250+ TOP MCQs on Shell Heat Exchangers – Shell Side Effective Diameter

Heat Transfer Operations online test on “Shell Heat Exchangers – Shell Side Effective Diameter”.

1. The pitch of a baffle arrangement in a shell and tube HE is the distance between centres of two tubes in the shell side arrangement.
a) True
b) False

2. What is the pitch of a shell and tube HE if the diameter of the tubes is 30mm and clearance is 10mm?
a) 35mm
b) 38mm
c) 40mm
d) 45mm

Answer: c
Clarification: The pitch of the setup can be defined as Pt = D_t + C = 30 + 10 = 40mm where C is the Clearance.

3. What is the flow area if the pitch of a shell and tube HE is 40mm, the inner diameter of the tubes is 20mm, clearance is 10mm and baffle space is 10mm?
a) 50 sq mm
b) 60 sq mm
c) 55 sq mm
d) 53 sq mm

Answer: a
Clarification: Flow area in a shell and tube HE is represented as: A = (frac{ID×C×B}{Pt} = frac{20×10×10}{40} )= 50 sq mm.

4. In the calculation of flow area for a shell and tube heat exchanger, which is the correct formula?
Where C = Clearance, Pt = Pitch, ID = Inner Diameter, B = Baffle spacing.
a) A = (frac{ID×Pt×C}{B})
b) A = (frac{ID×Pt×B}{C})
c) A = (frac{Pt×C×B}{ID})
d) A = (frac{ID×C×B}{Pt})

5. Which of the following formula is correct for Equivalent diameter for a triangular pitch?
a) D = (frac{4(frac{1}{2}πPt^2+frac{frac{1}{2}πOD^2}{8})}{frac{1}{2}πOD^2})
b) D = (frac{4(frac{1}{2}πPt^2+frac{frac{1}{2}πOD^2}{4})}{frac{1}{2}πOD^2})
c) D = (frac{4(0.86frac{1}{2}πPt^2-frac{frac{1}{2}πOD^2}{4})}{frac{1}{2}πOD^2})
d) D = (frac{4(frac{0.86}{2}πPt^2-frac{frac{1}{2}πOD^2}{4})}{frac{1}{2}πOD})

Answer: d
Clarification: The Equivalent diameter for a triangular pitch is correctly given by the formula,
D = (frac{4(frac{0.86}{2}Pt^2-frac{frac{1}{2}πOD^2}{4})}{frac{1}{2}πOD})

6. What is the correct Entity X in the equivalent diameter of a Triangular pitch?
D = (frac{4(frac{X}{2}Pt^2-frac{frac{1}{2}πOD^2}{4})}{frac{1}{2}πOD})
a) 0.23
b) 0.027
c) 0.86
d) 0.086

Answer: c
Clarification: The area of equilateral triangle is (frac{sqrt{3}}{2}Pfrac{t^2}{2}) hence,(frac{sqrt{3}}{2})=0.86.

7. What is the equivalent diameter of a triangular pitch shell and tube HE if pitch is 40mm, the outer diameter of the tubes is 30mm?
a) 29.4 mm
b) 28.4 mm
c) 26.2 mm
d) 24.1 mm

Answer: b
Clarification: The Equivalent diameter for a triangular pitch is correctly given by the formula,
D = (frac{4(frac{0.86}{2}Pt^2-frac{frac{π}{2}OD^2}{4})}{frac{π}{2}OD}) = (frac{4(frac{0.86}{2}40^2-frac{frac{1}{2}π30^2}{4})}{frac{1}{2}π30})=28.43.

8. What is the equivalent diameter of a square pitch shell and tube HE if pitch is 40mm, the outer diameter of the tubes is 30mm?
a) 39.4 mm
b) 37.9 mm
c) 36.2 mm
d) 34.1 mm

Answer: b
Clarification: The Equivalent diameter for a square pitch is correctly given by the formula,
D = ( frac{4(Pt^2-frac{πOD^2}{4})}{πOD}=frac{4(40^2-frac{π30^2}{4})}{π30}) = 37.92 mm.

9. Which one of the following is the correct formula for the equivalent diameter for square pitch?
a) D = ( frac{4(Pt^2-frac{πOD^2}{4})}{πOD})
b) D = ( frac{4(0.86×Pt^2-frac{πOD^2}{4})}{πOD})
c) D = ( frac{(frac{0.86}{2}Pt^2-frac{πOD^2}{4})}{frac{π}{2}OD})
d) D = ( frac{4(frac{0.86}{2}Pt^2-frac{πOD^2}{4})}{frac{π}{2}OD})

Answer: a
Clarification: The equivalent diameter for a square pitch is approximately the radius of the flow area available which is D = ( frac{4(Pt^2-frac{πOD^2}{4})}{πOD}). Whereas the following formula is for a triangular pitch: D = ( frac{4(frac{0.86}{2}Pt^2-frac{πOD^2}{4})}{frac{π}{2}OD}).