91Ó°ÊÓ

The term 'isentropic efficiency' describes how well a pump, compressor, or turbine operates compared to an ideal version of that device.
An ideal pump or compressor increases the pressure of a fluid in a perfectly efficient manner. Real-world devices, however, are not perfectly efficient.
The formula for isentropic efficiency (\( \eta_S \)) is:
\[ \eta_S = \frac{W_{\text{ideal}}}{W_{\text{actual}}} \]
Here, \( W_{\text{ideal}} \) represents the work an ideal pump would do, and \( W_{\text{actual}} \) represents the actual work done, considering losses due to friction and heat.
In this exercise, the given isentropic efficiency is 75%, meaning the pump operates at 75% of the efficiency of an ideal pump.

To understand the work done by the pump, let's first look at the formula provided:
\[ W_{\text{pump,ideal}} = \frac{(P_2 - P_1)}{\rho} \]
This formula calculates the work done by an ideal pump per unit mass of fluid.
In our problem, \( P_1 \) is 1 bar and \( P_2 \) is 4 bar.
After converting these pressures to Pascals and using the density of water (\( \rho = 1000 \, \text{kg/m}^3 \)), we find:
\[ W_{\text{pump,ideal}} = \frac{(4 \times 10^5 \text{ Pa} - 10^5 \text{ Pa})}{1000 \text{ kg/m}^3} = 300 \text{ J/kg} \]
However, the actual work done by the pump is higher because of the pump's efficiency.
The formula to find the actual work done is:
\[ W_{\text{pump}} = \frac{W_{\text{pump,ideal}}}{\eta_S} \]
Substituting the values gives us:
\[ W_{\text{pump}} = \frac{300 \text{ J/kg}}{0.75} = 400 \text{ J/kg} \]
This adjustment takes into account the pump’s isentropic efficiency.

The cost of operating an electrically driven pump is influenced by the electricity rate and the pump's power consumption.
The power required can be calculated using the actual work done by the pump and the flow rate:
\[ P = W_{\text{pump}} \times \frac{dm}{dt} \]
With an actual work of 400 J/kg and flow rate of 50 kg/s:
\[ P = 400 \text{ J/kg} \times 50 \text{ kg/s} = 20000 \text{ W} = 20 \text{ kW} \]
The hourly cost of running the pump is then calculated by multiplying the power consumption by the electricity rate and the operating time.
Here, the electricity cost is given as 8.5 cents per kWh.
Therefore, the hourly cost is:
\[ \text{cost} = P \times \text{rate per kW} \times h \]
\[ \text{cost} = 20 \text{ kW} \times 8.5 \text{ cents/kW} \times 1 \text{ hour} \]
Converting to dollars:
\[ \text{cost} = 20 \times 8.5 \times 0.01 = 1.70 \text{ dollars/hour} \]
This means it costs $1.70 per hour to run the pump under the given conditions.