Consider the process in Example 13.4. Can you reduce the required area substantially by changing the coolant

Question:

Consider the process in Example 13.4. Can you reduce the required area substantially by changing the coolant flow rate?

Example 13.4

We must reduce the temperature of a small process steam flowing at 2 kg/min from 90◦C to 30◦C. The stream has the physical properties of water. Cooling water is available at 20◦C and may be discharged at a maximum temperature of 50◦C. What are the specifications of a double-pipe heat exchanger if h_T = 400 W/m²K? The heat load is obtained from Equation 13.34a, with c_pt = 4,180 J/kg and ρ_tq_t = 2 kg/m = 1/30 kg/s, to be 8,360 W. If we assume that the cooling water is discharged at the maximum allowable temperature, 50^◦C, then it follows from Equation 13.34b that ρ_sq_s = 1/15 kg/s = 4 kg/min. ΔT_L = 30−20 = 10◦C and ΔT_o = 90 – 50 = 40◦C, hence ΔT_lm = (10−40)/ln (10/40) = 21.65◦C. From Equation 13.33, 8,360 W = (400 W/m2K)πDL(21.65 K), and πDL = 0.965 m². If D= 2.5 cm, say, then L = 12.3 m. This is unrealistically long, but if ten parallel tubes of the same diameter were used then the length of the exchanger would be only 1.23 m.