Calculate energy required to heat water using a steam coil

[sci0x]

Homework Statement

Cold water at 15°C is heated to 85°C in a hot water tank by means of a steam coil
immersed in the water at the bottom of the tank. The combined (empty) tank and steam
coil weigh 1000 kg and are made of 304 stainless steel with a specific heat of 1.8 kJ kg-1 K
-1
.
Using the table below for water and steam properties data, calculate the amount of
energy required to heat 200 hL of water from 15°C to 85°C assuming the tank is well
insulated and there are negligible thermal losses to the environment.

Relevant Equations

Q = m c dT

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
= (199800 kg)(4.18 kJ kg-1K-1)((85-15)
=58461480 kJ

Am I correct here?

 

[Chestermiller]

20 m^3

 

[Steve4Physics]

Homework Statement:: Cold water at 15°C is heated to 85°C in a hot water tank by means of a steam coil
immersed in the water at the bottom of the tank. The combined (empty) tank and steam
coil weigh 1000 kg and are made of 304 stainless steel with a specific heat of 1.8 kJ kg-1 K
-1
.
Using the table below for water and steam properties data, calculate the amount of
energy required to heat 200 hL of water from 15°C to 85°C assuming the tank is well
insulated and there are negligible thermal losses to the environment.
Relevant Equations:: Q = m c dT

View attachment 278494

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
= (199800 kg)(4.18 kJ kg-1K-1)((85-15)
=58461480 kJ

Am I correct here?

In addition to @Chestermiller's (admirably concise) observation, note that in heating the water, you will unavoidably have to heat the container. You are probably expected to include the energy required to do this. The wording of the question doesn't make this clear, but I guess you are given the mass of the container and its specific heat capacity for this reason.

Also '58461480 kJ' has 7 significant figures. (Lose 1 mark in an exam'!) You have given your answer with a precision of about 1 part in about 6 million. But the data you are using are far less precise.

EDIT: Although it will only change the answer slightly, you might consider using the average specific heat capacity of water over the temperature range.

 

[sci0x]

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)
= 126,000 kg

Total heat req to heat water and stainless steel:
5,860,134 + 126,000 = 5,986,134 kJ

The next part of the Q is:
If the steam feeding the heating tube is at 1 bar gauge pressure and has a dryness fraction of 90%, estimate the amount of steam required to perform this heating task.

1 bar G = 10^5 Pa or 100 kPa
At 100 kPa
Enthalpy of steam = 2676 kJ kg-1
Enthalpy of water = 419 kJ kg-1
Change of enthalpy of steam = 2676 - 419 = 2257 kJ kg-1
Dryness factor is 90%: (2257)(0.9) = 2031.3 kj kg-1

5,986,134 kJ = Mass of Steam (2031.3 kj kg-1)
2946.94 kg = Mass of Steam

 

[Steve4Physics]

Total heat req to heat water and stainless steel:
5,860,134 + 126,000 = 5,986,134 kJ

You've given the answer to 7 significant figures. What would you write if this were a value required in an exam' and you didn't want to lose a mark for an inappropiate number of significant figures?

1 bar G = 10^5 Pa or 100 kPa
At 100 kPa
Enthalpy of steam = 2676 kJ kg-1
Enthalpy of water = 419 kJ kg-1
Change of enthalpy of steam = 2676 - 419 = 2257 kJ kg-1
Dryness factor is 90%: (2257)(0.9) = 2031.3 kj kg-1

5,986,134 kJ = Mass of Steam (2031.3 kj kg-1)
2946.94 kg = Mass of Steam

Why state '100kPa' when the table explicitly uses 101kPa?

Edit - the above pressure is wrong.
See @Chestermiller's post #6 below.

No need to say "Change of enthalpy of steam = 2676 - 419" when this value has already been calculated for you (penultimate column of table).

You are asked for an *estimate* of the steam mass. An 'estimate' is an approximate value. You have given you answer to 6 significant figures!

Can you suggest why your value (suitably rounded) might be a reasonable rough estimate - given that you have ignored the water from the steam cooling from 100ºC to 85ºC?

 

[Chestermiller]

I bar gauge pressure is 2 bars absolute pressure. So the temperature is not 100 C.

 

[Chestermiller]

Can you suggest why your value (suitably rounded) might be a reasonable rough estimate - given that you have ignored the water from the steam cooling from 100ºC to 85ºC?

The steam is not cooled to 85 C. The streams are not mixed. It is a heat exchanger. The steam exits as saturated liquid at 2 bars.

 

[Steve4Physics]

The steam is not cooled to 85 C. The streams are not mixed. It is a heat exchanger. The steam exits as saturated liquid at 2 bars.

OK but I don't get why the condensed liquid leaving the heat exchanger cannot have cooled nearer to 85ºC. Mixing is not needed to achieve this.

There is nothing in the question which requires input and output temperatures of the heat exchanger to be equal.

My gut feel is that there could be a temperature-drop along the heat exchanger but its contribution to heating the water is negligible (compared to the contribution from the condensation process). So it need not be calculated and this is the basis for the question asking only for an 'estimate'.

 

[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 am i gone wrong here

 

[Chestermiller]

Looks OK