Math Problem Statement

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

flows through the tubes at 65Β°C while steam at 75Β°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 π‘˜π½β„π‘˜π‘” Properties of water at 70Β°C: Density =

977.8 π‘˜π‘” π‘š3 ⁄ , π‘˜ = 0.668 π‘Šβ„π‘šπΎ, 𝐢𝑝 = 4.187 π‘˜π½β„π‘˜π‘”πΎ, πœ‡ =

0.415 Γ— 10βˆ’6 π‘˜π‘”π‘šβ„π‘ 

Solution

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 AA of all the tubes is calculated using the formula for the lateral surface area of a cylinder:

A=n×π×DΓ—LA = n \times \pi \times D \times L

Where:

  • nn is the number of tubes,
  • DD is the diameter of the tube (25 mm = 0.025 m),
  • LL is the length of the tube (120 cm = 1.2 m),
  • Ο€\pi is 3.1416.

You have 16 tubes in total.

A=16×π×0.025Γ—1.2A = 16 \times \pi \times 0.025 \times 1.2

Aβ‰ˆ16Γ—3.1416Γ—0.025Γ—1.2=1.507 m2A \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=hΓ—AΓ—Ξ”TQ = h \times A \times \Delta T

Where:

  • QQ is the heat transfer rate,
  • hh is the heat transfer coefficient,
  • AA is the surface area (calculated above),
  • Ξ”T=Tsteamβˆ’Twater=75Β°Cβˆ’65Β°C=10Β°C=10 K\Delta T = T_{steam} - T_{water} = 75Β°C - 65Β°C = 10Β°C = 10 \, \text{K}.

The heat transfer coefficient hh 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 hh based on typical values in condensation on a tube surface, say h=5000 W/m2Kh = 5000 \, \text{W/m}^2\text{K}.

Thus, the heat transfer rate QQ becomes:

Q=5000Γ—1.507Γ—10Q = 5000 \times 1.507 \times 10

Qβ‰ˆ75350 W=75.35 kWQ \approx 75350 \, \text{W} = 75.35 \, \text{kW}

3. Determine the Rate of Condensation:

The rate of condensation mΛ™\dot{m} can be found using the relation:

mΛ™=QLv\dot{m} = \frac{Q}{L_v}

Where:

  • mΛ™\dot{m} is the rate of condensation (kg/s),
  • LvL_v is the latent heat of vaporization (2300 kJ/kg = 2300 Γ— 10^3 J/kg).

Now substitute the values:

mΛ™=753502300Γ—103\dot{m} = \frac{75350}{2300 \times 10^3}

mΛ™β‰ˆ0.0328 kg/s\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!

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Math Problem Analysis

Mathematical Concepts

Heat Transfer
Condensation
Thermodynamics

Formulas

A = n Γ— Ο€ Γ— D Γ— L
Q = h Γ— A Γ— Ξ”T
ṁ = Q / Lv

Theorems

Heat Transfer Rate
Rate of Condensation

Suitable Grade Level

College Level