# Heat Transfer Operations Questions and Answers – Condensers – Heat Transfer Coefficients – 2

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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?
hVER=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

Answer:a
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.
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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 Re0.4[K3ρ2g ⁄ μ2]1/3
b) h=0.0077 Re0.4[K3ρ2g ⁄ μ2]1/4
c) h=0.0077 Re0.4[K4ρ2g ⁄ μ2]1/3
d) h=0.0077 Re0.4[K3ρ3g ⁄ μ2]1/3
View Answer

Answer:a
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 Re0.4[K3 ρ2g ⁄ μ2]1/3

3. When do we apply the Kirkbride equation for determination of heat transfer coefficient in a condenser?
h=0.0077 Re0.4[K3 ρ2g ⁄ μ2]1/3
a) Re < 1800
b) Re > 1800
c) Re < 1000
d) Re < 18000
View Answer

Answer:b
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 –
hHOR=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 Re0.4[K3 ρ2g ⁄ μ2]1/3
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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 Tsat= 41.5℃, density = 995kg/m3, hfg = 2402 KJ/m hr K , K = 2.22 KJ/mK, μ = 8.01x 10–4 Ns/m2
a) 0.0364 m
b) 0.024 m
c) 0.0046 m
d) 0.0056 m
View Answer

Answer:a
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 Tsat = 41.5℃, T = 18.5℃, density = 995kg/m3, hfg = 2.4 KJ/m2K, K = 2.22 KJ/mK, μ = 8.01x 10–4 N/m2s. Hence we get δ=0.0364
δ=[(4×2.22×103 x (41.5-18.5)x8.01x 10-4 x 0.25)/995×9.8×2.4×103]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 Tsat = 41.5℃, density = 995kg/m3, hfg = 2402 KJ/m2K, K = 2.22 KJ/mK, μ = 8.01 x 10–4 N/m2s
a) 160×105 J/m2K
b) 3.5×105 J/m2K
c) 450×105 J/m2K
d) 150×105 J/m2K
View Answer

Answer:b
Explanation: We can apply the equation/relation for the value to be h =$$\frac{k}{\Delta}$$ = 2.22×103/0.0064 = 3.5×105 J/m2K.
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=[4×2.22×103 x(41.5-18.5)x8.01 x 10-4 x0.25 ⁄ 995×9.8x2402x103]1/4=0.0064m
Given Tsat = 41.5℃, density = 995kg/m3, hfg = 2402 KJ/m hr K, K = 2.22 KJ/mK, μ = 8.01x 10–4 Ns/m2 Hence we get δ=0.0064m.
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6. What is the value of havg 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 Tsat= 41.5℃, density = 995kg/m3, hfg = 2402 KJ/m hr K, K = 2.22 KJ/mK, μ = 8.01x 10–4 Ns/m2
a) 50×105 J/m2K
b) 10×105 J/m2K
c) 350×105 J/m2K
d) 26 x105 J/m2K
View Answer

Answer:d
Explanation: We know the relation between them to be
hVER=0.943$$[\frac{K^3 p^2 g h_{fg}}{\mu x(T_{sat} – T_L)}]^{0.25}$$
hVER=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 Tsat = 41.5℃, T = 18.5℃, density = 995kg/m3, hfg = 2402 KJ/m2K, K = 2.22 KJ/mK, μ = 8.01 x 10–4 N/m2s. Hence we get δ=0.000148 and h = 200 x105 J/m2K.

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 Tsat = 41.5℃, density = 995kg/m3, hfg = 2402 KJ/m hr K, K = 2.22 J/mK, μ = 8.01x 10–4 Ns/m2
a) 0.09 m/s
b) 0.9 m/s
c) 0.6m/s
d) 169.6m/s
View Answer

Answer:a
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)⁄ 995×9.8x2402x103]1/4=0.000148m
Given Tsat= 41.5℃, density = 995kg/m3, hfg = 2402 KJ/m hr K, K = 2.22 J/mK, μ= 8.01x 10 – 4 Ns/m2 Hence we get δ=1.48×10-4 m.
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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 Tsat = 41.5℃, density = 995kg/m3, hfg = 2402 KJ/m hr K, K = 2.22 J/mK, μ = 8.01x 10–4 Ns/m2
a) 0.135 Kg/hr
b) 22.39 Kg/hr
c) 0.135 kg/sec
d) 22.39 kg/s
View Answer

Answer:d
Explanation: M = VρA = 0.09x995x0.25 = 22.39 kg/s where
v = ρgδ2 / 3μ = (995x10x0.0001482)/(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 = ρgδ2 / μ
c) v = ρgδ2 / 3
d) v = ρgδ2 / 3μ
View Answer

Answer:d
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 = ρgδ2 / 3μ.
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10. What is the expression for total heat transferred in a condenser of length L and diameter D?
a) havg A Δ Tsat-L
b) havg A Δ TLMTD
c) hX A ΔTsat-L
d) hX A ΔTLMTD
View Answer

Answer:a
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 havg as the heat transfer coefficient and LMTD is not applicable here.

11. What is the expression for total steam condensate rate?
a) M = V / hfg
b) M = hfg/V
c) M = V / havg
d) M = VϸA
View Answer

Answer:d
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

Answer:a
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.

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