Water, initially saturated vapor at 4 bar, fills a closed, rigid container. The water is heated until its temperature is 360°C. For the water, determine the heat transfer, in kJ per kg of water. Kinetic and potential energy effects can be ignored.

Explanation:

Using table A-3, we will obtain the properties of saturated water as follows.

Hence, pressure is given as p = 4 bar.

= 2553.6 kJ/kg

At state 2, we will obtain the properties. In a closed rigid container, the specific volume will remain constant.

Also, the specific volume saturated vapor at state 1 and 2 becomes equal. So,

According to the table A-4, properties of superheated water vapor will obtain the internal energy for state 2 at and temperature so that it will fall in between range of pressure p = 5.0 bar and p = 7.0 bar.

Now, using interpolation we will find the internal energy as follows.

                   = 2963.2 - 2.005

                   = 2961.195 kJ/kg

Now, we will calculate the heat transfer in the system by applying the equation of energy balance as follows.

      Q - W = ......... (1)

Since, the container is rigid so work will be equal to zero and the effects of both kinetic energy and potential energy can be ignored.

            = 0

Now, equation will be as follows.

           Q - W =

           Q - 0 =

           Q =

Now, we will obtain the heat transfer per unit mass as follows.

                      = (2961.195 - 2553.6)

                      = 407.595 kJ/kg

Thus, we can conclude that the heat transfer is 407.595 kJ/kg.

acceleration = (change in speed) / (time for the change)

change in speed = (20-120)= -100 m/s.

1.5 min = 90 sec

acceleration = -100/90 = -1.11 m/s^2.

we don't care about the 150kg or the 2,500 meters. they don't make any difference.

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