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Tpy6a [65]
3 years ago
15

The interactive activities in this course

Engineering
1 answer:
MariettaO [177]3 years ago
3 0

Answer:

A

Explanation:

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A fluid has a dynamic viscosity of 0.048 Pa.s and a specific gravity of 0.913. For the flow of such a fluid over a flat solid su
sattari [20]

Answer:

Explanation:

First we should recall how Newton's laws relates shear stress to a fluid's velocity profile:

\tau = \mu \cfrac{\partial v}{\partial y}

where tau is the shear stress, mu is viscosity, v is the fluid's velocity and y is the direction perpendicular to flow.

Now, in this case we have a parabolic velocity profile, and also we know that the fluid's velocity is zero at the boundary (no-slip condition) and that the vertex (maximum) is at y=75 \, mm and the velocity at that point is 1.125 \, m/s

We can put that in mathematical terms as:

v(y)= A+ By +Cy^2 \\v(0) = 0\\v(75 \, mm) = 1.125 \, m/s\\v'(75 \, mm) = 0\\

From the no-slip condition, we can deduce that A=0 and so we are left with just two terms:

v(y) = By + C y ^2 \\

We know that the vertex is at y= 75 \, mm and so we can rewrite the last equation as:

v(y) = k(y-75 \, mm) ^2+h

where k and h are constants to be determined. First we check that v( 75 \, mm) = 1.125 \,  m/s :

v( 75 \, mm) = k(75 \, mm -75 \, mm) ^2+h = h = 1.125 \, m/s\\\\h= v_{max} = 1.125 \,  m/s

So we found that h was the maximum velocity for the fluid, now we have to determine k, for that we need to make use of the no-slip condition.

v( 0) = k( -75 \, mm) ^2+  1.125 \,  m/s= 0 \quad (no \, \textendash slip)  \\\\k= - \cfrac{ 1.125 \, m/s }{(75 \, mm ) ^2} = - \cfrac{ 1125 \, mm/s }{(75 \, mm ) ^2}\\\\k= -  \cfrac{0.2}{mm \times s}

And thus we find that the final expression for the fluid's velocity is:

v( y) = 1125-  0.2 ( y -75 ) ^2

where v is in mm/s and y is in mm.

In SI units it would be:

v( y) = 1.125-  200 ( y -0.075 ) ^2

To calculate the shear stress, we need to take the derivative of this expression and multiply by the fluid's viscosity:

\tau = \mu \cfrac{\partial v}{\partial y}

\tau =0.048\,   \cdot  (-400) ( y-0.075   )

for y= 0.050 \, m we have:

\tau =0.048\,   \cdot  (-400) ( 0.050 -0.075   ) = 0.48\, Pa

Which is our final result

5 0
4 years ago
The surface of an object is the<br> texture value appearance or function
maksim [4K]
Texture value appearance
4 0
3 years ago
Why a brass lid ring on a glass canning jar will loosen when heated.
Zolol [24]

hi here is your answer hope it helps or then sry

Explanation:

A brass lid on a glass canning jar will loosen when heated because brasshas the greater coefficient of thermal expansion 

4 0
3 years ago
An overhead 25m-long, uninsulated industrial steam pipe of 100-mm diameter, is routed through a building whose walls and air are
wariber [46]

Answer:

1) q=18414.93 W

2) C=12920$

Explanation:

Given data:

pipe length L=25m

pipe diameter D=100mm =0.1 m

air temperature T_{s1}=T_{\infty1} }=25 °C.....= 298.15k

pipe surface temp T_{s2}=150 °C.....=423.15k

surface emissivity e= 0.8

boiler efficiency η=0.90

natural gas price Cg=$0.02 per MJ

1) Total heat loss and radiation heat loss combined

          q=q_{conv} +q_{rad}

          q=A[h(T_{s2}-T_{s1})+eб(T_{s2}^4-T_{s1}^4)]....... (1)

б=5.67×10^-8 W/m^2K^4 (boltzmann constant)

area A =L.Dπ=25×0.1π=7.85 m^2

putting all these values in eq (1)

q=18414.93 W

2) suppose boiler is operating non stop annual energy loss will be

               E=q.t

                  =18414.93.3600.24.365

                  =5.81×10^11 J

   to find furnace energy consumption

               Ef =E/η

                  =6.46×10^5 MJ

   annual cost

                  C=Cg. Ef

                    =12920$

8 0
3 years ago
Poached eggs are cooked in bath of boiling water at 100°C. Over time they reach thermal equilibrium with the bath. They are then
andrew11 [14]

Answer:

a) hmax = 13334.65 w/m²k

b) t = 0.288 s

c) t = 0.693 s

d) Q = 6032.37 w = 6.032 kw

e) Transmit Heat transfer Rate into the egg is asked so it is heat of poached egg in boiling watch.

Explanation:

7 0
3 years ago
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