Heat Transfer Questions & Answers  
Question by Student #201427132
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While you give this example, you missed the assumptions and let assumptions as blank.
So, i try to fill it out and here's my idea :
1. Steady-state flow
2. Plate should be flat and thin plate
3.ρ, c, k are constant
4. $E_c$<<$\frac{1}{Pr}$
I wonder whether my guess is right or wrong.
sorry to bothering you
Here $\rho$, $c$, $k$ are evaluated at the film temperature. So you should say that they are assumed constant at the film temperature. Also, you should mention S-S and no radiation. It's also better to say negligible viscous dissipation rather than ${\rm Ec} \ll 1/{\rm Pr}$. 1.0 point bonus.
Question by Student #201427132
In previous lecture at 5/16, while you are teaching the inline tube bank, you said that $U_{max}$ = $ \frac{{U_{\infty}}{A_{\infty}}}{A_{min}}$ = $ \frac{{U_{\infty}}{S_{n}}}{S_{n}-D}$.
I think it should be corrected to $U_{max}$ = $ \frac{{U_{\infty}}{A_{\infty}}}{A_{min}}$=$ \frac{{U_{\infty}}{S_{n}^2\Pi}}{({S_{n}-D})^2\Pi}$=$ \frac{{U_{\infty}}{S_{n}^2}}{({S_{n}-D})^2}$.
Is there something specific reason?
I'm not sure why you want to correct it this way... You need to explain better what your idea is, because I can not understand.
Question by Student #201427132
Dear professor, i rewrite my question.
In previous lecture at 5/16, while you are teaching the inline tube bank, you said that $U_{max}$=$\frac{{U_\infty} {A_\infty}}{A_{min}}$=$\frac{{U_\infty} {S_n}}{{S_n}-D}$.
I guess that shape of cross-section of the path of air is circle.
So i think it should be corrected to $U_{max}$=$\frac{{U_\infty} {A_\infty}}{A_{min}}$ = $\frac{{U_\infty} {S_n}^2\Pi}{({S_n}-D)^2\Pi}$ = $\frac{{U_\infty} {S_n}^2}{({S_n}-D)^2}$.
If shape of cross-section of the path of air is not circle, what the shape should be?
Well, how can the cross-sectional area be a circle of radius $S_n$? this doesn't make sense.. We are dealing with a tube bank where $S_n$ is a spacing between the tubes, and not a radius of anything. 0.5 point bonus boost.
Question by Student #201427102
For bulk temp in ducts, v was not 0. Because you didn't get rid of v at enregy eq. But if L is very very long, does flow becomes fully developed?? And can we assume v=o for all y direction in energy eq???? And can get rid of v at energy eq.??
No, be careful here. When deriving $\dot{m}(c_{p2}T_{b2}-c_{p1}T_{b1})=q_{\rm w}$, we did not assume that the flow is fully developed. Thus the equation $\dot{m}(c_{p2}T_{b2}-c_{p1}T_{b1})=q_{\rm w}$ can be applied for any internal flow either fully developed or not. 1 point bonus boost.
Question by Student #201527118
I have a question about #1 of assignment 7.
: 10 rows deep and 50 tubes high. $$S_{\rm p}=S_{\rm n}=1.9cm, \ D = 6.33mm $$ $$T_{\rm s}=90^{\circ}C$$ $$T_{\rm \infty}=20^{\circ}C, \ u_{\rm \infty}=4.5m/s, \ p = 1atm$$ Wanted : $q/L, \ T_{\rm 2}$
: S-S, No radiation, No V.D., $\ T_{\rm \infty}=T_{\rm 1}, \ \rho = const$ $$Sol)$$ $$u_{\rm max}=u_{\rm \infty}\frac{S_{\rm n}}{S_{\rm n}-D}=6.75m/s$$ $$T_{\rm f}=\frac{T_{\rm \infty}+T_{\rm s}}{2}=328K$$ Table : Properties of air with $T_{\rm f}=328K$
$$ \rightarrow \rho_{\rm f}=1.08 \ kg/m^3, \ \mu_{\rm f}=1.96 \times 10^{-5} kg/m\cdot s ,\ Pr = 0.702, \\ k = 0.028 \ W/m \cdot ^{\circ} C $$ $$Then, $$ $$Re_{\rm D}=\frac{\rho_{\rm f}u_{\rm max}D}{\mu_{\rm f}} = 2354$$ From table, $ \ {Nu}_{\rm D}=C{Re_{\rm D}}^n{Pr}^{1/3}$
$$and \ {S_{\rm N}}/D={S_{\rm P}}/D=3 \rightarrow C=0.317, \ n=0.608$$ Then,
$${Nu}_{\rm D}=C{Re_{\rm D}}^n{Pr}^{1/3}=31.6$$ $$h = \frac{k}{D}{Nu}_{\rm D} = 140 \ W/m^2 \cdot ^{\circ}C$$ and, $ \ A_{\rm s}=49 \times 10 \times \pi D \times L$
$$ \therefore q/L=h \frac{A_{\rm s}}{L}(T_{\rm s}-T_{\rm \infty}) = h \times 490 \times \pi D \times (T_{\rm s}-T_{\rm \infty}) = 95.5 kW/m$$ So, I think the answer 54.9 kW/m is wrong.
You are right that there was a mistake. But it is not in the answer but in the question formulation: it should be 6 rows deep, not 10 rows deep. I fixed the mistake in the assignment: please check the latest version. 2 points bonus.
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