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alexgriva [62]
2 years ago
14

Help me, iv been having problems with ads going in my phones storage files, what can i do to stop this?

Engineering
2 answers:
Viefleur [7K]2 years ago
7 0

Answer:

so you can do this to avoid paying but you can just try to rest set your phone it sound crazy like you will loose all your contacts but you wont just re- set it. i mean u can try it other wise i dont know :)

Explanation:

hodyreva [135]2 years ago
7 0

Answer:

You can try pluging your phone into your computor and backing up your data onto itunes and factory resetting it then install your data back and see if that works.

Explanation:

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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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