This set of Heat Transfer Operations Problems focuses on “Condensers – Heat Transfer Coefficients – 2”.

1. For the given equation of heat transfer coefficient, which one of the following condition should hold true?

h_{VER}=0.943 \([\frac{K^3 p^2 g h_{fg}}{\mu L(T_{sat} – T_L )}]^{0.25}\)

a) Re < 1800

b) Re > 1800

c) Re < 1000

d) Re < 18000

View Answer

Explanation: The expression is valid only when the Reynolds number is less than the required 1800 value otherwise the constant term in the given equation would have to be changed according to the new value of forces and mass balances.

2. If the Reynolds Number is more than 1800, then what is the equation for the heat transfer coefficient of the condenser?

a) h=0.0077 Re^{0.4}[K^{3}ρ^{2}g ⁄ μ^{2}]^{1/3}

b) h=0.0077 Re^{0.4}[K^{3}ρ^{2}g ⁄ μ^{2}]^{1/4}

c) h=0.0077 Re^{0.4}[K^{4}ρ^{2}g ⁄ μ^{2}]^{1/3}

d) h=0.0077 Re^{0.4}[K^{3}ρ^{3}g ⁄ μ^{2}]^{1/3}

View Answer

Explanation: The expression is valid only when the Reynolds number is less than the required 1800 value otherwise the constant term in the given equation would have to be changed according to the new value of forces and mass balances. If the Reynolds Number increases to a higher value, then we use the Kirkbride formula to derive the given equation:

h=0.0077 Re

^{0.4}[K

^{3}ρ

^{2}g ⁄ μ

^{2}]

^{1/3}

3. When do we apply the Kirkbride equation for determination of heat transfer coefficient in a condenser?

h=0.0077 Re^{0.4}[K^{3} ρ^{2}g ⁄ μ^{2}]^{1/3}

a) Re < 1800

b) Re > 1800

c) Re < 1000

d) Re < 18000

View Answer

Explanation: The expression is valid only when the Reynolds number is less than the required 1800 value otherwise the constant term in the given equation would have to be changed according to the new value of forces and mass balances. If the Reynolds Number increases to a higher value, then we use the Kirkbride formula. For below 1800 value, Re < 1800, the expression is –

h

_{HOR}=0.943 \([\frac{K^3 p^2 g h_{fg}}{\mu L(T_{sat} – T_L )}]^{0.25}\)

And thus for answer when Re > 1800

h=0.0077 Re

^{0.4}[K

^{3}ρ

^{2}g ⁄ μ

^{2}]

^{1/3}

4. A 0.5m square plate is to be used to condense dry saturated steam at 0.08bar, if the wall surface is to be maintained at 18.5℃, calculate the film thickness at a distance of 25cm.

Given T_{sat}= 41.5℃, density = 995kg/m^{3}, h_{fg} = 2402 KJ/m hr K , K = 2.22 KJ/mK, μ = 8.01x 10^{–4} Ns/m^{2}

a) 0.0364 m

b) 0.024 m

c) 0.0046 m

d) 0.0056 m

View Answer

Explanation: The equation for Film thickness is given by Nusselt theory of condensation as

\(\delta = [\frac{4K(T_{sat}-T_L)\mu x}{\rho gh_{fg}}]^{1/4}\)

Here T

_{sat}= 41.5℃, T = 18.5℃, density = 995kg/m

^{3}, h

_{fg}= 2.4 KJ/m

^{2}K, K = 2.22 KJ/mK, μ = 8.01x 10

^{–4}N/m

^{2}s. Hence we get δ=0.0364

δ=[(4×2.22×10

^{3}x (41.5-18.5)x8.01x 10

^{-4}x 0.25)/995×9.8×2.4×10

^{3}]

^{1/4}=0.036m.

5. A 0.5m square plate is to be used to condense dry saturated steam at 0.08bar, if the wall surface is to be maintained at 18.5℃, calculate the Convective heat transfer coefficient h at a distance of 25cm.

Given T_{sat} = 41.5℃, density = 995kg/m^{3}, h_{fg} = 2402 KJ/m^{2}K, K = 2.22 KJ/mK, μ = 8.01 x 10^{–4} N/m^{2}s

a) 160×10^{5} J/m^{2}K

b) 3.5×10^{5} J/m^{2}K

c) 450×10^{5} J/m^{2}K

d) 150×10^{5} J/m^{2}K

View Answer

Explanation: We can apply the equation/relation for the value to be h =\(\frac{k}{\Delta}\) = 2.22×10

^{3}/0.0064 = 3.5×105 J/m

^{2}K.

The equation for Film thickness is given by Nusselt theory of condensation as

\(\delta = [\frac{4K(T_{sat}-T_L)\mu x}{\rho gh_{fg}}]^{1/4}\)

δ=[4×2.22×10

^{3}x(41.5-18.5)x8.01 x 10

^{-4}x0.25 ⁄ 995×9.8x2402x10

^{3}]

^{1/4}=0.0064m

Given T

_{sat}= 41.5℃, density = 995kg/m

^{3}, h

_{fg}= 2402 KJ/m hr K, K = 2.22 KJ/mK, μ = 8.01x 10

^{–4}Ns/m

^{2}Hence we get δ=0.0064m.

6. What is the value of h_{avg} of a condenser at a distance of 25cm if a 0.5m square plate is to be used to condense dry saturated steam at 0.08bar, if the wall surface is to be maintained at 18.5℃?

