POLYMER ENGINEERING - QUESTION 24.2 : Let C% be the fractional crystallinity, Rs

POLYMER ENGINEERING - QUESTION 24.2 : Let C% be the fractional crystallinity, Rs = density of sample, Ra = density of amorphous form and Rc = density of crystalline form. In a polymer, these unknowns could be related by the equation C% = (Rc / Rs) (Rs - Ra) / (Rc - Ra). (a) Find the equation of Rc as a function of C%, Rs and Ra. (b) Two samples of a polymer, C and D exist. For sample C, C% = 0.513 when Rs = 2.215 unit. For sample D, C% = 0.742 when Rs = 2.144 unit. Both samples C and D have the same values of Ra and Rc. Find the values of Ra and Rc in 6 decimal places.

Question Posted / kangchuentat

1 Answers
2197 Views
I also Faced
E-Mail Answers

POLYMER ENGINEERING - QUESTION 24.2 : Let C% be the fractional crystallinity, Rs = density of sample..

Answer / kangchuentat

POLYMER ENGINEERING - ANSWER 24.2 : (a) C% = (Rc / Rs) (Rs - Ra) / (Rc - Ra), C% (Rc - Ra) Rs = Rc Rs - Rc Ra = Rc C% Rs - Ra C% Rs, Rc C% Rs - Rc Rs + Rc Ra = Ra C% Rs = Rc (C% Rs - Rs + Ra), Rc = Ra C% Rs / (C% Rs - Rs + Ra). (b) For sample C, Rc = Ra C% Rs / (C% Rs - Rs + Ra) = Ra (0.513) (2.215) / [ (0.513) (2.215) - 2.215 + Ra ] = 1.136295 Ra / (Ra - 1.078705). For sample D, Rc = Ra C% Rs / (C% Rs - Rs + Ra) = Ra (0.742) (2.144) / [ (0.742) (2.144) - 2.144 + Ra ] = 1.590848 Ra / (Ra - 0.553152). Then Rc = 1.136295 / (Ra - 1.078705) = 1.590848 / (Ra - 0.553152), 1.136295 (Ra - 0.553152) = 1.590848 (Ra - 1.078705), 1.136295 Ra - 0.628544 = 1.590848 Ra - 1.716056, Ra (1.590848 - 1.136295) = 1.716056 - 0.628544 = 1.087512 = 0.454553 Ra. Then Ra = 1.087512 / 0.454553 = 2.392487. When Ra = 2.392487, Rc = 1.136295 / (Ra - 1.078705) = 1.590848 / (Ra - 0.553152) = 0.864904 with 6 decimal places. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

Is This Answer Correct ?

0 Yes

0 No

Post New Answer

More Chemical Engineering Interview Questions

QUANTUM COMPUTING - EXAMPLE 32.7 : If | ± > = 0.707 ( | 0 > ± | 1 > ), prove that | Ψ (t = 0) > = | 0 > = 0.707 ( | + > + | - > ).

1 Answers

What is difference between overall heat transfer coefficient & individual heat transfer coefficient?

0 Answers

Hi, Please give me chemical engineering paper for IOCL exam for entire written exam, GD and personel interview model questions to my email id. (chemistnathan@rediffmail.com) Rgds, Ragu

0 Answers HAL, IOCL,

In Galvonised iron and stainsteel pipes, in which pipe friction losses will more and why ?

0 Answers

What is the Import Procurement Cycle ? and what are the customization steps in SAP ?

0 Answers Siemens,

Question 31 – For liquid benzene, the CP constants are : a = 129440, b = - 169.5, c = 0.64781. Reference temperature is 298 K. The temperature of benzene is 60 degree Celsius. Calculate the enthalpy of benzene by using the formula H = a (DT) + (b/2) (T^2 – TREF^2) + (c/3) (T^3 – TREF^3) where ^ is power, DT is temperature difference with TREF = 298 K. H is in J / kmol. DT = T – TREF.

1 Answers

What is the difference between coagulation and flocculation in water treatment plant?

1 Answers

why u opted for chemical engg?

2 Answers

Explain what are the disadvantages of using gear pumps?

0 Answers

What are the some common problems associated with dilute phase pneumatic conveying?

0 Answers

Explain how can one determine if a particular solid can be fluidized as in a fluidized bed?

0 Answers

Hi, This is Naveen, plz if anybody have experience of IOCL interview, send me question.

0 Answers

For more Chemical Engineering Interview Questions Click Here