## Heat Exchanger

In the diagram below, water flows into an inner pipe at 25°C. An outer pipe encapsulates the inner water pipe with oil entering at 100°C and 3 kg/s. If water exits the heat exchanger at 65°C, what is the mass flow rate (kg/s) of the water? Note the specific heat (kJ/kg·K) for both water and oil are 4.186 and 3.5 respectively.

##
__
__**Hint**

**Hint**

In a heat exchanger, the rate of heat transfer associated with either steam, an incompressible fluid, or ideal gas with constant specific heat flows is

$$$\dot{Q}=\dot{m}c_p(\Delta T)$$$

where
$$c_p$$
is the specific heat (at constant pressure),
$$\dot m$$
is the mass flow rate, and
$$\Delta T$$
is the change in temperature.

##
__
__**Hint 2**

**Hint 2**

Because the heat transfer between the water & oil occurs within the same system,
$$\dot{Q}_{oil}=\dot{Q}_{water}$$
.

In a heat exchanger, the rate of heat transfer associated with either steam, an incompressible fluid, or ideal gas with constant specific heat flows is

$$$\dot{Q}=\dot{m}c_p(\Delta T)$$$

where
$$c_p$$
is the specific heat (at constant pressure),
$$\dot m$$
is the mass flow rate, and
$$\Delta T$$
is the change in temperature. Per the problem statement, we can write a rate of heat transfer equation for both the oil and water:

$$$\dot{Q}_{water}=\dot{m}_{water}\times c_{p,water}(\Delta T_{water})$$$

$$$\dot{Q}_{oil}=\dot{m}_{oil}\times c_{p,oil}(\Delta T_{oil})$$$

Because the heat transfer between the water & oil occurs within the same system,
$$\dot{Q}_{oil}=\dot{Q}_{water}$$
:

$$$\dot{m}_{water}\times c_{p,water}(\Delta T_{water})=\dot{m}_{oil}\times c_{p,oil}(\Delta T_{oil})$$$

$$$\dot{m}_{water}=\dot{m}_{oil}\times \frac{c_{p,oil}}{c_{p,water}}\times \frac{(\Delta T_{oil})}{(\Delta T_{water})}$$$

$$$\dot{m}_{water}=3\frac{kg}{s}\times \frac{3.5kJ/kg\cdot K}{4.186kJ/kg\cdot K}\times \frac{[(100+273K)-(65+273K)]}{[(65+273K)-(25+273K)]}$$$

$$$=3\frac{kg}{s}\times 0.836\times \frac{35}{40}=2.19\:kg/s$$$

2.19 kg/s