Advanced Thermodynamics: Lecture 4 Shivasubramanian Gopalakrishnan - - PowerPoint PPT Presentation

Advanced Thermodynamics: Lecture 4

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

Specific heats for solids and liquids

The specific heats of incompressible substances depend on temperature only. Enthalpy change for a solid or an incompressible liquid. h = u + Pv dh = du + vdP +

• *0

Pdv ∆h = ∆u + v∆P ≈ cavg∆T + v∆P Constant–pressure processes, as in heaters (∆P = 0) ∆h = ∆u = cavg∆T

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

Specific heats for solids and liquids

The specific heats of incompressible substances depend on temperature only. Enthalpy change for a solid or an incompressible liquid. h = u + Pv dh = du + vdP +

• *0

Pdv ∆h = ∆u + v∆P ≈ cavg∆T + v∆P Constant–pressure processes, as in heaters (∆P = 0) ∆h = ∆u = cavg∆T Constant–temperature processes, as in pump (∆T = 0) ∆h = v∆P

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

Cooling of an Iron Block by Water

A 50–kg iron block at 80C is dropped into an insulated tank that contains 0.5 m3 of liquid water at 25C. Determine the temperature when thermal equilibrium is reached.

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

Temperature Rise due to Slapping

If you ever slapped someone or got slapped yourself, you probably remember the burning sensation. Recall the infamous slapgate incident between Harbhajan Singh and Sreesanth. Assuming the temperature of the affected area of Sreesanth’s face to rose by 1.8OC (ouch!) and that the Bhajji’s slapping hand has a mass of 1.2 kg and about 0.150 kg of the tissue on the face and the hand is affected by the incident, estimate the velocity of Bhajji’s hand just before impact. Take the specific heat of the tissue to be 3.8 kJ/kg

OC.

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

Analysis of open systems

The conservation of mass principle for a control volume can be expressed as: The net mass transfer to or from a control volume during a time interval ∆t is equal to the net change (increase or decrease) in the total mass within the control volume during ∆t. min − mout = ∆mCV where ∆mCV = mfinal − minitial is the change in the mass of the control volume during the process. In the rate form this is given as ˙ min − ˙ mout = dmCV dt

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

Conservation of mass

Rate of change of mass with CV dmCV dt = d dt Z

CV

ρdV Differential mass flow rate δ ˙ m = ρVndA = r(Vcosθ)dA = ρ( ¯ V ·¯ n)dA Net mass flow rate δ ˙ m = Z

CS

˙ m = Z

CS

ρ( ¯ V · ¯ n)dA General COM d dt Z

CV

ρdV + Z

CS

ρ( ¯ V · ¯ n)dA = 0

Image source: Thermodynamics An Engineering Approach, Cengel and Boles, 7th edition Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

Flow work and Flow energy

Control volumes involve mass flow across their boundaries, and some work is required to push the mass into or out of the control

• volume. This work is known as the flow work, or flow energy, and

is necessary for maintaining a continuous flow through a control volume.

Image source: Thermodynamics An Engineering Approach, Cengel and Boles, 7th edition

Wflow = FL = PAL = PV On a per unit mass basis wflow = Pv KJ/Kg

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

Total Energy of a flowing fluid

The total energy of a simple compressible system consists of three parts: internal, kinetic, and potential energies e = u + ke + pe = u + V 2 2 + gz The fluid entering or leaving a control volume possesses an additional form of energy–the flow energy Pv, as already discussed. Then the total energy of a flowing fluid on a unit-mass basis (denoted by θ) becomes θ = Pv + e = Pv + (u + ke + pe) θ = h + ke + pe = h + V 2 2 + gz

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

Steady Flow Process

A process during which a fluid flows through a control volume

• steadily. That is, the fluid properties can change from point to

point within the control volume, but at any point, they remain constant during the entire process. Mass balance for Steady Flow system X

in

˙ m = X

• ut

˙ m ˙ Ein − ˙ Eout =

• *0

dEsystem dt ˙ Qin + ˙ Win + X

in

˙ mθ = ˙ Qout + ˙ Wout + X

• ut

˙ mθ ˙ Qin+ ˙ Win+ X

in

˙ m ✓ h + V 2 2 + gz ◆ = ˙ Qout+ ˙ Wout+ X

• ut

˙ m ✓ h + V 2 2 + gz ◆

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

Water Flow through a Garden Hose Nozzle

A garden hose attached with a nozzle is used to fill a 10-gal bucket. The inner diameter of the hose is 2 cm, and it reduces to 0.8 cm at the nozzle exit. If it takes 50 s to fill the bucket with water, determine (a) the volume and mass flow rates of water through the hose, and (b) the average velocity of water at the nozzle exit.

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

Discharge of Water from a Tank

A 4-ft-high, 3-ft-diameter cylindrical water tank whose top is open to the atmosphere is initially filled with water. Now the discharge plug near the bot- tom of the tank is pulled out, and a water jet whose diameter is 0.5 in streams out. The average velocity of the jet is given by V = √2gh, where h is the height of water in the tank measured from the center of the hole (a variable) and g is the gravitational acceleration. Determine how long it will take for the water level in the tank to drop to 2 ft from the bottom.

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

Energy Transport by Mass

Steam is leaving a 4-L pressure cooker whose operating pressure is 150 kPa. It is observed that the amount of liquid in the cooker has decreased by 0.6 L in 40 min after the steady operating conditions are established, and the cross-sectional area of the exit opening is 8 mm2. Determine (a) the mass flow rate of the steam and the exit velocity, (b) the total and flow energies of the steam per unit mass, and (c) the rate at which energy leaves the cooker by steam.

