Chapter 4: Modeling Fluid and

Thermal Systems
• Pressurized fluids (liquids and gases) are used by mechanical engineers in the
design of devices that deliver forces and torques to mechanical loads
– Hydraulic systems use a liquid as the working fluid
– Pneumatic systems use air or other gases

• Like electromechanical systems (Chapter 3), fluid systems convert energy from a
power source (pressurized fluid) to mechanical energy

• Thermal systems involve the transfer of heat energy, and temperature is typically the
dynamic variable of interest
 4.2 Hydraulic Systems
• A hydraulic fluid system is composed of a pump that provides high-pressure fluid, a fluid capacitance
(reservoir) that stores energy due to pressure, and hoses/valves that connect the various elements

• The fundamental variables are pressure P (N/m2 or Pa), mass-flow rate w (kg/s), and volumetric
flow rate Q (m3/s).

• The fluid-bulk modulus b measures the fluid’s resistance to compressibility:

Typical hydraulic fluid: b = 109 Pa = 1 GPa and r0 = 860 kg/m3

Therefore, for large fluid pressures (dP = 20 MPa) fluid density only changes by about 2% (hydraulic fluids
are very “stiff”)
 Fluid Resistance: Laminar Flow
• A fluid resistance element >> component that resists flow and dissipate energy >> electrical resistors
• For laminar pipe flow, the pipe flow is “smooth” and “streamlined” >> Large diameter, low velocity

• Consider a long capillary flow tube with low volumetric flow rate (Q) or low pressure drop (P1 – P2) so
that the flow is dominated by viscous forces

RL = “laminar fluid resistance” (Pa-s/m3)

• Laminar flow: linear relationship between the pressure drop DP = P1 – P2 and the volumetric flow rate Q

• This relation is analogous to Ohm’s law (e = RI) where e (voltage) is “pressure”, and I (current) is “flow.”
• Laminar flow exists when the Reynolds number Re < 2300; [Re is nondimensional and is the ratio of
inertial to viscosity forces in the fluid]
 Fluid Resistance: Laminar Flow

RL = “laminar fluid resistance” (Pa-s/m3)

• For laminar pipe flow where the pipe length is L is significantly larger than the diameter d, the laminar
fluid resistance can be computed from Hagen-Poiseuille Law

μ : dynamic/absolute viscosity of the fluid

 Fluid Resistance: Turbulent Flow
• Liquid flowing through valves, orifices (sharp-edge opening), and certain pipes is usually turbulent

• For turbulent flow (rule of thumb: Re > 2300), the flow is “rough” and no longer smooth but “swirls”

• The fluid resistance relation for turbulent flow is nonlinear:

or,

RT = “turbulent fluid resistance”

• Components such as valves, elbow bends, and couplings resist flow and usually induce turbulent flow at
nominal pressures

• In most cases, turbulent fluid resistance RT (or KT) must be determined experimentally.
 Fluid Resistance: Orifice Flow
• An orifice is an opening through which fluid can flow (e.g., hole in a tank, a valve opening, etc).
The figure below shows fluid flow through a sharp-edged orifice
• The classical orifice flow equation can be derived using Bernoulli’s equation

Bernoulli’s Equation

Idealized flow rate

without any frictional
losses

Cd = “discharge coefficient” = 0.62

(for sharp-edged orifice)
“orifice flow equation”
A0 = orifice area

ρ = fluid density, kg/m3

 Fluid Resistance: Valve Flow
• A 4-way spool valve is shown below:

Valve position y
alters orifice area Av
and hence meters
flows Q1 and Q2

• The generic symbol for fluid resistance (lumped fluid resistance) is shown below:
 Fluid Capacitance
• Fluid capacitance C is ability to store energy due to pressure

Fluid mass:
units: m /Pa
3

Hydrostatic
pressure:

Capacitance of tank (A = constant):

Divide by dt to get similar to electric

capacitor!
 Fluid Inertance (Fluid Inductance)
• The effect due to the fluid’s inertia as it accelerate along a pipe

  units: Pa-s2/m3

  like electrical inductance!  

