You need to cool 8.0 kg/s of superheated steam at 1000 kPa from 325oC to 200oC. You can either use R-134a that is available as a saturated liquid at 0oC, or air that is available at 15oC and 100 kPa. Due to other operational factors, the R-134a must not have an exit quality of greater than 0.60, and the air can not have an exit temperature greater than 75oC. Design a system that will meet the constraint for each substance, and comment on the relative practicality of each system.

Question

Consider State 1 to be the steam inlet and State 2 to be the steam outlet. Consider State 3 to be the coolant inlet and State 4 to be the coolant outlet.
Given: P1 = P2 = 1000 kPa; T1 = 325oC; m?s = 8.0 kg/s; T2 = 200oC
Assume: Q? = 0 (insulated). Also, given no other information regarding the heat exchanger, make the following common heat exchanger assumptions: W? = ?KE = ?PE = 0
Also, assume the heat exchanger is a multiple-inlet, multiple-outlet, steady-state, steady-flow device.

What will be an ideal response?

Answer 1

It is probably easiest to find the amount of heat to be removed from the steam, and then use that to determine the necessary mass flow of the coolant options.
For the steam, the First Law reduces to Q?s = m?s(h2 ? h1)
h1 = 3104.5 kJ/kg
h2 = 2827.9 kJ/kg
Solving: Q?s =-2213 kW

Then, the heat out of the steam is into the coolant (so the sign changes as the system changes): Q?cool = ?Q?s = 2213 kW
For each coolant, to minimize the mass flow rate, we need to maximize the change in enthalpy. This is done with using the maximum quality for the R-134a, and the maximum temperature for the air.
R-134a: T3 = T4 = 0oC ; x3 = 0.0; x4 = 0.60 (assumes a constant pressure process) Then, h3 = 50.02 kJ/kg; h4 = 168.35 kJ/kg
For the heat input to the R-134a, the First Law yields: m?R134a =Q?cool/(h4?h3) = 18.7 kgs
Air: T3 = 15oC, P3 = 100 kPa; T4 = 75oC
Consider the air to be an ideal gas with constant specific heats: cp = 1.005 kJ/kg-K

While the flow rates may not seem much different at first, the fact that the air is a gas while the volume of the R-134a is primarily a liquid means that the R-134a system is more practical from a size-standpoint. One would need approximately 30 m3/s of air to go through the system, but only approximately 0.015 m3/s of R-134a to enter the system.