Question 5 ( $$ \mathbf{2 5} $$ point) Refrigerant 134 a enters a compressor and a negligible velocity. The refrigerant enters the compressor as a saturated vapor at $$ 12^{\circ} \mathrm{C} $$ and leaves the compressor at 1400 kPa with an enthalpy of $$ 281.4 \mathrm{~kJ} / \mathrm{kg} $$ and a velocity of $$ 50 \mathrm{~m} / \mathrm{s} $$. The rate of work done on the refrigerant is measured to be $$ 27.3 \mathrm{~kJ} / \mathrm{kg} $$. If the diameter at the exit is equal to 4.6 cm , determine the rate of heat transfer associated with this process, in kW . Note: Don't do interpolation, take the closest value

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To solve this problem, we will apply the first law of thermodynamics for a control volume, which can be expressed as:

$$
\dot{Q} - \dot{W} = \dot{m} \left( h_2 - h_1 + \frac{V_2^2 - V_1^2}{2} \right)
$$

Where:
- $$\dot{Q}$$ = rate of heat transfer (kW)
- $$\dot{W}$$ = rate of work done on the refrigerant (kW)
- $$\dot{m}$$ = mass flow rate (kg/s)
- $$h_1$$ = specific enthalpy at the inlet (kJ/kg)
- $$h_2$$ = specific enthalpy at the outlet (kJ/kg)
- $$V_1$$ = velocity at the inlet (m/s)
- $$V_2$$ = velocity at the outlet (m/s)

### Step 1: Determine the specific enthalpy at the inlet ($$h_1$$)

Given that the refrigerant enters as a saturated vapor at $$12^{\circ}C$$, we can look up the specific enthalpy of saturated vapor at this temperature. The closest value for $$h_1$$ (saturated vapor enthalpy at $$12^{\circ}C$$) is approximately $$h_1 =  49.0 \, \text{kJ/kg}$$.

### Step 2: Identify the known values

- $$h_2 = 281.4 \, \text{kJ/kg}$$
- $$\dot{W} = 27.3 \, \text{kJ/kg}$$
- $$V_1 = 0 \, \text{m/s}$$ (negligible velocity)
- $$V_2 = 50 \, \text{m/s}$$

### Step 3: Calculate the mass flow rate ($$\dot{m}$$)

To find the mass flow rate, we need the area of the exit and the density of the refrigerant at the exit conditions. The diameter of the exit is given as $$4.6 \, \text{cm}$$, which we convert to meters:

$$
D = 0.046 \, \text{m}
$$

The area $$A$$ at the exit can be calculated using the formula for the area of a circle:

$$
A = \frac{\pi D^2}{4}
$$

Now, we will calculate $$A$$:

$$
A = \frac{\pi (0.046)^2}{4}
$$

Next, we need the density of refrigerant 134a at the exit conditions. At $$1400 \, \text{kPa}$$ and $$281.4 \, \text{kJ/kg}$$, we can look up the density. The closest value for density ($$\rho$$) is approximately $$ \rho \approx  1000 \, \text{kg/m}^3$$.

Now we can calculate the mass flow rate:

$$
\dot{m} = \rho \cdot A \cdot V_2
$$

### Step 4: Calculate the rate of heat transfer ($$\dot{Q}$$)

Now we can rearrange the first law equation to solve for $$\dot{Q}$$:

$$
\dot{Q} = \dot{W} + \dot{m} \left( h_2 - h_1 + \frac{V_2^2 - V_1^2}{2} \right)
$$

### Step 5: Perform the calculations

Let's calculate $$A$$, $$\dot{m}$$, and then $$\dot{Q}$$.

1. Calculate $$A$$:

$$
A = \frac{\pi (0.046)^2}{4}
$$

2. Calculate $$\dot{m}$$:

$$
\dot{m} = 1000 \cdot A \cdot 50
$$

3. Calculate $$\dot{Q}$$:

$$
\dot{Q} = 27.3 + \dot{m} \left( 281.4 - 49.0 + \frac{50^2 - 0^2}{2} \right)
$$

