Q1. Q = hA(T2-T1) = hw*(2*.6*.75)*(90-15) where hw is convective heat transfer coeff of water.
we find hw from the Nusselt number relation given below
here L = .60m and T film = (90+15)/2 = 52.5 oC = 325.5 K
Pr is the Prandtl number given by
From tables, Pr = 3.42
Gr is Grashof number which is given by the relation:
Where Ts is surface temperature and water temp is the other temp required. Prandtl number can be found out by looking up the properties (evaluated at film temp.)table for water at back of book.
Where Film Temp. = (Ts+T8)/2Here:
µ : viscosity, (SI units : Pa s)
k : thermal conductivity, (SI units : W/(m K) )
cp : specific heat, (SI units : J/(kg K) )
Q= 43.503 KW
Q2. 1820 = hCO*3.14*D*(370-45) D is diameter
For a long horizontal cylinder:
Where h is the hCO required. C and n are constants which in the question are
Ra C n
Now Ra is Rayleigh number given by:
Where RaL = RaD
Ra = 3733775200*D*D*D
Hence, C =.480 n = .250 (assuming diameter of the order of cm)
The formula for Grashof number and Prandtl number is already given in the previous question.
L = D
D = 30 cm
Q3. Q = hair*(190-15)*4*3.14*(7/200)*(7/200)
Nusselt number for a sphere is given by:
Ra = 202566.06
Pr = .695
Nu = 11.62
h = 5.3 W/m2K
Here Ra and Pr number are found as shown earlier and properties (measured at film temperature already mentioned above) looked up from the tables given at the back of the book.
Nu = hD/k so we can find h after finding the Nusselt number.
Q4. Q’’= hnitrogen*0.5*3.14*40*580/100
Re = 4592.6 C =.193 m =.618
Pr = .702
h = 3.24 W/m2K
Pr number’s relationship is already given above.
From Nusselt No. We can find h and hence find the convective heat transfer.
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