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A tank contains 50 kg of water, which is stirred by a paddle wheel at 300 rpm while transmitting a torque of 0.2 kNm. After the tank achieves steady state, determine (a) the rate of heat transfer, (b) the rate of entropy transfer, and (c) the rate of entropy generation in the tank's universe. Assume the atmospheric temperature to be 25C.

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Answer:

To determine the rate of heat transfer, we need to use the first law of thermodynamics, which states that the rate of change of energy in a system is equal to the rate of heat transfer into the system minus the rate of work done by the system. At steady state, the rate of change of energy is zero, so we have:

Rate of heat transfer = Rate of work done by the paddle wheel

The rate of work done by the paddle wheel can be calculated using the formula:

Rate of work = torque x angular velocity

where torque is given as 0.2 kNm and angular velocity is given as 300 rpm. We need to convert angular velocity to radians per second:

Angular velocity = 300 rpm x (2π/60) = 31.4 rad/s

Rate of work = 0.2 kNm x 31.4 rad/s = 6.28 kW

Therefore, the rate of heat transfer is 6.28 kW.

To determine the rate of entropy transfer, we need to use the formula:

Rate of entropy transfer = Rate of heat transfer / Temperature

The temperature is given as 25°C, which is equivalent to 298 K.

Rate of entropy transfer = 6.28 kW / 298 K = 0.0211 kW/K

Therefore, the rate of entropy transfer is 0.0211 kW/K.

To determine the rate of entropy generation, we need to use the formula:

Rate of entropy generation = Rate of heat transfer / Temperature - Rate of entropy transfer

Rate of entropy generation = 6.28 kW / 298 K - 0.0211 kW/K = 20.97 kW/K

Therefore, the rate of entropy generation is 20.97 kW/K.

Step-by-step explanation:

Hope this helps!

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