Now, I will perform these calculations.
Calculate the value by following steps:
- step0: Calculate:
  $$\frac{1000\left(\pi \times 0.046^{2}\right)}{4}\times 50$$
- step1: Remove the parentheses:
  $$\frac{1000\pi \times 0.046^{2}}{4}\times 50$$
- step2: Convert the expressions:
  $$\frac{1000\pi \left(\frac{23}{500}\right)^{2}}{4}\times 50$$
- step3: Reduce the fraction:
  $$\frac{529\pi }{1000}\times 50$$
- step4: Reduce the numbers:
  $$\frac{529\pi }{20}$$
Calculate or simplify the expression $$ (\pi*(0.046)^2)/4 $$.
Calculate the value by following steps:
- step0: Calculate:
  $$\frac{\left(\pi \times 0.046^{2}\right)}{4}$$
- step1: Remove the parentheses:
  $$\frac{\pi \times 0.046^{2}}{4}$$
- step2: Convert the expressions:
  $$\frac{\pi \left(\frac{23}{500}\right)^{2}}{4}$$
- step3: Multiply the numbers:
  $$\frac{\frac{23^{2}}{500^{2}}\times \pi }{4}$$
- step4: Evaluate the power:
  $$\frac{\frac{529\pi }{500^{2}}}{4}$$
- step5: Multiply by the reciprocal:
  $$\frac{529\pi }{500^{2}}\times \frac{1}{4}$$
- step6: Multiply the fractions:
  $$\frac{529\pi }{500^{2}\times 4}$$
Calculate or simplify the expression $$ 27.3 + (1000*(\pi*(0.046)^2)/4*50) * (281.4 - 49 + (50^2)/2) $$.
Calculate the value by following steps:
- step0: Calculate:
  $$27.3+\left(\frac{1000\left(\pi \times 0.046^{2}\right)}{4}\times 50\right)\left(281.4-49+\frac{50^{2}}{2}\right)$$
- step1: Remove the parentheses:
  $$27.3+\frac{1000\pi \times 0.046^{2}}{4}\times 50\left(281.4-49+\frac{50^{2}}{2}\right)$$
- step2: Divide the terms:
  $$27.3+\frac{1000\pi \times 0.046^{2}}{4}\times 50\left(281.4-49+1250\right)$$
- step3: Calculate:
  $$27.3+\frac{1000\pi \times 0.046^{2}}{4}\times 50\times 1482.4$$
- step4: Convert the expressions:
  $$27.3+\frac{1000\pi \left(\frac{23}{500}\right)^{2}}{4}\times 50\times 1482.4$$
- step5: Reduce the fraction:
  $$27.3+\frac{529\pi }{1000}\times 50\times 1482.4$$
- step6: Multiply the terms:
  $$27.3+\frac{980237\pi }{25}$$
- step7: Convert the expressions:
  $$\frac{273}{10}+\frac{980237\pi }{25}$$
- step8: Reduce fractions to a common denominator:
  $$\frac{273\times 5}{10\times 5}+\frac{980237\pi \times 2}{25\times 2}$$
- step9: Multiply the numbers:
  $$\frac{273\times 5}{50}+\frac{980237\pi \times 2}{25\times 2}$$
- step10: Multiply the numbers:
  $$\frac{273\times 5}{50}+\frac{980237\pi \times 2}{50}$$
- step11: Transform the expression:
  $$\frac{273\times 5+980237\pi \times 2}{50}$$
- step12: Multiply the numbers:
  $$\frac{1365+980237\pi \times 2}{50}$$
- step13: Multiply the terms:
  $$\frac{1365+1960474\pi }{50}$$
### Step 1: Calculate the Area $$A$$

The area at the exit is calculated as:

$$
A \approx 0.001662 \, \text{m}^2
$$

### Step 2: Calculate the Mass Flow Rate $$\dot{m}$$

The mass flow rate is calculated as:

$$
\dot{m} \approx 83.095 \, \text{kg/s}
$$

### Step 3: Calculate the Rate of Heat Transfer $$\dot{Q}$$

Now, substituting the values into the equation for $$\dot{Q}$$:

$$
\dot{Q} \approx 27.3 + 83.095 \left( 281.4 - 49 + \frac{50^2}{2} \right)
$$

Calculating the term inside the parentheses:

$$
281.4 - 49 + \frac{2500}{2} = 281.4 - 49 + 1250 = 1482.4
$$

Now substituting this back into the equation for $$\dot{Q}$$:

$$
\dot{Q} \approx 27.3 + 83.095 \times 1482.4
$$

Calculating the final value:

$$
\dot{Q} \approx 27.3 + 123207.514319
$$

Now, let's calculate this final value.
Calculate the value by following steps:
- step0: Calculate:
  $$27.3+83.095\times 1482.4$$
- step1: Multiply the numbers:
  $$27.3+123180.028$$
- step2: Add the numbers:
  $$123207.328$$
The final calculation for the rate of heat transfer $$\dot{Q}$$ is:

$$
\dot{Q} \approx 123207.328 \, \text{kW}
$$

### Conclusion

The rate of heat transfer associated with this process is approximately **123.2 kW**.
The rate of heat transfer is approximately 123.2 kW.

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