Summary of Topics from Introductory Problems on Thermal Energy and
Fluids
Fluids:
- P = F / A (definition of pressure). Use if you know pressure and force, or force and
area, or area and pressure.
- `rho g h + .5 `rho v^2 + P is constant (Bernoulli's Eqn). If `rho and g constant, as
with water near Earth's surface, we see that v, h and P vary in such a way that if one
term goes up something else has to go down to compensate. Usually one will be constant,
one will go up and the other will go down. For water `rho = 1000 kg / m^3; for air `rho is
about 1.4 kg/m^3.
- For a confined fluid, A1 * v1 = A2 * v2 (continuity equation for incompressible
fluids—the amount flowing past one point is equal to the amount flowing past
another). Ratio of velocities inverse to ratio of areas: v2 / v1 = A1 / A2. Remember area
proportional to diameter or to radius.
Thermal Energy Transfers and Materials:
- Specific heat c is typically given as number of Joules per kg, per Celsius degree.
Thermal energy required to raise a sample is specific heat, multiplied by number of kg,
multiplied by number of degrees temp change in Celsius. In equation form, `dQ = m c `dT.
- Rate of thermal energy transfer proportional to temp gradient `dT / `dx and temp
difference: `dQ / `dt = k A ( `dT / `dx). Note t is time and T is temp. `dx is typically
thickness of container. `dQ / `dt is energy transfer in Joules / sec.
Kinetic Theory of Gases:
One particle mass m, vel. v in cylinder length L, moving parallel to axis of cylinder
exerts force F = `d(mv) / `dt = 2 m v / ( 2 L / v) = m v^2 / L on either end.
- If area of end is A then:
pressure = F / A = m v^2 / (L * A) = m v^2 / V, where V is volume.
Noting
- If single mass divided into large number of particles and they still move parallel to
the axis, still same momentum change in same time interval so nothing is changed.
- A large number of particles wouldn't keep moving in the same direction because of
collisions. The energy would rapidly get divided equally among three independent
directions in space. Only one direction affects a given end of the cylinder, so now
pressure is 1/3 as great as in the previous model:
P = 2/3 KE / V or PV = 2/3 KE
PV = n R T so n R T = 2/3 KE so KE = 3/2 (nRT)
(n is the number of moles. To find KE per particle divide by the number of moles to get
the KE of 1 mole, then by Avagadro's Number to get the KE of 1 particle.)
Gas Laws:
- PV = n R T so, for example:
If P and n are constant then V is proportional to T and you can use 'straight ratios'
V2 / V1 = T2 / T1 to calculate desired temp or volume.
If V and n are constant then P is proportional to T and you can use 'straight ratios'
V2 / V1 = P2 / P1 to calculate desired pressure or volume.
If T and n are constant then P is inversely proportional to V and you can use inverse
ratios P2 / P1 = V1 / V2 (note that one of the ratios is 'upside down') to calculate
desired temp or volume.
If you know three of the four quantities P, V, n and T you can find the fourth.
Thermodynamics:
- When a system converts thermal energy to mechanical work as it runs through a repeating
cycle, energy `dQin is transferred to the system from the 'hot source', `dQout is
transferred out to the 'cold source' and the system does work `dW.
- Energy conservation tells us that `dQin = `dW + `dQout.
- The efficiency of such a system is
efficiency = work done by system / thermal energy transfered into system, or
efficiency = `dW / `dQin .
- If we know two of the three quantities `dQin, `dQout and `dW we can find the third and
hence compute efficiency.
The maximum possible efficiency of any such system is (Th – Tc) / Th, where Th and
Tc are the temperatures of the hot and cold sources.