Question

A flat plate is 2 m long, 0.8 m wide and 3 mm thick. Density of plate = 3000
kg/m3
. Specific heat of plate material = 700 J / kg K. Its initial temperature is
800C. A stream of air at 200C is blown over surface of the plate along its width,
at a velocity 2 m /s. Calculate.
i) Rate of heat dissipation from the plate.
ii) Initial rate of cooling of the plate.
Properties of air: ρ = 1.09 kg / m3 , k = 0.028 W / mK, Cp = 1.007 kJ
/ kg k.
µ = 2.03 X 10-5 kg / ms, Nu = 0.664 Re 0.5 Pr 0.3333

Answer 1

Answer:

Rate of heat dissipation from the plate is 580.73 w and initial rate of cooling of the plate is 403.2MW

Explanation:

Step 1: Given : $L=2 \mathrm{~m}, w=0.8 \mathrm{~m}, t=3 \mathrm{~mm}=0.003 \mathrm{~m}$

$\begin{aligned}& \rho_P=3000 \mathrm{~kg} / \mathrm{m}^3, C_p=700 \mathrm{~J} / \mathrm{kg}-\mathrm{K},\left(t_P\right)_i=80^{\circ} \mathrm{C}, \rho_a=1.09 \mathrm{~kg} / \mathrm{m}^3, \\& k=0.028 \mathrm{~W} / \mathrm{m}-{ }^{\circ} \mathrm{C}, C=1007 \mathrm{~J} / \mathrm{kg}-\mathrm{K}, \mu=2.03 \times 10^{-5} \mathrm{~kg} / \mathrm{m}-\mathrm{s} \\& P r=0.698, \text { For air, } t_{\infty}=20^{\circ} \mathrm{C}, U=2 \mathrm{~m} / \mathrm{s}\end{aligned}$

To Find : i. Heat dissipation rate.

ii. Initial rate of cooling of plate.

Step 2: Air blows over both surfaces of plate along width.

Hence, characteristic length, $L_c=w=0.8 \mathrm{~m}$

Now, Reynolds number,

$\begin{aligned}& R e=\frac{\rho_a U L_c}{\mu} \\& =\frac{1.09 \times 2 \times 0.8}{2.03 \times 10^{-5}}=8.59 \times 10^4 \\& \because \quad R e < 5 \times 10^5 \text {, hence flow is laminar. } \\& N u_a=0.664(R e)^{0.5}(P r)^{0.33} \\& =0.664\left(8.59 \times 10^4\right)^{0.5}(0.698)^{0.33} \\& =172.837 \\&\end{aligned}$

Since,

$\begin{aligned}& N u_a=\frac{h \times 0.8}{0.028} \\& \Rightarrow \quad \frac{h \times 0.8}{0.028}=172.837 \\& h=6.0493 \mathrm{~W} / \mathrm{m}^2{ }^{\circ} \mathrm{C} . \\& \text { 4. Now, heat transfer rate for one surface, } \\& Q=h A \Delta t \\& Q=6.0493 \times 2 \times 0.8 \times(80-20) \\& Q=580.7328 \mathrm{~W}\end{aligned}$

For both surfaces,

$Q_T=2 \times 580.7328=1161.4656 \mathrm{~W}$

Step 3: Now, initial cooling rate of plate for one surface,

$\begin{aligned}Q_P & =\dot{m} C_p \Delta t \\& =\rho_P A U C_p \Delta t \\& =3000 \times 2 \times 0.8 \times 2 \times 700 \times(80-20) \\& =403.2 \mathrm{MW}\end{aligned}$

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