20 long tons of ice per 12 hrs at -15°C were produced from steam at 190°C. Allow 12% for the
losses. How many TR are required? ((Use 2257 and 334 as your reference values for the latent
heats and 4.187 for the value of specific heat of water).

516.7229 TR

Step-by-step explanation

given M = 20 long tons 20X 1016.05 kg = 0. 47039 kg 12 S hys 12 x 3600 Heat required to Produce ice(19from steam at 190'c is ce = sensible Heat required to bring steam from /go'c to loo'c. Q, + Latent heat required to convert steam to water atleo's 2 + Sensible heat required to bring water from lovic to o'c ly + 1 Latent heat required to Convert water into ice at "o'c Qy + Sensible heat require to bring ice from octo - 15 c , Q , = mcp [AT] = 0. 47039 x 4.187 [190 - 100] Q 1 = 177. 258 KW Q 2 = mhig = 0. 476 3 9 x 2257 = 1061.67 KW Q3 = mcp ( 1 00 - 0 ) = 0 . 470 3 9 x 4. 187 x 100 0 3 = 196. 952 KW dy = mhig = 0. 47039 X 834 = 157. 11 KW as = my (0 - (- 15) ) = 0.470 39 x 4.187 x 15 as = 29. 543 RW Total heat required is " = Q , + 2 2 + 03 + Qy tas .

Q = 177.25 8 + 1061 . 67 + 196. 952 + 157. 11 + 29.543 Q = (622.533 KW Considering 12.1. loss , Total Required heat will become Q = (622. 533 x 1. 12 Q = 1817. 237 KW Convert kw into TR ITR = 3.5 1685 KW Bop da C = 1817. 237 TR 3.51685 CQ = 516. 72 29 TR 516. 72 29 TR is required. to Produce l'ce, at -15'0 from 1soc of Steam at rate of 20 long long / his 72