136906.78 N/m

and

damping is express as

damping = 2∈ √(mk)

put here all value

damping = 2(0.08869) √(120×136906.78)

so damping constant is 718.96 N.s/m

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5 0
3 years ago
A gas tank is known to have a thickness of 0.5 inches and an internal pressure of 2.2 ksi. Assuming that the maximum allowable s
sergiy2304 [10]

Answer:

D_o=11.9inch

Explanation:

From the question we are told that:

Thickness T=0.5

Internal PressureP=2.2Ksi

Shear stress \sigma=12ksi

Elastic modulus \gamma= 35000

Generally the equation for shear stress is mathematically given by

 \sigma=\frac{P*r_1}{2*t}

Where

r_i=internal Radius

Therefore

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Generally

 r_o=r_1+t

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Generally the equation for outer diameter is mathematically given by

 D_o=2r_o

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Therefore

Assuming that the thin cylinder is subjected to integral Pressure

Outer Diameter is

 D_o=11.9inch

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3 years ago
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