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blagie [28]
3 years ago
13

A long cylindrical steel rod is heat treated in an oven that is 7 meters long and is maintained at a temperature of 900°C. The r

od is 10-cm in diameter and is drawn at a velocity of 3 m/min. The rod enters the oven at 30°C and leave at 700°C. The rod has a density of 7800 kg/m³ and an average heat capacity of 0.465 kJ/kg °C. Calculate the rate of heat transfer to the rods in the oven.
Physics
1 answer:
Maru [420]3 years ago
6 0

Answer:

Q' = 954.28 KJ/s

Explanation:

We are given;

Density of rod = 7800 kg/m³

Length of steel rod; L= 3m

Diameter; D = 10cm = 0.1m

Initial temperature;T1 = 30°C

Exit temperature; T2 = 700°C

Average heat capacity; c = 0.465 kJ/kg °C

Area = πD²/4 = π x 0.1²/4 = 0.0025π

Now,we know that volume = Area x Length

We also know that density = mass/volume.

Thus, mass = density x volume

So, mass = 7800 x 0.0025π x 3 = 183.78 kg

Formula for heat transfer is;

Q = m•c•(T2 - T1)

Q = 183.78 x 0.465(700 - 30)

Q = 57256.66 KJ

Rate of heat transfer is given as;

Q' = Q/t

Question says velocity at 3m/minutes. So, for every 3m it's i minute.. Thus, t = 1 minute = 60 seconds

Thus, Q' = 57256.66/60 = 954.28 KJ/s

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A heat engine with 0.300 mol of a monatomic ideal gas initially fills a 1000 cm3 cylinder at 500 K . The gas goes through the fo
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Complete Question:

A heat engine with 0.300 mol of a monatomic ideal gas initially fills a 1000 cm3 cylinder at 500 K . The gas goes through the following closed cycle: - Isothermal expansion to 5000 cm3. - Isochoric cooling to 400 K . - Isothermal compression to 1000 cm3. - Isochoric heating to 500 K .

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b) what is the thermal efficiency

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a) Work done for 1 cycle = 402.13

b) Thermal efficiency = 20%

Explanation:

Number of moles, n = 0.300 mol

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Temperature, T = 500 K

Isothermal expansion to 5000 cm³

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Isochoric cooling

In an Isochoric process, volume is constant i.e. V₂ = V₁ = V

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But  ln(V/V) = ln 1 = 0

Work done, W₂ = 0 Joules

Isothermal Compression to 1000 cm³

V₂ = 1000 cm³

V₁ = 5000 cm³

W = nRT ln(V₂/V₁)

W = 0.3 * 8.314 * 400 ln(1000/5000)

W₃ = -1605 J

Isochoric heating to 500 K

Since there is no change in volume, no work is done

W₄ = 0 J

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W = W₁ + W₂ + W₃ + W₄

W = 2007.13 + 0 + 0 -1605+0

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Thermal efficiency = (Net workdone for 1 cycle)/(Heat absorbed)

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Thermal efficiency = 402.13/2007.13

Thermal efficiency = 0.2

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