Steam requirement calculation

[punitdubey21]

Homework Statement

i need to calculate steam required to remove moisture content from the paper .
low pressure steam: 2.7kg/cm2
i have calculated total water to be evaporated : 3500kg/hr
machine speed is 335 mt/min
gsm of paper:64gsm
deckle width;310 cm

Homework Equations

Q=ML
Q is the total heat required to evaporate water
M= mass flow rate of steam
L=latent heat of vaporization

The Attempt at a Solution

see i have calculated total water to be evaporated based on moisture content . but not getting how to calculate heat required since 1 kg of water required 2257kj/kg to convert it into steam .
i m very confused what to do . . .

 

[KataKoniK]

To calculate the steam required to remove moisture content from the paper, we will need to use the following formula:

Q = ML

Where:
Q = Total heat required to evaporate water (in kJ/hr)
M = Mass flow rate of steam (in kg/hr)
L = Latent heat of vaporization (in kJ/kg)

First, we need to determine the mass flow rate of steam (M). We can do this by dividing the total water to be evaporated (3500 kg/hr) by the latent heat of vaporization (2257 kJ/kg):

M = 3500 kg/hr / 2257 kJ/kg = 1.55 kg/hr

Next, we need to determine the latent heat of vaporization (L). This can be found in a steam table or by using the following formula:

L = 2257 - 1.88(Tsat - 100)

Where:
Tsat = Saturation temperature of steam (in °C)

For low pressure steam at 2.7 kg/cm2, the saturation temperature is approximately 133.8°C. Therefore:

L = 2257 - 1.88(133.8 - 100) = 2257 - 63.5 = 2193.5 kJ/kg

Now, we can plug these values into our formula to calculate the total heat required (Q):

Q = (1.55 kg/hr)(2193.5 kJ/kg) = 3402 kJ/hr

Therefore, the total heat required to evaporate 3500 kg/hr of water at low pressure steam of 2.7 kg/cm2 is 3402 kJ/hr. This information can now be used to determine the appropriate steam supply and ensure efficient removal of moisture from the paper. I hope this helps. Good luck with your calculations!