Respuesta :
Answer:
heat rate = 7.38 W
Explanation:
Given Data:
Pressure = 1atm
diameter (D) = 5mm = 0.005m
length = 2
mass flow rate (m) = 140*10^-6 kg/s
Exit temperature = 160°C,
At 400K,
Dynamic viscosity (μ) = 22.87 *10^-6
Prandtl number (pr) = 0.688
Thermal conductivity (k) = 33.65 *10^-3 W/m-k
Specific heat (Cp) = 1.013kj/kg.K
Step 1: Calculating Reynolds number using the formula;
Re = 4m/πDμ
= (4*140*10^-6)/(π* 0.005*22.87 *10^-6)
= 5.6*10^-4/3.59*10^-7
= 1559.
Step 2: Calculating the thermal entry length using the formula
Le = 0.05*Re*Pr*D
Substituting, we have
Le = 0.05 * 1559 * 0.688 *0.005
Le = 0.268
Step 3: Calculate the heat transfer coefficient using the formula;
Nu = hD/k
h = Nu*k/D
Since Le is less than given length, Nusselt number (Nu) for fully developed flow and uniform surface heat flux is 4.36.
h = 4.36 * 33.65 *10^-3/0.005
h = 0.1467/0.005
h = 29.34 W/m²-k
Step 4: Calculating the surface area using the formula;
A = πDl
=π * 0.005 * 2
=0.0314 m²
Step 5: Calculating the temperature Tm
For energy balance,
Qc = Qh
Therefore,
H*A(Te-Tm) = MCp(Tm - Ti)
29.34* 0.0314(160-Tm) = 140 × 10-6* 1.013*10^3 (Tm-100)
0.921(160-Tm) = 0.14182(Tm-100)
147.36 -0.921Tm = 0.14182Tm - 14.182
1.06282Tm = 161.542
Tm = 161.542/1.06282
Tm = 151.99 K
Step 6: Calculate the rate of heat transferred using the formula
Q = H*A(Te-Tm)
= 29.34* 0.0314(160-151.99)
= 7.38 W
the Prandtl number using the formula
Following are the response to the given question:
Surface temperature of the tube at the exit [tex]T_{s,0} = 160^{\circ}\ C[/tex].
Properties of air at [tex]T= 400\ K[/tex] are as follows
Air Density, [tex]\rho =0.8711 \ \frac{kg}{m^3}\\\\[/tex]
Specific heat of air, [tex]c_p, =1.014 \times 10^3 \ \frac{J}{kg \ K} \\\\[/tex]
Dynamie viscosity of air, [tex]\mu = 230.1\times 10^{-7} \ \frac{N-s}{m^2}\\\\[/tex]
Thermal conductivity of air,[tex]k = 33.8\times 10^{-3} \ \frac{W}{m-K}\\\\[/tex]
Prandtl number of air, [tex]Pr = 0.690[/tex]
Reynolds number is given by
[tex]Re_{D}=\frac{4m}{\pi D \mu} \\\\=\frac{4 \times 135\times 10^{-6}}{\pi \times 5\times 10^{-3} \times 230.1 \times 10^{-7}}\\\\= 1494.02[/tex]
Since the Reynolds number is less than 2300 the flow is considered to be laminar
Thermal entry length is given by,
[tex]X_{fd , \lambda}= 0.05 Re_{D} Pr D \\\\= 0.05 \times 1494.02 \times 0.690 \times 5 \times 10^{-3}\\\\= 0.257 (< < L)\\\\[/tex]
Since thermal entry length is less than given length the flow is considered to be fully developed.
For fully developed laminar flow with constant surface heat flux, Nusselt number is given by.
[tex]Nu_{D}, = 4.36 \\\\\frac{h_0 D}{k}=4.36\\\\h_0=\frac{4.36 \ k}{D}\\\\[/tex]
[tex]=\frac{4.36 \times 33.8 \times 10^{-3}}{5 \times 10^{-3 }}\\\\= 29.47 \frac{W}{m^2 \ K}[/tex]
Apply energy balance.
[tex]mc_p,(T_{m,o}- T_{m,j})=h_0\pi DL(T_{s,0}-T_{m,0})\\\\(T_{m,o}- T_{m,j})= \frac{h_0\pi DL}{mc_p} (T_{s,0}-T_{m,0}) \\\\(T_{m,o}- 100)= \frac{29.47\times \pi \times 5x10^{-3} \times 2}{135 \times 10^{-6} \times 1.014 \times 10^{3}} (160-T_{m,0})\\\\(T_{m,o}- 100)= \frac{925.82\times 10^{-3}}{136.89 \times 10^{-3} } (160-T_{m,0})\\\\(T_{m,o}- 100)= 6.76 (160-T_{m,0})\\\\T_{m,o}=152.32^{\circ} \ C[/tex]
Heat rate entering the tube is given by,
[tex]q=h_o \pi DL(T_{s,o} -T_{m, o}) \\\\q= 29.47\times \pi 5 \times 10^{-3} \times 2(160-152.26) \\\\q=7.165 \ W[/tex]
Therefore, the heat transfer is[tex]q = 7.165 W.[/tex]
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