• Fluid inertia effects are usually insignificant and can be ignored in modeling fluid systems.
 Conservation of Mass
• Fluid system models are obtained by applying the conservation of mass to a control volume (CV)

• If we consider a fixed control volume (CV) as shown below, fluid mass may be entering the CV, leaving the
CV, or accumulating in the CV

Control
volume Mass leaves CV
(CV)
Mass enters CV

• From the conservation of mass, we can compute the net or accumulated mass flow rate by assigning a sign
convention dm/dt > 0 for mass entering CV, and dm/dt < 0 for mass for mass leaving CV
• Therefore, conservation of mass yields:

• If mass does not accumulate in the CV (i.e., steady flow through CV), then the net mass-flow rate in the CV
is zero, or
 Modeling a Hydraulic Tank: Example 4.1

Figure shows a single hydraulic tank with input volumetric-flow rate Q in.

(a) Derive the mathematical model of the hydraulic system assuming laminar flow through the valve.

(b) Derive the mathematical model of the hydraulic system assuming turbulent flow through the valve.

(c) Repeat problems (a) and (b) with the model expressed in terms of liquid height h.
 Modeling a Hydraulic Tank:
Example 4.1a (Laminar Flow)
• Consider a storage tank containing a liquid, with in-flow Qin and laminar out-flow through the valve

Pressure at base (hydrostatic Eq):

Laminar out-flow through valve:

• Apply conservation of mass: Divide by ρ,

we get

From fluid capacitance

• Substitute for volumetric out-flow, Qout recall

Tank model with

Laminar output
 Modeling a Hydraulic Tank:
Example 4.1b (Turbulent Flow)
• Consider a storage tank containing a liquid, with in-flow Qin and laminar out-flow through the valve

Pressure at base (hydrostatic Eq):

Laminar out-flow through valve:

• Apply conservation of mass: Divide by ρ,

we get

From fluid capacitance

Tank model with

turbulent output
 Modeling a Hydraulic Tank:
Example 4.1c (In terms of h)

Tank model with Tank model with

Laminar output turbulent output

Divide by ρg

• Hydraulic tank systems require a first-order ODE for each fluid capacitance
(reservoir)
• ODE can be expressed with either pressure P or liquid height h as the
dynamic variable
• For two interconnected tanks, the complete mathematical model will involve
two first-order ODEs.
Modeling a Hydraulic Tank:
Practice Problems

Other related problem to the

topic from exercise
 Modeling Hydromechanical Systems
• Hydromechanical systems are created by combining hydraulic and mechanical components,
• Used to convert the energy stored in the pressurized fluid to mechanical energy

Fluid mass:

• Fluid volumetric flow-rate Q is entering the CV from the left, and no flow is leaving the CV. The
piston moves (due to pressure), and therefore the size of the CV is not fixed but rather varies with
time
• Applying conservation of mass to the CV:

0
 Modeling Hydromechanical Systems (2)

Fluid mass:

Bulk modulus

• The time-rate of fluid density is

• Substituting this expression into the previous mass-continuity (CV) equation and rearranging we
obtain
Instantaneous CV,

Fundamental pressure-rate ODE for hydraulic cylinder with

compressible fluid >> Nonlinear (Inverse of x in 1/V)
Hydraulic Actuator
 Hydromechanical Actuator:
Example 4.2
• Derive the mathematical model of the hydraulic actuator

• Start with the pressure-rate equation for the cylinder chamber:

V0 >> No deflection of spring, x = 0

where instantaneous volume is

 The fluid model is Nonlinear!

 Hydromechanical Actuator:
Example 4.2 (2)

• Next, use the FBD of the mechanical subsystem:

• Rearranging we obtain the mechanical model:

(linear)
 Hydromechanical Actuator:
Example 4.2 (3)

System model
Fluid ODE:
3rd-order nonlinear
system

Mechanical ODE:

Dynamic Variables of interest: P, x;

Input variables: Qin, Patm, FL
 Hydromechanical Systems
• Reading/Practice Assignment
• Example 4.3: Hydraulic accumulator
• Example 4.4: Fluid capacitance of Hose
• Problems on similar concepts from the exercise

Hydraulic accumulator
 4.3 Pneumatic Systems
• Pneumatic systems involve compressible fluids (gas) such as air, where density is not constant

• Analysis of pneumatic systems is complex due to thermodynamic effects and the fact that gas flow is more
complicated that liquid flow
– Compressibility effects may cause oscillations in the system response

– Flow can become “choked” at the throat (sonic conditions, or Mach 1) of a valve or orifice, resulting in highly a
nonlinear relationship between pressure and flow rate

