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: Read the flow arrangement from the figure.
Fluid 2 flows along the central axis of the inner tube, entering at \(20^{\circ}\)C and leaving at \(30^{\circ}\)C, while Fluid 1 connects to the annulus around the inner tube, entering at \(60^{\circ}\)C and leaving at \(40^{\circ}\)C, so the two streams flow counter to each other.

Step 2: Assign the convective coefficients to the correct surfaces.
Fluid 2 (inside the inner tube) has \(h_i = 80\) W/(m\(^2\)K), acting on the tube's inner surface of radius \(r_i=0.010\) m. Fluid 1 (in the annulus) has \(h_o = 50\) W/(m\(^2\)K), acting on the tube's outer surface of radius \(r_o = 0.011\) m.

Step 3: Compute the three resistances in the tube wall, for length \(L=1\) m.
Inner convection: \(A_i=2\pi r_iL=0.06283\) m\(^2\), \(R_i=\dfrac{1}{h_iA_i}=\dfrac{1}{80(0.06283)}=0.1989\) K/W.
Wall conduction: \(R_{wall}=\dfrac{\ln(r_o/r_i)}{2\pi kL}=\dfrac{\ln(1.1)}{2\pi(386)}=\dfrac{0.0953}{2425.3}=0.0000393\) K/W, very small since the tube conducts well.
Outer convection: \(A_o=2\pi r_oL=0.06912\) m\(^2\), \(R_o=\dfrac{1}{h_oA_o}=\dfrac{1}{50(0.06912)}=0.2894\) K/W.

Step 4: Add the resistances and get \(UA\).
\(R_{total}=0.1989+0.00004+0.2894=0.4884\) K/W, so \(UA=1/R_{total}=2.048\) W/K.

Step 5: Find the LMTD for the counter-flow arrangement.
At one end, \(\Delta T_A = 60-30=30^{\circ}\)C; at the other end, \(\Delta T_B=40-20=20^{\circ}\)C. \(\text{LMTD}=\dfrac{\Delta T_A-\Delta T_B}{\ln(\Delta T_A/\Delta T_B)}=\dfrac{10}{\ln(1.5)}=\dfrac{10}{0.4055}=24.66^{\circ}\)C.

Step 6: Get the heat transfer rate.
\(Q=UA\times\text{LMTD}=2.048(24.66)=50.5\) W.

Final Answer:
Adding the three resistances in series to build \(UA\), then multiplying by the counter-flow LMTD, gives the total heat duty. \[ \boxed{Q = 50.5 \ \text{W}} \]