ME 24-221 THERMODYNAMICS I Solutions to extra problems in









bar psi kPa/MPa Kg/cm2 Mbar

Page 1. bar psi. kPa/MPa. Kg/cm2. Mbar. 0.025. 0.363. 2.5 kPa. 0.025. 25. 0.04. 0.58 5. 72.516. 500 kPa. 5.098. 5000. 6. 87.020. 600 kPa. 6.117. 6000.
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(a) the specific volume at T = 240°C p = 1.25 MPa
ThermoHmwk Soln


OEM Perspective on Cryogenic H2 Storage

15 kg/min. (3 MW). CGH2. 40 g/L. 69-65 g/L. 1
compressed hydrogen brunner


ME 24-221 THERMODYNAMICS I Solutions to Assignment 11

December 1 2000. J. Murthy. 11.4 A steam power plant operating in an ideal Rankine cycle has a high pressure of 5. MPa and a low pressure of 15 kPa.
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Table 1. Saturation (Temperature)

Table 1. Saturation (Temperature) (continued). 5 t °C p
NISTIR Tab


ME 24-221 THERMODYNAMICS I Solutions to extra problems in

Nov 9 2000 1. 2 s = C. 8.29 A mass and atmosphere loaded piston/cylinder contains 2 kg of water at 5 MPa
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KNOWN: Water is the working fluid in an ideal Rankine cycle. The

5. Saturated liquid exits the open feedwater heater and saturated liquid exits State 5: Ps= P2 = 1 MPa (10 bar)
HW


Thermo 1 (MEP 261) Thermodynamics An Engineering Approach

5–7 Air enters a nozzle steadily at 2.21 kg/m until the density in the tank rises to 7.20 kg/m ... steadily at 1 MPa and 500°C with a mass flow rate.
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AREN 2110 SOLUTIONS FALL 2006 HOMEWORK ASSIGNMENTS

5-45 A number of brass balls are to be quenched in a water bath at a specified rate. kJ/kg. 3240.9. /kg m. 0.02975. C. 450. MPa. 10. 1. 3. 1. 1. 1.
SOLUTIONS HOMEWORKS F


212073 ME 24-221 THERMODYNAMICS I Solutions to extra problems in

ME 24-221

THERMODYNAMICS I

Solutions to extra problems in chapter 8

November 9, 2000

Fall 2000

J. Murthy

8.5Water is used as the working fluid in a Carnot cycle heat engine, where it changes

from saturated liquid to saturated vapor at 200°C as heat is added. Heat is rejected in a constant pressure process (also constant T) at 20 kPa. The heat engine powers a Carnot cycle refrigerator that operates between -15°C and +20°C. Find the heat added to the water per kg water. How much heat should be added to the water in the heat engine so the refrigerator can remove 1 kJ from the cold space?

Solution:

Carnot cycle:

q H = T H (s 2 - s 1 ) = h fg = 473.15 (4.1014) = 1940 kJ/kg T L = Tsat (20 kPa) = 60.06 oC ref = QL / W = T L / (T H - T L ) = (273 - 15) / (20 - (-15)) = 258 / 35 = 7.37 W = Q L

β = 1 / 7.37 = 0.136 kJ

W = η

HE Q H H2O HE = 1 - 333/473 = 0.29 Q H H2O = 0.136 / 0.296 = 0.46 kJ

8.12A cylinder fitted with a piston contains ammonia at 50°C, 20% quality with a

volume of 1 L. The ammonia expands slowly, and during this process heat is transferred to maintain a constant temperature. The process continues until all the liquid is gone. Determine the work and heat transfer for this process.

C.V. Ammonia in the cylinder.1 2 T

S 50 C
3

NHTable B.2.1: T

1 = 50°C, x 1 = 0.20, V 1 = 1 L v 1 = 0.001777 + 0.2 × 0.06159 = 0.014095 s 1 = 1.5121 + 0.2 × 3.2493 = 2.1620 m = V 1 /v 1 = 0.001/0.014095 = 0.071 kg v 2 = v G = 0.06336, s 2 = s G = 4.7613

Process: T = constant to x2

= 1.0, P = constant = 2.033 MPa 1 W 2 = ⌡⌠PdV = Pm(v 2 - v 1 ) = 2033 × 0.071 × (0.06336 - 0.014095) = 7.11 kJ 1 Q 2 = ⌡⌠TdS = Tm(s 2 - s 1 ) = 323.2 × 0.071(4.7613 - 2.1620) = 59.65 kJ or 1 Q 2 = m(u 2 - u 1 1 W 2 = m(h 2 - h 1

ME 24-221

THERMODYNAMICS I

Solutions to extra problems in chapter 8

November 9, 2000

Fall 2000

J. Murthy

8.5Water is used as the working fluid in a Carnot cycle heat engine, where it changes

from saturated liquid to saturated vapor at 200°C as heat is added. Heat is rejected in a constant pressure process (also constant T) at 20 kPa. The heat engine powers a Carnot cycle refrigerator that operates between -15°C and +20°C. Find the heat added to the water per kg water. How much heat should be added to the water in the heat engine so the refrigerator can remove 1 kJ from the cold space?

Solution:

Carnot cycle:

q H = T H (s 2 - s 1 ) = h fg = 473.15 (4.1014) = 1940 kJ/kg T L = Tsat (20 kPa) = 60.06 oC ref = QL / W = T L / (T H - T L ) = (273 - 15) / (20 - (-15)) = 258 / 35 = 7.37 W = Q L

β = 1 / 7.37 = 0.136 kJ

W = η

HE Q H H2O HE = 1 - 333/473 = 0.29 Q H H2O = 0.136 / 0.296 = 0.46 kJ

8.12A cylinder fitted with a piston contains ammonia at 50°C, 20% quality with a

volume of 1 L. The ammonia expands slowly, and during this process heat is transferred to maintain a constant temperature. The process continues until all the liquid is gone. Determine the work and heat transfer for this process.

C.V. Ammonia in the cylinder.1 2 T

S 50 C
3

NHTable B.2.1: T

1 = 50°C, x 1 = 0.20, V 1 = 1 L v 1 = 0.001777 + 0.2 × 0.06159 = 0.014095 s 1 = 1.5121 + 0.2 × 3.2493 = 2.1620 m = V 1 /v 1 = 0.001/0.014095 = 0.071 kg v 2 = v G = 0.06336, s 2 = s G = 4.7613

Process: T = constant to x2

= 1.0, P = constant = 2.033 MPa 1 W 2 = ⌡⌠PdV = Pm(v 2 - v 1 ) = 2033 × 0.071 × (0.06336 - 0.014095) = 7.11 kJ 1 Q 2 = ⌡⌠TdS = Tm(s 2 - s 1 ) = 323.2 × 0.071(4.7613 - 2.1620) = 59.65 kJ or 1 Q 2 = m(u 2 - u 1 1 W 2 = m(h 2 - h 1