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Vaselesa [24]
3 years ago
6

R 134a enters a air to fluid heat exchanger at 700 kPa and 50 oC. Air is circulated into the heat exchanger to cool the R134a to

saturated liquid at 700 kPa. Ambient air at 20 oC and 100 kPa is circulated into the heat exchanger to cool the R134a. If The mass flow rate of R134a is 0.099 kg/sec and the circulated air increases in temperature by 5 oC, what is the volumetric flow rate of air through the heat exchanger
Engineering
1 answer:
OLEGan [10]3 years ago
4 0

Answer:

The volumetric flow rate of air through the heat exchanger is 3.36 m³/s.

Explanation:

Here we have,

-\dot {m_1} c_1dt_1 = \dot {m_2} c_2dt_2

The properties of R 134a at 700 kPa Saturated temperature = 26.7 °C and enthalpy = 88.8 kJ/kg

While at super heated temperature and pressure of 70 kPa and 50 °C the enthalpy is 288.53 kJ/kg

Therefore, we have, the heat lost per kg  = 288.53 kJ/kg - 88.8 kJ/kg = 199.73 kJ/kg

For air we have at 20 ° and 100 kPa, enthalpy = 294 kJ/kg

At 25 °  c_p = 1.012 J·g⁻¹·K⁻¹

    20 ° C  c_p = 1.006 kJ/(kg·K)

Therefore,  c_p air = 1.006 kJ/(kg·K)

Plugging in the values into the above equation, we have

-\dot {m_1} c_1dt_1 = \dot {m_2} c_2dt_2 = 0.099 kg/sec × 199.73 kJ/kg  = \dot {m_2} \times 1.006 \times 5  \textdegree

\dot {m_2} = 19.77/5.03 = 3.93 kg/s

At 100 kPa  and  25° C the density of air is 1.16843 kg/m³

Therefore the volume flow rate = 3.93/1.16843 =3.36 m³/s.

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2 years ago
Create a C language program that can be used to construct any arbitrary Deterministic Finite Automaton corresponding to the FDA
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Answer:

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Explanation:

/* C Program to construct Deterministic Finite Automaton */

#include <stdio.h>

#include <DFA.h>

#include <stdlib.h>

#include <math.h>

#include <string.h>

#include <stdbool.h>

struct node{

struct node *initialStateID0;

struct node *presentStateID1;

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printf("Please enter the total number of states:");

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3 years ago
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7 0
3 years ago
Using the following data, determine the percentage retained, cumulative percentage retained, and percent passing for each sieve.
vekshin1

Solution :

<u>Sieve Size</u> (in)                   <u>Weight retain</u><u>(g)</u>

3                                         1.62

2                                         2.17

$1\frac{1}{2}$                                       3.62

$\frac{3}{4}$                                        2.27

$\frac{3}{8}$                                        1.38

PAN                                    0.21

Given :

Sieve       weight       % wt. retain    % cumulative       % finer

size        retained                               wt. retain

No. 4        59.5            10.225%          10.225%            89.775%

No. 8        86.5            14.865%          25.090%           74.91%

No. 16       138              23.7154%        48.8054%         51.2%

No. 30      127.8           21.91%              70.7154%          29.2850%

No. 50      97               16.6695%         87.3849%         12.62%

No. 100     66.8            11.4796%         98.92%              1.08%

Pan          <u>  6.3    </u>           1.08%              100%                   0%

                581.9 gram

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0.149 mm, N 100, % finer 1.08

0.297, N 50 , % finer 12.62%

x  ,   10%

$y-1.08 = \frac{12.62 - 1.08}{0.297 - 0.149}(x-0.149)$

$(10-1.08) \times \frac{0.297 - 0.149}{12.62 - 1.08}+ 0.149=x$

x = 0.2634 mm

Effective size, $D_{10} = 0.2643 \ mm$

Now, N 16 (1.19 mm)  ,  51.2%

N 8 (2.38 mm)  ,  74.91%

x,  60%

$60-51.2 = \frac{74.91-51.2}{2.38-1.19}(x-1.19)$

x = 1.6317 mm

$\therefore D_{60} = 1.6317 \ mm$

Uniformity co-efficient = $\frac{D_{60}}{D_{10}}$

   $Cu= \frac{1.6317}{0.2643}$

Cu = 6.17

Now, fineness modulus = $\frac{\Sigma \text{\ cumulative retain on all sieve }}{100}$

$=\frac{\Sigma (10.225+25.09+48.8054+70.7165+87.39+98.92+100)}{100}$

= 4.41

which lies between No. 4  and No. 5 sieve [4.76 to 4.00]

So, fineness modulus = 4.38 mm

7 0
2 years ago
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Answer:

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Explanation:

Thinking process:

The relation N_{o} = N_{i} * \frac{E_{f}-E_{i}  }{KT}

With the expression Ef - Ei = 0.36 × 1.6 × 10⁻¹⁹

and ni = 1.5 × 10¹⁰

Temperature, T = 300 K

K = 1.38 × 10⁻²³

This generates N₀ = 1.654 × 10¹⁶ per cube

Now, there are 10¹⁶ per cubic centimeter

Hence, N_{d}  = 1.65*10^{16}  - 10^{16} \\           = 6.5 * 10^{15} per cm cube

5 0
3 years ago
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