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14 days ago

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Water is boiled at 300 kPa pressure in a pressure cooker. The cooker initially contains 3 kg of water. Once boiling started, it

is observed that half of the water in the cooker evaporated in 30 minutes. Assume negligible heat loss from the cooker.

1 answer:

inna [2.2K]14 days ago

3 0

Answer:

The cooker receives an average energy transfer rate of 1.80 kW.

Explanation:

Given that,

Pressure of the boiling water = 300 kPa

Mass of water = 3 kg

Duration = 30 min

Dryness fraction of the water = 0.5

What is the average energy transfer rate to the cooker?

We know that,

The specific enthalpy of vapor at 300 kPa is

$h_{f}=561.47\ kJ/kg$

$h_{fg}=2163.8\ kJ/kg$

We need to determine the initial state enthalpy of water

$h_{1}=h_{f}$

$h_{1}=561.47\ kJ/kg$

We must find the enthalpy of water in the final state

Using the enthalpy formula

$h_{2}=h_{f}+xh_{fg}$

Insert the value into the formula

$h_{2}=561.47+0.5\times2163.8$

$h_{2}=1643.37\ kJ/kg$

Next, we calculate the energy transfer rate to the cooker

Using the energy rate formula

$Q=\dfrac{m(h_{2}-h_{1})}{t}$

Substituting the value into the formula

$Q=\dfrac{3\times(1643.37-561.47)}{30\times60}$

$Q=1.80\ kW$

Therefore, the average energy transfer rate to the cooker stands at 1.80 kW.

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