Problem 14A.6

Problem 14A.9

 

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restart;

We used the following properties of air to solve both problems 14A.6 and 14A.9 in Bird, Stewart, & Lightfoot.  The properties include viscosity, density, specific heat, and thermal conductivity.

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mu:=0.0190*cp;rho:=.0723*lb/ft^3;Cphat:=0.241*Btu/lb/R;k:=0.0152*Btu/hr/ft/R;

mu := 0.190e-1*cp

rho := 0.723e-1*lb/ft^3

Cphat := .241*Btu/(lb*R)

k := 0.152e-1*Btu/(hr*ft*R)

Problem 14A.6 – Forced-convection heat transfer from an isolated sphere.  (a) A solid sphere 1 in. in diameter is placed in an otherwise undisturbed air stream, which approaches at a velocity of 100 ft/s, a pressure of 1 atm, and a temperature of an imbedded electric heating coil.  What must be the rate of electrical heating in cal/s to maintain the stated conditions?  Neglect radiation and use Eq. 14.4-5.  (b) Repeat the problem in (a), but use Eq. 14.4-6.

part a

start out with the given parameters in the problem

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l0:=0.083*ft;Tinf:=560*R;T0:=660*R;r:=.0415*ft;v0:=100*ft/s;

l0 := 0.83e-1*ft

Tinf := 560*R

T0 := 660*R

r := 0.415e-1*ft

v0 := 100*ft/s

first convert the velocity to feet per hour and convert the viscosity to the correct English units

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v:=v0*3600*s/hr;mu:=mu*2.4191*lb/ft/hr/cp;

v := 360000*ft/hr

mu := 0.4596290e-1*lb/(ft*hr)

now calculate the Reynolds number and the Prandtl number

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Rey:=l0*v*rho/mu;Pr:=Cphat*mu/k;

Rey := 47001.47293

Pr := .7287538750

Now use equation 14.4-5 to solve for the Nusselt Number

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Num:=2+0.60*Rey^.5*Pr^(1/3);

Num := 119.0578680

Use the definition of the heat transfer coefficient

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hm:=Num*k/l0;

hm := 21.80336860*Btu/(hr*ft^2*R)

Use the definition of the heat flow to find the necessary heat

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Q:=hm*4*3.14157*r^2*(T0-Tinf);

Q := 47.18745151*Btu/hr

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Q:=Q*hr/3600/s*252*cal/Btu;

Q := 3.303121606*cal/s

part b

instead of equation 14.4-5 we will now use equation 14.4-6 to solve the problem

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muinf:=.051796*lb/ft/hr;

muinf := 0.51796e-1*lb/(ft*hr)

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Num2:=2+(0.4*Rey^.5+0.06*Rey^(2/3))*Pr^.4*(muinf/mu)^.25;

Num2 := 151.6672930

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hm2:=Num2*k/l0;

hm2 := 27.77521511*Btu/(hr*ft^2*R)

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Q2:=hm2*4*3.14157*r^2*(T0-Tinf);

Q2 := 60.11188638*Btu/hr

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Q2:=Q2*hr/3600/s*252*cal/Btu;

Q2 := 4.207832047*cal/s

Problem 14A.9 – The ice-fisherman on Lake Mendota.  Compare the rates of heat loss of an ice-fisherman, when he is fishing in calm weather (wind velocity zero) and when the wind velocity is 20 mph out of the north.  The ambient air temperature is -10oF.  Assume that a bundled-up ice-fisherman can be approximated as a sphere 3 ft. in diameter.

the poor fisherman, using the same properties of air

For the first situation, the wind velocity is zero

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l20:=3*ft;r2:=1.5*ft;T20:=558*R;T2inf:=449.4*R;Num3:=2;hm3:=Num3*k/l20;

l20 := 3*ft

r2 := 1.5*ft

T20 := 558*R

T2inf := 449.4*R

Num3 := 2

hm3 := 0.1013333333e-1*Btu/(hr*ft^2*R)

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Q3:=hm3*4*3.14157*r2^2*(T20-T2inf);

Q3 := 31.11511457*Btu/hr

for the second situation there is an wind velocity of 20mph

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v20:=20*mi/hr;v2:=v0*5280*ft/mi;

v20 := 20*mi/hr

v2 := 105600*ft/hr

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Rey2:=rho*l20*v2/mu;

Rey2 := 498328.8696

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Num4:=2+0.60*Rey2^.5*Pr^(1/3);

Num4 := 383.1561223

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hm4:=Num4*k/l20;

hm4 := 1.941324353*Btu/(hr*ft^2*R)

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Q4:=hm4*4*3.14157*r2^2*(T20-T2inf);

Q4 := 5960.973323*Btu/hr

Thus, we can see that there is a heat loss of 190X greater when there is a wind velocity of 20mph.  Considering that the average human who is seated and quiet produces 360 Btu/hr, our fisherman friend may be in trouble!

                                

http://ebusscottcity.ebusbuilder.com/images/ice%20fishing.jpg

 

 

 

 

 

 

 

 

 

 

 

 

Jones, Esplin 2005