Heat Transfer Operations Questions and Answers – Condensers – Heat Transfer Coefficients – 1

«
»

This set of Heat Transfer Operations Multiple Choice Questions & Answers (MCQs) focuses on “Condensers – Heat Transfer Coefficients – 1”.

1. Which one of the following is not an assumption of condensation heat regime taken to calculate the heat transfer coefficient?
a) Presence of linear temperature profile
b) Absence of high pressure
c) Absence of viscous shear of the vapour
d) Thickness of the film is too small to create a temperature difference

Explanation: The three assumptions of the of condensation heat regime taken to calculate the heat transfer coefficient are –

1. Absence of viscous shear of the vapour
2. Thickness of the film is too small to create a temperature difference
3. Presence of linear temperature profile

2. What is the expression for the flow velocity of the falling film in a vertical condenser?
a) U=$$\frac{\rho}{\mu}g(\delta-\frac{y^2}{2})$$
b) U=$$\rho g(\delta y -\frac{y^2}{2})$$
c) U=$$\frac{\rho}{\mu}g(\delta y-y^2)$$
d) U=$$\frac{\rho}{\mu}g(\delta y-\frac{y^2}{2})$$

Explanation: U=$$\frac{\rho}{\mu}g(\delta y-\frac{y^2}{2})$$ is the correct equation obtained by deriving the Mass balance equation from the Nusselt’s theory of condensation.
Weight of fluid = Bouyancy + Shear
g(ρ-φ)(δ-y)=$$\mu \frac{du}{dy}$$
∫ g(ρ -φ)(δ-y)dy=∫μ du
U = $$\frac{\rho -\phi}{mu}g(\delta y – \frac{y^2}{2})$$
When the vapour density is negligible, $$\phi$$ = 0, we get –
U=$$\frac{\rho}{\mu}g(\delta y-\frac{y^2}{2})$$

3. What is the term y in the expression for velocity of condensate flow?
U=$$\frac{\rho}{\mu}g(\delta y-\frac{y^2}{2})$$
a) Film thickness
b) Film thickness at y
c) Distance from the wall at x
d) Film thickness at x

Explanation: The term y in the expression for velocity of condensate flow is the distance from the cooling wall inside the film at a distance x from the top. It would be clear from the diagram –

Sanfoundry Certification Contest of the Month is Live. 100+ Subjects. Participate Now!

4. What is the expression for mass flow rate of condensate in a condenser?
a) M°=$$\frac{\rho^2}{\mu}g(\delta y-\frac{y^2}{2})$$
b) M°=$$\frac{\rho^2}{\mu}g(\delta y-\frac{y^3}{2})$$
c) M°=$$\frac{\rho^2}{\mu}g\delta(\frac{y^3}{3})$$
d) M°=$$\frac{\rho^2}{\mu}g(\frac{\delta^3}{3})$$

Explanation: Weight of fluid = Bouyancy + Shear
g(ρ-φ)(δ-y)=$$\mu \frac{du}{dy}$$
∫ g(ρ -φ)(δ-y)dy=∫μ du
U = $$\frac{\rho -\phi}{\mu}g(\delta y – \frac{y^2}{2})$$
Integrating the equation with φ = 0, M° = $$\frac{\rho^2}{\mu}g(\delta y-\frac{y^2}{2}) = \frac{\rho^2}{\mu}g(\frac{\delta^3}{3})$$ we get the required expression.

5. What is the term delta in the expression for velocity of condensate flow?
M°=$$\frac{\rho^2}{\mu}g(\frac{\delta^3}{3})$$
a) Final stable film thickness
b) Film thickness at y
c) Average film thickness
d) Film thickness at x

Explanation: The term delta in the expression for velocity of condensate flow is the film thickness at a distance x from the top.

