Need help with this engineering problem. ## Solar Power Plant Operating on a Rankine Cycle with Refrigerant 134a### DescriptionThe figure below provides steady-state operating data for a solar power plant that operates on a Rankine cycle with Refrigerant 134a as its working fluid. The turbine and pump operate adiabatically. The rate of energy input to the collectors from solar radiation is:\[ \frac{\dot{Q}_{in}}{A} = 0.3 \text{ kW per } m^2 \]of collector surface area, with 60% of the solar input to the collectors absorbed by the refrigerant as it passes through the collectors.### Diagram ExplanationThe accompanying diagram illustrates the process:1. **Solar Collectors**: - The solar collectors receive solar radiation energy (\(\dot{Q}_{in}\)) at a rate of 0.3 kW per m^2 of the collector's surface area. - 60% of this solar input is absorbed by the refrigerant as it flows through the collectors. - The refrigerant, heated up, moves to the turbine.2. **Turbine**: - The refrigerant vapor enters the turbine at state 1 and expands to produce work (\(W_{t}\)). - The refrigerant exits the turbine at state 2.3. **Condenser**: - The refrigerant, still in vapor form, passes through the condenser. - It releases heat (\(\dot{Q}_{out}\)) and condenses into a liquid. - The refrigerant exits the condenser at state 3.4. **Pump**: - The liquid refrigerant is then pressurized by the pump, absorbing work (\(W_{p}\)). - The pressurized liquid refrigerant exits the pump at state 4 and re-enters the solar collectors, completing the cycle.### Key Points- **Working Fluid**: Refrigerant 134a- **Operating Mode**: Adiabatic for both turbine and pump- **Energy Input**: \(\frac{\dot{Q}_{in}}{A} = 0.3 \text{ kW per } m^2\)- **Energy Absorption**: 60% of solar input absorbed by refrigerantThis system is an example of how solar energy can be utilized in a Rankine cycle to generate power efficiently using a refrigerant as the working fluid. ### Thermodynamic State Table for Solar Collector System#### Table of States:| State | \( p \) (bar) | \( h \) (kJ/kg) | \( x \) ||-------|:------------:|:---------------:|:------:|| 1 | 24 | 279.1 | 1 || 2 | 7 | 259.1 | 0.9845 || 3 | 7 | 86.78 | 0 || 4 | 24 | 88.55 | -- |- **State:** Represents the different stages in the thermodynamic process within the solar collector system.- **\( p \) (bar):** Pressure in bars at each state.- **\( h \) (kJ/kg):** Specific enthalpy in kilojoules per kilogram at each state.- **\( x \):** Quality of the steam at each state, representing the proportion of the mass that is vapor (0 to 1 scale). A value of 1 indicates pure vapor, while 0 indicates pure liquid. A value of "--" signifies that the quality is not applicable or not provided.#### Task:Determine the solar collector surface area, in \(m^2\) per kW of power developed by the plant.This table, alongside the specified task, is useful for analyzing the cyclic processes of the working fluid in solar thermal power plants and determining necessary collector characteristics for efficient energy transfer.

Given data: - The rate of energy input is Qin/A = 0.3 kW/m2. The enthalpy at state 1 is h1 = 279.1…

The figure below provides steady-state operating data for a solar power plant that operates on a Rankine cycle with Refrigerant 134a as its working fluid. The turbine and pump operate adiabatically. The rate of energy input to the collectors from solar radiation is Qin /A = 0.3 kW per m2 of collector surface area, with 60% of the solar input to the collectors absorbed by the refrigerant as it passes through the collectors. Turbine W Solar collectors Condenser Pump W. wwww