Question

Very cold water (5°C) is heated at atmospheric pressure (1 atm) until saturated vapor is formed.

(a)

Draw an enthalpy path for the water.

(b)

Calculate the specific enthalpy change (kJ/mol) for each step in the path, and calculate the total specific enthalpy change (kJ/mol).

(c)

Compare the total enthalpy change from the enthalpy path to the total enthalpy change using the steam tables (see zyExample in another section).

(d)

Compute the ratio of latent heat to sensible heat.

(e)

The process is completed in reverse, cooling saturated water vapor to very cold water. Will the magnitude of the specific enthalpy change increase, decrease, or stay the same?

A heat exchanger is a piece of equipment that allows heat to be transferred from one stream to another while preventing the streams from mixing. Here, a heat exchanger has one stream containing an ice slurry entering at 0°C that exits as water (H₂O) at 20°C. The second stream enters at 8.8 kg/min as hot ammonia (NH₃) vapor at 12.2 atm and 144°C. The ammonia stream leaves the exchanger as a vapor at 12.2 atm and 34°C. Two heat capacities of ammonia are provided as: C_p (NH₃, vapor, P=12.2 atm) = 8.0 J/mol-K and C_p (NH₃, liquid, P=12.2 atm) = 82.3 J/mol-K.

(a)

Draw the enthalpy path for each component.

(b)

Find the mass flow rate (kg/s) of the water.

(c)

What percentage of the enthalpy change of the water is due to the phase change?

(d)

The water enters as liquid water at 0^oC instead of an ice slurry. Will the enthalpy change of the water stream increase, decrease, or stay the same?

Answer 1

a)

Assuming constant heat capacity; sensible heat is given by

dh = C.dT

where

h = specific enthalpy in kJ / kg

C = specific heat capacity of water

dT = temperature difference

Latent heat is given by

dh = L

where

L = latent heat of vaporisation of water.

For water; C = 4.2 kJ / kg.oC

L = 2260 kJ / kg

From 5 oC < T < 100 oC

dh = 4.2 dT

At T = 100 oC

dh = 2260

The enthalpy path thus is

b)

For the slanting part of the enthalpy diagram:

dh1 = 4.2 X (100 - 5) = 399 kJ / kg

For the vertical part of the enthalpy diagram:

dh2 = 2260 kJ / kg

Total specific enthalpy change:

dh = dh1 + dh2

dh = 399 + 2260 = 2659 kJ / kg

c)

From the steam table:

Enthalpy of liquid water at 5 oC = 21.02 kJ / kg

Enthalpy of vapor at 100 oC = 2675.6 kJ / kg

Therefore; specific enthalpy change:

dH = 2675.6 - 21.02 = 2654.58 kJ / Kg

The two values of enthalpy change are very close to each other.

Percentage difference = 100 X (2659 - 2654.58) / 2659 = 0.17 %

which is a very small value.

d)

Ratio of latent heat to sensible heat = 2260 / 399 = 5.66

e)

The enthalpy with remain the same as it is a state function and depends only on the intial and final state of the system.