250+ TOP MCQs on Condensers – Heat Transfer Coefficients – 2 and Answers

Heat Transfer Operations Problems 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
Answer:a
Clarification: 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[K³ρ²g ⁄ μ²]^1/3
b) h=0.0077 Re^0.4[K³ρ²g ⁄ μ²]^1/4
c) h=0.0077 Re^0.4[K⁴ρ²g ⁄ μ²]^1/3
d) h=0.0077 Re^0.4[K³ρ³g ⁄ μ²]^1/3
Answer:a
Clarification: 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[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 Re^0.4[K³ ρ²g ⁄ μ²]^1/3
a) Re < 1800
b) Re > 1800
c) Re < 1000
d) Re < 18000
Answer:b
Clarification: 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[K³ ρ²g ⁄ μ²]^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/m³, h_fg = 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
Answer:a
Clarification: 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/m³, h_fg = 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 T_sat = 41.5℃, density = 995kg/m³, h_fg = 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
Answer:b
Clarification: 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 T_sat = 41.5℃, density = 995kg/m³, h_fg = 2402 KJ/m hr K, K = 2.22 KJ/mK, μ = 8.01x 10^–4 Ns/m² 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/m³, h_fg = 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