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s2008m [1.1K]
4 years ago
8

Calculate the convective heat-transfer coefficient for water flowing in a round pipe with an inner diameter of 3.0 cm. The water

flow rate is 2 L/s and the water temperature is 30 °C
Engineering
1 answer:
olasank [31]4 years ago
6 0

Answer:

h = 10,349.06 W/m^2 K

Explanation:

Given data:

Inner diameter = 3.0 cm

flow rate  = 2 L/s

water temperature 30 degree celcius

Q = A\times V

2\times 10^{-3} m^3 = \frac{\pi}{4} \times (3\times 10^{-2})^2 \times velocity

V = \frac{20\times 4}{9\times \pi} = 2.83 m/s

Re = \frac{\rho\times V\times D}{\mu}

at 30 degree celcius = \mu = 0.798\times 10^{-3}Pa-s , K  = 0.6154

Re = \frac{10^3\times 2.83\times 3\times 10^{-2}}{0.798\times 10^{-3}}

Re = 106390

So ,this is turbulent flow

Nu = \frac{hL}{k} = 0.0029\times Re^{0.8}\times Pr^{0.3}

Pr= \frac{\mu Cp}{K} = \frac{0.798\times 10^{-3} \times 4180}{0.615} = 5.419

\frac{h\times 0.03}{0.615}  = 0.0029\times (1.061\times 10^5)^{0.8}\times 5.419^{0.3}

SOLVING FOR H

WE GET

h = 10,349.06 W/m^2 K

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Answer:

The temperature T= 648.07k

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T=output temperature of the first heat engine and input temperature of the second heat engine= unknown

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but carnot efficiency of heat engine =1 - \frac{Tl}{Th} \\

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since both engines have the same thermal capacities <em>n_{th} </em> therefore n_{th} =n_{th1} =n_{th2}\\n_{th }=1-\frac{T1}{T}=1-\frac{T}{T3}\\ \\= 1-\frac{1400}{T}=1-\frac{T}{300}\\

We have now that

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multiplying through by T

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multiplying through by 300

-420000+ T^{2} =0\\T^2 =420000\\\sqrt{T2}=\sqrt{420000}  \\T=648.07k

The temperature T= 648.07k

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3 years ago
Given a square matrix [A], write a single line MATLAB command that will create a new matrix [Aug] that consists of the original
Liono4ka [1.6K]

Answer:

Consider A is square matrix of order 4 x 4 generated using magic function. Augmented matrix can be generated using:

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The modulus of elasticity for a ceramic material having 6.0 vol% porosity is 303 GPa. (a) Calculate the modulus of elasticity (i
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Answer:

modulus of elasticity for the nonporous material is 340.74 GPa

Explanation:

given data

porosity = 303 GPa

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solution

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In an experiment, the local heat transfer over a flat plate were correlated in the form of local Nusselt number as expressed by
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Answer:

R= 1.25

Explanation:

As given the local heat transfer,

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But we know as well that,

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Replacing the values

h_x=Nu_x \frac{k}{x}\\h_x= 0.035Re^{0.8}_xPr^{1/3} \frac{k}{x}

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<em>*Note that A is just a 'summary' of all of that constat there.</em>

<em>That is A=0.035(\frac{V}{\upsilon})^{0.8}Pr^{1/3}k</em>

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