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tatuchka [14]
3 years ago
5

A refrigerated space is maintained at -15℃, and cooling water is available at 30℃, the refrigerant is ammonia. The refrigeration

capacity is 105 kJ/h. If the compressor is operated reversibly:
(1) What is the value of ε for Carnot refrigerator?

(2) Calculate the ε for the vapor-compression cycle;

(3) Calculate the circulation rate for the refrigerant;

(4) Calculate the rating power of the compressor.​
Engineering
1 answer:
Illusion [34]3 years ago
6 0

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.

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Answer:

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Explanation:

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mass of sludge produced = 32909.09 kg

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and as we know that

\frac{100}{S sludge} = \frac{solid percentage}{S solid} = \frac{water percentage}{S water}

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S sludge = 1.152

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6 0
3 years ago
At many intersections, under certain circumstances, a right turn may be made while the traffic light is red. Before turning righ
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A lake with an area of 525 [acre] was monitored during a one-month period. The average inflow was 30 [cfs] for the month and the
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Answer:

\Delta S = 1581663.5ft^3

Explanation:

We need to calculate the change in storage through the changes given,

That is,

\Delta S = P+I-O-O_{seepage}-E

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\Delta S = P+(I-0)-O_{seepage}-E

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The values inside the are parenthesis need to be konverted as I note here.

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That is,

\Delta S = 0.83in\Delta S= (0.83in*1ft/12in)(525acres)\\\Delta S=36.31 acres.ft\\\Delta S=36.31acres.ft*(43560ft^3/acrees.ft)\\\Delta S = 1581663.5ft^3

3 0
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Q1. A truck traveling at 40 mph is approaching a stop sign. At time ????0 and at a distance of 80ft, the truck begins to slow do
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Answer:

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a = Deceleration = -12 ft/s² (it'll have a negative sign, since it is negative acceleration

v² = u² + 2ax

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x = 143.41 ft

143.41 ft > 80 ft; hence, the truck will not stop in time. The truck passes the stop sign by about 63.41 ft before it stops.

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2 years ago
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