Consider a cold fluid is flowing over a warmer flat surface, at a velocity low e

ID: 2994691 • Letter: C

Question

Consider a cold fluid is flowing over a warmer flat surface, at a velocity low enough for the

flow to be laminar. Assume plate temperature does not change. a) For the same bulk velocity of

the fluids, how many times faster would water cool (i.e., remove heat from) the plate as

compared to air? b) Keeping all other parameters the same, if the bulk velocity (flow rate)

doubles (still laminar flow), what happens to average h? Water and air properties are given in

the Appendices (Tables C.8, C.11, and C.12). Consider a film temperature of 350 K.

Average Nu = 0.664*Re^0.5*Pr^0.33

Re = V*L / neu

Nu = 0.664*(V*L/neu)^0.5 *Pr^0.33

At 350 K, For air, neu = 20.92*10^-6 m^2/s and Pr = 0.7

At 350 K, For water, neu = 365*10^-6 m^2/s and Pr = 2.29

Since V and L are constant for both fluids,

Nu_water / Nu_air = (Pr_water / Pr_air)^0.33 / (neu_water / neu_air)^0.5

= (2.29 / 0.7)^0.33 / (365 / 20.92)^0.5

= 0.355

Now Nu = hL/k

At 350 K, for water k = 668*10^-3 W/mK

At 350 K, for air k = 30*10^-3 W/m-K

h_water / h_air = (Nu_water / Nu_air)*(k_water / k_air)

= 0.355*668/30

= 7.91

Hence, water will cool 7.91 times faster.

When V doubles, Nu will change by V^0.5 times = 2^0.5 = 1.41 times.

Hence, h will also increase by 1.41 times.

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