QUESTION 1
In a Thermal Power Station, the steam input to a turbine operating on a fully condensing
mode is 100 TPH. The heat rejection requirement of the steam turbine condenser is 555
kcal/kg of steam condensed. The temperature of cooling water at the inlet and outlet of the
turbine condenser is 27 0C and 37 0C respectively. Find out the circulating cooling water
flow.
ANSWER 1
The quantum of heat rejected in the turbine condenser
= Quantum of steam condensed (kg) x heat rejection (kcal/kg)
= 100,000 x 555 = 55.5 Million kcal/h.
Heat gained by circulating cooling water = Heat rejected in the condenser
Circulating cooling water flow
= 100,000 x 555 / (37-27) x specific heat (1)
= 5550 m3/hr
QUESTION 2
A medium sized engineering industry has installed two 480 CFM screw compressors, A &
B. Compressor-A is operating at full load and Compressor-B is running in load – unload
condition. The load power of both the compressor is 74 kW and the unload power of the
Compressor-B is 26 kW. Both the compressors are operated during working day.The
percentage loading of the Compressor-B during working day is 64 %. After arresting the
leakage in the system, the loading of the compressor was found to be 35 %. Estimate the
energy savings per day.
ANSWER 2
Existing Case:
Energy consumed per hour by Compressor -A= 74 kW
Energy consumed per hour by Compressor -B= 0.64 x 74 + 0.36 X 26 = 56.72 kW
Total energy consumed (Compressor A& B) = 74 + 56.72 = 130.72 kW/hr
Energy consumed per day= 130.72 X 24 hrs = 3137.3 kWh/day
Leakage Calculation:
Energy consumed per hour by Compressor -B= 0.64 x 74 + 0.36 X 26 = 56.72 kW
Energy consumed per hour by Compressor -B= 0.35 x 74 + 0.65 X 26 = 42.8 kW
Difference in power consumption = 56.72 - 42.8 = 13.92 kW/hr
Savings by arresting leakage per day= 13.92 X 24 = 334 kWh/day
QUESTION 3
A plant is operating a chilled water system always at full load. The chilled water inlet and
outlet temperatures are 12 0C and 7 0C respectively. The chilled water pump discharge
pressure is 3.6 kg/cm2g and the suction is 5 meters above the pump centreline. The power
drawn by the chilled water pump’s motor is 70 kW and an efficiency of 90 %. The chilled
water pump efficiency at the operating point from pump characteristic curve is 60 %. Find
out the operating refrigeration load in TR.
ANSWER 3
Total head 36 – 5 = 31 m
Pump shaft power 70 x 0.9
63 kW
Flow rate (63 x 1000) X 0.6 / 31 x 1000 x 9.81
0.124297 m3/s
447.5 m3/hr
Refrigeration load (447500 x 5) / 3024
740 TR
QUESTION 4
The operating data of an induced draft-cooling tower is as follows:
Observed range : 8 0C.
Cooling water flow rate : 12,500 m3/hr
Drift loss : 0.1 % of circulation rate
Wet Bulb Temperature : 27 0C
Ambient Dry Bulb Temperature : 35 0C
Effectiveness : 67 %
Cycle of Concentration : 3
Estimate the evaporation loss; make up water requirement and TR load of cooling tower.
ANSWER 4
Evaporation loss = 0.00085 X 1.8 X 12500 X 8 = 153 m3/hr
Blow Down =153 /(3-1)= 76.5 m3/hr
Make up =153 + 76.5 + (12500*0.001)= 242 m3/hr
Heat load =12500*1000*8/3024 = 33069 TR
QUESTION 5
In an air washer of a textile humidification system with an airflow of 3000 m3/h at 25 0C
and 10 % relative humidity is humidified to 60 % relative humidity by adding water
through spray nozzles. The specific humidity of air at inlet and outlet are 0.002 kg/kg of
dry air and 0.0062 kg/kg of dry air respectively. The density of air at 25 0C is 1.184
kg/m3. Calculate the amount of water required in kg/hr.
ANSWER 5
The amount of water required:
mw =v ρ ( Wout - Win)
= 3000 X 1.184 X (0.0062- 0.002)
= 14.9 kg/h
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