Calculate the flow of steam to heat wort (in the brewing process)

[sci0x]

Homework Statement

A brewery requires wort to undergo 5% by mass evaporation in one hour. The wort is boiled at atmospheric. Using the data below, calculate flow of steam required.

Saturated steam pressure = 3 bar gauge
Condensate pressure in wort boiler = 3 bar gauge
Enthalpy of steam, at 3 bar gauge = 2739 kg kj-1
Enthalpy of water at 3 bar g = 605 kj kg-1
Initial wort volume at 100 deg c = 500 hl
Enthalpy of vapourisation of water at 100 deg C and atmospheric pressure = 2257 kj kg-1
Density of water at 100 deg C = 958 kg m-3

Relevant Equations

Q = U A dT

I need help solving this please. Its a past exam paper.

requires wort to undergo 5% by mass evaporation in one hour
does this mean starting vol is 500hl and after one hour vol should be 475hl

Wort volume = 500 hl = 50,000 L / hr
1 L = 10^-3 m^3
50,000 L = 50 m^3 /hr
= 0.833 m^3 / min
= 0.01388 m^3 / sec

Final Enthalpy
= (0.01388)(2739)
= 38.0173 kJ

Initial Enthalpy
= (0.01388)(605)
= 8.3974 kJ

Final - Initial = 29.6199 kJ

Am i on the right track here?

 

[Chestermiller]

What are the thermodynamic states of the wort and steam entering the device?

What are the thermodynamic states of the wort and steam exiting the device?

 

[BvU]

If it says 5% by mass, doesn't that mean 5% by mass and not by volume ?
And that you evaporate 5%, so not 100 % ?Try to get your units straightened out. 50,000 L = 50 m^3 /hr hurts my eyes !

E.g: m3/h * kJ/kg is not kJ

Advice: take a break, get some rest .

 

[sci0x]

Okay wort and steam enter the boiler and wort -5% mass and water leave

We have steam enthalpy but not mass = (2739)(x)
We have water enthalpy and density = (605kJ kg-1)(958 kg m-3) = 6227 kJ m-3

Can I say (2739 kJ kg-1)(x) = 6227 kJ m-3
X = 2.27 kg m-3 = steam density

Right direction?
Examiners notes say answer is 0.704 kg/s
Can you show me how to get to this Chester

 

[Chestermiller]

Okay wort and steam enter the boiler and wort -5% mass and water leave

We have steam enthalpy but not mass = (2739)(x)
We have water enthalpy and density = (605kJ kg-1)(958 kg m-3) = 6227 kJ m-3

Can I say (2739 kJ kg-1)(x) = 6227 kJ m-3
X = 2.27 kg m-3 = steam density

Right direction?
Examiners notes say answer is 0.704 kg/s
Can you show me how to get to this Chester

You still haven't answered my questions.

 

[sci0x]

Wort entering the boiler: Cool
Steam entering boiler: Superheated

Wort leaving boiler: heated to 100 deg C with evaporation
Steam leaving boiler: heat energy supplied to wort, condense to water

Thanks for your help

 

[Chestermiller]

Wort entering the boiler: Cool
Steam entering boiler: Superheated

Wort leaving boiler: heated to 100 deg C with evaporation
Steam leaving boiler: heat energy supplied to wort, condense to water

Thanks for your help

What is the temperature and pressure of the steam leaving? What makes you think that the entering steam is superheated?
What is the temperature and pressure of the steam entering?

What is the temperature and pressure of the wort leaving?

What is the temperature and pressure of the wort entering?

 

[sci0x]

What is the temperature and pressure of the steam leaving?
Temp ? // Pressure: 3 bar g (1 bar = 100,000 Pa) so 300,000 Pa

What is the temperature and pressure of the steam entering?
Temp: 100 deg C // Pressure: 3 bar g = 300,000 Pa

What is the temperature and pressure of the wort leaving?
Temp: 100 deg C // Pressure: Atm pressure - 10^5 Pa

What is the temperature and pressure of the wort entering?
Temp: ? // Pressure: Atm pressure

 

[Chestermiller]

I guess the steam is flowing through tubes in the boiler. This was not clear to me before.

