HTandPI_Problem11_Soln

7/30/2019 HTandPI_Problem11_Soln

1/5

Heat Transfer and Process Integration

2011-2012

Semester 2

Problem Solving Session 11 Solution - The Grand Composite curve

The problem table for a chemical process is shown in Table 1.

Tmin = 20.0000 [C]

Minimum Hot Utility = 935.000 [kW]

Minimum Cold Utility = 695.000 [kW]

Interval Temperature* Enthalpy

[C] [kW]

---------------------------------------------------------------------

1 335.00000 935.000002 195.00000 235.00000

3 180.00000 310.00000

4 170.00000 200.00000

5 120.00000 0.0000000 Pinch

6 95.000000 300.00000

7 80.000000 555.00000

8 60.000000 695.00000

---------------------------------------------------------------------

Table 1

It is proposed to use the exhaust from a gas turbine to meet the external heating

requirements of the process. The gas turbine exhaust is at 510OC. The specific heat

capacity of the exhaust can be taken as 1.2kJ/kg.OC. Ambient temperature is 20

OC.

a) Use the graph on reverse to produce the grand composite curve.

b) What is the mass flowrate of the gas turbine exhaust needed to provideexternal heating for the process?

c) What is the stack loss of the gas turbine exhaust if the stack temperaturewas the same as the pinch temperature?

d) What would be the mass flowrate of the gas turbine exhaust and the stackloss if the stack temperature was 160

OC?


2/5

)H (kW)

T (oC)*600

550

500

450

400

350

300

250

200

150

100

50

0


3/5

The Grand Composite curve for the process with a Tmin of 20OC is shown

in Fig. 1.

The hot utility requirement needs to be met with a utility that provides the necessary

heat to the process. It is envisaged that this will be carried out by the exhaust from a

gas turbine. The gas turbine exhaust is 510O

C. The amount of heat that the GTexhaust has to supply is the QHmin value of 935kW. If this was matched exactly

against the grand composite curve we would have the situation shown in Fig. 2.

The supply temperature of the GT exhaust is shown at 500OC (real temperature

510OC), and the target temperature is shown at the pinch temperature of 120

OC (the

actual target temperature of the exhaust is 130OC). As the GT exhaust is a hot utility

the temperatures shown on the grand composite have to be reduced by 0.5Tmin.

What then is the mass flowrate of the GT exhaust?

We can use the general expression;H = mCpT

Where = H = enthalpy or heat load (kW)

T = temperature change (OC)

m = mass flowrate (kg/s)

Cp = specific heat capacity (kJ/kg.OC)

We knew H = 935kW, and T = 510 130 = 380OC real temperatures, or 500 -120

shifted temperatures (grand composite curve).

The Cp of the exhaust gas can be taken as 1.2 kJ/kg.OC (same as air which it is mostly

composed of).

The mass flowrate, m, is then ;


4/5

Fig. 1

Fig. 2


5/5

2.05kg/s.

However, in reality the gas turbine exhaust has to be cooled down to ambient

temperature, say 20OC. This is done by the exhaust being ejected through the stack.

The mass flowrate of the exhaust does not change, and so consequently the total heat

content of the exhaust is;H = mCpT

Which is;

H = 2.05*1.2*(510-20) = 1205.4kW (using real temperatures)

(Note if shown on Grand composite the temperatures would be 500OC and 10

OC).

The stack loss (heat that cant be used) is therefore 1205.4 935 = 270.4kW.

In the above case we have set the stack temperature (or dew point temperature) equal

to the pinch temperature. Normally the stack temperature is around 160OC. If this wasapplied in the above case, what would be the mass flowrate of the exhaust and the

stack loss?

So, the mass flowrate of the exhaust is now determined by the same heat load

requirement (935kW) but a temperature change of 510-170OC (or 500 160

OC if

reading from Grand composite). So now mass flowrate = 2.3kg/s. The flowrate of the

exhaust has increased. The total heat content of the exhaust down to ambient

temperature is now 1352.4kW. The stack loss is now 1352.4-935 = 417.4kW. These

results are shown in Fig. 3.

Ambient temperaure (20OC real,10OC interval)

Mass flowrate = 2.05kg/s.Stack loss = 270.4kW.Mass flowrate = 2.3kg/s.

Stack loss = 417.4kW.

Fig. 3

Documents

HTandPI_Problem11_Soln