Determine the amount of heat required to produce 1Kg of steam at a pressure of 7 bar, at a temperature 29 degree Celsius, under the following . (i) When the steam is wet having dryness fraction 0.87 (ii) When the steam is dry saturated Assume specific heat = 2.35KJ/KgK Answer : P 7bar T 29ᵒC x=0.9 m =1kg C p =2.35 KJ/KgK At pressure 7 bar saturation Temp T sat =165 ᵒC h f =697.2 KJ/Kg h fg= 2064.8 KJ/Kg i) Heat required when steam is wet ... ) When the stem is dry saturated hg=hf+hfg =2762 kj Heat actually required = 2762-121.8 = 2640.2 kj ...
Represent generation of steam on H-S diagram. Show constant dryness fraction lines, constant temperature lines, saturated line and superheated region on the same. Answer : Represent generation of steam on H-S diagram. Show constant dryness fraction lines, constant temperature lines, saturated line and superheated region on the same. ...
In a constant pressure process steam is generated from 10 bar and 0.8 dry condition till it become dry and saturated. Determine amount of heat added per kg of steam. From steam table at 10 bar Tsat = 179.9°C hf = 762.6 kl/kg, hg = 2776.2 kJ/kg Answer : Constant pressure process Final condition of steam is dry saturated Enthalpy of dry steam = hg= hf + hfg =2776.2 KJ/Kg .Given Latent heat of vaporization = hfg = hg - hf =2776.2-762.6 =2013.6 KJ/Kg Initial condition ... hfg = 762.6+0.8 X 2013.6 =2373.48 KJ/Kg Heat added = hg-hw = 402.72 KJ/kg...
In a boiler enthalpy of water supplied was 2000 kJ/kg. Enthalpy being added by fuel combustion is 3200 kJ/kg. Using first law of thermodynamics, find amount of heat supplied if steam generation rate is 10 tons per hour. Answer : Heat supplied Q = m (h2-h1) …….(Steady flow energy equation for boiler) ; h2=3200 KJ/kg h1=2000 KJ/kg m = 10 Tons /hour = (10 x 1000)/3600 kg/sec = 2.77 Kg/sec Q = 2.77(3200-2000) = 3324 KJ/sec heat supplied for 10 Tons/hr steam generation...
An aqueous solution of sodium chloride is prepared by dissolving 10kg of sodium chloride in 50 kg of water find (i) Weight % (ii) Mole % of solution. [Atomic weight of Na = 23, Cl = 35.5] Answer : Weight of NaCl = 10 kg Weight of H2O = 50 kg Total weight = 60 kg Weight % of NaCl = (10/ 60) * 100 =16.67% Weight % of H2O = (50/ 60) * 100 = 83.33% Molecular weight of NaCl = 58.5 k moles of NaCl ... 100 = 5.79% Mol % of H2O = (moles of H2O / Total moles)*100 = (2.78/ 2.949)*100 = 94.26%...
50.5k questions
47.1k answers
240 comments
7.0k users
