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vivado [14]
3 years ago
12

In the final stages of production, a pharmaceutical is sterilized by heating it from 25 to 75°C as it moves at 0.2 m/s through a

straight thin-walled stainless steel tube of 12.7-mm diameter. A uniform heat flux is maintained by an electric resistance heater wrapped around the outer surface of the tube. If the tube is 10 m long, what is the required heat flux?
Engineering
1 answer:
stepan [7]3 years ago
3 0

Answer:

The required heat flux = 12682.268 W/m²

Explanation:

From the given information:

The initial = 25°C

The final = 75°C

The volume of the fluid = 0.2 m/s

The diameter of the steel tube = 12.7 mm = 0.0127 m

The fluid properties for density \rho = 1000 kg/m³

The mass flow rate of the fluid can be calculated as:

m = pAV

m = \rho \dfrac{\pi}{4}D^2V

m = 1000 \times \dfrac{\pi}{4} \times ( 0.0127)^2 \times 0.2

m = 0.0253 \ kg/s

To estimate the amount of the heat by using the expression:

q = mc_p(T_{final}-T_{initial})

q = 0.0253 × 4000(75-25)

q = 101.2 (50)

q = 5060 W

Finally, the required heat of the flux is determined by using the formula:

q" = \dfrac{q}{A_s}

q" = \dfrac{q}{\pi D L}

q" = \dfrac{5060}{\pi \times 0.0127 \times 10}

q" =  12682.268 W/m²

The required heat flux = 12682.268 W/m²

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A) W' = 15680 KW

B) W' = 17113.87 KW

Explanation:

We are given;

Temperature at state 1; T1 = 290 K

Temperature at state 3; T3 = 1100 K

Rate of heat transfer; Q_in = 35000 kJ/s = 35000 Kw

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A) The power assuming constant specific heats at room temperature is gotten from;

W' = [1 - ((T4 - T1)/(T3 - T2))] × Q_in

Now, we don't have T4 and T2 but they can be gotten from;

T4 = [T3 × (r_p)^((1 - k)/k)]

T2 = [T1 × (r_p)^((k - 1)/k)]

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T4 = [(1100 × (8^((1 - 1.4)/1.4)]

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T2 = [290 × (8^((1.4 - 1)/1.4)]

T2 = 525.32 K

Thus;

W' = [1 - ((607.25 - 290)/(1100 - 525.32))] × 35000

W' = 0.448 × 35000

W' = 15680 KW

B) The power accounting for the variation of specific heats with temperature is given by;

W' = [1 - ((h4 - h1)/(h3 - h2))] × Q_in

From the table attached, we have the following;

At temperature of 607.25 K and by interpolation; h4 = 614.64 KJ/K

At T3 = 1100 K, h3 = 1161.07 KJ/K

At T1 = 290 K, h1 = 290.16 KJ/K

At T2 = 525.32 K, and by interpolation, h2 = 526.12 KJ/K

Thus;

W' = [1 - ((614.64 - 290.16)/(1161.07 - 526.12))] × 35000

W' = 17113.87 KW

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