• For compressible fluids, mass-flow rate (w) and volume-flow rate (Q) are not readily interchangeable since
density can change significantly

Mass of gas in volume V

Mass-flow rate (w)
 Pneumatic Systems (2)
• We must include thermodynamic effects when modeling pneumatic systems

• Pneumatic systems exhibit a functional relationship between pressure, density, and temperature:

Perfect/Ideal gas law

R = gas constant
T = absolute temperature in kelvin (K)
 Resistance of Pneumatic Systems

• In rare cases the flow in pneumatic systems is incompressible (low-speed flow)

• In these rare cases we may model resistance using either the laminar or turbulent models and
obtain the corresponding coefficients RL or RT via experimental data, such as mass-flow rate w
vs ΔP
Laminar flow (linear)

Turbulent flow (nonlinear)

• Note that resistance coefficients RL and RT have different units for pneumatic systems since we are using
mass-flow rate (w) instead of Q
 Resistance of Pneumatic Systems (2)
• For most pneumatic applications, flow through valves or orifices is turbulent, compressible, and
complex
– We need to consider “choked” vs. “unchoked” flow at the minimum area (throat), where “choked” flow
involves sonic conditions, or Mach = 1

Mass-flow rate:
nonlinear function
for unchoked flow

P1 > P2
w
upstream downstream
pressure, P1 pressure, P2
throat area, A
upstream temperature, T
 Resistance of Pneumatic Systems (3)
• Mass-flow rate w for pneumatic systems:

Flow is NOT choked

Cd: Discharge coefficient
(subsonic):

P2/P1 > Cr γ: ratio of specific heats

(1.4 for air)

Flow IS choked (Sonic)

(Mach = 1 at throat)
(linear in P1)
P2/P1 < Cr

Critical pressure ratio:

P1 > P2 w

P1 P2 (air)

throat area, A
temperature, T
Resistance of Pneumatic Systems (3)

• Downstream pressure P2 is varied between

a very small value (near vacuum) and an
upper limit equal to P1

P1 =6 *105 Pa
A0 =4 mm2
T1 =298 K,
Cd =0.8

P1 > P2 w

P1 P2

throat area, A
temperature, T
 Mass-Flow Rate for Pneumatic Systems
3

area A = 20.3 cm2 2.5 unchoked

T = 298 deg K
P2 = Patm = 1.013(10)5 N/m2

Mass-flow rate, kg/s

2
choked
g = 1.4 (air)
R = 287 m2/s2-K 1.5 Linear in P1
Cd = 0.8
1

Upstream (input) pressure is varied 0.5

while downstream pressure P2 is
held constant at atmospheric pressure 0
1 2 3 4 5 6 7 8
2 5
Input pressure, N/m x 10
P1 > P2 w
Critical
P1 P2 pressure P1 = 1.92(105) N/m2
 P2/P1 = 0.528
throat area, A
temperature, T
 Pneumatic Capacitance
• Pneumatic fluid capacitance is defined as

• Since mass is m = ρV, for gases filling a constant-volume (rigid) vessel we have

 ?

• Next, apply thermodynamics concepts: assume gas undergoes a polytropic expansion:

= constant, or

where n = 1 for an isothermal process;

specific heats of an ideal gas at constant
For an adiabatic process (no heat transfer): pressure and at constant volume
 Pneumatic Capacitance (2)
• Take differentials of both sides of the polytropic process

• Substitute for a using and substitute for P using the perfect gas law P = ρRT to obtain

• Substitute the above expression into the capacitance equation

Pneumatic capacitance of a fixed vessel can vary depending on the

Capacitance for fixed volume
gas temperature, type of gas, and thermodynamic process (n)
 Modeling Pneumatic Systems
• Pneumatic system models are based on conservation of mass:

Time-rate of mass in CV:

• Compute the time-rate of density ρ by taking the time derivative of the polytropic expansion process

• Substitute the polytropic process equation and solve for dρ/dt

 Modeling Pneumatic Systems (2)
• Substitute the previous equation for dρ/dt and the perfect gas law for density ρ into the mass-
conservation equation

• Because we want a dynamic equation for gas pressure, we solve the above expression for dP/dt

Fundamental modeling
equation for a pneumatic
capacitance with pressure P

Note: for a fixed volume, we get

 Pneumatic System: Example 4.5
Supply tank has pressure PS

Assume compressible flow

through valve with orifice area A0

Rigid chamber (V = constant)

