condenser consists of 16 vertical tubes in 4x4 array. Water 

flows through the tubes at 65°C while steam at 75°C condenses 

over the tubes. Find the rate of condensations, Take Length of 

tube = 120 cm, Diameter of tube =25 mm Latent heat of steam 

2300 𝑘𝐽⁄𝑘𝑔 Properties of water at 70°C: Density = 

977.8 𝑘𝑔 𝑚3 ⁄ , 𝑘 = 0.668 𝑊⁄𝑚𝐾, 𝐶𝑝 = 4.187 𝑘𝐽⁄𝑘𝑔𝐾, 𝜇 =

0.415 × 10−6 𝑘𝑔𝑚⁄𝑠


To find the rate of condensation for this setup, we'll use principles from heat transfer and thermodynamics. Here's the step-by-step breakdown:

### 1. **Determine Surface Area of the Tubes**:
The total surface area \( A \) of all the tubes is calculated using the formula for the lateral surface area of a cylinder:

\[
A = n \times \pi \times D \times L
\]

Where:
- \( n \) is the number of tubes,
- \( D \) is the diameter of the tube (25 mm = 0.025 m),
- \( L \) is the length of the tube (120 cm = 1.2 m),
- \( \pi \) is 3.1416.

You have 16 tubes in total.

\[
A = 16 \times \pi \times 0.025 \times 1.2
\]

\[
A \approx 16 \times 3.1416 \times 0.025 \times 1.2 = 1.507 \, \text{m}^2
\]

### 2. **Determine the Heat Transfer Rate**:
The heat transfer due to condensation is given by:

\[
Q = h \times A \times \Delta T
\]

Where:
- \( Q \) is the heat transfer rate,
- \( h \) is the heat transfer coefficient,
- \( A \) is the surface area (calculated above),
- \( \Delta T = T_{steam} - T_{water} = 75°C - 65°C = 10°C = 10 \, \text{K}\).

The heat transfer coefficient \( h \) can be estimated using empirical correlations for condensation, but this depends on further data such as the flow regime of the steam. For simplicity, we will assume a reasonable value for \( h \) based on typical values in condensation on a tube surface, say \( h = 5000 \, \text{W/m}^2\text{K} \).

Thus, the heat transfer rate \( Q \) becomes:

\[
Q = 5000 \times 1.507 \times 10
\]

\[
Q \approx 75350 \, \text{W} = 75.35 \, \text{kW}
\]

### 3. **Determine the Rate of Condensation**:
The rate of condensation \( \dot{m} \) can be found using the relation:

\[
\dot{m} = \frac{Q}{L_v}
\]

Where:
- \( \dot{m} \) is the rate of condensation (kg/s),
- \( L_v \) is the latent heat of vaporization (2300 kJ/kg = 2300 × 10^3 J/kg).

Now substitute the values:

\[
\dot{m} = \frac{75350}{2300 \times 10^3}
\]

\[
\dot{m} \approx 0.0328 \, \text{kg/s}
\]

So, the rate of condensation is approximately **0.0328 kg/s**.

### Summary:
- **Rate of condensation:** 0.0328 kg/s
- **Heat transfer rate:** 75.35 kW

Would you like more details on any specific part of the solution? 

---

Here are 5 related questions to explore further:
1. How does changing the diameter of the tubes affect the rate of condensation?
2. What would happen to the rate of condensation if the steam temperature were lower?
3. How would the calculation change if the number of tubes was different?
4. What are the effects of different water flow rates inside the tubes?
5. How can we estimate the heat transfer coefficient more accurately?

**Tip:** Always ensure that all units are consistent when performing heat transfer calculations!

condenser consists of 16 vertical tubes in 4x4 array. Water flows through the tubes at 65°C while steam at 75°C condenses over the tubes. Find the rate of condensation. Take Length of tube = 120 cm, Diameter of tube = 25 mm. Latent heat of steam 2300 kJ/kg. Properties of water at 70°C: Density = 977.8 kg/m³, k = 0.668 W/m·K, Cp = 4.187 kJ/kg·K, μ = 0.415 × 10−6 kg/m·s.

This detailed solution explains how to find the rate of condensation for a system with 16 vertical tubes arranged in a 4x4 array. Learn the heat transfer principles, formulas, and thermodynamic properties used in the calculation. Suitable for college-level thermodynamics courses.

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