riaology
Thermal and Fluid Science / Formula & reasoning reference

A compact map.
Keep the meaning.

Find a relationship, check its assumptions, and connect it to the model. Each entry includes a distinction worth remembering.

Includes every question in the available lessons. Future lessons will join this bank when published.

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26 lesson references found.

Foundations of thermodynamics

Small systems. Big ideas.

Full visual lesson ↗

Coffee cools, pumps move water, and rooms warm up. Thermal and fluid science turns those familiar changes into predictions: what moves, what accumulates, and how quickly a system responds.

An amount is not a rate.

Etransfer=W˙ Δt(constant power)E_{transfer}=\dot W\,\Delta t\quad\text{(constant power)}

Accumulation connects a balance to change.

dEdt=E˙in−E˙out\frac{dE}{dt}=\dot E_{in}-\dot E_{out}

Assumptions and scope

Distinguish an amount of energy from a rate of transfer. · Predict change from competing inputs and outputs. · Connect thermodynamics, fluid mechanics, and heat transfer.

Foundations of thermodynamics

A useful model starts with a boundary.

Full visual lesson ↗

Before choosing an equation, choose the question. A tank can be a container, a region of space, or a collection of water particles. Each description makes some questions easier to answer.

Mass accumulation becomes a level change.

dmdt=ρ(V˙in−V˙out),dhdt=V˙in−V˙outA\frac{dm}{dt}=\rho(\dot V_{in}-\dot V_{out}),\qquad\frac{dh}{dt}=\frac{\dot V_{in}-\dot V_{out}}{A}

Assumptions and scope

Choose a system that matches a question. · State assumptions before calculating. · Translate a mass balance into a changing liquid level.

Foundations of thermodynamics

Closed is not the same as isolated.

Full visual lesson ↗

A sealed gas can gain energy. An insulated pump can exchange mass and shaft work. The words closed, open, and isolated describe exchanges across your chosen boundary.

This topic starts with a physical interpretation. Use its lesson to practice the setup.

Assumptions and scope

Classify systems by actual boundary crossings. · Recognize a moving closed-system boundary. · Distinguish steady flow from no flow.

Foundations of thermodynamics

What changes when you split a system?

Full visual lesson ↗

Divide a uniform sample in half. Its mass halves; its temperature does not. This simple experiment separates properties that depend on system size from those that describe local state.

Specific properties remove the mass dependence.

v=Vm=1ρ,u=Umv=\frac Vm=\frac1\rho,\qquad u=\frac Um

Assumptions and scope

Distinguish intensive, extensive, and specific properties. · Separate a state property from a process quantity. · Recognize the conditions needed for equilibrium.

Foundations of thermodynamics

The zero on a gauge is a choice.

Full visual lesson ↗

A pressure gauge can read zero in a pressurized atmosphere. A temperature of 20°C is not twice as hot as 10°C. Reference points matter whenever a measurement enters an equation.

Pressure is force per area.

F=Δp A(uniform pressure difference)F=\Delta p\,A\quad\text{(uniform pressure difference)}

Gauge pressure subtracts a reference.

pabs=pg+patmp_{abs}=p_g+p_{atm}

Assumptions and scope

Convert gauge and absolute pressure. · Use absolute temperature in gas-law ratios. · Distinguish a temperature value from a temperature difference.

Foundations of thermodynamics

Energy has more than one place to hide.

Full visual lesson ↗

A lifted weight, a speeding cart, and warmer water store energy in different ways. Decide which stores change before simplifying an energy balance.

Bulk motion stores kinetic energy.

KE=12mV2KE=\frac12mV^2

Elevation stores gravitational potential energy.

ΔPE=mg(z2−z1)\Delta PE=mg(z_2-z_1)

Internal energy is a separate store.

E=U+KE+PEE=U+KE+PE

Assumptions and scope

Calculate kinetic and gravitational potential energy changes. · Distinguish bulk motion from internal energy. · Recognize energy conversion when mechanical motion stops.

Foundations of thermodynamics

Energy crosses. Heat and work describe how.

Full visual lesson ↗

A warmer wall, a rotating shaft, and a moving piston can all transfer energy. The mechanism at the boundary determines whether we call the transfer heat or work.

Work has several mechanisms.

