Process Engineering Division – Instrumentation and Automation Section – Daily Topics – Topic No. 2020-12-23
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Process Engineering Division – Instrumentation and Automation Section – Daily Topics – Topic No. 2020-12-23: Discussion topic: 335. When using a single-flange level gauge to measure the liquid level in an open container, the distances from the gauge to the highest and lowest liquid levels are respectively h1 = 1 m and h2 = 3 m. If the density of the medium being measured is ρ=980 kg/m3, then: 2. If the height of the liquid surface is h=2.5 m, what will be the output of the electro-hydraulic level gauge? (5 points are given only to those who provide the correct answer; additional points are awarded for a correct calculation process or detailed explanation.)
If the level gauge height H is 2.5 m, then (2.5–1)/(3–1) = 0.75; therefore, the output is 16 mA.
A single-flange level gauge is used to measure the liquid level in an open container, with the distances from the gauge to the highest and lowest liquid levels being h1=1m and h2=3m, respectively. If the density of the medium being measured is ρ=980 kg/m3, then: 2. If the height of the liquid surface is h=2.5 m, what will be the output of the electro-liquid level gauge? Solution: I = (h – h1) / (h2 – h1) × (20 – 4) + 4 = (2.5 – 1) / (3 – 1) × 16 + 4 = 16(mA). Answer: The height of the liquid level is h=2.5m, and the output of the electro-liquid level gauge is 16mA.
I=*(2.5-1)+4=16mA
I=(h-h1)/(h2-h1)×(20-4)+4=(2.5-1)/(3-1)×16+4=16(mA).
4+(2.5-1)/(3-1)*(20-4)=16ma
If the level gauge height H is 2.5 m, then (2.5–1)/(3–1) = 0.75; therefore, the output is 16 mA.
A single-flange level gauge is used to measure the liquid level in an open container, with the distances from the gauge to the highest and lowest liquid levels being h1=1m and h2=3m, respectively. If the density of the medium being measured is ρ=980 kg/m3, then: 2. If the height of the liquid surface is h=2.5 m, what will be the output of the electro-liquid level gauge? Solution: I = (h – h1) / (h2 – h1) × (20 – 4) + 4 = (2.5 – 1) / (3 – 1) × 16 + 4 = 16(mA). Answer: The height of the liquid level is h=2.5m, and the output of the electro-liquid level gauge is 16mA.
A single-flange level gauge is used to measure the liquid level in an open container, with the distances from the gauge to the highest and lowest liquid levels being h1=1m and h2=3m, respectively. If the density of the medium being measured is ρ=980 kg/m3, then: 2. If the height of the liquid surface is h=2.5 m, what will be the output of the electro-liquid level gauge? Solution: I = (h – h1) / (h2 – h1) × (20 – 4) + 4 = (2.5 – 1) / (3 – 1) × 16 + 4 = 16(mA). Answer: The height of the liquid level is h=2.5m, and the output of the electro-liquid level gauge is 16mA.