26 - Question 20 - Boiler Deaerator energy balance - Post by Kambiz Ehsani - 5/12/2018

A boiler generates 50,000 lb/hr of saturated steam at 300 psia, out of which 10,000 lb/hr is taken for process and is returned to the deaerator as condensate at 180 degree F, the remainder is consumed. Makeup water enters the DA tank at 70 degree F and steam is available at 300 psia for deaeration. The DA tank operates at 25 psia. The blow down has TDS of 1,500 ppm by weight and make up has a TDS of 100 ppm. Evaluate the blow down and deaeration steam quantities.

Solution:

If boiler concentration is 0.5 ppm solid and water hardness concentration is 2,500 ppm: then the percent moisture in steam = Steam Purity, ppm / Boiler water concentration, ppm or

percent moisture in steam = (0.5 / 2,500) * 100 = 0.02 so the steam quality = 100 – 0.02 = 99.98 %

DA Energy balance:

The mass balance around two control volumes gives us:

 If M is the make up water to DA, D is the steam for deaeration to DA, F the BFW and B is the Blowdown

For the boiler 50,000 + B = F and for DA: 10,000 + M + D = F                         Eq (1)

by equating this two we will have: 50,000 + B = 10,000 + M + D  

 Energy Balance around the two control volumes gives:

10,000 * 148 + 1,202.8 * D + M * 38 = 209 * F = 209 * (50,000 + B)           Eq (2)

Balance of the solids in the makeup and blowdown gives 100 * M = 1,500 B   Eq (3)

25 - Question 19 - General Physics - gym lifting energy burned - Post by Kambiz Ehsani - 5/9/2018 - Solved

You are at gym and working out. You are lifting weight. What is the work for lifting 30 pounds weight for 1 foot? How much energy or calories will you burn if you do this exercise for 10 minutes and you do it 500 times? Although it looks impossible :-)

Solution:

Work for lifting a weight @ gym :
30 lbf lifted 1 ft = 30 * 1 = 30 lbf-ft = 30 * 1.3558 = 41 J

Total work including 500 times lifting = 41 * 500 = 20,500 J

Power = 34 J/sec = Watts  or 5 kcal

Assuming the human body efficiency in converting the Fat to exercise of 0.25

Power converted to to burned fat = 5 / 0.75 = approximately 7 Kcal

24 - Question 18 - Entropy Calculation - posted by Kambiz Ehsani - 5/8/2018

A peice of metal with the weight of 100 kg and temperature of 500 K falls into a big lake and gets cooled down with the lake temperature of 285 K. Find the entropy change of the lake water.