Thermodynamics Problems Chapter 1

Revision of Thermodynamic Concepts

S. Y. B. Tech. Prod Engg.ME0223 SEM-IV Applied Thermodynamics & Heat Engines

Revision of

Thermodynamic Concepts

Examples

Applied Thermodynamics & Heat Engines

S.Y. B. Tech.

ME0223 SEM - IV

Production Engineering



Example 1The e.m.f. in a thermocouple with the test junction at t ⁰C on gas thermometer scale and reference junction at Ice Point is given by;

ε = 0.20 t – 5 X 10-4 t2 mVThe millivoltmeter is calibrated at Ice and Steam Points. What will this thermocouple read when the gas thermometer reads 50 ⁰C ?

At Ice Point, when t = 0 ⁰C, ε = 0 mV.

At Steam point, when t = 100 ⁰C, ε = 0.20 X 100 – 5 X 10-4 X (100)2

= 15 mV

Thus, ΔT = 100 ⁰C → Δε = 15 mV.

At t = 50 ⁰C, ε = 0.20 X 50 – 5 X 10-4 X (50)2

= 8.75 mV

Hence, when gas thermometer reads 50 ⁰C (corresponding to 8.75 mV); the thermocouple will read;

100 15

X 8.75 = 58.33 ⁰C …ANS


S. Y. B. Tech. Prod Engg.

Example 2A water in a tank is pressurised by air and a pressure is measured by a multifluid manometer. The tank is located on a mountain at 1400 m altitude, where the atmospheric pressure is 85.6 kPa. Determine the air pressure in the tank if h1 = 0.1 m, h2 = 0.2 m and h3 = 0.35 m. Take the densities of water, oil and Mercury to be 1000 kg/m3, 850 kg/m3 and 13600 kg/m3 respectively.

P1 + h1ρwaterg + h2ρoilg - h3ρmercuryg = Patm

P1 = Patm - h1ρwaterg - h2ρoilg + h3ρmercuryg

P1 = Patm + g(h3ρmercury - h1ρwater - h2ρoil)

= (85.6)+(9.81)[(13600*0.35)-(1000*0.1)-(850*0.2)]

kPa Kg/m3

m/sec2 mKg/m3

mKg/m3

m

= 130 kPa…ANS

Oil

Air

Water

h1

h2h3

1

2

Mercury



Example 3

500 kJ

U1 = 800 kJU2 = ?

Wsh,in = 100 kJ

A rigid tank contains a hot fluid that is cooled while stirred by a paddle wheel. Initially, the internal energy of the fluid is 800 kJ. During the cooling process, the fluid loses 500 kJ of heat, and the paddle wheel does 100 kJ of work on the fluid. Determine the final internal energy of the fluid.

Tank is stationary → ∆KE = ∆PE = 0

Thus, ∆E = ∆U

Applying the First Law of Thermodynamics (i.e. Energy Balance);

Ein – Eout = ∆Esystem

Wsh,in – Qout = ∆U = U2 – U1

100 – 500 = U2 - 800

kJ kJ kJ kJ

U2 = 400 kJ….ANS



Example 4A room is initially at the outdoor temperature of 25ºC. A large fan that consumes 200 W of electricity is turned on. The Heat transfer rate between the room and the outdoor air is given by Q = UA(Ti – To) where U = 6 W/m2.ºC overall Heat transfer coefficient. A= 30 m2 exposed surface area of the room. Ti – To are the indoor and outdoor temperatures. Determine the indoor air temperature when steady conditions are established.

The electricity consumed by a fan is the Energy input to the room, i.e. 200 W.

This increases room air temperature.

However, at steady state, rate of Heat transfer from the room = Heat input to the room.

Applying the First Law of Thermodynamics (i.e. Energy Balance);

Ein – Eout = dEsystem / dt = 0 Steady State

Ein – Eout

Welect,in = Qout = UA(Ti – To)

200 (W) = (6 W/m2ºC).(30 m2).(Ti – To) Ti = 26.1 ºC…ANS



Example 5A frictionless piston-cylinder device contains 5 kg of steam at 400 kPa and 200 C. ⁰Heat is transferred to the steam until the temperature reaches 250 C. If the piston ⁰is not attached to the shaft, and its mass is constant, determine the Work done by the steam during the Process . Use the data given in FIG. for Sp. Volume.

Constant Pressure, Quasi-Static Process.

Pre

ssur

e (P

)

Volume (V)

P=Const

v1 = 0.53434m3/kg

v2 = 0.59520 m3/kg

)( 1221

2

1

VVPPdVWV

V

W1-2 = mP(v1 – v2)…..V = mv

W1-2 = (5)(400)[(0.59520 – 0.53434)

kg kPa m3/kg

W1-2 = 121.7 kJ…ANS



Example 6

Pre

ssur

e

T0 = 80 C = Const.⁰

Volume

State 2

State 1P1

P2

V2 V1

A piston-cylinder device initially contains 0.4 m3 of air at 100 kPa and 80 C. The air ⁰is now compressed to 0.1 m3 in such a way that the temperature remains constant. Determine the Work done during this Process.

