An air compressor requires 11.5 kW to deliver air at the rate of 2.3 kg/min. The compressor is cooled by water whose specific enthalpy rise during the process is not to exceed 35 kJ/kg. Assuming that the change in the kinetic energy of the air is negligible and that the heat energy loss to the atmosphere is negligible, determine the flow of cooling water required. The following data relate to the state of the air at inlet to and outlet from the compressor: Specific volume Specific internal Pressure 10 237 0,185 (208 kg h)
The correct answer and explanation is :
To solve this problem, we need to determine the flow rate of cooling water required to absorb the heat from the compressor, given the air compressor’s power input, mass flow rate, and the permissible enthalpy rise of the water.
Step-by-Step Calculation:
Step 1: Determine the Heat Energy Added to the Air
The total power input to the compressor is 11.5 kW. This power is essentially used to compress the air, and some of it is converted into heat that needs to be carried away by the cooling water. The mass flow rate of the air is given as 2.3 kg/min.
To convert this into kg/s (SI unit of mass flow rate):
[
\dot{m}_{\text{air}} = \frac{2.3 \, \text{kg/min}}{60} = 0.03833 \, \text{kg/s}
]
Step 2: Determine the Energy to Be Absorbed by the Cooling Water
The air compressor requires 11.5 kW, which is the power input, and a portion of this power will be released as heat that must be absorbed by the cooling water. Assuming the change in kinetic energy of the air is negligible and that heat loss to the atmosphere is also negligible, the entire energy input is converted to heat energy.
Since the heat energy transferred to the water is equal to the heat energy added to the air, the heat absorbed by the cooling water, in watts (joules per second), is:
[
Q_{\text{water}} = 11.5 \, \text{kW} = 11,500 \, \text{W}
]
Step 3: Apply the Heat Absorption Equation for Water
The heat absorbed by the water is given by the formula:
[
Q_{\text{water}} = \dot{m}{\text{water}} \times \Delta h{\text{water}}
]
Where:
- ( Q_{\text{water}} ) is the heat energy transferred to the cooling water in watts.
- ( \dot{m}_{\text{water}} ) is the mass flow rate of water in kg/s.
- ( \Delta h_{\text{water}} ) is the enthalpy rise of the water, which is 35 kJ/kg (or 35,000 J/kg).
Rearranging the formula to solve for the mass flow rate of water:
[
\dot{m}{\text{water}} = \frac{Q{\text{water}}}{\Delta h_{\text{water}}}
]
Substitute the known values:
[
\dot{m}_{\text{water}} = \frac{11,500 \, \text{W}}{35,000 \, \text{J/kg}} = 0.3286 \, \text{kg/s}
]
Step 4: Convert to kg/min
Finally, to convert the mass flow rate of the cooling water into kg/min:
[
\dot{m}_{\text{water}} = 0.3286 \, \text{kg/s} \times 60 = 19.716 \, \text{kg/min}
]
Final Answer:
The flow rate of cooling water required is approximately 19.7 kg/min.
Explanation:
The compressor uses 11.5 kW of power, which is mostly converted into heat energy. The cooling water must absorb this heat to maintain the compressor’s temperature. By applying the principle of energy conservation and the given enthalpy rise for the water, we can calculate the required flow of water to ensure that it can absorb the heat without exceeding the enthalpy rise limit.