ΔE=Q−W(closed system)\Delta E=Q-W\quad\text{(closed system)}

Moving-boundary work depends on the path.

Wb=∫V1V2p dV,Wb=p(V2−V1)(constant pressure)W_b=\int_{V_1}^{V_2}p\,dV,\qquad W_b=p(V_2-V_1)\quad\text{(constant pressure)}

Assumptions and scope

Identify heat, electrical work, shaft work, and boundary work. · Apply a consistent work sign convention. · Recognize process dependence through a pressure-volume diagram.

Foundations of thermodynamics

Every joule needs an account.

Full visual lesson ↗

Choose the boundary, name the transfers, and identify the energy stores. The first law makes the story quantitative; efficiency asks how much of the input serves your intended purpose.

Write the balance before inserting numbers.

ΔU=Q−W−ΔKE−ΔPE\Delta U=Q-W-\Delta KE-\Delta PE

Useful is a design decision.

η=Euseful,outEin\eta=\frac{E_{useful,out}}{E_{in}}

Assumptions and scope

Apply ΔE=Q−W\Delta E=Q-W with a stated sign convention. · Include kinetic and potential energy changes when needed. · Distinguish conservation from useful-output efficiency.

Properties & energy balances

What happens during a phase change?

Full visual lesson ↗

Temperature can pause while energy keeps entering.

Saturation ties pressure to temperature.

T=Tsat(p)T=T_{sat}(p)

The plateau is an energy conversion.

Q=m(h2−h1),hfg=hg−hfQ=m(h_2-h_1),\quad h_{fg}=h_g-h_f

Assumptions and scope

One kilogram at constant 100 kPa; equilibrium heating with only pressure-volume work. Liquid heat capacity is approximated as 4.18 kJ/(kg K); saturated properties use IAPWS-IF97.

Properties & energy balances

How do we read property tables?

Full visual lesson ↗

Choose the region before choosing a row.

Classify before lookup.

T<Tsat: liquid;T>Tsat: vaporT<T_{sat}:\text{ liquid};\quad T>T_{sat}:\text{ vapor}

Read the headings and units.

V=mv,U=mu,H=mhV=mv,\quad U=mu,\quad H=mh

Interpolation is a local approximation.

h(T)=ha+T−TaTb−Ta(hb−ha)h(T)=h_a+\frac{T-T_a}{T_b-T_a}(h_b-h_a)

Assumptions and scope

Superheated water at 200 kPa. Interpolation is linear between the displayed IAPWS-IF97 table rows; values are not interpolated across phase boundaries.

Properties & energy balances

What does vapor quality mean?

Full visual lesson ↗

A little vapor can occupy most of the space.

Quality measures mass.

x=mgmf+mgx=\frac{m_g}{m_f+m_g}

Derive the mixture relation.

v=(1−x)vf+xvg=vf+xvfgv=(1-x)v_f+xv_g=v_f+xv_{fg}

Invert when another property is given.

x=v−vfvg−vf,h=hf+xhfgx=\frac{v-v_f}{v_g-v_f},\quad h=h_f+xh_{fg}

Volume fractions tell a different story.

ϕg=xvg(1−x)vf+xvg\phi_g=\frac{xv_g}{(1-x)v_f+xv_g}

Assumptions and scope

Equilibrium saturated water at 100 kPa. Quality is defined only inside the two-phase region, including its saturated endpoints.

Properties & energy balances

How does a closed system store energy?

Full visual lesson ↗

A rigid wall stops boundary work, not every energy transfer.

Start with a material system.

ΔU+ΔKE+ΔPE=Q−W\Delta U+\Delta KE+\Delta PE=Q-W

Name each work mechanism.

W=Wb+Ws+WeW=W_b+W_s+W_e

Connect storage to state.

ΔU=mcv(T2−T1)\Delta U=m c_v(T_2-T_1)

Check the outcome.

Q=0,Wb=0⟹ΔU=Ws,inQ=0,\quad W_b=0\quad\Longrightarrow\quad\Delta U=W_{s,in}

Assumptions and scope

Rigid sealed tank with 2 kg ideal gas, cv=0.718kJcv = 0.718 \mathrm{kJ}/(kg K); unchanged bulk kinetic and potential energy. Rigid means Wb=0,W_b = 0, not zero shaft work.

Properties & energy balances

What does a moving boundary do?

