Department of Chemical Engineering IIT Bombay

CL 356, Process Plant UtilitiesSpring Semester 2014

Quiz

March 20th, 2014

This is a closed book examinationAnswer all questionsTotal marks: 15

Question 1

5 marks

Stack gases at a heater outlet were found to have the composition given below:

Component Wt%Carbon di oxide 19.38Water 7.82Sulphur di oxide 0.37Nitrogen 70.47Oxygen 1.96Total 100.00

Ambient air has the following characteristics (dry basis):

Component Wt %Nitrogen 76Oxygen 23Carbon di oxide 01Total 100

Assuming that the fuel had no appreciable concentrations of Nitrogen or Oxygen, determine the percentage of excess air used to fire the fuel

Solution

Basis 100 kg of flue gas

From ambient air analysis, 76 kg of nitrogen is associated with 23 kg of OxygenSo, 70.47 kg of nitrogen is associated with 70.47 * 23/76 = 21.3 kg of oxygen (1.5)Oxygen consumed = 21.3 – 1.96 = 19.34 kg (1.5)Thus, for complete combustion, 19.34 kg of oxygen are requiredExcess oxygen = 100 * 1.96/19.34 = 10.00 % (2)

Question 2

5 marks

Page 1 of 3

A pump delivers 40 m3/h at a head of 60 mlc when coupled to a 4 pole motor operating on %0 Hz. 440 V, three phase supply. Determine the synchronous speed of the pump at full load if the slip is 3%. What would be the flow rate and head when the pump is coupled to a 2 pole motor with the same electrical supply, if the slip increases to 5%.

Solution:

Synchronous speed (rpm) = 120*f/N Where, f is frequency in Hertz and N is number of poles (0.5)

Hence for case 1, 4 pole motor synchronous speed;= 120 x 50/4 = 1500 rpmActual speed, with 3% slip = 1500 x 0.97 = 1455 rpm (1.5)

For case 2, 2 pole motorsynchronous speed = 120*50/2 = 3000 rpmActual speed with 5% slip = 3000 x 0.95 = 2850 rpm (1.5)

From pump laws, flow rate is proportional to rpmHead is proportional to rpm^2

Hence;New flow rate = 40 x (2850/1455) = 78.3 m3/hNew head = 60x (2850/1455) 2 = 230.2 mlc (1.5)

Page 2 of 3

Question 3

5 marks

A steam turbine is fed two levels of steam (HP and LP) with one exhaust. If the mechanical losses are 5%, what is the power output of the turbine.

Schematic, properties of steam and feed rates are as under:

Solution:

H in = 50,000 * 682.1 + 80,000 * 778.4 = 9.6e7 kcal/h (1.5)Hout = (80, 000 + 50, 000) * 559.3 = 7.2e7 kcal/h (1.5) H in - Hout = 2.36e7 kcal/h = 27.5 MW (1 kW = 860 kcal/h) (1) Deducting 5% losses, gives generator output as 26.1 MW (1)

ooOoo

Page 3 of 3

50 t/h, 5 bar, 200 oC, H = 682.1kcal/kg

80 t/h, 50 bar, 425 oC, H = 778.4 kcal/kg

0.139 bar, 68 oC, H = 559.3 kcal/kg