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just olya [345]
3 years ago
7

Air enters a compressor operating at steady state at 1.05 bar, 300 K, with a volumetric flow rate of "84" m3/min and exits at 12

bar, 400 K. Heat transfer occurs at a rate of 14 kW from the compressor to its surroundings. Assuming the ideal gas model for air and neglecting kinetic and potential energy effects, determine the power input, in kW.
Engineering
1 answer:
Nina [5.8K]3 years ago
8 0

Answer:

W = - 184.8 kW

Explanation:

Given data:

P_1 = 1.05 bar

T_1 = 300K

\dot V_1 = 84 m^3/min

P_2 = 12 bar

T_2 = 400 K

We know that work is done as

W = - [ Q + \dor m[h_2 - h_1]]

forP_1 = 1.05 bar,  T_1 = 300K

density of air is 1.22 kg/m^3 and h_1 = 300 kJ/kg

for P_2 = 12 bar, T_2 = 400 K

h_2 = 400 kj/kg

\dot m = \rho \times \dor v_1 = 1.22 \frac{84}{60} =1.708 kg/s

W = -[14 + 1.708[400-300]]

W = - 184.8 kW

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A counter-flow double-piped heat exchange is to heat water from 20oC to 80oC at a rate of 1.2 kg/s. The heating is to be accompl
lawyer [7]

Answer:

110 m or 11,000 cm

Explanation:

  • let mass flow rate for cold and hot fluid = M<em>c</em> and M<em>h</em> respectively
  • let specific heat for cold and hot fluid = C<em>pc</em> and C<em>ph </em>respectively
  • let heat capacity rate for cold and hot fluid = C<em>c</em> and C<em>h </em>respectively

M<em>c</em> = 1.2 kg/s and M<em>h = </em>2 kg/s

C<em>pc</em> = 4.18 kj/kg °c and C<em>ph</em> = 4.31 kj/kg °c

<u>Using effectiveness-NUT method</u>

  1. <em>First, we need to determine heat capacity rate for cold and hot fluid, and determine the dimensionless heat capacity rate</em>

C<em>c</em> = M<em>c</em> × C<em>pc</em> = 1.2 kg/s  × 4.18 kj/kg °c = 5.016 kW/°c

C<em>h = </em>M<em>h</em> × C<em>ph </em>= 2 kg/s  × 4.31 kj/kg °c = 8.62 kW/°c

From the result above cold fluid heat capacity rate is smaller

Dimensionless heat capacity rate, C = minimum capacity/maximum capacity

C= C<em>min</em>/C<em>max</em>

C = 5.016/8.62 = 0.582

          .<em>2 Second, we determine the maximum heat transfer rate, Qmax</em>

Q<em>max</em> = C<em>min </em>(Inlet Temp. of hot fluid - Inlet Temp. of cold fluid)

Q<em>max</em> = (5.016 kW/°c)(160 - 20) °c

Q<em>max</em> = (5.016 kW/°c)(140) °c = 702.24 kW

          .<em>3 Third, we determine the actual heat transfer rate, Q</em>

Q = C<em>min (</em>outlet Temp. of cold fluid - inlet Temp. of cold fluid)

Q = (5.016 kW/°c)(80 - 20) °c

Q<em>max</em> = (5.016 kW/°c)(60) °c = 303.66 kW

            .<em>4 Fourth, we determine Effectiveness of the heat exchanger, </em>ε

ε<em> </em>= Q/Qmax

ε <em>= </em>303.66 kW/702.24 kW

ε = 0.432

           .<em>5 Fifth, using appropriate  effective relation for double pipe counter flow to determine NTU for the heat exchanger</em>

NTU = \\ \frac{1}{C-1} ln(\frac{ε-1}{εc -1} )

NTU = \frac{1}{0.582-1} ln(\frac{0.432 -1}{0.432 X 0.582   -1} )

NTU = 0.661

          <em>.6 sixth, we determine Heat Exchanger surface area, As</em>

From the question, the overall heat transfer coefficient U = 640 W/m²

As = \frac{NTU C{min} }{U}

As = \frac{0.661 x 5016 W. °c }{640 W/m²}

As = 5.18 m²

            <em>.7 Finally, we determine the length of the heat exchanger, L</em>

L = \frac{As}{\pi D}

L = \frac{5.18 m² }{\pi (0.015 m)}

L= 109.91 m

L ≅ 110 m = 11,000 cm

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11. Wet-cell batteries are commonly referred to as lead-acid batteries.<br> A) O True<br> B) O False
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3 years ago
Create a file named students containing the following data in your current directory. Each line in this file represents a studen
Ostrovityanka [42]

Answer:

#!/bin/bash

# Simple line count example, using bash

# Usage: ./line_count.sh file

# -----------------------------------------------------------------------------

# Link filedescriptor 10 with stdin

exec 10<&0

# stdin replaced with a file supplied as a first argument

exec < $1

# remember the name of the input file

in=$1

# init

file="current_line.txt"

let count=0

# this while loop iterates over all lines of the file

while read LINE

do

   # increase line counter  

   ((count++))

   # write current line to a tmp file with name $file (not needed for counting)

   echo $LINE > $file

   # this checks the return code of echo (not needed for writing; just for demo)

   if [ $? -ne 0 ]  

    then echo "Error in writing to file ${file}; check its permissions!"

   fi

done

echo "Number of lines: $count"

echo "The last line of the file is: `cat ${file}`"

# Note: You can achieve the same by just using the tool wc like this

echo "Expected number of lines: `wc -l $in`"

# restore stdin from filedescriptor 10

# and close filedescriptor 10

exec 0<&10 10<&-

Explanation:

4 0
4 years ago
An estimated 60% of annual precipitation in a watershed (drainage area = 20000 acres) is evaporated. If the average annual river
Nastasia [14]

Answer:

The answer is 2.715 In

Explanation:

An estimated 60% of annual precipitation in a watershed (drainage area = 20000 acres) is evaporated. If the average annual river flow at the outlet of the basin has been observed to be 2.5 cfs, determine the annual precipitation (inches) in the basin

The annual precipitation in inches in the basin is 2.715 inches.

The solution and steps is explained in the attachment.

I hope i have been able to help.

3 0
3 years ago
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