2017 Real Exam Questions 19-21

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Question: Air at 20°C flows through pipes ABC at a rate of 20,000 m3/h and enters a device at constant pressure at point C, as shown in the diagram below. The specifications of the pipes are φ530 × 7 mm. Due to changes in production requirements, it is necessary to reduce the amount of air delivered to point C by 8,000 m3/h; moreover, a branch pipe should be installed at point B so that the excess air can be sent to another device at constant pressure at point D. The tube in the BD section is extremely short, so its straight-line resistance can be ignored. If the local resistance coefficients between BC and BD (excluding the outlet resistance coefficient) are 7 and 4 respectively, with the pipe friction coefficient λ = 0.014, which of the following values is the calculated inner diameter of branch BD in millimeters? (A) 517 (B) 412 (C) 363 (D) 394. The solution involves calculating pressure loss. Before the change, the gas volume at point C was such that the pressure difference drove the flow of gas between A and C. After the change, the gas volume at C became 12,000, and the corresponding pressure difference decreased to 36% of its original value. Since D is also at atmospheric pressure, the pressure difference between A and D is the same as that between A and C; in other words, the pressure difference between B and D is the same as that between B and C. The formula for calculating the pressure difference is: https://www.cryogeny.cn/wp-content/ql-cache/quicklatex.com-37f8d08f65f7dea5433d698ba5867c23_l3.svg https://www.cryogeny.cn/wp-content/ql-cache/quicklatex.com-46c4a85605ee96c08051c45f71b28c9f_l3.svg https://www.cryogeny.cn/wp-content/ql-cache/quicklatex.com-bc285ccd7a1de84b61c7f7296ca9355c_l3.svg https://www.cryogeny.cn/wp-content/ql-cache/quicklatex.com-dda08a88338944c602c1a1e5741f657c_l3.svg https://www.cryogeny.cn/wp-content/ql-cache/quicklatex.com-d2197b67a96c9375f59f8ed26b74e891_l3.svg For question C: A factory uses saturated steam at a temperature of 133.3 °C to heat a certain liquid. The temperature of the liquid rises from 105 °C to 115 °C. It is known that the flow rate of the liquid is 360 m³/h, its density is 1080 kg/m³, and its specific heat capacity is 2.93 kJ/(kg·K). The area of the heat exchanger is 180 m². If heat losses are ignored, what should be the value of the heat transfer coefficient? https://www.cryogeny.cn/wp-content/ql-cache/quicklatex.com-bb5269470d9cb82f3b1b1325c42d07cf_l3.svg (A) 1780 (B) 1287 (C) 923 (D) 754. The solution involves calculating heat. https://www.cryogeny.cn/wp-content/ql-cache/quicklatex.com-5d3cc48019e0e5654bb78714da4c0901_l3.svg https://www.cryogeny.cn/wp-content/ql-cache/quicklatex.com-6708f1a49f3f29a4d54494f1740bc531_l3.svg To determine K, it is first necessary to obtain the temperature difference. https://www.cryogeny.cn/wp-content/ql-cache/quicklatex.com-e3658fe72e1d7963837b071e799e642c_l3.svg The logarithmic temperature difference method is used. https://www.cryogeny.cn/wp-content/ql-cache/quicklatex.com-09a5ba97e6fdb1cc9e4c7a3d96b5f54d_l3.svg https://www.cryogeny.cn/wp-content/ql-cache/quicklatex.com-2731d2abcecb5c8db9e425afc529b798_l3.svg https://www.cryogeny.cn/wp-content/ql-cache/quicklatex.com-5894e0a64dcbbc2e8ee5fc0519222b06_l3.svg https://www.cryogeny.cn/wp-content/ql-cache/quicklatex.com-91a174664a8df28fbb0fd0b53eaf64f8_l3.svg https://www.cryogeny.cn/wp-content/ql-cache/quicklatex.com-3692b440bf9024ae2fade05ffbc3203a_l3.svg K = 360 * 1080 * 2.93 * 10 / 180 / 22.94 * 1000 / 3600 = 766. Unit conversion is also important; otherwise, the results will be confusing and none of them will match those given in the problem. A certain oil refinery uses two shell-and-tube heat exchangers with identical heat transfer areas and a heat transfer coefficient of K=350 each. https://www.cryogeny.cn/wp-content/ql-cache/quicklatex.com-ec9d7407ca0d0f2044ab804455dca032_l3.svg The average temperature difference correction factor is as follows: https://www.cryogeny.cn/wp-content/ql-cache/quicklatex.com-3fc71d70bf7ac22f6d69d527d42f5c49_l3.svg. The relationship between R and S is as follows: | R | S | https://www.cryogeny.cn/wp-content/ql-cache/quicklatex.com-3fc71d70bf7ac22f6d69d527d42f5c49_l3.svg | |——|——-|——| | 2.42 | 0.33 | 0.92 | | 3.06 | 0.15 | 0.98 | | 3.06 | 0.292 | 0.88 | What is the heat transfer area of each heat exchanger? (https://www.cryogeny.cn/wp-content/ql-cache/quicklatex.com-7bd22cca2802b4314e8f06fb217359f5_l3.svg) (A) 43 (B) 86 (C) 113 (D) 226. The solution requires an understanding of the concepts and calculations related to https://www.cryogeny.cn/wp-content/ql-cache/quicklatex.com-3fc71d70bf7ac22f6d69d527d42f5c49_l3.svg. The calculated values are R = (208–113)/(133–102) = 3.06 and S = (133–102)/(208–102) = 0.2897. According to the table at https://www.cryogeny.cn/wp-content/ql-cache/quicklatex.com-3fc71d70bf7ac22f6d69d527d42f5c49_l3.svg, the value is 0.88. Treating these two heat exchangers as one large heat exchanger, the logarithmic mean temperature difference can be calculated using the formulas at https://www.cryogeny.cn/wp-content/ql-cache/quicklatex.com-9fc9780d473566fd6ec9b308f30668aa_l3.svg, https://www.cryogeny.cn/wp-content/ql-cache/quicklatex.com-76b37193aceb1d4349d269d56cb053d9_l3.svg, https://www.cryogeny.cn/wp-content/ql-cache/quicklatex.com-df96b1f56bc57141760b7985eccaec34_l3.svg, https://www.cryogeny.cn/wp-content/ql-cache/quicklatex.com-5b29fd5d740731e9004b7ff3d5944bf2_l3.svg, and https://www.cryogeny.cn/wp-content/ql-cache/quicklatex.com-5ae404d68ffdb3c3d8ae18864c58a661_l3.svg. Using the formula for heat calculation at https://www.cryogeny.cn/wp-content/ql-cache/quicklatex.com-522c3aac55876583549078ec38a98577_l3.svg, https://www.cryogeny.cn/wp-content/ql-cache/quicklatex.com-8ed650956f2f8f399aeee36f8547f489_l3.svg, https://www.cryogeny.cn/wp-content/ql-cache/quicklatex.com-047eb7775aede47b095d5f355006fef9_l3.svg, and https://www.cryogeny.cn/wp-content/ql-cache/quicklatex.com-142ec0ffad47f3a3715b7609088ab38d_l3.svg, the area of each individual heat exchanger is 113; therefore, the answer is C