Give step-by-step solution with explanation and final answer:434 . EdaE | [1 5 i x i i H H H g - IE £ Hm i
Question:
Give step-by-step solution with explanation and final answer:
434 .
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This is AI generated Response.
Asked by: Test User 100251
Created at: 2025-05-20 22:28:42
Asked by: Test User 100251
Created at: 2025-05-20 22:28:42
Final Answer:
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Let's break down the given problem step by step:
**Problem Restatement:**
> In a Rankine cycle, steam expands from 3 MPa and 400°C to 10 kPa. What is the turbine work per kg if \( h_1 = 3214 \) kJ/kg and \( h_2 = 2200 \) kJ/kg?
Options:
- 1000 kJ/kg
- 1114 kJ/kg
- 714 kJ/kg
- 1344 kJ/kg
---
### **Step 1: Understanding the Rankine Cycle**
In a Rankine cycle, the turbine work per kg (\( w_t \)) is the difference in enthalpy between the inlet and exit of the turbine:
\[
w_t = h_1 - h_2
\]
Where:
- \( h_1 \) = Enthalpy at turbine inlet (kJ/kg)
- \( h_2 \) = Enthalpy at turbine exit (kJ/kg)
---
### **Step 2: Substitute the Given Values**
Given:
- \( h_1 = 3214 \) kJ/kg
- \( h_2 = 2200 \) kJ/kg
\[
w_t = 3214 - 2200
\]
---
### **Step 3: Calculate the Turbine Work**
\[
w_t = 3214~\text{kJ/kg} - 2200~\text{kJ/kg}
\]
\[
w_t = 1014~\text{kJ/kg}
\]
---
### **Step 4: Match with the Options**
The closest value to 1014 kJ/kg from the options is **1000 kJ/kg**.
---
### **Final Answer:**
\[
\boxed{1000~\text{kJ/kg}}
\]
**Explanation:**
The turbine work per kg in a Rankine cycle is simply the enthalpy drop across the turbine, \( h_1 - h_2 \). Given values lead directly to 1014 kJ/kg, best matching the option 1000 kJ/kg.
Question:
i need detailed calculations along with the final Answer
This is AI generated Response.
Asked by: Test User 100251
Created at: 2025-05-20 22:30:17
Asked by: Test User 100251
Created at: 2025-05-20 22:30:17
Final Answer:
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Did you like this answer?
Certainly! Let's go through the problem step-by-step with detailed calculations and explanations.
---
### **Problem Restatement:**
In a Rankine cycle, steam expands from a pressure of **3 MPa and 400°C** to **10 kPa**.
Given:
- Enthalpy at turbine inlet, \( h_1 = 3214 \text{ kJ/kg} \)
- Enthalpy at turbine exit, \( h_2 = 2200 \text{ kJ/kg} \)
**Question:**
Calculate the turbine work per kg of steam, i.e., \( w_{t} = h_1 - h_2 \).
---
### **Step 1: Understand what is being asked**
The turbine work per unit mass in a Rankine cycle is given by the difference in enthalpy between the inlet and outlet of the turbine:
\[
w_{t} = h_1 - h_2
\]
---
### **Step 2: Plug in the known values**
\[
w_{t} = 3214 \text{ kJ/kg} - 2200 \text{ kJ/kg}
\]
---
### **Step 3: Perform the subtraction**
\[
w_{t} = 3214 - 2200 = 1014 \text{ kJ/kg}
\]
---
### **Step 4: Match with options**
Given options:
- 1000 kJ/kg
- 1114 kJ/kg
- 714 kJ/kg
- 1344 kJ/kg
The calculated value, **1014 kJ/kg**, is very close to **1000 kJ/kg**.
In practical engineering problems, slight differences can occur due to rounding or approximation; hence, **1000 kJ/kg** is the best match.
---
### **Final Answer:**
\[
\boxed{\textbf{1000 kJ/kg}}
\]
---
### **Summary:**
- The enthalpy drop across the turbine determines the work output.
- Calculated \( w_{t} = 1014 \text{ kJ/kg} \), approximate to 1000 kJ/kg.
- Therefore, **the turbine work per kg of steam is approximately 1000 kJ**.
---
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