3. Steam enters an adiabatic turbine at 4MPA and 350°C and expands to 75 kPa with an isentropic efficiency of 85 percent. Determine the power output of this turbine when the mass flow rate is 1.7kg/s.
3. Steam enters an adiabatic turbine at 4MPA and 350°C and expands to 75 kPa with an isentropic efficiency of 85 percent. Determine the power output of this turbine when the mass flow rate is 1.7kg/s.
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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![**Problem Statement:**
Steam enters an adiabatic turbine at 4 MPa and 350°C and expands to 75 kPa with an isentropic efficiency of 85 percent. Determine the power output of this turbine when the mass flow rate is 1.7 kg/s.
**Solution Explanation:**
To determine the power output of the turbine, you will need to perform energy calculations using properties of steam and the given conditions. The process includes the following steps:
1. **Obtain Inlet and Outlet States:**
- You will use the steam tables or a thermodynamics software tool to find the specific enthalpies at the inlet and outlets of the turbine.
2. **Calculate Isentropic Enthalpy:**
- Identify the isentropic (ideal) enthalpy at the outlet pressure (75 kPa).
3. **Account for Isentropic Efficiency:**
- Use the isentropic efficiency to determine the actual enthalpy at the outlet.
4. **Determine Change in Enthalpy:**
- Calculate the enthalpy change, which is the difference between the inlet and outlet enthalpies.
5. **Calculate Power Output:**
- Use the mass flow rate (1.7 kg/s) and the change in enthalpy to find the power output of the turbine.
**Example Calculation:**
For educational purposes, let’s assume hypothetical values (students should use actual values from the steam tables):
- Specific enthalpy at inlet (h1) = 3200 kJ/kg
- Specific enthalpy at isentropic outlet (hs2) = 2500 kJ/kg
Given:
- Isentropic Efficiency (η) = 85% = 0.85
\[ \eta = \frac{h1 - h2}{h1 - hs2} \]
Step-by-step calculations:
- Find actual outlet enthalpy (h2):
\[ h2 = h1 - \eta \times (h1 - hs2) \]
\[ h2 = 3200 - 0.85 \times (3200 - 2500) \]
\[ h2 = 3200 - 0.85 \times 700 \]
\[ h2 = 3200 - 595 \]
\[ h2 = 2605 \text{ kJ/kg} \]
- Calculate](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F74ab977b-b4a0-40ea-97fa-cc243f1e86a7%2F0888c6d6-70ab-43f4-9c12-03fc930da02c%2F5lmxgtp_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Steam enters an adiabatic turbine at 4 MPa and 350°C and expands to 75 kPa with an isentropic efficiency of 85 percent. Determine the power output of this turbine when the mass flow rate is 1.7 kg/s.
**Solution Explanation:**
To determine the power output of the turbine, you will need to perform energy calculations using properties of steam and the given conditions. The process includes the following steps:
1. **Obtain Inlet and Outlet States:**
- You will use the steam tables or a thermodynamics software tool to find the specific enthalpies at the inlet and outlets of the turbine.
2. **Calculate Isentropic Enthalpy:**
- Identify the isentropic (ideal) enthalpy at the outlet pressure (75 kPa).
3. **Account for Isentropic Efficiency:**
- Use the isentropic efficiency to determine the actual enthalpy at the outlet.
4. **Determine Change in Enthalpy:**
- Calculate the enthalpy change, which is the difference between the inlet and outlet enthalpies.
5. **Calculate Power Output:**
- Use the mass flow rate (1.7 kg/s) and the change in enthalpy to find the power output of the turbine.
**Example Calculation:**
For educational purposes, let’s assume hypothetical values (students should use actual values from the steam tables):
- Specific enthalpy at inlet (h1) = 3200 kJ/kg
- Specific enthalpy at isentropic outlet (hs2) = 2500 kJ/kg
Given:
- Isentropic Efficiency (η) = 85% = 0.85
\[ \eta = \frac{h1 - h2}{h1 - hs2} \]
Step-by-step calculations:
- Find actual outlet enthalpy (h2):
\[ h2 = h1 - \eta \times (h1 - hs2) \]
\[ h2 = 3200 - 0.85 \times (3200 - 2500) \]
\[ h2 = 3200 - 0.85 \times 700 \]
\[ h2 = 3200 - 595 \]
\[ h2 = 2605 \text{ kJ/kg} \]
- Calculate
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