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2 years ago

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A two-fluid heat exchanger has inlet and outlet temperatures of 65 and 40 degree C for the hot fluid and 15 and 30 degree C for

the cold fluid. Can you tell whether this exchanger is operating under counterflow or parallel-flow conditions? Determine the effectiveness of the heat exchanger.

1 answer:

Darya [45]2 years ago

5 0

Answer:

Parallel flow

$\epsilon =0.5$

Explanation:

Given that

For hot fluid

Inlet temperature=65 C

Outlet temperature=40 C

For cold fluid

Inlet temperature=15 C

Outlet temperature=30 C

Here is only temperature is given and did not provide the information about heat capacity of fluids so by only seeing temperature we can not predict the types of heat exchanger. But there is not too much difference in temperature of exits fluids so we can say that this is parallel flow heat exchanger.

Effectiveness of heat exchanger given as

$\epsilon =\dfrac{\Delta T}{\Delta T_{max}}$

$\epsilon =\dfrac{65-40}{65-15}$

$\epsilon =0.5$

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2 years ago

Read 2 more answers

An adiabatic gas turbine expands air at 1300 kPa and 500◦C to 100 kPa and 127◦C. Air enters the turbine through a 0.2-m2 opening

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Given:

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Temperature, $T_{1}$ = $500^{\circ}$

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velocity, v = 40 m/s

A = 1 $m^{2}$

Solution:

For air propertiess at

$P_{1}$ = 1300 kPa

$T_{1}$ = $500^{\circ}$

$h_{1}$ = 793kJ/K

$v_{1}$ = $0.172\frac{m^{3}}{kg}$

and also at

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$T_{2} = 127^{\circ}$

$h_{2}$ = 401 KJ/K

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$m' = \frac{Av}{v_{1}}$

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2 years ago

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eimsori [14]

Answer:

Here is the JAVA program:

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for (int i = 0; i < arrowBaseHeight; i++) {

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System.out.print("*"); } //displays stars

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for (int i = 1; i <= arrowHeadWidth; i++)

{ for(int j = k; j > 0; j--) {//inner loop to print stars

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k = k - 1;

System.out.println(); } } } // continues to add more asterisks for new line

Explanation:

The program asks to enter the height of the arrow base, width of the arrow base and the width of arrow head. When asking to enter the width of the arrow head, a condition is checked that the arrow head width arrowHeadWidth should be less than or equal to width of arrow base arrowBaseWidth. The while loop keeps iterating until the user enters the arrow head width larger than the value of arrow base width.

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The last nested loop is used to output an arrow head of width arrowHeadWidth. The inner loop forms the arrow head and prints the stars needed to form an arrow head. So (3) is satisfied.

The value of temporary variable k is decreased by 1 so the next time it enters the nested for loop it will be one asterisk lesser.

The screenshot of output is attached.

3 0

2 years ago

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

Lera25 [3.4K]

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

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

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

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

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

8 0

2 years ago