Posted: November 16th, 2014
A gas used in a combustion process has molecular weight 16 g/mol. The universal gas constant R = 8.314462 J/mol K. A sample of this gas (totaling 262.5 mol) is heated in a combustion process is from 20oC to 200oC at constant pressure.
(Question 18)
A gas used in a combustion process has molecular weight 16 g/mol. The universal gas constant R = 8.314462 J/mol K. A sample of this gas (totaling 262.5 mol) is heated
in a combustion process is from 20oC to 200oC at constant pressure.
(i) Determine the specific heat capacity at constant pressure, cp, if the sample experiences a change in enthalpy of 945 kJ.
(ii) Determine the specific heat capacity at constant volume, cv, given that the adiabatic index for the gas is 4/3.
(Question 2)
(Question 3)
A rectangular hollow beam of length 3m is simply supported at its ends. It has width 100mm and depth 200mm. The beam is subjected to a uniformly distributed load of 2
tonnes/m and a point load of 200N at its centre position.
Determine;
(i) The maximum stress due to the bending (occurring at the centre point)
(ii) The value of the radius of curvature of the neutral layer
(iii) The factor of safety if the maximum allowable stress of the material is 100 Mpa
Name;
Instructor’s name;
Course:
Date:
Question 2)
2 i. Stress = frorce / area = 80000 * 2 / (100 * 5 * 2) = 160000 / 1000 = 160N/mm2
ii. strain= stress/ Modulus of elasticity = 160 * 1000000 / 71 * 1,000,000,000
= 0.0022535211267606
iii. Given that the force is acting in the direction along the length L, the strain shall act only to lengthen the length L.
The change in length shall be: 200 * 0.0022535211267606 = 0.45070422535212
The thickness t changes by 0.45070422535212 * 0.34 = 0.1532394366197208
iv. Assuming force F is equal to force P; the strain from P is equal to thet from F. The change in width W shall be: 100 * 0.0022535211267606 = 0.22535211267606
The change in length L shall be * 0.0022535211267606 = 0.45070422535212
The change in thickness shall be 0.45070422535212 * 0.34 = 0.1532394366197208
(Question 3)
A rectangular hollow beam of length 3m is simply supported at its ends. It has width 100mm and depth 200mm. The beam is subjected to a uniformly distributed load of 2
tonnes/m and a point load of 200N at its centre position.
Determine;
(i) The maximum stress due to the bending (occurring at the centre point)
(200/2) + (3 * 2) = 106 N/mm2
(ii) The value of the radius of curvature of the neutral layer
106*7/22 *200 = 0.1686363636363636 M
(iii) The factor of safety if the maximum allowable stress of the material is 100 Mpa
100 / 40 = 2.5
(Question 4)
i. Velocity = 3000 * 22/ 7 * 60 = 1260000 mm/s
ii. 20 * 3000 / 90 = 6.666666666666667
(Question 5)
i. For simple harmonic motion to take place, the piston s poised to move in a manner that only lets the piston to draw out and in a regular fashion that is
consistent and repeated.
ii. As can be seen from the graph, it indeed exudes simple harmonic motion.
(Question 6)
600 * 60 * 7/22 * 1/60 *1 * 2 * 0.4 = 152.7272727272727rpm
(Question 10)
i. 25 / 2 * 60 * 1000 * 1 / 500 = 1500N / mm2
ii. 1500 / 90 = 16.666666666667 degrees
(Question 11)
i. O.3 sin 30 = 0.33
ii. 14 * 0.02 + 4 = 4.28
iii. 0.02 * 4.28 = 0.0856
(Question 12)
i. T = 30 /10 = 3
Circular frequency = 8*10 = 80
Natural frequency = 80 / 9.81 = 8.154943934760449
ii. 30 * 8/10 = 24
iii. 24 * 22/7 = 75.42857142857143
(Question 13)
i. f = w/80 T= w * 80
ii. 6/3 * w * 80 = 160w
(Question 18)
A gas used in a combustion process has molecular weight 16 g/mol. The universal gas constant R = 8.314462 J/mol K. A sample of this gas (totaling 262.5 mol) is heated
in a combustion process is from 20oC to 200oC at constant pressure.
