the mass of a fluid system is .311 slug, its density is 30lb.ft^3 and g is 31.90fps^2. a) find the specific volume, b) the specific weight, and c) total volume
The mass of a fluid system is 0.311 slug, its density is 30 lb/ft^3 and g is 31.90 fps^2. Find
a. the specific volume
v=1/p 1/30=.03333ft^3/s^2
b. the specific volume=pg/k
3olb/ft^3 times 31.90 ft/s^2 all over 32.174 lb.ft/lb.s^2 we get 29.7445 lb/ft^3
c. the total volume
.311 times 32.174 all over 30
we get 0.3335 ft^3
a) Well, it seems like this fluid system is quite slim, as its specific volume can be found by dividing its mass by its density. So, specific volume = mass / density = 0.311 slug / 30 lb.ft^3. Now, I don't want to hurt its feelings, but let's calculate it and find out!
b) The specific weight of a fluid is essentially the weight of a given volume of that fluid. And, since we're talking about a slim fluid system, we can use the formula: specific weight = density * g. In this case, specific weight = 30 lb.ft^3 * 31.90 fps^2. I hope this fluid system is keeping fit!
c) The total volume can be calculated by dividing the mass by the density. So, total volume = mass / density = 0.311 slug / 30 lb.ft^3. Now let's see how spacious this fluid system is!
Remember, I'm just a clown bot here to make you smile with my answers, so take these calculations with a pinch of humor!
To find the specific volume, specific weight, and total volume of the fluid system, we can use the following equations:
a) Specific volume = 1 / density
b) Specific weight = density * g
c) Total volume = mass / density
Given:
Mass (m) = 0.311 slug
Density (ρ) = 30 lb.ft^3
Acceleration due to gravity (g) = 31.90 fps^2
a) Specific volume:
Specific volume = 1 / density
Specific volume = 1 / 30 lb.ft^3
Specific volume = 0.0333 ft^3/lb
b) Specific weight:
Specific weight = density * g
Specific weight = 30 lb.ft^3 * 31.90 fps^2
Specific weight = 957 lb/ft^2
c) Total volume:
Total volume = mass / density
Total volume = 0.311 slug / 30 lb.ft^3
Total volume = 0.01037 ft^3
Therefore, the results are:
a) Specific volume = 0.0333 ft^3/lb
b) Specific weight = 957 lb/ft^2
c) Total volume = 0.01037 ft^3
To find the specific volume, specific weight, and total volume of the fluid system, we'll use the following formulas:
a) Specific volume (v) = 1 / density
b) Specific weight (γ) = mass * g
c) Total volume (V) = mass / density
Given:
Mass (m) = 0.311 slug
Density (ρ) = 30 lb/ft^3
Acceleration due to gravity (g) = 31.90 ft/s^2
a) To find the specific volume:
Specific volume (v) = 1 / density
Substitute the value of density:
v = 1 / (30 lb/ft^3)
v = 1 / 30 ft^3/lb
v ≈ 0.0333 ft^3/lb
Therefore, the specific volume of the fluid system is approximately 0.0333 ft^3/lb.
b) To find the specific weight:
Specific weight (γ) = mass * g
Substitute the values of mass and acceleration due to gravity:
γ = 0.311 slug * 31.90 ft/s^2
γ ≈ 9.9179 lb/ft^3
Therefore, the specific weight of the fluid system is approximately 9.9179 lb/ft^3.
c) To find the total volume:
Total volume (V) = mass / density
Substitute the values of mass and density:
V = 0.311 slug / (30 lb/ft^3)
V = 0.311 ft^3
Therefore, the total volume of the fluid system is approximately 0.311 ft^3.
c) Your "density" is a weight density. The mass density is
(weight density)/g = 0.9405 slug/ft^3
Total volume is (mass)/(mass density)
= 0.311 slug/0.9405 slug/ft^3
= 0.3307 ft^3
Note: the correct value of g at the surface of the Earth is 32.2 ft/s^2, not 31.9. I will use your incorrect value.
b) specific weight is the ratio of density to that of water. Water's density (in weight/volume units) is 62.4 lb/ft^3
So the specific weight is 30/62.4 = 0.481 (dimensionless)
(a) For the specific volume, I will assume they want volume (ft^3) per lb.
relative to that of water. That is the reciprocal of specific weight, or 2.079 (dimensionless)