Non-Adiabatic Steam–Air Heat Exchanger: 04-BS-10, December 2019, Q8

Question 8: Non-Adiabatic Steam–Air Heat Exchanger (15 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Check — reconstructed missing values. The pressure (0.07 MPa), air inlet temperature (30°C), the "exits as saturated liquid at the same pressure" condition, and $c_p=1.005$ for air are all confirmed legible. Clean representative values are substituted: steam enthalpy $h_1=2600$ kJ/kg, steam flow $1.5$ kg/min (the one surviving flow-rate figure), air flow $85$ kg/min, air exit $65^\circ$C — chosen so the resulting heat-loss fraction is physically reasonable (a modest loss to surroundings rather than the dominant term). The method below applies unchanged to the real printed values if recovered.

Given. Steam: $P=0.07$ MPa, $h_1=2600$ kJ/kg (assumed), $\dot m_{steam}=1.5$ kg/min, exits saturated liquid at 0.07 MPa. Air: $\dot m_{air}=85$ kg/min (assumed), $T_{air,1}= 30^\circ$C, $T_{air,2}=65^\circ$C (assumed), $c_p=1.005$ kJ/kg·K. KE/PE negligible.

Find. (a) $x_1$; (b) $\dot Q_{surr}$ [kW].

Approach

Get the saturated liquid/vapor enthalpies at 0.07 MPa to back out the entering steam's quality from its given enthalpy, then close an overall energy balance: heat released by the condensing steam minus heat absorbed by the warming air stream equals the (unbalanced) heat lost to the surroundings.

Quality of the entering steam (part a). At 0.07 MPa, $h_f=376.75$, $h_g=2659.42$ kJ/kg: $$x_1=\frac{h_1-h_f}{h_g-h_f}=\frac{2600-376.75}{2659.42-376.75}=\boxed{0.9740}.$$
Heat released by the condensing steam. $$\dot Q_{steam}=\dot m_{steam}(h_1-h_f)=\frac{1.5}{60}(2600-376.75)=\boxed{55.58\ \text{kW}}.$$
Heat absorbed by the air stream. $$\dot Q_{air}=\dot m_{air}\,c_p\,(T_{air,2}-T_{air,1})=\frac{85}{60}(1.005)(65-30)=\boxed{49.83\ \text{kW}}.$$
Heat transfer to the surroundings (part b). The steam releases more heat than the air absorbs; the difference is lost to the surroundings: $$\dot Q_{surr}=\dot Q_{steam}-\dot Q_{air}=55.58-49.83=\boxed{5.75\ \text{kW}}.$$

Quantity	Result
(a) $x_1$	0.9740
(b) $\dot Q_{surr}$	5.75 kW

Concept & Theory

Topic. Non-adiabatic two-stream heat exchanger: condensing steam heating a separate air stream, with an explicit heat-loss term to the surroundings.

Key relations. Two-phase quality from $x=(h-h_f)/(h_g-h_f)$; overall energy balance $\dot Q_{steam,released}=\dot Q_{air,absorbed}+\dot Q_{surr}$, since the two streams don't exchange exactly matched duties when the exchanger isn't perfectly insulated.

Why this works. If the exchanger WERE adiabatic, the steam's released heat would exactly equal the air's absorbed heat; asking for $\dot Q_{surr}$ signals that a portion of the duty genuinely escapes (or enters) through the casing, which the overall energy balance isolates as whatever residual keeps the balance closed.

Common pitfall. Assuming the exchanger is adiabatic and trying to solve for an unknown flow rate from an equality of the two streams' duties — here the problem explicitly asks for the heat LOST, so the two duties are not expected to match.

Source. Moran, Shapiro, Boettner & Bailey, Fundamentals of Engineering Thermodynamics, 8th ed., Ch. 4 (Control Volume Energy Analysis, Heat Exchangers).

Question 8 of 9: Non-Adiabatic Steam–Air Heat Exchanger

Question 8: Non-Adiabatic Steam–Air Heat Exchanger (15 marks)

Approach