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Agata [3.3K]
3 years ago
13

The Hubble Space Telescope is an optical imaging telescope with extremely good angular resolution. Someone discovers an object t

hat Hubble cannot see at all. How might they have found it?
Engineering
1 answer:
HACTEHA [7]3 years ago
8 0

Answer:

<h2><em><u>Through</u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>use</u></em><em><u> </u></em><em><u>of</u></em><em><u> </u></em><em><u>Hubble</u></em><em><u> </u></em><em><u>Space</u></em><em><u> </u></em><em><u>Telescope</u></em><em><u> </u></em><em><u>that's</u></em><em><u> </u></em><em><u>why</u></em><em><u> </u></em><em><u>they</u></em><em><u> </u></em><em><u>might</u></em><em><u> </u></em><em><u>find</u></em><em><u> </u></em><em><u>it</u></em><em><u> </u></em><em><u>or</u></em><em><u> </u></em><em><u>enlarging</u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>resolution</u></em><em><u> </u></em><em><u>of</u></em><em><u> </u></em><em><u>a</u></em><em><u> </u></em><em><u>telescope</u></em><em><u>.</u></em></h2>

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3 years ago
A refrigerated space is maintained at -15℃, and cooling water is available at 30℃, the refrigerant is ammonia. The refrigeration
Illusion [34]

Answer:

(1) 5.74

(2) 5.09

(3) 3.05×10⁻⁵ kg/s

(4) 0.00573 kW

Explanation:

The parameters given are;

Working temperature, T_C  = -15°C = 258.15 K

Temperature of the cooling water, T_H = 30°C = 303.15 K

(1) The Carnot coefficient of performance is given as follows;

\gamma_{Max} = \dfrac{T_C}{T_H - T_C}  =  \dfrac{258.15}{303.15 - 258.15}   = 5.74

(2) For ammonia refrigerant, we have;

h_2 = h_g = 1466.3 \ kJ/kg

h_3 = h_f = 322.42 \ kJ/kg

h_4 = h_3 = h_f = 322.42 \ kJ/kg

s₂ = s₁ = 4.9738 kJ/(kg·K)

0.4538 + x₁ × (5.5397 - 0.4538) = 4.9738

∴ x₁ = (4.9738 - 0.4538)/(5.5397 - 0.4538) = 0.89

h_1 = h_{f1} + x_1 \times h_{gf}

h₁ = 111.66 + 0.89 × (1424.6 - 111.66) = 1278.5 kJ/kg

\gamma = \dfrac{h_1 - h_4}{h_2 - h_1}

\gamma = \dfrac{1278.5 - 322.42}{1466.3 - 1278.5} = 5.09

(3) The circulation rate is given by the mass flow rate, \dot m as follows

\dot m = \dfrac{Refrigeration \ capacity}{Refrigeration \ effect \ per \ unit \ mass}

The refrigeration capacity = 105 kJ/h

The refrigeration effect, Q = (h₁ - h₄) = (1278.5 - 322.42) = 956.08 kJ/kg

Therefore;

\dot m = \dfrac{105}{956.08}  = 0.1098 \ kg/h

\dot m = 0.1098 kg/h = 0.1098/(60*60) = 3.05×10⁻⁵ kg/s

(4) The work done, W = (h₂ - h₁) = (1466.3 - 1278.5) = 187.8 kJ/kg

The rating power = Work done per second = W×\dot m

∴ The rating power = 187.8 × 3.05×10⁻⁵ = 0.00573 kW.

6 0
3 years ago
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