Question

Convective heat transfer coefficients for Fluid 1 and Fluid 2 in a heat exchanger, as shown in the figure below, are 50 W/(m²K) and 80 W/(m²K), respectively. The inner tube is made of a material which has a thermal conductivity 386 W/(m K) for the given range of temperatures in the heat exchanger. The length of the heat exchanging surface, the inside radius of the inner tube, and thickness of the inner tube are 1 m, 10 mm, and 1 mm, respectively. Considering no heat exchange between the outer tube and the surrounding, the heat transfer rate for the heat exchanger is ________ W (rounded off to 1 decimal place).

Hint:

Work out which fluid flows inside the inner tube from the figure, then build up the overall resistance from the two convective and one conductive term.

Answer 1

Correct Answer: 50.5

Step 1: Fix which fluid touches which surface.
The figure shows Fluid 2 running straight through the inner tube's core between $20^{\circ}$C and $30^{\circ}$C, and Fluid 1 connecting to the annulus around it, between $60^{\circ}$C and $40^{\circ}$C, so heat moves from Fluid 1 in the annulus into Fluid 2 in the core, in a counter-flow pattern.

Step 2: Build the overall coefficient from the individual resistances.
With $r_i=0.010$ m, $r_o=0.011$ m, $L=1$ m, $h_i=80$ W/(m$^2$K) for Fluid 2 and $h_o=50$ W/(m$^2$K) for Fluid 1: $\dfrac{1}{UA} = \dfrac{1}{h_iA_i} + \dfrac{\ln(r_o/r_i)}{2\pi k L} + \dfrac{1}{h_oA_o}$.

Step 3: Compute each term.
$A_i = 2\pi(0.010)(1) = 0.06283$ m$^2$, so $1/(h_iA_i) = 1/(80\times0.06283) = 0.1989$ K/W.
$\ln(r_o/r_i)/(2\pi kL) = \ln(1.1)/(2\pi\times386) = 0.0000393$ K/W, negligible next to the convective terms.
$A_o = 2\pi(0.011)(1) = 0.06912$ m$^2$, so $1/(h_oA_o) = 1/(50\times0.06912) = 0.2894$ K/W.

Step 4: Sum and invert.
Total $= 0.1989+0.0000393+0.2894 = 0.4884$ K/W, so $UA = 2.048$ W/K.

Step 5: Bring in the LMTD for the counter-flow ends.
End differences are $60-30=30$ K and $40-20=20$ K, so LMTD $= (30-20)/\ln(30/20) = 10/0.4055 = 24.66$ K.

Step 6: Multiply to get the duty.
$Q = UA \times \text{LMTD} = 2.048 \times 24.66 = 50.5$ W.

Final Answer:
Whether the resistance is tracked from the inside out or folded straight into $UA$, the exchanged heat comes out the same. \[ \boxed{Q = 50.5 \ \text{W}} \]