Question

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.

Answer 1

Answer:

The average rate of energy transfer to the cooker is 1.80 kW.

Explanation:

Given that,

Pressure of boiled water = 300 kPa

Mass of water = 3 kg

Time = 30 min

Dryness friction of water = 0.5

Suppose, what is the average rate of energy transfer to the cooker?

We know that,

The specific enthalpy of evaporate at 300 kPa pressure

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

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

We need to calculate the enthalpy of water at initial state

$h_{1}=h_{f}$

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

We need to calculate the enthalpy of water at final state

Using formula of enthalpy

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

Put the value into the formula

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

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

We need to calculate the rate of energy transfer to the cooker

Using formula of rate of energy

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

Put the value into the formula

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

$Q=1.80\ kW$

Hence, The average rate of energy transfer to the cooker is 1.80 kW.