Engineering Math and Equations Flashcards
Determine Pressure when a force of 200 lb is applied to an area of 4 sq in.
P = F/A P = 2OOlb / 4 sq in. P = 50 psi
What is the area a 1500 lb force is acting on if it exerts a pressure of 100 psi?
A = F/P A = 707lb / 100 psi A = 7.07 sq in.
What is the force on the bottom of a tank that has base dimensions of 6’ by 12’ and contains a water depth of 7” Note: feet must be converted to inches. There is a pressure of 0.433 psi per vertical foot (depth) of water.
F = P x A F = (0.433 x 7) psi x [(6' x 12) x ( 12' x 12)} sq in F = 3.031 psi x (72 x 144) sq in. F = 3.031 psi x 10,368 sq in. F = 31,425 lb
What is the force on a flat head of a steam boiler drum (tube sheet) with a pressure of 100 psi and an area of 1000 sq in.
F = P x A F = 100 psi x 1000 sq in. F = 100,000 lb
Define Work and it’s unit of measurement.
Work is the movement of an object by a constant force to a specific distance. The units of distance are stated in linear measurement (feet). Force is a push or pull measured in pounds (weight).
Write the equation for work and unit of measurement
Work = Force x distance W = F (in lb) x D (in ft) W = F x D W = ft-lb
Write equation for distance and force.
Distance = work/force D = W/F Force = Work/Distance F = W/D
How many ft-lb of work are done if a force of 80 lb is applied and moves an object 100’?
W = F x D W = 80 lb x 100' W = 8000 ft-lb
How high is 2500 lb load lifted by a hoist performing 50,000 ft-lb work?
D = W/F D = 50,000 ft-lb / 2500 lb D = 20'
What is the force applied to move an object if 450 ft-lb of work is done while moving this object 15’?
F = W/D F = 450 ft-lb F = 30 lb
Define Power and its unit of measurement.
Power is the rate at which worked is performed.
The unit of measurement is ft-lb per unit of time.
Write the equation for Power.
Power = Work / Time P = ft-lb / T Power = Force x Distance / Time
Write the equation for Work.
Work = Force / Distance
What is the power output of an engine that performs 140,000 ft-lb of work in 4 minutes.
P = W / T P = ft-lb / T P = 140,000 ft-lb / 4 min P = 35,000 ft-lb / min
Define mechanical HP.
1 mechanical HP = 33,000 ft-lb of work / minute
Horsepower = Power / 33,000 ft-lb / minute.
How much horsepower will an engine develop if it does 168,000 ft-lb of work in 2.5 minutes?
HP = W / T x 33,000 HP = 168,000 ft-lb / 2.5 min x 33,000 ft-lb/min HP = 168,000d ft-lb / 82,500 ft-lb HP = 2.04 HP
A pump delivers 2500 lb of water per minute and lifts it 280’. What is the horsepower of the pump?
HP = F x D / T x 33,000 ft-lb/min HP = 2500 lb x 280' / 1 min x 33,000 ft-lb/min HP = 700,000 ft-lb / 33,000 ft-lb HP = 21.21 HP
Write the equation do determine the quantity of latent added or removed from a substance.
Quantity of heat = Weight of substance x latent heat.
Q (in btu) = W (in lb) x Lh (in Btu/lb to change it state
Q = W x Lh
How much heat is required to change 1000 lb of water to steam at 212 deg. F.
Q = W x Lh Q = 1000 lb x 970.3 Btu/lb Q = 970,300 Btu
Write the equation to determine the quantity of sensible heat added to or removed from a substance.
Quantity of heat = Weight of substance x specific heat x (Final temp - Int. temp)
Q (in Btu) = W (in lb) x Sph (in Btu/lb/F x {T1 (in F) -T2 (in F)}
Q = W x Sph x (T1 - T2)
How much heat transfer is required to raise the of 1000 lb of water from 92°F to 212°F
Q = W x Sph (T1 - T2) Q = 1000 lb x 1 Btu/ln/°F x (212 - 92)°F Q = 1000 lb x 120 Btu/lb Q = 120,000 Btu
Convert F to C
(1.8 x °C) to 32
Convert ˚C to ˚F
˚F - 32 / 1.8
1 Btu is equal to how ft-lb of mechanical energy
1 btu is equal to 778 ft-lb of mechanical energy.
Write Eq. for Charles first laws
V1/T1 = V2/T2
How many Btu/min are in 1 mechanical HP
42.4 Btu/min
Write equation for Charles 2nd law.
P1/T1 = P2/T2
What is the final volume of 10 cu ft. of gas @ 90˚F when heated to 200˚F at a constant pressure?
Answer is 12 cu ft. (pg. 14)
What is the pressure on a gas tank with an initial temp. of 90˚F and a press. of 100 psi (85.3 psi) when heated to 200˚F.
P2 = 120 psia ( 105.3 psi)
Pg. 15
Eq. for Boyle’s law
P1 x V1 = P2 x V2
What is the final press. of 1 cu ft of air with an initial pressure of 100 psi that is expanded at a constant temperature to 2 cu. ft?
P2 = 50 psi (35.3 psig)
pg. 16
