Question

A coal-fired power plant is burning bituminous coal that has an energy content of 12,000 Btu/lb. The power plant is burning the coal at a rate of 110 lb/s. The efficiency of the powerplant is 33%, which means that 67% of the total energy from the coal is lost as waste heat.

a. A new heat exchanger is installed that is capable of using 0.12% of the waste heat to heat an air stream for use in another process in the plant. The air enters the heat exchanger at a volumetric flow rate of 2,000 acfm at 298 K and 1 atm. Assuming the heat exchanger is 90% efficient (i.e., 90% of the heat supplied to the heat exchanger is transferred to the air stream), at what temperature will the air leave the heat exchanger, assuming an exit pressure of 1 atm? Use an average specific heat of 0.244 Btu/lb-°R.
b. What will be the volumetric flow rate of the gas leaving the heat exchanger (in units of acfm)?
c. If the air exiting the heat exchanger has a moisture content of 12% (v/v), what is the moisture-corrected dry flow rate (in units of dry standard cubic meters per minute) at STP (1 atm, 298 K)?

Answer 1

Answer:

a) $T_{out} = 2190.455 ^{\textdegree}R$ b) $\dot V_{air,out} = 8352.941 acfm$  c) $\dot V_{air,out,corr} = 7350.588 afcm$

Explanation:

a) The heat lost by the power plant is:

$\dot Q_{loss} = (0.67) \cdot (110 \frac{lbm}{s} ) \cdot (12000 \frac{BTU}{lbm} )\\\dot Q_{loss} = 884400 BTU$

The waste heat used by heat exchanger is:

$\dot Q_{used} = (0.0012) \cdot \dot Q_{loss}\\\dot Q_{used} = 1061.28 BTU$

Assuming that air behaves as an ideal gas, density is given by following expression:

$\rho_{air} = \frac{P \cdot M_{air}}{R_{u} \cdot T}$

$\rho_{air} = \frac{(101.325 kPa) \cdot (28 \frac{kg}{kmol})}{(8.314 \frac{kPa\cdot m^{3}}{kmol \cdot K} )(298 K)}\\\rho_{air} = 1.145 \frac{kg}{m^{3}}$

The density unit is converted to pounds per cubic feet:

$\rho_{air} = 1.145 \frac{kg}{m^{3}} \cdot (\frac{1 lb}{0.453 kg}) \cdot (\frac{0.304 m}{1 ft} )^{3} \\\rho_{air} = 0.071 \frac{lb}{ft^{3}}$

The heat received by air flow through heat exchanger is:

$\dot Q_{air} = (0.90) \cdot \dot Q_{used}\\\dot Q_{air} = 955.152 BTU$

Outlet temperature can isolated from the following formula:

$\dot Q_{air} = \rho_{air} \cdot \dot V_{air} \cdot c_{p,air} \cdot (T_{out} - T_{in})$

$T_{out} =T_{in} + \frac{\dot Q_{air}}{\rho_{air} \cdot \dot V_{air} \cdot c_{p,air}}$

$T_{out} = 536.4 ^{\textdegree}R + \frac{955.152 BTU}{(0.071\frac{lb}{ft^{3}})\cdot (33.333 \frac{ft}{s} )\cdot(0.244 \frac{BTU}{lb ^{\textdegree}R})}\\T_{out} = 2190.455 ^{\textdegree}R$

b) Due to the compressibility of air, density has to be calculated by using the approach from section a).

$\rho_{air,out} = \frac{(101.325 kPa) \cdot (28 \frac{kg}{kmol})}{(8.314 \frac{kPa\cdot m^{3}}{kmol \cdot K} )(1217 K)}\\\rho_{air,out} = 0.280 \frac{kg}{m^{3}}$

$\rho_{air,out} = 0.017 \frac{lb}{ft^{3}}$

Volumetric flow can be found by Principle of Mass Conservation:

$\dot V_{air, out} = \frac{\rho_{air}}{\rho_{air,out}}\cdot \dot V_{air}$

$\dot V_{air,out} = 8352.941 acfm$

c) The moisture component indicates that 88 percent of volume is occupied by dry-air. The moisture-corrected dry flow rate is:

$\dot V_{air,out,corr} = 0.88 \cdot (8352.941 afcm)\\\dot V_{air,out,corr} = 7350.588 afcm$