Full visual lesson ↗

The shaded area is a process, not a property.

Derive moving-boundary work.

δWb=pA ds=p dV,Wb=∫V1V2p dV\delta W_b=pA\,ds=p\,dV,\quad W_b=\int_{V_1}^{V_2}p\,dV

Integrate a polytropic path.

Wb=C1−n(V21−n−V11−n)=p2V2−p1V11−nW_b=\frac{C}{1-n}(V_2^{1-n}-V_1^{1-n})=\frac{p_2V_2-p_1V_1}{1-n}

The n=1 path needs a logarithm.

Wb=p1V1ln⁡ ⁣(V2V1)(n=1)W_b=p_1V_1\ln\!\left(\frac{V_2}{V_1}\right)\quad(n=1)

Assumptions and scope

Quasi-equilibrium ideal-gas path; p1=200kPa1 = 200 \mathrm{kPa}, V1=0.011 = 0.01 m³. Work by the gas is positive. kPa m3=kJ\mathrm{m}^3 = \mathrm{kJ}.

Properties & energy balances

How much energy changes the temperature?

Full visual lesson ↗

A heat capacity is a slope you can integrate.

Define the slopes.

cv=(∂u∂T)v,cp=(∂h∂T)pc_v=\left(\frac{\partial u}{\partial T}\right)_v,\quad c_p=\left(\frac{\partial h}{\partial T}\right)_p

Derive the ideal-gas difference.

h=u+RT⟹cp−cv=Rh=u+RT\quad\Longrightarrow\quad c_p-c_v=R

Integrate before approximating.

Δu=∫T1T2cv(T) dT≈cv,avg(T2−T1)\Delta u=\int_{T_1}^{T_2}c_v(T)\,dT\approx c_{v,avg}(T_2-T_1)

Liquids and solids have another approximation.

Δh≈∫T1T2c(T) dT+v(p2−p1)\Delta h\approx\int_{T_1}^{T_2}c(T)\,dT+v(p_2-p_1)

Assumptions and scope

One kilogram of a teaching-model ideal gas with cv(T)=0.70+0.00010.70+0.0001T kJ/(kg K), R=0.287kJR=0.287 \mathrm{kJ}/(kg K). This illustrative heat-capacity model is not a property correlation for a named gas.

Properties & energy balances

What enters, leaves, and accumulates?

Full visual lesson ↗

A steady inlet does not guarantee a steady tank.

Conservation starts with accumulation.

dmCVdt=∑inm˙−∑outm˙\frac{dm_{CV}}{dt}=\sum_{in}\dot m-\sum_{out}\dot m

Convert a port measurement.

m˙=ρV˙=ρAV\dot m=\rho\dot V=\rho AV

Steady flow constrains the sum.

m˙1=m˙2,ρ1A1V1=ρ2A2V2\dot m_1=\dot m_2,\quad\rho_1A_1V_1=\rho_2A_2V_2

Unsteady tanks need integration.

h(t)=h0+V˙in−V˙outAth(t)=h_0+\frac{\dot V_{in}-\dot V_{out}}{A}t

Assumptions and scope

Water density 1000 kg/m³, tank area 0.5 m², initial height 0.5 m, rim at 1 m. Prescribed constant flow rates; prediction stops at empty/full.

Properties & energy balances

Why does flowing fluid carry enthalpy?

Full visual lesson ↗

Enthalpy packages the energy needed to move mass.

A port needs pressure work.

wflow=pAℓm=pvw_{flow}=\frac{pA\ell}{m}=pv

Combine two terms once.

h=u+pvh=u+pv

Add motion and elevation.

eflow=h+V22+gze_{flow}=h+\frac{V^2}{2}+gz

Assumptions and scope

Illustrative parcel with u=250kJu = 250 \mathrm{kJ}/kg; pressure and volume controls explore the energy identity, not a real-fluid equation of state. Flow work is already included in enthalpy.

Properties & energy balances

How do we balance energy in a flowing system?

Full visual lesson ↗

Mass carries energy through the boundary.

Write the complete control-volume balance.

dECVdt=Q˙−W˙s+∑inm˙(h+V22+gz)−∑outm˙(h+V22+gz)\frac{dE_{CV}}{dt}=\dot Q-\dot W_s+\sum_{in}\dot m\left(h+\frac{V^2}{2}+gz\right)-\sum_{out}\dot m\left(h+\frac{V^2}{2}+gz\right)

Nozzle speed comes from an enthalpy drop.

