Atmospheric air with free flow velocity U_∞= 30 m/s flows through a circular cylinder at T_∞= 250 K and D = 2.5 cm diameter. The surface of the circular cylinder is kept at a uniform temperature T_w = 350 K. The CD drag coefficient is 1.1. (D=1 cm- 3 cm and U_∞= 10 m/sec 30 m/s) (a) Calculate the average heat transfer coefficient ( h ̅_m,W/m2K) (b) Determine the amount of heat transfer (Q, W) per 1 m length of the cylinder (c) Calculate the drag force (FD, N) acting on the 1 m length of the cylinder FD=0.1 A=0.34 D=2.9 cd=1.2
Atmospheric air with free flow velocity U_∞= 30 m/s flows through a circular cylinder at T_∞= 250 K and D = 2.5 cm diameter. The surface of the circular cylinder is kept at a uniform temperature T_w = 350 K. The CD drag coefficient is 1.1. (D=1 cm- 3 cm and U_∞= 10 m/sec 30 m/s) (a) Calculate the average heat transfer coefficient ( h ̅_m,W/m2K) (b) Determine the amount of heat transfer (Q, W) per 1 m length of the cylinder (c) Calculate the drag force (FD, N) acting on the 1 m length of the cylinder FD=0.1 A=0.34 D=2.9 cd=1.2
Refrigeration and Air Conditioning Technology (MindTap Course List)
8th Edition
ISBN:9781305578296
Author:John Tomczyk, Eugene Silberstein, Bill Whitman, Bill Johnson
Publisher:John Tomczyk, Eugene Silberstein, Bill Whitman, Bill Johnson
Chapter29: Troubleshooting And Typical Operating Conditions For Commercial refrigeration
Section: Chapter Questions
Problem 16RQ
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Question
Atmospheric air with free flow velocity U_∞= 30 m/s flows through a circular cylinder at T_∞= 250 K and D = 2.5 cm diameter. The surface of the circular cylinder is kept at a uniform temperature T_w = 350 K.
The CD drag coefficient is 1.1. (D=1 cm- 3 cm and U_∞= 10 m/sec 30 m/s)
(a) Calculate the average heat transfer coefficient ( h ̅_m,W/m2K)
(b) Determine the amount of heat transfer (Q, W) per 1 m length of the cylinder
(c) Calculate the drag force (FD, N) acting on the 1 m length of the cylinder
FD=0.1
A=0.34
D=2.9
cd=1.2
Transcribed Image Text:u,D C =
puž /2
FD. Re
Q = hA (T, - T)
A
St =
Pr-2/3
ST V
h = Nu, kD
Umax
ST -D
1/4
Pr
Nup = C,C, Re5max Pro.36
Prs.
To = T (Ts - T;) exp
max
Ap = NLX
P = W = PVNTSTL Ap/p
Q= PVNTSTL Cp (To-Ti)
k
Num
= (0.4 Re0.5 + 0.06 Re2/3) Pr0.4
hm
Num
D
pu
2 = hm(TDLX(T, - T) F,= cp LD Po
2
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