6. What is the expression for flow velocity of the condensate if the density of the vapour is not zero?
a) U=$$\frac{\rho\rho v}{\mu}g(\delta-\frac{y^2}{2})$$
b) U=$$\rho g(\delta y -\frac{y^2}{2})$$
c) U=$$\frac{\rho}{\mu\rho v}g(\delta y-y^2)$$
d) U=$$\frac{(\rho-\rho v)}{\mu}g(\delta y-\frac{y^2}{2})$$

Explanation: U=$$\frac{(\rho-\rho v)}{\mu}g(\delta y-\frac{y^2}{2})$$
The derivation shown is –
Weight of fluid = Bouyancy + Shear
g(ρ-φ)(δ-y)=$$\mu \frac{du}{dy}$$
∫ g(ρ -φ)(δ-y)dy=∫μ du
U = $$\frac{\rho-\phi}{\mu}g(\delta y – \frac{y^2}{2})$$

7. What is the expression for mass flow rate of condensate in a condenser?
a) M°=$$\frac{\rho^3}{\mu}g(\delta y-\frac{y^2}{2})$$
b) M°=$$\frac{(\rho-\rho v)^2}{\mu}g(\delta y-\frac{y^3}{2})$$
c) M°=$$\frac{\rho^2(\rho-\rho v)^2}{\mu}g\delta(\frac{y^3}{3})$$
d) M°=$$\frac{(\rho-\rho v)\rho}{\mu}g(\frac{\delta^3}{3})$$

Explanation: Integrating the equation M°=$$\int_0^\delta\frac{(\rho-\rho v)}{\mu}g(\delta y-\frac{y^2}{2})dy = \frac{(\rho-\rho v)\rho}{\mu}g(\frac{\delta^3}{3})$$

8. What is the expression for the laminar film thickness of the condensate at a distance of x from the top of the condenser?
a) δ=[4K(Tsat – TL)μ x)/(ρ ghfg)]1/2
b) δ=[4K(Tsat – TL)μ x)/(ρ ghfg)]1/4
c) δ=[4K(Tsat – TL)μ x)/(ρ ghfg)]1/8
d) δ=[K(Tsat – TL)μ x)/(ρ ghfg)]1/4

Explanation: The derivation considering vapour density is –
$$d\dot{m}=\frac{\rho(\rho-\rho_v)g\delta^2d\delta}{\mu}$$
θ = mλ=heat transfer
$$d\dot{m} \lambda= \frac{kdx}{\delta}(T_{sat}-T_L)$$
∴ equating
$$\delta^3\frac{ds}{dx} = \frac{K(T_{sat}-T_L)\mu}{\rho(\rho-\rho_v)g\lambda}$$
$$\int\delta^3d\delta = \int \frac{K(T_{sat}-T_L)\mu dx}{\rho(\rho-\rho_v)g\lambda}$$
$$\frac{\delta^4}{4} = \int \frac{K(T_{sat}-T_L)\mu x}{\rho(\rho-\rho_v)g\lambda}$$
$$\delta = [\frac{4K(T_{sat}-T_L)\mu x}{\rho(\rho-\rho_v)g\lambda}]^{1/4}$$

9. What is the term TL in the Nusselt theory of condensation equation for film thickness?
$$\delta = [\frac{4K(T_{sat}-T_L)\mu x}{\rho gh_{fg}}]^{1/4}$$
a) Liquid temperature
b) Gas temperature
c) Wall temperature
d) Bulk temperature

Explanation: The term TL is the liquid temperature of the film-

10. What is the expression for Averaged convective heat transfer coefficient for a vertical condenser?
a) hVER=0.943 $$[\frac{K^3 p^2 g h_{fg}}{\mu L(T_{sat} – T_L )}]^{0.25}$$
b) hVER=0.943 $$[\frac{K^3 p^2 g h_{fg}}{\mu L(T_{sat} – T_L )}]^{0.33}$$
c) hVER=0.943 $$[\frac{K^3 p^2 g h_{fg}}{\mu L(T_{sat} – T_L )}]^{0.35}$$
d) hVER=0.943 $$[\frac{K^3 p^2 g h_{fg}}{\mu L(T_{sat} – T_L )}]^{25}$$

