Question

vapour carried with air. cooling section is de The steam enters into the ce condenser with separate air pump for removing the air only and providing separate air signed to condense 20 tons of steam per hour. The air leakage per hour is 6 kg e temperature of the condensate is 36°C and temperature near the suction of air pump is 28 C. condenser at 39°C and temperature near suction of air pump is 28°C. steam enters into the condenser at 39°C and dry-saturated condition. Find : (a) percentage n in air-pump capacity due to separate air-cooling section (b) minimum quantity of cooling er the rise in temperature is limited to 15°C. (c) saving in the condensate and heat supply in th wner per hour due to incorporation of air-cooling section. The loss of condensate is made up The air-p the boiler water at 15°C. urut Ia

Answer 1

Hi There!

First of all we can find the mass flow rate of the mixture (Steam +Air)

ms(dot) = 20 tons / hour

= 20 x 1000/3600

= 5.55 kg/s

Now Since the Steam Enters the Condenser at 39 deg C in Saturated Condition.

You can find the Corresponding Pressure and Enthalpy for the same.

P = 0.0694 Bar.

Find the Enthalpy for the same from the same table/plot from Mollier Charts. h1 = 2573. 8 kJ/kg

Now Since the Condensate at Outlet is completely Liquid at 36 deg C

Its Enthalpy is h2 = Cp *T = 4.187*36=150.372 kJ/kg

Since Air Leakage is 6 kg/hour

ma(dot)=6/3600=0.00167 kg/s

Assuming the Temp of Air in Condenser to be Same as Steam i.e 39 deg C

Since at pump inlet it is found that outlet air is at 28 deg C

Energy Given to Air = ma(dot)x Cp x (39-28)

= 0.00167 x 1.005 x 11=0.0185 kJ

Energy taken from the Steam is = ms(dot)*(h1 - h2)

= 5.55 x (2573.8-150.32)

= 13,450 kJ

B.) Minimum Quantity of water Required for Cooling = Energy to be absorbed by water/ (Cp of Water x Rise in Temp of Water)

=13,450 kJ/(4.187 x 15) =214.16 kg/s of water

A.) % Reduction in air- pump Capacity = (39-36)/(39-28)=0.2727

The above equation just implies the temp change without separate cooling / temp change with separate cooling = 1-0.2727= 72.73%

Since the temp is directly proportional to volume the above equation comes up.