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saul85 [17]
3 years ago
6

Bananas are to be cooled from 28°C to 12°C at a rate of 1140 kg/h by a refrigerator that operates on a vapor-compression refrige

ration cycle. The power input to the refrigerator is 9.8 kW. Determine (a) the rate of heat absorbed from the bananas, in kJ/h, and the COP, (b) the minimum power input to the refrigerator, and (c) the second-law efficiency and the exergy destruction for the cycle. The specific heat of bananas above freezing is 3.35 kJ/kg·°C.
Engineering
1 answer:
Lera25 [3.4K]3 years ago
8 0

Answer:

A) COP = \frac{16.97}{9.8} = 1.731

B) P_{IN} = 0.4763

C) Second law efficiency 4.85%

exergy destruction for the cycle = 9.3237 kW

Explanation:

Given data:

T_1 = 28 degree celcius

T_2 = 12 degree celcius

\dot m = 1140 kg/h

Power to refrigerator = 9.8 kW

Cp = 3.35 kJ/kg degree C

A) Q = \dot m Cp \Delta T

        = 1140 \times 3.35\times (28-12) = 61,104 kJ/h

Q_{abs} = 61,104 kJ/h = 16.97 kJ/sec

COP = \frac{16.97}{9.8} = 1.731

b)

COP ∝ \frac{1}{P_{in}}

P_{in} wil be max when COP maximum

taking surrounding temperature T_H = 20 degree celcius

COP_{max} = \frac{T_L}{T_H- T_L} = \frac{285}{293 - 285} = 35.625

we know that

COP = \frac{heat\ obsorbed}{P_{in}}

P_{IN} = \frac{16.97}{35.62} = 0.4763

c) second law efficiency

\eta_{11} = \frac{COP_R}{(COP)_max} = \frac{1.731}{35.625} = 4.85\%

exergy destruction os given as X = W_{IN} - X_{Q2}

                                                         = 9.8 - 0.473 = 9.3237 kW

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Answer:

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void add()

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  cout << "\nmatrix1 # of columns: ";

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      {

          cout << "\nEnter element (" << i << "," << j << "): ";

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          cout << "\n";

      }

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  cin >> rows2;

 

  cout << "\nmatrix2 # of columns: ";

  cin >> cols2;

 

  int m2[rows2][cols2];

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          cout << "\nEnter element (" << i << "," << j << "): ";

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          cout << m2[i][j] << " ";

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  if(rows1 == rows2 && cols1 == cols2)

  {

      cout << "\n";

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  int rows1,cols1,i,j,k,rows2,cols2;

  cout << "\nmatrix1 # of rows: ";

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  cin >> cols1;

 

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  cout << "\nmatrix1 # of columns: ";

  cin >> cols1;

 

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  cin >> rows2;

 

  cout << "\nmatrix2 # of columns: ";

  cin >> cols2;

 

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      {

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5 0
3 years ago
The heat required to raise the temperature of m (kg) of a liquid from T1 to T2 at constant pressure is Z T2CpT dT (1) In high sc
a_sh-v [17]

Answer:

(a)

<em>d</em>Q = m<em>d</em>q

<em>d</em>q = C_p<em>d</em>T

q = \int\limits^{T_2}_{T_1} {C_p} \, dT   = C_p (T₂ - T₁)

From the above equations, the underlying assumption is that  C_p remains constant with change in temperature.

(b)

Given;

V = 2L

T₁ = 300 K

Q₁ = 16.73 KJ    ,   Q₂ = 6.14 KJ

ΔT = 3.10 K       ,   ΔT₂ = 3.10 K  for calorimeter

Let C_{cal} be heat constant of calorimeter

Q₂ = C_{cal} ΔT

Heat absorbed by n-C₆H₁₄ = Q₁ - Q₂

Q₁ - Q₂ = m C_p ΔT

number of moles of n-C₆H₁₄, n = m/M

ρ = 650 kg/m³  at 300 K

M = 86.178 g/mol

m = ρv = 650 (2x10⁻³) = 1.3 kg

n = m/M => 1.3 / 0.086178 = 15.085 moles

Q₁ - Q₂ = m C_p' ΔT

C_p = (16.73 - 6.14) / (15.085 x 3.10)

C_p = 0.22646 KJ mol⁻¹ k⁻¹

6 0
3 years ago
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