Question

team with a quality of 0.7, pressure of 1.5 bar, and flow rate of 10 kg/s enters a steam separator operating at steady state. Sat-urated vapor at 1.5 bar exits the separator at state 2 at a rate of 6.9 kg/s while saturated liquid at 1.5 bar exits the separator at state 3. Neglect-ing kinetic and potential energy effects, determine the rate of heat transfer, in kW, and its associated direction.

Answer 1

Answer:

$\dot Q = - 194.295\,kW$

Explanation:

The model for the separator is obtained after applying the First Law of Thermodynamics:

$\dot Q + \dot m_{1}\cdot h_{1} - \dot m_{2} \cdot h_{2} - \dot m_{3} \cdot h_{3} = 0$

The rate of the heat transfer is:

$\dot Q = \dot m_{3}\cdot h_{3} + \dot m_{2}\cdot h_{2} - \dot m_{1}\cdot h_{1}$

The specific enthalpies of each state are, respectivelly:

$h_{1} = 844.55\,\frac{kJ}{kg} + 0.7\cdot (1946.4\,\frac{kJ}{kg} )$

$h_{1} = 2207.03\,\frac{kJ}{kg}$

$h_{2} = 2791.0\,\frac{kJ}{kg}$

$h_{3} = 844.55\,\frac{kJ}{kg}$

$\dot Q = (3.1\,\frac{kg}{s})\cdot (844.55\,\frac{kJ}{kg} )+ (6.9\,\frac{kJ}{kg} )\cdot (2791.0\,\frac{kJ}{kg} ) - (10\,\frac{kg}{s} )\cdot (2207.03\,\frac{kJ}{kg} )$

$\dot Q = - 194.295\,kW$

Answer 2

Answer:

-222.649 kW

Explanation:

The first step to solve this problem is to use the first law of thermodynamics:

$\frac{dE}{dt}=Q-W+\sum{m_{i}*h_{i}}-\sum{m_{e}*h_{e}}$

Note that Q, W, and m represent rate of heat transfer, rate of work, and mass flow rate.

This reduces to:

$0=Q-0+m_{1}h_{1}-m_{2}h_{2}-m_{3}h_{3}$

This equation can be rearranged to solve for rate of heat transfer (Q):

$Q=m_{3}h_{3}+m_{2}h_{2}-m_{1}h_{1}$

The enthalpy values for this problem can be found from tables. From Fundamentals of Engineering Thermodynamics 9th Edition, the table used was Table A-3: Properties of Saturated Water (Liquid-Vapor): Pressure Table.

The value for $h_{1}$ is found by using the equation $h_{1}=h_{f}+x(h_{g}-h_{f})$

$h_{1}=467.11+0.7(2693.6-467.11)=2025.653$ kJ/kg

The value for $h_{2$ is the value for $h_{g}$ at 1.5 bar due to it being saturated vapor.

$h_{2}$ = 2693.6 kJ/kg

The value for $h_{3}$ is the value for $h_{f}$ at 1.5 bar due to it being saturated liquid.

$h_{3}$ = 467.11 kJ/kg

Now these values can be substituted back into the equation solving for Q:

$Q=m_{3}(467.11)+m_{2}(2693.6)-m_{1}(2025.653)$

Note that $m_{3}=m_{1}-m_{2}=10-6.9=3.1$ kg/s

The equation solving for Q becomes:

$Q=(3.1)(467.11)+(6.9)(2693.6)-(10)(2025.653)$

$Q=1448.041+18585.84-20256.53=-222.649$ kW