V2=V12+2000(h1−h2)V_2=\sqrt{V_1^2+2000(h_1-h_2)}

Turbines and throttles differ.

W˙turbine≈m˙(h1−h2),h1≈h2  (throttle)\dot W_{turbine}\approx\dot m(h_1-h_2),\quad h_1\approx h_2\;\text{(throttle)}

Assumptions and scope

Steady, single-inlet/single-outlet adiabatic nozzle; no shaft work and negligible elevation change. Enthalpy drop in kJ/kg requires factor 1000 to combine with m²/s².

Fluid mechanics

What makes a fluid a fluid?

Full visual lesson ↗

Viscosity connects motion to shear stress.

Fluids deform under shear.

ρ=mV,τ=μdudy\rho=\frac m{\mathcal V},\quad\tau=\mu\frac{du}{dy}

No slip supplies boundary values.

u(0)=0,u(H)=U,u(y)=UyHu(0)=0,\quad u(H)=U,\quad u(y)=U\frac yH

Find stress and force.

τ=μUH,F=τA\tau=\mu\frac UH,\quad F=\tau A

Kinematic viscosity is a ratio.

ν=μρ\nu=\frac\mu\rho

Assumptions and scope

Steady laminar Newtonian Couette flow; no pressure gradient, no slip, fixed lower plate and moving upper plate. Edge effects neglected.

Fluid mechanics

How does pressure change with depth?

Full visual lesson ↗

Depth sets pressure; density sets its slope.

Derive a small-column balance.

p(z)A−p(z+dz)A−ρgA dz=0⟹dpdz=−ρgp(z)A-p(z+dz)A-\rho gA\,dz=0\quad\Longrightarrow\quad\frac{dp}{dz}=-\rho g

Integrate layer by layer.

p−ps=g∑iρiΔhip-p_s=g\sum_i\rho_i\Delta h_i

Gauge is relative to the surface reference.

pabs=patm+pgp_{abs}=p_{atm}+p_g

Pressure loads a submerged wall.

F=∫Apg dA=12ρgbH2,hCP=2H3F=\int_A p_g\,dA=\frac12\rho gbH^2,\quad h_{CP}=\frac{2H}{3}

Assumptions and scope

Static open tank; upper layer 1 m deep, lower layer water at 1000 kg/m³, g=9.81m/sg = 9.81 \mathrm{m}/\mathrm{s}². No surface-tension pressure jump at the flat interface.

Fluid mechanics

How do we describe fluid motion?

Full visual lesson ↗

A steady flow can accelerate its particles.

Eulerian and material descriptions.

V=V(x,t)\mathbf V=\mathbf V(\mathbf x,t)

Continuity fixes the mean speed.

A1V1=A2V2A_1V_1=A_2V_2

Acceleration has two contributions.

DVDt=∂V∂t+V∂V∂x\frac{DV}{Dt}=\frac{\partial V}{\partial t}+V\frac{\partial V}{\partial x}

Assumptions and scope

Steady, incompressible, one-dimensional mean flow; area decreases linearly over 1 m. Particle paths are not the same concept as instantaneous streamlines in unsteady flow.

Fluid mechanics

How does mechanical energy move with a fluid?

Full visual lesson ↗

Pressure, speed, and height share a mechanical-energy account.

State the mechanical-energy relation.

pρg+V22g+z=C\frac p{\rho g}+\frac{V^2}{2g}+z=C

Combine continuity with energy.

V2=V1(D1/D2)2V_2=V_1(D_1/D_2)^2

Turn head into measurable pressure.

p1−p2=ρ2(V22−V12)p_1-p_2=\frac\rho2(V_2^2-V_1^2)

Assumptions and scope

Steady incompressible water, same elevation, negligible losses, no machines, along a streamline; uniform-speed approximation. Pressure-drop result does not itself check cavitation.

Fluid mechanics

When can we use Bernoulli?

Full visual lesson ↗

An equation earns its cancellations.

The ideal reservoir prediction.

Videal=2gΔzV_{ideal}=\sqrt{2g\Delta z}

Add the missing physics.