(i) Determine the specific heat capacity at constant pressure, cp, if the sample experiences a change in enthalpy of 945 kJ.
8.314462 * 945 / 262.5 – 16
=13.9320632
(ii) Determine the specific heat capacity at constant volume, cv, given that the adiabatic index for the gas is 4/3.
= 13.9320632 * 4/3
= 18.57608426666667
Name;
Instructor’s name;
Course:
Date:
Question 2)
2 i. Stress = force / area = 80000 * 2 / (100 * 5 * 2) = 160000 / 1000 = 160N/mm2
ii. strain= stress/ Modulus of elasticity = 160 * 1000000 / 71 * 1,000,000,000
= 0.0022535211267606
iii. Given that the force is acting in the direction along the length L, the strain shall act only to lengthen the length L.
The change in length shall be: 200 * 0.0022535211267606 = 0.45070422535212
The thickness t changes by 0.45070422535212 * 0.34 = 0.1532394366197208
iv. Assuming force F is equal to force P; the strain from P is equal to thet from F. The change in width W shall be: 100 * 0.0022535211267606 = 0.22535211267606
The change in length L shall be * 0.0022535211267606 = 0.45070422535212
The change in thickness shall be 0.45070422535212 * 0.34 = 0.1532394366197208
v. change in area shall be : 0.22535211267606 * 0.45070422535212 = 0.1015671493751273mm2
(Question 3)
A rectangular hollow beam of length 3m is simply supported at its ends. It has width 100mm and depth 200mm. The beam is subjected to a uniformly distributed load of 2
tonnes/m and a point load of 200N at its centre position.
Determine;
(i) The maximum stress due to the bending (occurring at the centre point)
(200/2) + (3 * 2) = 106 N/mm2
(ii) The value of the radius of curvature of the neutral layer
106*7/22 *200 = 0.1686363636363636 M
(iii) The factor of safety if the maximum allowable stress of the material is 100 Mpa
100 / 40 = 2.5
(Question 4)
i. Velocity = 3000 * 22/ 7 * 60 = 1260000 mm/s
ii. 20 * 3000 / 90 = 6.666666666666667
(Question 5)
i. For simple harmonic motion to take place, the piston s poised to move in a manner that only lets the piston to draw out and in a regular fashion that is
consistent and repeated.
ii. As can be seen from the graph, it indeed exudes simple harmonic motion.
(Question 6)
600 * 60 * 7/22 * 1/60 *1 * 2 * 0.4 = 152.7272727272727rpm
(Question 10)
i. 25 / 2 * 60 * 1000 * 1 / 500 = 1500N / mm2
ii. 1500 / 90 = 16.666666666667 degrees
iii. 90 * 2 / 9.81 = 18.34862385321101
(Question 11)
i. O.3 sin 30 = 0.33
ii. 14 * 0.02 + 4 = 4.28
iii. 0.02 * 4.28 = 0.0856
(Question 12)
i. T = 30 /10 = 3
Circular frequency = 8*10 = 80
Natural frequency = 80 / 9.81 = 8.154943934760449
ii. 30 * 8/10 = 24
iii. 24 * 22/7 = 75.42857142857143
(Question 13)
i. w = 160 * 2/60 = 5.333333333333333 rpm f = w/80 T= w * 80
ii. 6/3 * w * 80 = 160w
(Question 18)
A gas used in a combustion process has molecular weight 16 g/mol. The universal gas constant R = 8.314462 J/mol K. A sample of this gas (totaling 262.5 mol) is heated
in a combustion process is from 20oC to 200oC at constant pressure.
(i) Determine the specific heat capacity at constant pressure, cp, if the sample experiences a change in enthalpy of 945 kJ.
C= q/(m*delta T)
8.314462 * 945 / 262.5 – 16
=13.9320632
(ii) Determine the specific heat capacity at constant volume, cv, given that the adiabatic index for the gas is 4/3.
Cv = C/ I
= 13.9320632 * 4/3
= 18.57608426666667
PLACE THIS ORDER OR A SIMILAR ORDER WITH US TODAY AND GET AN AMAZING DISCOUNT 🙂
Expert paper writers are just a few clicks away
Place an order in 3 easy steps. Takes less than 5 mins.