Can anyone tell me the formula and what the symbols mean for the following problems?
1. You have a 1/2 inch internal diameter hose at your home and water moves through it at 2.5 feet/sec when the faucet is fully turned on. If you leave the water running while you wash your car (15 min) how much water have you just used?
2. You are given a hydraulic jack which you'd like to keep in your car in case you have a flat. Your vehicle weighs 4500lbs, and you know the jack will have to lift 1500 lbs when you put it under the axle. The hand end cylinder of the jack is 1/2inch in diameter and the "work" end of the cylinder is 3.5 inches in diameter. It has an 8 inch long handle; the distance from the attachment point of the handle to the small piston of the jack is 1.5 inches. How much force will you have to apply to the end of the handle to lift the car and change a tire?
Q (volume flow rate) = V * A, where
A = pi * (D^2/4) is the cross sectional area of the hose
D is the hose diameter (0.5 inch = 1/24 foot)
V is the average flow velocity in the hose. (It is higher in the middle than at the wall)
If you have A in square feet and V in ft/s, your answer will be in cubic feet per second. Convert that to gallons/minute and multiply by the washing time (15 minutes) to get the answer in gallons.
i did
A=pi(1/48)^2=.0014
2.5 ft/sec * .0014ft
.0035 ft^3
450 galmin/1 ft^3/sec * .0035 ft^3/sec*15 min
= 23.625 gpm
Sure! Let's start with the first problem:
1. In order to calculate the amount of water used, we need to use the formula for volume flow rate, which is given by Q = V * A, where Q is the volume flow rate, V is the average flow velocity, and A is the cross-sectional area of the hose.
2. The cross-sectional area of the hose can be calculated using the formula A = π * (D^2 / 4), where A is the area and D is the diameter of the hose.
3. In this case, the given diameter of the hose is 1/2 inch, which is equivalent to 1/24 foot. Therefore, D = 1/24 foot.
4. Now, let's calculate the cross-sectional area A using the formula:
A = π * (1/24)^2 / 4 = 0.0014 square feet.
5. The given flow velocity is 2.5 feet per second. Let's use this value to calculate the volume flow rate using the formula Q = V * A:
Q = 2.5 ft/s * 0.0014 ft = 0.0035 cubic feet per second.
6. To convert cubic feet per second to gallons per minute, we need to multiply the value by a conversion factor of 450 gallons per minute per 1 cubic foot per second:
0.0035 ft^3/sec * 450 gal/min = 1.575 gallons per minute.
7. Finally, to find out how much water you have used during a 15-minute car wash, multiply the flow rate by the washing time:
1.575 gal/min * 15 min = 23.625 gallons.
So, you have used approximately 23.625 gallons of water during the car wash.
Now, let's move on to the second problem:
1. To calculate the force required to lift the car and change a tire, we need to use the principle of Pascal's law in hydraulics.
2. According to Pascal's law, the pressure is transmitted equally in all directions in an enclosed fluid. Therefore, the pressure on the smaller piston (hand end) will be the same as the pressure on the larger piston (work end) of the jack.
3. The force exerted by the hand end can be calculated using the formula F = P * A, where F is the force, P is the pressure, and A is the cross-sectional area of the hand end cylinder.
4. The given diameter of the hand end cylinder is 1/2 inch, which is equivalent to 1/24 foot. Therefore, D = 1/24 foot.
5. The cross-sectional area A can be calculated using the formula A = π * (D^2 / 4):
A = π * (1/24)^2 / 4 = 0.0014 square feet.
6. Now, let's calculate the force exerted by the hand end using the formula F = P * A. However, to calculate the pressure P, we need to find the force exerted on the larger piston (work end) and consider the forces and distances involved.
7. The force exerted on the work end can be calculated using the formula F_work = W_vehicle * (A_work / A_hand), where F_work is the force on the work end, W_vehicle is the weight of the vehicle, A_work is the cross-sectional area of the work end cylinder, and A_hand is the cross-sectional area of the hand end cylinder.
8. The given weight of the vehicle is 4500 lbs.
9. The given diameter of the work end cylinder is 3.5 inches, which is equivalent to 3.5/12 feet. Therefore, D_work = 3.5/12 feet.
10. The cross-sectional area A_work can be calculated using the formula A_work = π * (D_work^2 / 4):
A_work = π * (3.5/12)^2 / 4 = 0.079 square feet.
11. Now, let's calculate the force exerted on the work end using the formula F_work = W_vehicle * (A_work / A_hand):
F_work = 4500 lbs * (0.079 / 0.0014) = 254,285.71 lbs.
12. Since the pressure is transmitted equally in all directions, the pressure on the hand end is the same as the pressure on the work end. Therefore, we can equate the forces on both ends:
F_work = P * A_hand.
13. Rearranging the equation, we can solve for the pressure P:
P = F_work / A_hand = 254,285.71 lbs / 0.0014 square feet.
14. Now, let's calculate the force required to lift the car by considering the lever arm of the handle. The distance from the attachment point of the handle to the small piston is given as 1.5 inches, which is equivalent to 1.5/12 feet.
15. The force required to lift the car can be calculated using the formula F_required = P * L, where F_required is the force required, P is the pressure, and L is the lever arm of the handle.
16. The given length of the handle is 8 inches, which is equivalent to 8/12 feet.
17. Now, let's calculate the force required to lift the car using the formula F_required = P * L:
F_required = P * L = (254,285.71 lbs / 0.0014 square feet) * (8/12) feet.
By plugging in the values, you should be able to calculate the force required to lift the car and change a tire.
Please note that the final calculation involves unit conversions to ensure the consistency of units throughout the calculations.