Q1. Q = hA(T2-T1) = h_{w}*(2*.6*.75)*(90-15) where h_{w} is convective heat transfer coeff of water.

we find h_{w} from the Nusselt number relation given below

here L = .60m and T film = (90+15)/2 = 52.5 ^{o}C = 325.5 K

Pr is the Prandtl number given by

From tables, Pr = 3.42

and

f =.84

Gr is Grashof number which is given by the relation:

Gr=3.28*10^11

h=644 W/m^{2}K

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) )

c_{p} : __specific heat__, (SI units : J/(kg K) )

Q= 43.503 KW

Q2. 1820 = h_{CO}*3.14*D*(370-45) D is diameter

For a long horizontal cylinder:

Where h is the h_{CO} required. C and n are constants which in the question are

Ra C n

Now Ra is Rayleigh number given by:

Where Ra_{L} = Ra_{D}

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 = h_{air}*(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/m^{2}K

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.

Q=14.27W

Q4. Q’’= h_{nitrogen}*0.5*3.14*40*580/100

Where

Re = 4592.6 C =.193 m =.618

Pr = .702

h = 3.24 W/m^{2}K

Pr number’s relationship is already given above.

From Nusselt No. We can find h and hence find the convective heat transfer.

Q=2364.8 W/m

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