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I'm not sure what data you have to answer your question specifically. For H2 with a MW of 2.016, the SG is 0.0696. This problem has been solved Youll get a detailed solution from a subject matter expert that helps you learn core concepts. Compute the density of water and the specific weight and density of mercury. The specific gravity of mercury is 13.55. CO has a MW of about 28, air is 28.96 so the SG for CO would be 0.967. The specific weight of water at ordinary pressure and temperature is 62.4 lb/ft3 (9.81 kN/m3).
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The weight density of water is 62.4 lb/ft3. Question: The weight density of water is 62.4 lb/ft3. Just to confuse things more, SG can also be used to describe the ratio of the MW of a gas to that of air. Youll get a detailed solution from a subject matter expert that helps you learn core concepts. Back the density, the specific volume is the reciprocal of that (1/density) or 0.01335 ft3/lb in this case.
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(Click on the graph for a larger version.) If the trough is full of water, find the force of the water on a triangular end. See this result also down below: The density of seawater 63.9262 lb/ft3. the trough in the figure below has width w3 ft, length L18 ft and height h5 ft. The average density of seawater is 63.9262 lb/ft3. If the block was in fresh water (density 1000 kg/cm3), what would be the buoy A cube of wood with mass m 1.0 kg floats 80 submerged in a cylinder filled with water (water density 1000 kg/m). If the trough is full of water, find the force of the water on a triangular end. That SG can be at temperature (which for geothermal can be pretty hot) or it can be at the same reference temperature as the water reference, you need to understand what you have and it's not something we can likely tell you what you have. It measures 6.00 cm wide by 8.00 cm high by L cm long. Properties and Definitions Useful constants, properties, and conversions gc 32.2 ft/sec2 lbm-ft/lbf-sec2 water 1.96 slugs/ft3 water 62.4 lb/ft3 1 ft3/sec 449 gpm 1 mgd 1.547 ft3/sec 1 foot of water 0.433 psi 1 psi 2. Since the specific weight of water is given as 62.4 lb/ft3 and we know that 1 gallon is equal to 1/7.48 cubic feet, we can substitute these values into the formula: Weight of 1 gallon of water 62.4 lb/ft3 x (1/7.48 ft3) Simplifying the equation: Weight of 1 gallon of water 62.4 lb/ft3 x 0.1337 ft3. So, if the brine has an SG of 1.2, its density is 1.2*62.4 = 74.9 lb/ft3. The density doesn't shift a lot with 'small' temperature changes but there are times where that can be a factor, say for custody transfer of fluids. The density of water is usually around 62.4 lb/ft3 or 8.34 lb/gal depending what temperature you use. In other words, the 62.4 lb/ft3 represents rho g, not just rho in the second equation. So it already has the gravitational constant built in. For example, suppose that in a setting similar to the problem posed in Preview Activity 6.As David and Latexman have said, SG is the ratio of the density of a fluid at some temperature to that of water at a reference temperature. begingroup Xyzar I think the key to understanding this is that density in lb/ft3 is not mass per unit volume, but rather gravitational force per unit volume. What is the meaning of the value you find? Why?īecause work is calculated by the rule \( W = F \cdot d\), whenever the force \( F\) is constant, it follows that we can use a definite integral to compute the work accomplished by a varying force. Remember the important points being made here: 1.
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Evaluate the definite integral \( \int^50_0 B(h) dh\). 62.4 lb of 1 ft 1 ft water 1 ft FIGURE 9.1 One cubic foot of water weighs 62.4 lb.= B(h)\Delta h\) measure for a given value of \( h\) and a small positive value of \( \Delta h\)?