Explanation: The derivation for the averaged heat transfer coefficient can be derived as –
$$\delta = [\frac{4K(T_{sat}-T_L)\mu x}{\rho(\rho-\rho_v)g\lambda}]^{1/4}$$
hx = $$\frac{K}{\delta}$$
havg = $$\frac{4}{3} \frac{K}{\delta}$$
hVER=0.943 $$[\frac{K^3 p^2 g \lambda}{\mu L(T_{sat} – T_L )}]^{\frac{1}{4}}$$
hVER=0.943 $$[\frac{K^3 p^2 g h_{fg}}{\mu L(T_{sat} – T_L )}]^{0.25}$$

11. What is the relation between the averaged heat transfer coefficient over the entire condenser length vs its value at a point x?
a) $$\frac{h_{avg}}{h_x}$$ =4
b) $$\frac{h_{avg}}{h_x} =\frac{2}{3}$$
c) $$\frac{h_{avg}}{h_x} =\frac{4}{3}$$
d) $$\frac{h_{avg}}{h_x} =\frac{3}{4}$$

Explanation: havg=$$\frac{4}{3}$$ K/δ and hx = K/δ hence dividing the two expressions we get $$\frac{h_{avg}}{h_x} = \frac{4}{3}$$.

12. What is the relation between the averaged heat transfer coefficient over the entire condenser length vs its film thickness value at a point x?
a) havg = $$\frac{4}{3} K/\delta$$
b) havg = $$\frac{8}{3} K/\delta$$
c) havg = $$\frac{8}{3} \delta/K$$
d) havg = $$\frac{4}{3} \delta/K$$

Explanation: The correct representation of the average value is 4K/3Δ because the given statement is a dimensional and calculation result obtained after the integration of hX as the basic equation.

13. What is the relation between the heat transfer coefficient vs its film thickness value at a point x?
a) hx = K/δ
b) havg = $$\frac{8}{3} K/\delta$$
c) havg = δ/K
d) havg = $$\frac{4}{3} \delta/K$$

Explanation: The term hx = K/δ is the dimensionally obtained term by minutely examining the dimensions of each of the given terms hx, k and δ, the diagram describing delta is –

14. What is the value of the constant C when the condenser is placed vertically?
hVER=C $$[\frac{K^3 p^2 g h_{fg}}{\mu L(T_{sat} – T_L )}]^{0.25}$$
a) 0.0943
b) 0.943
c) 0.725
d) 0.633

Explanation: The derivation can be presented as-
$$\delta = [\frac{4K(T_{sat}-T_L)\mu x}{\rho(\rho-\rho_v)g\lambda}]^{1/4}$$
hx = $$\frac{K}{\delta}$$
havg = $$\frac{4}{3} \frac{K}{\delta}$$
hVER=0.943 $$[\frac{K^3 p^2 g \lambda}{\mu L(T_{sat} – T_L)}]^{\frac{1}{4}}$$
hVER=0.943 $$[\frac{K^3 p^2 g h_{fg}}{\mu L(T_{sat} – T_L)}]^{0.25}$$
Here the term is 0.943

15. What is the value of the constant C when the condenser is placed horizontally?
hHOR=C $$[\frac{K^3 p^2 g h_{fg}}{\mu L(T_{sat} – T_L)}]^{0.25}$$
a) 0.942
b) 0.725
c) 0.325
d) 0.027

Explanation: The expression derived from the Equation, –
$$\delta = [\frac{4K(T_{sat}-T_L)\mu x}{\rho(\rho-\rho_v)g\lambda}]^{1/4}$$
hx = $$\frac{K}{\delta}$$
havg = $$\frac{4}{3} \frac{K}{\delta}$$
hVER=0.943 $$[\frac{K^3 p^2 g \lambda}{\mu L(T_{sat} – T_L)}]^{\frac{1}{4}}$$
hVER=0.943 $$[\frac{K^3 p^2 g h_{fg})}{\mu L(T_{sat} – T_L)}]^{0.25}$$
This equation is for vertical wall, for horizontal wall, we multiply with the factor of sin, we get the value 0.725.

Sanfoundry Global Education & Learning Series – Heat Transfer Operations.

To practice all areas of Heat Transfer Operations, here is complete set of 1000+ Multiple Choice Questions and Answers.