Thermodynamics/The First Law Of Thermodynamics/Example Problem 1

Steam phase water enters a heat exchanger at 0.5 bar with an inlet mass quality of 0.75. At the outlet, the fluid leaves at a saturated liquid state. The cooling fluid, which is water, enters the heat exchanger at 15° C and leaves with a temperature of 35° C. Pressure drop is assumed to negligible for both fluid streams of the heat exchanger. The heat exchanger is assumed to be perfectly insulated and at steady-state conditions.

Determine the following:

The mass flow rate ratio between the cooling fluid in and the steam in.
The heat transfer Q(dot) delivered or received within the heat exchanger, assuming a steam mass flow rate of 1 kg/s.

Given:

1 Cal = 4.184 J, 1 bar = 10.01325x10^5 Pa

Assumptions:

The first assumption, the system operates under steady-state conditions, is given. Thus we have ${\dot{m}}_{1} = {\dot{m}}_{2}$ and ${\dot{m}}_{3} = {\dot{m}}_{4}$ for mass flow balances for both sides of the heat exchanger. Also assume that the potential and kinetic energy is maintained from the inlet to the outlet of both sides of the heat exchanger

Part a)

Define the system: The system is the whole heat exchanger

First Law of Thermodynamics using the power formulation:

$δ E_{T} / δ t = 0 = \dot{Q} + \dot{W} + {\dot{m}}_{1} h_{1} - {\dot{m}}_{2} h_{2} + {\dot{m}}_{3} h_{3} - {\dot{m}}_{4} h_{4}$

$\Leftrightarrow 0 = 0 + 0 + {\dot{m}}_{1} h_{1} - {\dot{m}}_{2} h_{2} + {\dot{m}}_{3} h_{3} - {\dot{m}}_{4} h_{4}$ $\Leftrightarrow 0 = {\dot{m}}_{1} (h_{1} - h_{2}) + {\dot{m}}_{3} (h_{3} - h_{4})$

$\Leftrightarrow \frac{{\dot{m}}_{3}}{{\dot{m}}_{1}} = \frac{h_{1} - h_{2}}{- (h_{3} - h_{4}} = \frac{h_{1} - h_{2}}{h_{4} - h_{3}}$

To determine enthalpy into $h_{1}$ the quality of the inlet mass must be used. X = 0.95. Using the equation for the quality of a mixed state vapor:

$X = \frac{h_{x} - h_{l, s a t}}{h_{v, s a t} - h_{l, s a t}}$

Substituting $h_{1}$ for $h_{x}$ and rearranging the equation to solve for $h_{1}$ , results in:

$h_{1} = h_{l, s a t} + X (h_{v, s a t} - h_{l, s a t})$

Using a steam table dependent on pressure, the following values for $h_{l, s a t}$ and $h_{v, s a t}$ can be found and substituted into the equation above to solve for $h_{1}$ :

$h_{l, s a t} (0.1 b a r) = X k c a l / k g$

Thermodynamics/Tables/Saturated Steam - Properties v.s. Pressure - Bar