# Steam at 1 bar enters an adiabatic heat exchanger with a quality of 0.855 and exits as a...

## Question:

Steam at 1 bar enters an adiabatic heat exchanger with a quality of 0.855 and exits as a saturated liquid. The steam is being cooled by a 725 kg/s stream of water which enters the heat exchanger at 20C and exits at 40C. The two flows do not mix.

Determine the volumetric flow rate (m{eq}\displaystyle ^3 {/eq}/s) of the steam entering the heat exchanger.

## Heat Exchanger:

A device that is used to transfer heat between two fluids is called a heat exchanger. The fluids typically are not in contact with each other and can exist in two separate phases. Counter and parallel flow heat exchangers are the most common type of heat exchangers used.

The given conditions for steam are:

{eq}P_1 = 1 \ bar \\ x = 0.855 \\ {/eq}

The steam exits as saturated liquid.

From the steam tables,

Specific volume at inlet, {eq}\displaystyle v_1 = x * v_g = 0.855*1.694 = 1.45 \ m^3/kg {/eq}

Note: Since the term {eq}v_f {/eq} is much smaller than {eq}\displaystyle v_g {/eq} it can be ignored.

Specific enthalpy at inlet, {eq}h_1 = h_f + x * h_{fg} \\ h_1 = 417.46 + 0.855*2258 = 2348.05 \ kJ/kg {/eq}

Specific enthalpy at exit, {eq}h_2 = h_f = 417.46 \ kJ/kg {/eq}

Let the flow rate of steam be, {eq}\dot m_s {/eq}

Heat lost by steam = {eq}\dot m_s*(h_1 - h_2) = \dot m_s*(2348.05 - 417.46) = 1930.59* \dot m_s {/eq}

Heat gained by water = {eq}\dot m_w * C * \Delta T {/eq}

C is the specific heat of water = 4.186 \ kJ/kgK

Heat gained by water = {eq}725 * 4.186 * (40 - 20) = 60,697 \ KW {/eq}

Heat lost by steam = Heat gained by water

{eq}1930.59* \dot m_s = 60,697 \displaystyle m_s = \frac {60,697}{1930.59} = 31.44 \ kg/s {/eq}

Volumetric flow rate at inlet, {eq}\dot V_1 = \dot m_s * v_1 = 31.44 * 1.45 = 45.54 \ m^3/s {/eq}