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Give step-by-step solution with explanation and final answer:5. A sealed rigid vessel has a volume of 1 m* and contains 2 kg of water at 100 °C. The vessel
i

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Act like a helpful tutor and exlain me :
Give step-by-step solution with explanation and final answer:5. A sealed rigid vessel has a volume of 1 m* and contains 2 kg of water at 100 °C. The vessel
is now heated and the temperature and pressure both rise. A pressure safety valve is already
installed in the container wall. At what pressure should the valve be set so that the
maximum temperature reached in the container is 200 °C. [Total 10 points]

VIPSolutions · Accepted Answer

Let's solve the problem **step by step**:

---

## **Step 1: Known Data**

- Volume, $ V =  \, \text{m}^3 $
- Mass, $ m = 2 \, \text{kg} $
- Initial temperature, $ T_1 = 100^\circ \text{C} $
- Final temperature, $ T_2 = 200^\circ \text{C} $
- Rigid (constant volume) vessel

---

## **Step 2: Find Specific Volume**

The **specific volume** ($ v $) is:

$$
v = \frac{V}{m} = \frac{1 \, \text{m}^3}{2 \, \text{kg}} = .5 \, \text{m}^3/\text{kg}
$$

---

## **Step 3: Find Pressure at 200°C with $ v = .5 \, \text{m}^3/\text{kg} $**

We need to find **the pressure at 200°C** for water with a specific volume of $.5 \, \text{m}^3/\text{kg}$.

We use **superheated steam tables** for water at $T = 200^\circ C$:

- At 200°C, the **saturated vapor specific volume** $ v_g \approx .1272 \, \text{m}^3/\text{kg} $ (from steam tables).
- Our specific volume is much higher ($.5 > .1272$), so the water is in the **superheated region**.

Now, search the superheated steam tables for water at $T = 200^\circ C$:

| Pressure (MPa) | Specific Volume ($v$) ($\text{m}^3/\text{kg}$) |
|:--------------:|:--------------------------------------------------:|
| .1            | 1.694                                              |
| .2            | .8857                                             |
| .5            | .3745                                             |

We see that at .2 MPa, $v = .8857$, and at .5 MPa, $v = .3745$. Our $v = .5$ is between these values.

---

## **Step 4: Interpolate to Find Pressure**

Let's interpolate between .2 MPa and .5 MPa for $v = .5$:

Using the interpolation formula:

$$
P = P_1 + \frac{(v - v_1)}{(v_2 - v_1)} \cdot (P_2 - P_1)
$$
Where:
- $P_1 = .5\, \text{MPa}$, $v_1 = .3745$
- $P_2 = .2\, \text{MPa}$, $v_2 = .8857$
- $v = .5$

$$
P = .5 + \frac{(.5 - .3745)}{(.3745 - .8857)} \cdot (.5 - .2)
$$

Calculate numerator and denominator:
- $.5 - .3745 = .1255$
- $.3745 - .8857 = -.5112$
- $.5 - .2 = .3$

So,

$$
P = .5 + \frac{.1255}{-.5112} \times .3
$$
$$
= .5 - .0737
$$
$$
= .426 \, \text{MPa}
$$

---

## **Step 5: **Final Answer**

### **Set the safety valve at:**

$$
\boxed{.43 \, \text{MPa}}
$$

or (rounded):

$$
\boxed{.43 \, \text{MPa} \ \text{(430 kPa)}}
$$

---

### **Summary**

- Find specific volume: $.5 \, \text{m}^3/\text{kg}$
- At 200°C, use superheated steam tables to interpolate between pressures for specific volume.
- The safety valve should be set at **.43 MPa** so that the maximum temperature reached in the vessel is 200°C.

Let me know if you want to see the interpolation steps in more detail!

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