solution of test 1 sep13

Chemical Engineering Thermodynamics (CCB 2024) September 2013

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Test 1 Solution

And

Review Chapter 2 & 3 Problems


Replacement Tutorial Class

Date: 12th November 2013 (Tuesday)

Time: 8:00-10.00 am (For group 2)

12:00 noon -2.00 pm (For group 1)

Venue: 04.02.16

Review Class on lectures (From week 1 to week 7)

Date: Week 14

Time:

Venue:


Step 3: Assumptions and Approximations

PROBLEM-SOLVING TECHNIQUE

Step 7: Reasoning, Verification

and Discussion

Step 6: Calculations

Step 5: Properties

Step 4: Physical Laws

Step 2: Schematic

Step 1: Problem Statement


Air is compressed from an initial state of 1 bar to a final state of 5 bar by three

different mechanically reversible processes in a closed system:

(a)Heating at constant volume followed by cooling at constant pressure

(b) Isothermal compression

(c) Adiabatic compression followed by cooling at constant volume

Assume air to be an ideal gas. Calculate the changes in internal energy and

enthalpy of the air for each process.

Class example

Solution

Let system = 1 mol air.

For constant T (beginning and end of

process), dT = 0.

Thus,

= = 0

Example 3.2


Air is compressed from an initial state of 1 bar and 25oC (298.15 K) to a final state of

5 bar and 25oC (298.15 K) by three different mechanically reversible processes in a

closed system:

(a) Heating at constant volume followed by cooling at constant pressure

(b) Isothermal compression


Assume air to be an ideal gas with the constant heat capacities, CV = (5/2) R and CP =

(7/2) R. At 25oC (298.15 K) and 1 bar, the molar volume of air is 0.024 79 m3 mol-1.

Calculate the work required, heat transferred, and the changes in internal energy and

enthalpy of the air for each process.

Solution

Let system = 1 mol air. For R = 8.314 J mol-1 K-1,

CV = 20.785 J mol-1 K-1 and CP = 29.099 J mol

-1 K-1

For constant T (beginning and end of process), dT = 0.

Thus, = = for all three processes


Example 3.2 (Contd)

33

2

1

12

221121

2

22

1

11

m 958004.05

1 m 79024.0

,For

P

PVV

VPVPTTT

VP

T

VP

(b) Heating at constant volume followed by cooling at constant pressure

Fist step: heating at constant volume:

For constant volume:

Q = U = CV T = (20.78 J K-1)(1490.75 298.15)K = 24 788 J

Second step: cooling at constant pressure

Q = H = CP T = (29.10 J K-1)(298.15 1490.75) K = 34 703 J

U = H (PV) = H P V = 34 703 J ( 5 105 N m-2)(0.004 958 0.024 79) m3 = 24 788 J

Combined steps (complete process):

Q = 24 788 J 34 703 J = 9915 J U = 24 788J 24 788 J = 0 As U = Q + W , W = Q = 9915 J As H = U + (PV) and (PV) = 0 , H = U = 0

K 75.14901

5K 15.298

,For

1

2

1

21

1

11

P

PTT

PPVVT

VP

T

VP


Example 3.2 (Contd)

(b) Heating at constant volume followed by cooling

at constant pressure (Alternative method)

Fist step: heating at constant volume:

w = 0 J

Second step: cooling at constant pressure

w = P V = ( 5 10

5 N m-2)(0.004 958 0.024 79) m3

= 9915 J Total work W = 9915 J + 0 J = 9915 J Since, U is zero, Q =- W = - 9915 J

Combined steps (complete process):

H = U = 0 W = Q = 9915 J


(b) For the isothermal compression of ideal gas,

= = ln12

= 8.314 J K1 298.15 K ln1

5 = 3990 J

Alternative method

= = 11 ln12

= 1 x 105 Pa (0.02479 m3)ln1

5 = 3990 J


Fist step: adiabatic compression:

111 = 2

1

=

=29.099

20.785= 1.4

= 112

1

= 298.15 K0.02479 m3

0.004958 m3

0.4

= 567.57 K

For adiabatic process, Q = 0

= = 20.785 J K1 567.57 298.15 K = 5600 J

Example 3.2 (Contd)

Formulae:

1 = constant

=

W = U = CVT


Second step: cooling at constant

volume:

= 0 = = 2

= 20.785 J K1 (298.15 567.57)K

= 5600 J

Combined steps (overall process):

W = 5600 J

Q = 5600 J

Example 3.2 (Contd)


Test 1 Solution

1. a. An ideal gas at P1=1 bar, 25 C and V1=12 m3 is compressed to

P2= 5 bar, 25 C and V2= 1m3 by two different mechanically reversible

processes: (1) cooling at constant pressure followed by heating at constant

volume; (2) isothermal compression. For each path, calculate:

i. The enthalpy, H (J). [3 marks]

ii. The work, W (J). [3 marks]

iii. The heat, Q (J). [3 marks]


Test 1 Solution (Q. 1, a, Contd)


1. b. TWO (2) heat engines produce power of 90000 kW each:

(1) A Carnot engine operates between heat reservoirs at 750 K and 300 K.

(2) A practical engine operates between the same heat reservoirs but with

a thermal efficiency, = 0.30. Determine in each case the rates at which heat is absorbed from the hot reservoir and discarded to the cold

reservoir.

[6 marks]

Test 1 Solution (Q. 1, b)


Test 1 Solution (Q. 1, b, Contd)


Test 1 Solution (Q. 2, a)

2. a. Superheated steam at 1000 kPa and 250 C (523.15 K) expands isentropically to 200 kPa.

i. Find the enthalpy and entropy for the superheated steam at the

initial state (1000 kPa and 250 C). [4 marks] ii. Determine the down-stream state of the steam. [2 marks]

iii. What is the final specific enthalpy in kJ kg-1? [7 marks]



isentropic


2. b. A process is assumed to be reversible and adiabatic. What is

the entropy change of the process? Justify your answer.

[2 marks]

Test 1 Solution (Q. 2, b)

solution of test 1 sep13

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