V2=2g(Δz+hp)1+KV_2=\sqrt{\frac{2g(\Delta z+h_p)}{1+K}}

Assumptions and scope

Large open reservoir discharges to atmosphere 5 m lower; negligible reservoir speed. Ideal Bernoulli corresponds to K=0\mathrm{K} = 0 and no pump; extended energy accounting handles K and pump head.

Fluid mechanics

How do pumps and losses change the balance?

Full visual lesson ↗

A pump supplies the head the system requires.

Keep machine and loss terms visible.

p1ρg+α1V122g+z1+hp−ht=p2ρg+α2V222g+z2+hL\frac{p_1}{\rho g}+\alpha_1\frac{V_1^2}{2g}+z_1+h_p-h_t=\frac{p_2}{\rho g}+\alpha_2\frac{V_2^2}{2g}+z_2+h_L

Two large reservoirs simplify the endpoints.

hp=Δz+hLh_p=\Delta z+h_L

Convert head to shaft power.

W˙hyd=ρgV˙hp,W˙shaft,in=W˙hydηp\dot W_{hyd}=\rho g\dot Vh_p,\quad\dot W_{shaft,in}=\frac{\dot W_{hyd}}{\eta_p}

Assumptions and scope

Between two large open reservoirs with negligible surface speeds. Teaching system loss hL=0.2hL = 0.2 q² m with q measured in L/s; pump head is hydraulic energy added per unit weight.

Fluid mechanics

How do flow regimes differ?

Full visual lesson ↗

Reynolds number compares inertia with viscosity.

Form a dimensionless comparison.

Re=ρVDμ=VDν\mathrm{Re}=\frac{\rho VD}{\mu}=\frac{VD}{\nu}

Laminar flow has an exact profile.

u(r)=Δp4μL(R2−r2),umax=2Vu(r)=\frac{\Delta p}{4\mu L}(R^2-r^2),\quad u_{max}=2V

Assumptions and scope

Water ν = 10⁻⁶ m²/s in a circular pipe. Re < 2300 is typically laminar; 2300–4000 transitional; > 4000 typically turbulent. Thresholds depend on disturbances and entrance conditions.

Fluid mechanics

Why does pressure fall along a pipe?

Full visual lesson ↗

A small diameter can demand a large pressure.

Derive the laminar flow rate.

V˙=∫0Ru(r)2πr dr=πR4Δp8μL=πD4Δp128μL\dot V=\int_0^R u(r)2\pi r\,dr=\frac{\pi R^4\Delta p}{8\mu L}=\frac{\pi D^4\Delta p}{128\mu L}

Use a common pressure-loss form.

Δp=fDLDρV22,fD=64/Re\Delta p=f_D\frac LD\frac{\rho V^2}{2},\quad f_D=64/\mathrm{Re}

Turbulence introduces roughness.

1fD≈−1.8log⁡10[(ε/D3.7)1.11+6.9Re]\frac1{\sqrt{f_D}}\approx-1.8\log_{10}\left[\left(\frac{\varepsilon/D}{3.7}\right)^{1.11}+\frac{6.9}{\mathrm{Re}}\right]

State what is omitted.

hL=fDLDV22g+∑KV22gh_L=f_D\frac LD\frac{V^2}{2g}+\sum K\frac{V^2}{2g}

Assumptions and scope

Water ρ = 1000 kg/m³, μ = 0.001 Pa s; straight 10 m circular pipe, fully developed flow. Darcy factor, not Fanning factor. Haaland approximation used only above Re=4000Re = 4000; transition is bracketed.

Fluid mechanics

How do we model a pipe system?

Full visual lesson ↗

The operating point belongs to the pump and the pipe together.

Build the system curve.

hsystem=Δz+kq2h_{system}=\Delta z+kq^2

Use a pump characteristic.

hpump=20−0.1q2h_{pump}=20-0.1q^2

Solve the intersection.

20−0.1q2=Δz+kq2⟹q∗=20−Δz0.1+k20-0.1q^2=\Delta z+kq^2\quad\Longrightarrow\quad q_* =\sqrt{\frac{20-\Delta z}{0.1+k}}

Assumptions and scope

Illustrative pump and system curves; q is in L/s. Two large open reservoirs, negligible surface velocity. Loss multiplier includes pipe and fittings at a fixed configuration; not a universal friction law.