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

Deceleration of Air in a Diffuser

Air at 10C and 80 kPa enters the diffuser of a jet engine steadily with a velocity of 200 m/s. The inlet area of the diffuser is 0.4 m2. The air leaves the diffuser with a velocity that is very small compared with the inlet velocity. Determine (a) the mass flow rate

• f the air and (b) the temperature of the air leaving the diffuser.

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

Acceleration of Steam in a Nozzle

Steam at 250 psia and 700F steadily enters a nozzle whose inlet area is 0.2 ft2. The mass flow rate of steam through the nozzle is 10 lbm/s. Steam leaves the nozzle at 200 psia with a velocity of 900 ft/s. Heat losses from the nozzle per unit mass of the steam are estimated to be 1.2 Btu/lbm. Determine (a) the inlet velocity and (b) the exit temperature of the steam.

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

Compressing Air by a Compressor

Air at 100 kPa and 280 K is compressed steadily to 600 kPa and 400 K. The mass flow rate of the air is 0.02 kg/s, and a heat loss

• f 16 kJ/kg occurs during the process. Assuming the changes in

kinetic and potential energies are negligible, determine the necessary power input to the compressor.

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

Power Generation by a Steam Turbine

The power output of an adiabatic steam turbine is 5 MW, and the inlet and the exit conditions of the steam are as indicated in figure.

1 Compare the magnitudes of ∆h, ∆ke, and ∆pe. 2 Determine the work done per unit mass of the steam flowing

through the turbine.

3 Calculate the mass flow rate of the steam. Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

Expansion of Refrigerant-134a in a Refrigerator

Refrigerant-134a enters the capillary tube of a refrigerator as saturated liquid at 0.8 MPa and is throttled to a pressure of 0.12

• MPa. Determine the quality of the refrigerant at the final state

and the temperature drop during this process.

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

Mixing of Hot and Cold Waters in a Shower

Consider an ordinary shower where hot water at 140F is mixed with cold water at 50F. If it is desired that a steady stream of warm water at 110F be supplied, determine the ratio of the mass flow rates of the hot to cold water. Assume the heat losses from the mixing chamber to be negligible and the mixing to take place at a pressure of 20 psia.

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

Cooling of Refrigerant-134a by Water

Refrigerant-134a is to be cooled by water in a condenser. The refrigerant enters the condenser with a mass flow rate of 6 kg/min at 1 MPa and 70C and leaves at 35C. The cooling water enters at 300 kPa and 15C and leaves at 25C. Neglecting any pressure drops, determine (a) the mass flow rate of the cooling water required and (b) the heat transfer rate from the refrigerant to water.

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

Electric Heating of Air in a House

The electric heating systems used in many houses consist of a simple duct with resistance heaters. Air is heated as it flows over resistance wires. Consider a 15-kW electric heating system. Air enters the heating section at 100 kPa and 17C with a volume flow rate of 150 m3/min. If heat is lost from the air in the duct to the surroundings at a rate of 200 W, determine the exit temperature of air.

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

Heat engines

They receive heat from a high-temperature source (solar energy, oil furnace, nuclear reactor, etc.). They convert part of this heat to work (usually in the form of a rotating shaft). They reject the remaining waste heat to a low-temperature sink (the atmosphere, rivers, etc.). They operate on a cycle.

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

Heat engines

Wnet,out Low-temperature SINK Qout Qin HEAT ENGINE High-temperature SOURCE

Image source: Thermodynamics An Engineering Approach, Cengel and Boles, 7th edition Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

Schematic of a Power Plant

• System boundary

Boiler Pump Turbine Qout Win Wout Qin Energy source (such as a furnace) Energy sink (such as the atmosphere) Condenser

FIGURE 6–10

Total work output Wnet,out = Wout − Win

Image source: Thermodynamics An Engineering Approach, Cengel and Boles, 7th edition Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

Thermal Efficiency

Defined as Thermal Efficiency = Net work output Total heat input ηth = Wnet,out Qh Since Wnet,out = QH − QL ηth = 1 − QL QH

Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

Second Law: Kelvin Planck Statement

It is impossible for any device that operates on a cycle to receive heat from a single reservoir and produce a net amount of work.

HEAT ENGINE Wnet,out = 100 kW QH = 100 kW QL = 0 Thermal energy reservoir · · ·

In other words No heat engine can have a thermal efficiency of 100 percent, or as for a power plant to operate, the working fluid must exchange heat with the environment as well as the furnace.

Image source: Thermodynamics An Engineering Approach, Cengel and Boles, 7th edition Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

Refrigeration and heat pumps

• CONDENSER

EXPANSION VALVE 120 kPa –25°C 120 kPa –20°C 800 kPa 30°C 800 kPa 60°C COMPRESSOR QL QH Wnet,in Surrounding medium such as the kitchen air Refrigerated space EVAPORATOR

Image source: Thermodynamics An Engineering Approach, Cengel and Boles, 7th edition Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661

Refrigeration schematic

• Warm environment

at TH > TL Cold refrigerated space at TL R Wnet,in QH QL Required input Desired

• utput

Image source: Thermodynamics An Engineering Approach, Cengel and Boles, 7th edition Shivasubramanian Gopalakrishnan sgopalak@iitb.ac.in ME 661