Check the solution to "For laminar flow over a flat plate, the..."

High School
Algebra
Pre-Algebra
Ratios, rates, proportions

Donald Johnson

Answered question

2021-12-16

For laminar flow over a flat plate, the local heat transfer coefficient

h_{x}

is known to vary as

x^{- \frac{1}{2}}

, where x is the distance
from the leading edge (x=0) of the plate. What is the ratio of the average coefficient between the leading edge and some
location x on the plate to the local coefficient at x?

Answer & Explanation

Laura Worden

Beginner2021-12-17Added 45 answers

We are given following data:
$h_{x} \propto x^{\frac{1}{2}}$
$= K x^{\frac{1}{2}}$
Required:
$\times \frac{\overset{―}{h}}{h^{x}}$
Using the relation (6.14) of average coefficient:
$\overset{―}{h} = \frac{1}{x} \int_{0}^{x} h_{x} d x$
$= \frac{1}{x} \int_{0}^{x} K x^{\frac{- 1}{2}} d x$
$= \frac{1}{x} \times K {[\begin{matrix} \frac{x^{\frac{- 1}{2} + 1}}{\frac{- 1}{2} + 1} \end{matrix}]}_{0}^{x}$
$= \frac{K}{x} [\begin{matrix} 2 x \frac{1}{2} \end{matrix}]$
$h = 2 K x^{\frac{1}{2}}$
Calculating ratio of average coefficient between the leading edge and some locaion x on the plate to the local coefficient at x
$\frac{\overset{―}{h}}{h^{x}} = \frac{2 K x^{\frac{1}{2}}}{K x \frac{1}{2}}$
$\frac{\overset{―}{h}}{h^{x}} = 2$

Barbara Meeker

Beginner2021-12-18Added 38 answers

Solution: We have for laminar heat flow over a flat plate the local heat transfer coefficient

h_{x}

,
varies as

h_{x}, \sim C x^{- \frac{1}{2}}

, where

h_{x}

, is the coefficient at a distance x from the leading edge of the surface and the quantity C,
which depends on the fluid properties, is independent of x. Here, we
assume that

h_{x}, = C x^{- \frac{1}{2}}

.
The average heat transfer coefficient between the leading edge to some location x (say at point P) is piven by

\overset{―}{h_{x}} = \frac{\int_{0}^{x} h_{x} d x}{\int_{0}^{x} d x}

= \frac{1}{x} \int_{0}^{x} C x^{- \frac{1}{2}} d x

= \frac{2 C}{x} {[\begin{matrix} x^{\frac{1}{2}} \end{matrix}]}_{0}^{x} = 2 C x^{- \frac{1}{2}} = 2 h_{x}

Hence,

\frac{\overset{―}{h_{x}}}{h_{x}} = 2

nick1337

Expert2021-12-27Added 777 answers

Step1
The average heat transfer coefficient is given by as follow:-
$h_{x} o =$ average heat transfer coefficient
$h_{x = x} =$ local heat transfer coefficient
$\overset{―}{h_{x = 0}} = \frac{1}{x} \int_{0}^{x} h_{x} d x$
$\overset{―}{h_{x = 0}} = \frac{1}{x} \int_{0}^{x} C x^{- 1 / 2} d x$
$\overset{―}{h_{x = 0}} = \frac{1}{x} \int_{0}^{x} C \frac{x^{- 1 / 2 + 1}}{- 1 / 2 + 1} d x$
$\overset{―}{h_{x = 0}} = \frac{1}{x} \int_{0}^{x} C \frac{x^{1 / 2}}{1 / 2} d x$
$\overset{―}{h_{x = 0}} = \frac{2}{x} (C x^{1 / 2})_{0}^{x}$
Step2
after normal integration,
we get,
$\overset{―}{h_{x = 0}} = 2 C x^{- 1 / 2}$
$\overset{―}{h_{x = 0}} = 2 \times C \times h_{x}$
hence the final result shows that the average heat transfer coefficient is twice of local heat transfer coefficient.
Therefor the ratio of average and local heat transfer coefficient = 2

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