Question

Steam condenses at atmospheric pressures on the external surface of the tubes
of a steam condenser. The tubes are 12 in number and each is 30 mm in
diameter and 10 m long. The inlet and outlet temperatures of cooling water
flowing inside the tubes are 25 °C and 60 °C, respectively. If the flow rate is 1.1
kg/s, calculate the following: 0 the rate of condensation of steam, (ii) the mean
overall heat transfer coefficient based on the inner surface area and the
number of transfer units.​

Answer 1

Explanation:

Given Steam condenses at atmospheric pressures on the external surface of the tubes of a steam condenser. The tubes are 12 in number and each is 30 mm in  diameter and 10 m long. The inlet and outlet temperatures of cooling water  flowing inside the tubes are 25 °C and 60 °C, respectively. If the flow rate is 1.1 kg/s, calculate the following: 0 the rate of condensation of steam, (ii) the mean overall heat transfer coefficient based on the inner surface area and the number of transfer units.

• Given :  We need to find  
•               The rate of condensation of steam
•                  So heat lost by steam = heat gained by water
•                          So m1 x h = m2 x C (t2 – t1)    (c is the specific heat capacity)
•       So h is the latent heat of steam at atmospheric pressure = 2257 kJ / kg
•                        So m1 x 2257 = 1.1 x 4.187 x (60 – 25)  
•                                             m = 161.1995 / 2257
•                                                   = 0.0714 kg / s
•                                                    = 0.0714 x 3600
•                                                    = 257 kg / h
•         The mean overall hat transfer coefficient U:
•                             Q = m2 x c x (t2 – t1)
•                                 = 1.1 x 4.187 x 10^3 (60 – 25) (specific heat of water)
•                               Q  = 161199.5 J/s
•      Now Q = U A Theta m ------------------------- 1
•                           So theta m = theta 1 – theta 2 / log that 1 / theta 2
•                                              = 100 – 25 – (100 – 60) / log (100 – 25) / (100 – 60)
•                                              = 75 – 40 / log (75 / 40)
•                                              = 55.678 deg C
•                               So A = N x π d L
•                                        = 12 x 3.14 x 0.03 x 10 ( 30 mm = 0.03 m)
•                                      A = 11.304 sq m      
• Substituting the values in equation 1 we get
•                                 161199.5 = U x 11.304 x 55.678
•                                 161199.5 = U x 629.384112
•                                          U = 161199.5 / 629.384112
•                                             U = 256.122 W / m^2 deg C
•         The number of transfer units :
•      So C max is the hot fluid that stays at constant temperature and C min refers to water.
•           So C min = m x c (c is the specific heat capacity)
•                            = 1.1 x 4.187 x 10^3
•                           = 4605.7 W / deg C
•        Now number of transfer units will be = U A / C min
•                                                                           = 255.9 x 11.304 / 4605.7
•                                                                          = 0.628  

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https://brainly.in/question/22583268