Question

A boiler operates at steady state. The entering water is saturated liquid at P = 5 bar and has a flow rate of 10,000 kg/hr. The existing steam is also at P = 5 bar and T = 400o C. The pipe entering the boiler is 8 cm in diameter, and the pipe leaving the boiler is 30 cm in diameter. What is the rate at which heat is added in the boiler?

Answer 1

Answer:

The rate at which heat is added in the boiler is 8128.610 kilowatts.

Explanation:

A boiler is a device that works at steady state, whose function is pressurizing and heating water. All gravitational electric and magnetic effects can be neglected. From First Law of Thermodynamics the boiler is characterized by the following model:

$\dot Q_{in} + \dot H_{in}-\dot H_{out} + \dot K_{in} - \dot K_{out}= 0$ (Eq. 1)

Where:

$\dot Q_{in}$ - Heat transfer rate, measured in kilowatts.

$\dot H_{in}$, $\dot H_{out}$ - Inlet and outlet fluid enthalpy rates, measured in kilowatts.

$\dot K_{in}$, $\dot K_{out}$ - Inlet and outlet kinetic energy rate, measured in kilowatts.

And by the Principle of Mass Conservation we have that:

$\dot m_{in}-\dot m_{out} = 0$ (Eq. 2)

Where $\dot m_{in}$ and $\dot m_{out}$ are mass flows at entrance and exit of the boiler, measured in kilograms per second.

Then, we clear the heat transfer rate and expand (Eq. 1) by definitions of enthalpy and kinetic energy, as well as (Eq. 2):

$\dot Q_{in} = \dot H_{out}-\dot H_{in}+\dot K_{out}-\dot K_{in}$

$\dot Q_{in} = \dot m \cdot \left[(h_{out}-h_{in}\right)+\frac{1}{2}\cdot (v_{out}^{2}-v_{in}^{2})\left]$ (Eq. 3)

Speeds of fluid entering and exiting the boiler ($v$), measured in meters per second, are found by applying the following formula:

$\dot m = \left(\frac{\pi}{4} \right)\cdot \frac{D^{2}\cdot v}{\nu}$

$v = \frac{4\cdot \dot m\cdot \nu}{\pi\cdot D^{2}}$ (Eq. 4)

Where:

$\nu$ - Specific volume of fluid, measured in cubic meters per kilogram.

$D$ - Inner diameter of pipe, measured in meters.

Entrance ($\dot m = 2.778\,\frac{kg}{s}$, $\nu_{in} = 0.001093\,\frac{m^{3}}{kg}$ and $D_{in} = 0.08\,m$)

$v_{in} = \frac{4\cdot \left(2.778\,\frac{kg}{m^{3}} \right)\cdot \left(0.001093\,\frac{m^{3}}{kg} \right)}{\pi\cdot (0.08\,m)^{2}}$

$v_{in}\approx 0.604\,\frac{m}{s}$

Specific enthalpy at given conditions is:

$h_{in} = 640.09\,\frac{kJ}{kg}$ (From Saturated Water tables)

Exit ($\dot m = 2.778\,\frac{kg}{s}$, $\nu_{out} = 0.61731\,\frac{m^{3}}{kg}$ and $D_{out} = 0.30\,m$)

$v_{out} = \frac{4\cdot \left(2.778\,\frac{kg}{m^{3}} \right)\cdot \left(0.61731\,\frac{m^{3}}{kg} \right)}{\pi\cdot (0.30\,m)^{2}}$

$v_{out} \approx 24.261\,\frac{m}{s}$

Specific enthalpy at given conditions is:

$h_{out} = 3272.4\,\frac{kJ}{kg}$ (From Superheated Steam tables)

If we know that $\dot m = 2.778\,\frac{kg}{s}$, $h_{in} = 640.09\,\frac{kJ}{kg}$, $h_{out} = 3272.4\,\frac{kJ}{kg}$, $v_{in}\approx 0.604\,\frac{m}{s}$, $v_{out} \approx 24.261\,\frac{m}{s}$, the heat transfer rate is:

$\dot Q = \left(2.778\,\frac{kg}{s} \right)\cdot \left\{ \left(3272.4\,\frac{kJ}{kg}-640.09\,\frac{kJ}{kg} \right)+\frac{1}{2}\cdot \left[\left(24.261\,\frac{m}{s} \right)^{2}-\left(0.604\,\frac{m}{s} \right)^{2}\right] \right\}$

$\dot Q = 8128.610\,kW$

The rate at which heat is added in the boiler is 8128.610 kilowatts.