A steam turbine installation has a steam condenser that receives 34,100 kg of steam per hour with an enthalpy of 2565 kJ/kg. The steam condenses into water with an enthalpy of 160 kJ/kg. (a) Calculate the amount of heat removed from the steam to form water. (b) The condenser is cooled by seawater, which increases in temperature from 13°C to 24°C as it passes through the tubes. Find the amount of seawater passing through the tubes, assuming that 1 kg of water absorbs 4.19 kJ of heat for each degree Celsius increase.

1. Problem Description

A steam turbine installation has a steam condenser that receives 34,100 kg of steam per hour with an enthalpy of 2565 kJ/kg. The steam condenses into water with an enthalpy of 160 kJ/kg.

(a) Calculate the amount of heat removed from the steam to form water.

(b) The condenser is cooled by seawater, which increases in temperature from 13°C to 24°C as it passes through the tubes. Find the amount of seawater passing through the tubes, assuming that 1 kg of water absorbs 4.19 kJ of heat for each degree Celsius increase.

2. Solution Steps

(a) Heat removed from steam:

The mass flow rate of steam is $\dot{m} = 34100 \, \text{kg/hr}$. We need to convert this to kg/s for later calculations.

$$$$

\dot{m} = \frac{34100 \, \text{kg}}{1 \, \text{hr}} \times \frac{1 \, \text{hr}}{3600 \, \text{s}} = \frac{34100}{3600} \, \text{kg/s} \approx 9.472 \, \text{kg/s}

$$$$

The change in enthalpy is the enthalpy of the steam minus the enthalpy of the water:

$$$$

\Delta h = h_{\text{steam}} - h_{\text{water}} = 2565 \, \text{kJ/kg} - 160 \, \text{kJ/kg} = 2405 \, \text{kJ/kg}

$$$$

The total heat removed per second is:

$$$$

\dot{Q} = \dot{m} \Delta h = 9.472 \, \text{kg/s} \times 2405 \, \text{kJ/kg} = 22770.16 \, \text{kJ/s}

$$$$

The total heat removed per hour is:

$$$$

Q = \dot{Q} \times 3600 \frac{s}{hr} = 22770.16 \frac{kJ}{s} \times 3600 \frac{s}{hr} \approx 81972576 \, kJ/hr

$$$$

(b) Heat absorbed by seawater:

The temperature increase of the seawater is:

$$$$

\Delta T = 24^\circ \text{C} - 13^\circ \text{C} = 11^\circ \text{C}

$$$$

The heat absorbed by 1 kg of seawater is:

$$$$

q = 4.19 \, \text{kJ/kg} \cdot ^\circ\text{C} \times 11^\circ \text{C} = 46.09 \, \text{kJ/kg}

$$$$

Let $\dot{m}_{\text{seawater}}$ be the mass flow rate of seawater in kg/s. Then the total heat absorbed by the seawater per second is:

$$$$

\dot{Q} = \dot{m}_{\text{seawater}} \times q

$$$$

We know that the heat removed from the steam equals the heat absorbed by the seawater, so:

$$$$

22770.16 \, \text{kJ/s} = \dot{m}_{\text{seawater}} \times 46.09 \, \text{kJ/kg}

$$$$

Solving for $\dot{m}_{\text{seawater}}$:

$$$$

\dot{m}_{\text{seawater}} = \frac{22770.16}{46.09} \, \text{kg/s} \approx 494.037 \, \text{kg/s}

$$$$

Convert this to kg/hr:

$$$$

\dot{m}_{\text{seawater}} = 494.037 \, \text{kg/s} \times \frac{3600 \, \text{s}}{1 \, \text{hr}} \approx 1778533.2 \, \text{kg/hr}

$$$$

3. Final Answer

(a) The amount of heat removed from the steam is approximately 81972576 kJ/hr.

(b) The amount of seawater passing through the tubes is approximately 1778533.2 kg/hr.