If we place a condenser at an angle of α to the vertical, then the convective heat transfer coefficient can be represented as ____________

Question

TuteeHUB · Accepted Answer

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

TuteeHUB · Answer

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

TuteeHUB · Answer

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

TuteeHUB · Answer

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

TuteeHUB · Answer

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

1.	If we place a condenser at an angle of α to the vertical, then the convective heat transfer coefficient can be represented as ____________
A.	hINC = 0.943 \(0.943[\frac{K^3 p^2 g h_{fg}}{\mu L(T_{sat} – T_L)}sin⁡\alpha]^{0.25}\)
B.	hINC = 0.943 \(0.943[\frac{K^3 p^2 g h_{fg}}{\mu L(T_{sat} – T_L)}tan\alpha]^{0.25}\)
C.	hINC = 0.943 \(0.943[\frac{K^3 p^2 g h_{fg}}{\mu L(T_{sat} – T_L)}cos⁡\alpha]^{0.25}\)
D.	hINC = 0.943 \(0.943[\frac{K^3 p^2 g h_{fg}}{\mu L(T_{sat} – T_L)}cosec⁡\alpha]^{0.25}\)
Answer» B. hINC = 0.943 \(0.943[\frac{K^3 p^2 g h_{fg}}{\mu L(T_{sat} – T_L)}tan\alpha]^{0.25}\)

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