VIPSolutionsAct like a helpful tutor and exlain me : Give step-by-step solution with explanation and final answer:For questions 5.and 6 90 kg/min of refrigerant-134a at 1.2 MPa and 100°C is to be cooled to 1.2 MPa and 44°C in a condenser by air. The The mass flow rate of air required for this to be possible. Give your answer to two air enters at 100 kPa and 22°C and leaves at 95 kPa and decimal places. 62°C. Determine: Answer Air Py =100kPa 2 =22°C | R-134a — hy = 90 kg/min P,=12MPa T,=100°C — B= | Py=12MPa Ty=62C Ty=44C
Question:
Act like a helpful tutor and exlain me :
Give step-by-step solution with explanation and final answer:
For questions 5.and 6
90 kg/min of refrigerant-134a at 1.2 MPa and 100°C is to
be cooled to 1.2 MPa and 44°C in a condenser by air. The The mass flow rate of air required for this to be possible. Give your answer to two
air enters at 100 kPa and 22°C and leaves at 95 kPa and decimal places.
62°C. Determine:
Answer
Air
Py =100kPa
2 =22°C |
R-134a
—
hy = 90 kg/min
P,=12MPa
T,=100°C
—
B= | Py=12MPa
Ty=62C Ty=44C
This is AI generated Response.
Asked by: manish kumar
Created at: 2025-10-18 12:35:21
Asked by: manish kumar
Created at: 2025-10-18 12:35:21
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Let's solve the problem **step by step**:
## **Step1: List Given Data**
- **Refrigerant-134a (R-134a):**
- Mass flow rate, \(\dot{m}_1 = 90\, \text{kg/min} = 1.5\, \text{kg/s}\)
- Inlet: \(P_1 = 1.2\, \text{MPa},\ T_1 = 100^\circ \text{C}\)
- Outlet: \(P_2 = 1.2\, \text{MPa},\ T_2 = 44^\circ \text{C}\)
- **Air:**
- Inlet: \(P_3 = 100\, \text{kPa},\ T_3 = 22^\circ \text{C}\)
- Outlet: \(P_4 = 95\, \text{kPa},\ T_4 = 62^\circ \text{C}\)
---
## **Step 2: Find Heat Released by Refrigerant (\(\dot{Q}_{\text{out}}\))**
\[
\dot{Q}_{\text{out}} = \dot{m}_1 (h_1 - h_2)
\]
- You need **enthalpy values** (\(h_1, h_2\)) for R-134a at the given states (from refrigerant tables):
| State | \(P\) (MPa) | \(T\) (\(^\circ\)C) | \(h\) (kJ/kg) |
|-------|-------------|---------------------|---------------|
| 1 | 1.2 | 100 | \(h_1 = 272.63\) |
| 2 | 1.2 | 44 | \(h_2 = 251.93\) |
*(Values are approximate; please check your tables for accuracy.)*
\[
h_1 - h_2 = 272.63 - 251.93 = 20.70\ \text{kJ/kg}
\]
\[
\dot{Q}_{\text{out}} = 1.5\ \text{kg/s} \times 20.70\ \text{kJ/kg} = 31.05\ \text{kJ/s} = 31,050\ \text{W}
\]
---
## **Step 3: Find Heat Absorbed by Air (\(\dot{Q}_{\text{in}}\))**
Assuming all heat lost by refrigerant is gained by air (ignoring losses):
\[
\dot{Q}_{\text{in}} = \dot{m}_a c_p (T_4 - T_3)
\]
Where:
- \(c_p\) for air ≈ **1.005 kJ/kg·K**
- \(T_4 = 62^\circ \text{C} = 335.15\,\text{K}\)
- \(T_3 = 22^\circ \text{C} = 295.15\,\text{K}\)
- \((T_4 - T_3) = 40\,\text{K}\)
---
## **Step 4: Set Up the Energy Balance**
\[
\dot{Q}_{\text{out}} = \dot{Q}_{\text{in}}
\]
\[
31.05\ \text{kJ/s} = \dot{m}_a \times 1.005\ \text{kJ/kg·K} \times 40\ \text{K}
\]
\[
31.05 = \dot{m}_a \times 40.2
\]
\[
\dot{m}_a = \frac{31.05}{40.2} = .772\ \text{kg/s}
\]
---
## **Step 5: Final Answer (2 Decimal Places)**
\[
\boxed{.77\ \text{kg/s}}
\]
---
### **Summary Table**
| Quantity | Value |
|---------------|--------------|
| \(\dot{m}_1\) | 1.5 kg/s |
| \(h_1\) | 272.63 kJ/kg |
| \(h_2\) | 251.93 kJ/kg |
| \(\dot{Q}\) | 31.05 kJ/s |
| \(c_p\) (air) | 1.005 kJ/kg·K|
| \(T_4 - T_3\) | 40 K |
| \(\dot{m}_a\) | .77 kg/s |
---
## **Final Answer**
> **The mass flow rate of air required is \(\boxed{.77\ \text{kg/s}}\).**
Let me know if you'd like to see the refrigerant enthalpy table reference or further explanation!
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