Given T_{sat}= 41.5℃, density = 995kg/m^{3}, h_{fg} = 2402 KJ/m hr K, K = 2.22 KJ/mK, μ = 8.01x 10^{–4} Ns/m^{2}

a) 50×105 J/m^{2}K

b) 10×105 J/m^{2}K

c) 350×105 J/m^{2}K

d) 26 x105 J/m^{2}K

View Answer

Explanation: We know the relation between them to be

h

_{VER}=0.943\([\frac{K^3 p^2 g h_{fg}}{\mu x(T_{sat} – T_L)}]^{0.25}\)

h

_{VER}=0.943\([\frac{(2.22×1000)^3 x 995^2 x 9.8x2402x1000}{8.01×0.25(41.5 – 18.5)}]^{0.25}\)=2572000=2572000

Here T

_{sat}= 41.5℃, T = 18.5℃, density = 995kg/m

^{3}, h

_{fg}= 2402 KJ/m

^{2}K, K = 2.22 KJ/mK, μ = 8.01 x 10

^{–4}N/m

^{2}s. Hence we get δ=0.000148 and h = 200 x10

^{5}J/m

^{2}K.

7. A 0.5m square plate is to be used to condense dry saturated steam at 0.08bar, if the wall surface is to be maintained at 18.5℃, calculate the mean flow velocity at a distance of 25cm.

Given T_{sat} = 41.5℃, density = 995kg/m^{3}, h_{fg} = 2402 KJ/m hr K, K = 2.22 J/mK, μ = 8.01x 10^{–4} Ns/m^{2}

a) 0.09 m/s

b) 0.9 m/s

c) 0.6m/s

d) 169.6m/s

View Answer

Explanation: v = \(\frac{ρg\Delta^2}{3\mu} =\frac{995x10x0.000148^2}{3×8.01×10^{-4}}\)=0.09m/s

The equation for Film thickness is given by Nusselt theory of condensation as

\(\delta = [\frac{4K(T_{sat}-T_L)\mu x}{\rho gh_{fg}}]^{1/4}\)

δ=[4×2.22x(41.5-18.5)x8.01x 10

^{-4}x0.25)⁄ 995×9.8x2402x10

^{3}]

^{1/4}=0.000148m

Given T

_{sat}= 41.5℃, density = 995kg/m

^{3}, h

_{fg}= 2402 KJ/m hr K, K = 2.22 J/mK, μ= 8.01x 10 – 4 Ns/m

^{2}Hence we get δ=1.48×10

^{-4}m.

8. A 0.5m square plate is to be used to condense dry saturated steam at 0.08bar, if the wall surface is to be maintained at 18.5℃, calculate the mean mass flow at a distance of 25cm.

Given T_{sat} = 41.5℃, density = 995kg/m^{3}, h_{fg} = 2402 KJ/m hr K, K = 2.22 J/mK, μ = 8.01x 10^{–4} Ns/m^{2}

a) 0.135 Kg/hr

b) 22.39 Kg/hr

c) 0.135 kg/sec

d) 22.39 kg/s

View Answer

Explanation: M = VρA = 0.09x995x0.25 = 22.39 kg/s where

v = ρgδ

^{2}/ 3μ = (995x10x0.000148

^{2})/(3×8.01×10

^{-4})=0.09m/s.

9. What is the expression for mean flow velocity of the film fluid in a vertical condenser?

a) v = ρgδ / 3μ

b) v = ρgδ^{2} / μ

c) v = ρgδ^{2} / 3

d) v = ρgδ^{2} / 3μ

View Answer

Explanation: The mean flow velocity is the velocity with which the film of condensate moves down the wall of the condenser, hence the derived formula is, v = ρgδ

^{2}/ 3μ.

10. What is the expression for total heat transferred in a condenser of length L and diameter D?

a) h_{avg} A Δ T_{sat-L}

b) h_{avg} A Δ T_{LMTD}

c) h_{X} A ΔT_{sat-L}

d) h_{X} A ΔT_{LMTD}

View Answer

Explanation: The total heat transferred is the measurement done for the whole condenser setup and cannot be used for a particular point. Hence we use the h

_{avg}as the heat transfer coefficient and LMTD is not applicable here.

11. What is the expression for total steam condensate rate?

a) M = V / h_{fg}

b) M = h_{fg}/V

c) M = V / h_{avg}

d) M = VϸA

View Answer

Explanation: m = VϸA is the dimensionally sane equation among all the others which do not match the dimension of M = MT

^{-1}and hence VϸA is the correct option.

12. What is the expression for Reynolds number for the falling film in a condsenser?

a) Re= \(\frac{4\ddot{M}}{\rho u}\)

b) Re= \(\frac{4\ddot{M}}{\rho}\)

c) Re= \(\frac{4}{\rho u}\)

d) Re= \(\frac{\ddot{M}}{\rho u}\)

View Answer

Explanation: Re=\(\frac{4\ddot{M}}{\rho u}\) is the dimensionally sane equation among all the others which do not match the dimension of Reynolds number which is dimensionless and hence Re=\(\frac{4\ddot{M}}{\rho u}\) is the correct option.

**Sanfoundry Global Education & Learning Series – Heat Transfer Operations.**

To practice all areas of Heat Transfer Operations Problems, __here is complete set of 1000+ Multiple Choice Questions and Answers__.

**If you find a mistake in question / option / answer, kindly take a screenshot and email to [email protected]**

**Related Posts:**

- Practice Chemical Engineering MCQs
- Check Heat Transfer Operations Books
- Apply for Chemical Engineering Internship
- Check Chemical Engineering Books