Isothermal, Quasi-Static Process.

1

2111121

21

ln2

1

2

1

V

VVP

V

dVVPW

PdVW

V

V

V

V

W1-2 = -55.5 kJ…ANS

W1-2 = (100)(0.4) ln0.40.1

kPa m3

Negative sign indicates that the Work is done ON the system.



He gasm = 0.7 kg

T1 = 27 C⁰P1 = 350 kPa

Wsh

Example 7An insulated rigid tank initially contains 0.7 kg of Helium at 27 C and 350 kPa. A ⁰paddle wheel with a power rating of 0.015 kW is operated within the tank for 30 min. Determine (a) the final temperature, and (b) the final pressure of the Helium gas. Assume Cv = 3.1156 kJ/kg. C⁰

Tank is stationary → ∆KE = ∆PE = 0 Thus, ∆E = ∆U

Applying the Energy Balance; Ein – Eout = ∆Esystem

Wsh = ∆U = U2 – U1 = mCv(T2 - T1)

Wsh = (0.015 kW) (30 min) = 27 kJ

(a) 27 kJ = (0.7 kg)(3.1156 kJ/kg. C)(T⁰ 2 - 27 C)⁰

T2 = 39.4 C⁰ ….ANS

Applying the Ideal Gas Law;2

22

1

11

T

VP

T

VP

350 kPa (27+273) K

= P2 (39.4+273) K

(b)

P2 = 364.5 kPa…ANS



Example 8

Furnace

Wnet

QH = 80 kW

QL = 50 MW

Heat Engine

River

Heat is transferred to a Heat Engine from a furnace at the rate of 80 MW. If the rate of Waste Heat Rejection to a nearby river is 50 MW, determine the Net Power and the Thermal Efficiency of the Heat Engine.

QH = 80 MW and QL = 50 MW.

Net power of the Heat Engine is;

Wnet = QH – QL = (80 – 50) MW = 30 MW….ANS

Thermal Efficiency of the Heat Engine is;

ηth = Wnet,outQH

30 MW80 MW

=

= 0.375 or 37.5 %....ANS



Example 9

House,20 C⁰

Wnet = ?

QH = ?

QL = ?

Heat PumpCOP = 2.5

Outdoor Air, -2 C⁰

Heat Loss = 80,000 kJ/hr

A Heat Pump is used to used to meet the heating requirements of a house and maintained at 20 C on a day when the outdoor temperature drops to -2 C, the house is estimated to ⁰ ⁰lose Heat at the rate of 80,000 kJ/hr. If the Heat Pump is having COP of 2.5, determine (a) Power consumed by the Heat Pump (b) Rate at which Heat is absorbed from the cold outdoor air.

(a) Wnet,in = QH

COPHP

80,000 kJ/hr

2.5=

= 32,000 kJ/hr = 8.9 kW…ANS

The house is to be maintained at 20 C. ⁰Hence the Heat Pump has to deliver the SAME Heat to the house, i.e. 80,000 kJ/hr.

(b) QL = QH – Wnet,in = (80,000 – 32,000) kJ/hr

= 48,000 kJ/hr….ANS



T1 = 873 K

W1

Q1 = 2000 kJ

Q2 = ?

Heat Engine

Refrigerator

T2 = 313 K

Q4

T3 = 253 K

W2

W = 360 kJ Q3= Q4 + W2

A reversible Heat Engine operates between two reservoirs at temperatures of 600 C ⁰and 40 C. the Engine drives a reversible Refrigerator which operates between ⁰reservoirs of 40 C and -20 C. Heat transfer to the Engine is 2000 kJ and the net Work ⁰ ⁰output of the combined Engine is 360 kJ. Evaluate the Heat transfer to the refrigerant and the net Heat Transfer to the reservoir at 40 C.⁰

Example 10

ηmax = 1 - T2 = 1 - T1

313

873= 0.642

W1

Q1

= 0.642 W1 = 0.642 X 2000 = 1284 kJ

COPmax = T3

T2 – T3

253

313 - 253= = 4.22

COPmax = Q4

W2

= 4.22



W1 – W2 = W = 360 kJT1 = 873 K

W1

Q1 = 2000 kJ

Q2 = ?

Heat Engine

Refrigerator

T2 = 313 K

Q4

T3 = 253 K

W2

W = 360 kJ Q3= Q4 + W2

Example 10…contd

W2 = W1 - W = 1284 - 360 = 924 kJ

Q4 = COP X W2 = 4.22 X 924 = 3899 kJ

Q3 = Q4 + W2 = 3899 + 924 = 4823 kJ

Q2 = Q1 – W1 = 2000 - 1284 = 716 kJ

Thus, Heat Rejection to the 40 C reservoir⁰

= Q2 + Q3 = 716 + 4823 = 5539 kJ….ANS

Education

Thermodynamics Problems Chapter 1