Casing-and-Tube Water/Air Heat Exchanger: 17-Phys-B6 Applied Thermodynamics and Heat Transfer, May 2013, Q8

Question 8: Casing-and-Tube Water/Air Heat Exchanger (Part B, 20 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: part (b) as printed is self-contradictory — it asks for "10 kg/s of hot water" and in the same sentence says "the water flows at 6.5 kg/s ... as before"; one stream cannot have both flow rates. Per Note 1 both readings are worked below. Read literally (water still 6.5 kg/s), part (b) only restates part (a), whose own answer is already $42^{\circ}\text{C}$ — so the examiner almost certainly intends the exchanger to deliver 10 kg/s of water at $42^{\circ}\text{C}$ (water entering at $15^{\circ}\text{C}$ as before, air still 5.0 kg/s). That reading is taken as the answer to (b); the literal reading is shown as the check.

Given. $U=200$ W/m²K, $A=47.5$ m², 1 casing pass / 2 tube passes (135 tubes, double-pass); water (tube side) $\dot m_w=6.5$ kg/s, $T_{w,in}=15^{\circ}\text{C}$; air (casing side) $\dot m_a=5.0$ kg/s, $T_{a,in}=200^{\circ}\text{C}$.

Find. (a) $T_{w,out}$, $T_{a,out}$; (b) the air inlet temperature needed for $T_{w,out}=42^{\circ}\text{C}$ with the water flow raised to 10 kg/s (and, as a check, with it left at 6.5 kg/s).

Approach. Use the effectiveness–NTU method for a 1-casing-pass/2-tube-pass exchanger (unknown outlet temperatures rule out LMTD without iteration); part (b) then asks for the inlet condition that produces a STATED duty, re-evaluating $C_r$, $NTU$ and $\varepsilon$ for whichever water flow rate applies.

Capacity rates. With $c_{p,w}\approx4180$ J/kg·K, $c_{p,a}\approx1014$ J/kg·K (evaluated near the converged bulk means), $$C_w=\dot m_wc_{p,w}=27{,}170\ \text{W/K},\qquad C_a=\dot m_ac_{p,a}=5071\ \text{W/K} =C_{min},$$ $$C_r=C_{min}/C_{max}=5071/27{,}170=0.187,\qquad NTU=\frac{UA}{C_{min}}=\frac{200\times47.5}{5071}=1.873.$$
1–2 casing-and-tube effectiveness. $$\varepsilon=2\left\{1+C_r+\sqrt{1+C_r^{2}}\;\frac{1+\exp\!\left(-NTU\sqrt{1+C_r^{2}}\right)} {1-\exp\!\left(-NTU\sqrt{1+C_r^{2}}\right)}\right\}^{-1}=\boxed{0.781}.$$
Part (a) — outlet temperatures. $$q=\varepsilon\,C_{min}(T_{a,in}-T_{w,in})=0.781\times5071\times185=733.1\ \text{kW},$$ $$T_{w,out}=T_{w,in}+\frac{q}{C_w}=15+\frac{733{,}100}{27{,}170}=\boxed{42.0^{\circ}\text{C}},\qquad T_{a,out}=T_{a,in}-\frac{q}{C_a}=200-\frac{733{,}100}{5071}=\boxed{55.4^{\circ}\text{C}}.$$
Part (b) — 10 kg/s of water at $42^{\circ}\text{C}$. The water capacity rate rises to $C_w=10\times4180=41{,}800$ W/K; the air side stays $C_a=C_{min}\approx5108$ W/K ($c_{p,a}\approx1022$ J/kg·K at the new, hotter air bulk mean), so $C_r=0.122$ and $NTU=9500/5108=1.860$, giving from the same 1–2 relation $\varepsilon=0.802$. The duty needed is $$q_{req}=C_w(42-15)=41{,}800\times27=1128.6\ \text{kW},$$ and inverting $q=\varepsilon C_{min}(T_{a,in}-T_{w,in})$, $$T_{a,in}=T_{w,in}+\frac{q_{req}}{\varepsilon\,C_{min}}=15+\frac{1{,}128{,}600}{0.802\times5108} =\boxed{290.7^{\circ}\text{C}},$$ with the air leaving at $T_{a,out}=290.7-1{,}128{,}600/5108=69.7^{\circ}\text{C}$.
Check — the literal reading (water still 6.5 kg/s). $NTU$, $C_r$ and $\varepsilon=0.781$ are then those of Step 2, the duty is $q_{req}=27{,}170\times27=733.6$ kW, and $$T_{a,in}=15+\frac{733{,}600}{0.781\times5071}=200.1^{\circ}\text{C}\approx200^{\circ}\text{C},$$ i.e. part (a)'s own inlet: the exchanger already delivers 6.5 kg/s at $42^{\circ}\text{C}$ and nothing needs to change. That this reading returns the part-(a) data unchanged is why the 10 kg/s reading is taken as the intended question: raising the water flow by 54% at the same outlet temperature needs the air inlet raised by about $91^{\circ}\text{C}$.

Final results — Question 8
Quantity	Value
$NTU$, $\varepsilon$ (1–2 casing-and-tube)	$1.873$, $0.781$
(a) $T_{w,out}$	$42.0^{\circ}\text{C}$
(a) $T_{a,out}$	$55.4^{\circ}\text{C}$
(b) Required $T_{a,in}$, 10 kg/s water to $42^{\circ}\text{C}$	$290.7^{\circ}\text{C}$ (air leaves at $69.7^{\circ}\text{C}$)
(b) check, literal 6.5 kg/s reading	$200.1^{\circ}\text{C}\approx200^{\circ}\text{C}$ (unchanged)

Topic: Effectiveness–NTU analysis of a 1-casing-pass/2-tube-pass heat exchanger with unknown outlet temperatures.

Key relations: $NTU=UA/C_{min}$; $C_r=C_{min}/C_{max}$; 1–2 casing-and-tube $\varepsilon(NTU,C_r)$ (Incropera eq. 11.31a); $q=\varepsilon C_{min}\Delta T_{max}$.

Why this works: because $\varepsilon$ depends only on $NTU$ and $C_r$ (both fixed by geometry/flow rates, not by the inlet temperatures), $q=\varepsilon C_{min}(T_{a,in}-T_{w,in})$ is linear in the inlet temperature difference once the flow rates are set, so "what inlet temperature gives duty $q$" in part (b) is a one-line inversion after evaluating $\varepsilon$ for part (b)'s own flow rates.

Common pitfall: attempting LMTD directly when both outlets are unknown (circular, needs iteration) instead of the closed-form effectiveness–NTU route; or, in part (b), re-using part (a)'s $\varepsilon$ after the water flow has changed — a new flow rate changes $C_r$ (and here $C_{min}$'s partner), so $\varepsilon$ must be recomputed.

Source: Incropera & DeWitt, Fundamentals of Heat and Mass Transfer, Ch. 11 §11.4 (effectiveness–NTU method, casing-and-tube relations).

Question 8 of 8: Casing-and-Tube Water/Air Heat Exchanger

Question 8: Casing-and-Tube Water/Air Heat Exchanger (Part B, 20 marks)