The change in enthalpy of the steam passing through the tubes per kg of this steam is??

You have 50 m^3 / hr of wort entering the boiler and 5% of this evaporates. So the amount that evaporates is 2.5 m^3/hr. What is the mass flow rate in kg/hr of the wort water that evaporates? The rest of the wort does not change at all between inlet and outlet. What is the change in enthalpy per unit kg of the wort that evaporates? What heat transfer rate does this represent (in kJ/hr)?

 

[sci0x]

Hi Chester,

What is the mass flow rate in kg/hr of the wort water that evaporates?
Volume that evaporates is 2.5 m^3 / hr
Can I use density water here to calc mass..
Density = mass/vol
958 kg m-3 = mass / 2.5 m3
2395 kg / hr = mass
= 0.66 kg / sec

What is the change in enthalpy per unit kg of the wort that evaporates?
Change in Enthalpy = 2739 - 605 = 2135 kj kg-1
Can i say initial enthalpy = (2739)(0.66) = 2738.34
Final enthalpy = (605)(0.66) = 399.3
Difference = 2339.04 kJ

What heat transfer rate does this represent (in kJ/hr)
2339.04 kj = (enthalpy of vapourisation)(x)
2339.04 kj = x(2257 kj kg-1)
1.036 kg = x
1.036 kg per sec

Can you show me how to finish this off please?

 

[Chestermiller]

Hi Chester,

What is the mass flow rate in kg/hr of the wort water that evaporates?
Volume that evaporates is 2.5 m^3 / hr
Can I use density water here to calc mass..
Density = mass/vol
958 kg m-3 = mass / 2.5 m3
2395 kg / hr = mass

Correct

= 0.66 kg / sec

I asked for the mass flow rate in kg/hr, not kg/sec. Why are you not using an hourly basis, and insisting on using a seconds basis?

What is the change in enthalpy per unit kg of the wort that evaporates?
Change in Enthalpy = 2739 - 605 = 2135 kj kg-1

This is incorrect. The correct answer to this is given in the problem statement as 2257 kJ/kg

Can i say initial enthalpy = (2739)(0.66) = 2738.34
Final enthalpy = (605)(0.66) = 399.3
Difference = 2339.04 kJ

No.

What heat transfer rate does this represent (in kJ/hr)
2339.04 kj = (enthalpy of vapourisation)(x)
2339.04 kj = x(2257 kj kg-1)
1.036 kg = x
1.036 kg per sec

No. This is all wrong. I didn't ask you anything about the heating steam yet.

From the open system (control volume) version of the first law of thermodynamics, $$0=\dot{Q}-\dot{m}\Delta h$$
or $$0=\dot{Q}-(2395)(2257)$$or$$\dot{Q}=5.406\times\ 10^6\ kJ/hr$$
This is the heat transfer rate from the steam to the wort.

OK so far?

 

[sci0x]

Okay good with this Chester

 

[Chestermiller]

OK. For the steam flowing through the heat exchanger tubes immersed in the wort, we are going to do exactly the same thing.

What is the change in enthalpy (in kJ/kg) for the steam between the entrance and exit of the heat exchanger tubers?

In terms of this enthalpy change, the mass flow rate x of the steam in kg/hr, and the rate of heat transfer $\dot{Q}$ from the steam to the wort, what is the open system (control volume) version of the 1st law of thermodynamics applied to the steam passing through the heat exchanger tubes?

 

[sci0x]

Change in enthalpy of steam = 2739 - 605 = 2134 kj kg-1

5.406 x 10^6 kJ hr-1 = (2134 kJ kg-1)x
2533 kg hr-1 = x = flow of steam required ?