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Tutorial Condensation

This document provides information to calculate the steam condensation rate and heat transfer rate for saturated steam condensing on the outer surface of a vertical tube. It also calculates the water flow rate needed to maintain a certain temperature difference across the tube. Finally, it calculates similar parameters for a plate exposed to dry steam, including film thickness, heat transfer coefficient, and condensate rate.

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Kusmakar
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753 views5 pages

Tutorial Condensation

This document provides information to calculate the steam condensation rate and heat transfer rate for saturated steam condensing on the outer surface of a vertical tube. It also calculates the water flow rate needed to maintain a certain temperature difference across the tube. Finally, it calculates similar parameters for a plate exposed to dry steam, including film thickness, heat transfer coefficient, and condensate rate.

Uploaded by

Kusmakar
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
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Saturated steam at atm pressure condenses on the outer surface of a

vertical tube of length 1m and outer diameter 75mm. The tube wall is
maintained at a uniform surface temp of 40oC by the flow of cooling
water inside the tube. Estimate the steam condensation rate & the heat
transfer rate to the tube. What water flow rate will result in a 5oC temp
difference of water between the outlet & inlet of pipe? Also calculate the
flow Re No. to check the assumption of laminar flow condition.

• At atm pressure, saturation temp = 100oC,λ = 2258.76kJ/kg


• For saturated water, at mean film temp tf which is (100+40)/2 = 70
• Prop of fluid: density = 977.8kg/m3, k = 2.43kJ/m h K, viscosity = 4.06 x
10-4 kg/ms
1/4
4 𝜇 𝑘 𝑥 (𝑇𝑔 −𝑇𝑤 )
• 𝛿=
𝑔 𝜆 𝜌 (𝜌−𝜌𝑣 )
• = 2.354 x 10-4 m
3 1/4
𝜌 (𝜌 − 𝜌𝑣 )𝑔𝜆𝑘𝑓
ℎ = 0.943
𝐿𝜇𝑓 (𝑇𝑔 − 𝑇𝑤 )
• h = 13610 kJ/m2 h K
• Q = h A(Tg - Tw)
• = 13610 x π x 0.075 x 1(100-40)
• = 192321.66 kJ/h
• Steam condensation rate = m= Q/λ
• = 192321.66/2258.76
• = 85kg/h

• Re = 4m/μ P
• = 4x 85/3600x 4.06 x 10-4 x πx 0.075
• = 1187
• Heat released during condensation is picked by water in the tube
• Q = mw c Δ T
• mw = 192321.66/4.18 x 5
• = 9202 kg/h
A 0.5m square plate is exposed to dry steam at 0.08 bar. If
surface of the plate is to be maintained at 18.5oC, make
calculations for
(a) film thickness, local Heat tr. Coeff & mean flow vel of
condensate at 25cm from top of plate
(b) Av. Heat transfer coeff for entire plate
© total steam condensate rate & total heat tr. Rate to the plate
What change , if any, would result in av. Heat tr. Coeff. If the plate
is inclined at 60o to the vertical plate

• For sat vap at 0.08 bar,


• Saturation temp = 41.5oC, λ = 2402 kJ/kg
• At (41.5 + 18.5)/2 = 30oC, ρ = 995kg/m3
• µ= 8.01 x 10-4 N s/m2; k = 2.22 kJ/m hoC
1/4
4 𝜇 𝑘 𝑥 (𝑇𝑔 −𝑇𝑤 )
• 𝛿=
𝑔 𝜆 𝜌 (𝜌−𝜌𝑣 )
• At 25cm
• = 1.475x10-4 m = 0.1475mm
hx=k/δ
= 15051kJ/m2 h K
𝜌 𝑔𝛿 2
𝑣𝑒𝑙 =
3𝜇
= 0.088m/s
At bottom,i.e., x=0.5m
𝛿 = 1.775 x 10-4m
3 1/4
𝜌 (𝜌 − 𝜌𝑣 )𝑔𝜆𝑘𝑓
ℎ = 0.943
𝐿𝜇𝑓 (𝑇𝑔 − 𝑇𝑤 )
=24082kJ/m2 h K
• Q = hA ΔT
• = 24082 x (0.5x0.5)x (41.5 – 18.5)
• = 138471.5kJ/h

• m=Q/λ
• = 138471.5/2402
• = 57.65kg/h

• Re = 4m/µ b
• = 4x 57.65/(3600x8.01x10-4 x 0.5)
• = 160
• hinc= hcos60

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