A steam-heated single-pass tubular preheater is designed to raise 5.6 kg/s of air from 200C to 75°C, using saturated steam at 2.6 MPa(abs). It is proposed to double the flow rate of air and, in order to be able to use the same heat exchanger and achieve the desired temperature rise, it is proposed to increase the steam pressure. Calculate the steam pressure necessary for the new conditions and comment on the design characteristics of the new arrangement.

Question

GIVEN

• Steam-heated single-pass tubular preheated heating air

• Air flow rate ( m a)= 5.6 kg/s

• Air temperature

? Ta,in = 20°C

? Ta,out = 75°C

• Saturated steam pressure (ps) = 2.6 MPa = 2.6 × 106 N/m2

FIND

• The steam pressure necessary for a double m a with the same temperature rise

SKETCH



PROPERTIES AND CONSTANTS

the saturation temperature (Ts) of steam at 2.6 × 106 N/m2 = 225°C

Answer 1

The heat capacity rate of the steam is virtually infinite, therefore, Cmin/Cmax = 0. The effectiveness must be determined using Equation (10.22b)



Examination of e = e (NTU) when Cmin/Cmax = 0 or in and the solution of reveals that e = e (NTU) is the same of both counterflow and parallel flow when Cmin/Cmax = 0. For e = 0.268, gives NTU = (U A)/Cmin ? 0.25.

Doubling the flow rate of air doubles its heat capacity rate (Cmin). Therefore, the NTU is halved. For the new flow rate: NTU = 0.125. From Figure 10.18 for NTU = 0.125, Cmin/Cmax = 0 ? e = 0.15 The rate of heat transfer is



Solving for the steam temperature required.



The steam temperature required for the doubled air flow rate is 387°C for Ts = 387°C, the saturation pressure cannot be calculated as it is much larger than saturation temperature range of upto 3000C.

COMMENTS

Pressure increases rapidly with steam temperature, therefore, the solution is only practical if the equipment is designed to operate safely at high pressure.