Recent content by sci0x

Thanks Chester 1 bar G = 2 bar absolute 2 bar = 200kPa (so using 199 kPa data) Enthalpy of Steam - Enthalpy of water = Enthalpy of Evaporation = 2202 kJ kg-1 Dryness factor = (2202)(0.9) = 1981.84 kJ kg-1 5,986,134 kJ =(Mass of steam)(1981.8 kJ kg-1) Mass of steam = 3020 kg Would you agree or...

Thanks. Density = Mass/Volume 999 kg m^3 = Mass/(20 m^3) Mass of water = (20)(999) = 19980 kg Heat required to to heat 199800 kg water: Q =m C dT = (19980 kg)(4.19 kJ kg-1K-1)((85-15) = 5,860,134 kJ Also heating the stainless steel & Coil: Q = m C dT = 1000kg (1.8 kJ kg-1 K-1)(85-15) =...

I'm trying to complete this past exam paper Q. Water volume = 200hl = 20000L 1L=10^-3 m^3 20000L = 200 m^3 Density of water at 15 deg C = 999 kg m^3 Density = Mass/Volume 999 kg m^3 = Mass/(200 m^3) Mass of water = (200)(999) = 199800 kg Heat required to to heat 199800 kg water: Q =m C dT =...

Thanks for this

Thanks for your replies = 121000 = 0 = 4 m = 1.01 x 10^5 Pa = X = 0 P1+1/2ρ⋅V^21+ρ⋅g⋅h1= P2+1/2ρ⋅V^22+ρ⋅g⋅h2 121000 + 0 + (978)(9.81)(4) = 1.01x10^5 +1/2(978)V^2 + 0 159376.72 = 101,000 + 489V^2 58376.72 = 489 V^2 119.379 = V^2 10.926 m/s = V (hope this is right) If...

Hi there, I'm doing a past exam paper Q and i'd like some help. Assumptions are: The velocity in the tank is negligible and the hydrostatic head is 4m. Pressure in the vessel: Gauge pressure 1 bar g = 10^5 Pa 0.2 bar g = 20,000 Pa Hydrostatic Pressure: (4)(9.81)(978) = 38,376.72 Pa Absolute...

Can i ask where you got the the 9.9 and 266.7 in the Bernouilli Eq Also in the exam Q, they gave Darcys Eq, and use 4, not 1/2. Exam notes say answer for pump power is 7.1 kW

Hi Chester, this is my understanding of the Question My problem is that I am ending up with negative pump power

Great Chester. The next Q is: Set out the Bernouilli energy balance equation which describes the system at maximum pump power demand: My answer: Atmospheric pressure + Depth of beer to Transfer + Distance of pump below fermenting vessel + Pump Power = Pressure on maturation vessel + Pressure...

First calc reynolds no. Re Flow = 1.5ms-2 Re = (density x mean velocity x diameter) / viscosity = (1007 kgm-3)(1.5ms-1)(0.1m) / 0.002 Pa s = 75,525 = Turbulent flow I used this chart Here And I am getting friction factor of 0.02 Would you agree with this?

Abs Pressure: Gauge press: 1 bar G = 100,000 Pa = 0.987 atm 0.1 bar G = 0.0987 atm Atm press = 101.3 kPa = 1 atm Hydrostatic press: average CO2 occurs 2.5m up the vessel (9.81 ms-2)(1010kgm-3)(5/2) = 24,770.25 Pa 1 Pa = 9.869x10^-6 atm 24,770.25 Pa = 0.2445 atm Abs press= 0.0987 atm + 1 atm...

Yes i get you, the constant entropy lines are the lines going up to the right. So starting at inlet compressor following this line up to where it intersects with the outlet compressor has enthalpy of 700 kj/kg approx. (700kj/kg - 500 kj/kg) x 0.0923 kg/s = 18.46 kW 80% effecient so need 23.075...

Okay great, thanks. The next Q is to calc the electrical power required by the compressor set if it is 80% effecient Pressure and temp before compressor = 0.3 Mpa & 5 degrees superheat Enthalpy before compressor = 500 kj/kg Pressure and temp after compressor = 1.1Mpa & 5 deg superheat...

What is the equilibrium temperature at 300 kPa? Drawing line straight across from 0.3MPa gives me -16 deg C approx What is the inlet temperature and pressure? 5 degrees superheat and 0.3Mpa You asked me the equilibrium temp so I am guessing 5 deg superheat means inlet temp is -11 deg C What is...

This is after giving me more insight. Ive to draw the pressure bars 1.1 and 0.3 bar across to the temperature and use superheating and subcooling temperatures to calc h1, h2, h3, h4 like he does in the video.