0
Basic ODE: or

Input mass-flow rate win could be choked or unchoked depending on P/PS

Model

if (unchoked)

if (choked)
 Chapter 4: Summary
• We introduced a systematic approach for developing the mathematical model of fluid systems
• First, we presented the physical laws for resistance elements and energy-storage (capacitance)
elements
• Fluid-system models are derived by applying the conservation of mass to a control volume
– Each control volume (or fluid capacitance) in a fluid system will require a 1st-order ODE with pressure
as the dynamic variable
– Fluid models are typically nonlinear due to turbulent flow
 4.4 Thermal Systems
• Conservation of energy is used to model thermal systems;
thermal systems involve the storage and flow of heat energy

“Open” thermal system:

1) Boundary (insulation) encloses thermal capacitance C
2) Heat energy can enter/leave system (heat flow rate q)
3) Heat energy can enter/leave system due to mass transfer (e.g., fluid flow)
 Thermal Systems (2)
• Applying the conservation of energy to the thermal boundary (1st
Law of Thermodynamics expressed as time derivatives)

Units: J/s or watts (W)

Net rate of heat Time-rate of enthalpy Heat flow rates

energy stored due to mass transfer across boundary
across boundary
 Thermal Systems (3)
• In general, thermal systems are distributed systems
(temperatures are distributed throughout a body), and thus
they are modeled by partial differential equations (PDEs)
instead of ODEs.

• We will use a simplified lumped model approximation,

which means that we assume that all points in a body have
a single (average) temperature.

• For our lumped approximation, thermal systems can be

characterized by two passive thermal elements: thermal
resistance (R) and thermal capacitance (C)
 Thermal Resistance
• Heat can be transferred in three ways:
– Conduction (diffusion through a substance)
– Convection (fluid transport)
– Radiation (infrared waves like solar energy)

• We will only consider conduction and convection.

• For either conduction or convection heat transfer, the linear

model for heat flow rate is
DT = temperature difference, deg K
Heat flow rate,
R = thermal resistance, deg K-s/J
J/s or W

For , we have a perfect insulator (no heat transfer, q = 0)

 Thermal Resistance (2)
• Conduction:
x = thickness of material, A = area
k = thermal conductivity coefficient

For example, copper has thermal conductivity k that is 18x greater

than k for stainless steel (copper is a good conductor and poor insulator)

• Convection:

H = convection coefficient, A = area

For example, H for water is 50-100x greater than H for air (air is a
much better insulator than water)
 Thermal Capacitance
• Thermal capacitance C is a measure of a body’s ability to
store heat due to its mass and thermal properties

• Capacitance is the product of specific heat cp (at constant

pressure) and mass m :

Thermal capacitance
J/deg K

For example, 1 kg of water has more than 4x the thermal

capacitance of 1 kg of air
 Modeling Thermal Systems
• Lumped thermal models are based on the following heat
balance equation (i.e., energy balance):

heat stored in a body = net heat transfer to a body

incremental heat stored by incremental net incremental heat transferred in

temperature increase incremental time

 where

Divide by dt and
write as an ODE:
Basic thermal modeling Eq.
 Modeling Thermal Systems: Example 4.8
Double-pipe heat exchanger: transfer heat from “hot” fluid in tube to “cold” shell
COLD
flow

HOT
flow

Steady flow: in and out mass-flow rates are equal for both tube and shell

Tube: output temp T1,out = T1

Shell: output temp T2,out = T2
 Thermal Systems: Example 4.8 (2)

Thermal
boundaries:

Note we have two thermal capacitances and thus two temperatures

Plus, we have two heat-flow rates: q1 from “hot” tube to “cold” shell
and q2 from shell to ambient surroundings

Because we have an open system with mass-flow rate we include the

time-rates of enthalpy
 Thermal Systems: Example 4.8 (3)

Tube:

Shell:

Next, sub for heat flow

and enthalpy rates:
 Thermal Systems: Example 4.8 (4)

Tube:

Shell:

We may need two distinct values of cp if a chemical solution flows thru the tube
and water flows thru the shell

We can move all dynamic variables (T1 and T2) to the left-hand sides and
all input variables (Tin,1 , Tin,2 , and Ta ) to the right-hand sides (see textbook)

Mass-flow rates may also be considered as input variables or as constant

parameters.