C11. Consider a cylindrical reservoir 0.6 m high, with...

C11. Consider a cylindrical reservoir 0.6 m high, with inside diameter of 0.4 m and negligible wall thickness. The bottom and the top of the reservoir are perfectly insulated. The reservoir is exposed to an air current at 15 °C. In this reservoir, full of liquid, an electrical heating resistance is introduced dissipating continuously 200 W. The reservoir side wall is covered by an insulating material. Once the steady state is attained the liquid temperature stabilizes at 70 °C. The liquid side heat transfer coefficient is $h_i = 40 \text{ W/(m}^2 \text{.K)}$ and the air side heat transfer coefficient is $h_e = 10 \text{ W/(m}^2 \text{.K)}$. According to the given data and the operating conditions described, answer the following questions: a) Draw the equivalent thermal resistance circuit of the system presented. b) Determine the overall heat transfer coefficient. c) Determine the inside reservoir wall temperature. d) If the insulating material has 2 cm thickness, determine the value of the respective thermal conductivity to guarantee the operating conditions described above. e) Obtain the temperature profile equation along the insulating material wall. Sketch the temperature profile obtained, conveniently justifying. Note: if you did not solve d) use $k_{isol} = 0.2 \text{ W/(m.K)}$

Transcript

00:01 So let's first try to find a formula for the rate of heat transfer.

00:03 We know that the rate of heat transfer is going to be equal to ka times t sub 2 minus t sub 1 divided by the thickness l.

00:14 So we can say that this is going to be equal to negative ka times the change in temperature with respect to the radius.

00:23 We can say that h divided by 2 pi r times dr is going to be equal to negative kl.

00:32 Dt.

00:34 So at this point we can integrate and say from a to b h over 2 pi times d r over r this is going to equal integrating from t2 to t1 times negative kl d t and so we can say that h over 2 pi times l n of b over a is going to equal negative k l x times t t sub 1 minus t sub 2 now solve for h so h is going to be equal to 2 pi kl times t sub 2 minus t sub 1 divided by the natural log of b over a so this would be your answer for part a for part b we know that h over 2 pi r times d r again equals negative k l d t now we're trying to find a, essentially the temperature with respect to the radius.

01:49 So we can say that we can integrate from a to r of h over 2 pi times dr over r.

01:59 This is going to be equal to negative integral from t2 to t1 of kldt.

02:11 So we can say that h over 2 pi times ln of r over a.

02:18 Is going to be equal to negative kl t minus t 2.

02:24 Sorry, my apologies, this is actually just t because you're trying to find the temperature with respect to a radius.

02:33 So at this point, we can say that we can actually solve for t and say that t of r is going to be equal to t sub 2 minus h over 2.

02:57 Pi kl times l n of r over a however we need to substitute and we know that two pi k l equals h ln of b over a divided by t2 minus t sub 1 so we can actually substitute and say that t sub r a t of r is going to be equal to t sub 2 minus t sub 2 minus t sub 1 times ln of r over a divided by ln of b over a so this would be the temperature as a function of the radius that would be your answer for part b for part c however we're trying to be trying to find the rate of heat transfer and we can simplify it so we can first say that h again equals 2 pi kl times t sub 2 minus t sub 1 all divided by l and of b sub b divided by a so we can first make the i guess intuitively we can say that b over a is going to be equal to 1 plus b minus a over a of course this is simply algebraic manipulation and here a is approximately equal to b so we can say that ln of b over a is simply going to be equal to ln of 1 plus b minus a over a.

04:50 So at this point, and this is approximately going to be equal to b minus a divided by a.

04:58 So we can say that h now is going to be equal to 2 pi kl t sub 2 minus t sub 1, divided by b minus a, divided by a, and this is simply going to be equal to 2 pi a, kl times t sub 2 minus t sub 1, divided by b minus a.

05:29 This will equal h.

05:30 So this will be your answer for part c.

05:35 For part d, they want us to find the heat transfer in the cork.

05:39 This is again going to be equal to the 2 pi times the thermal conductivity of cork times the thickness times 140 degrees celsius minus t and this will be divided by ln of 4 over 2.

05:55 We know that the heat transfers for styrofoam is going to be equal to 2 pi times the thermoconductivity of styrofoam times the thickness times t minus 15 degrees celsius and then this will be divided by allen of 6 over 4...

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