Question

What will be an ideal response?

Answer 1

Thales

Answer 2

Ans: moisture, evaporation

Answer 3

Answer: A,C and F.

a) The velocity of the particle is different at different points on the circular path.

c) The magnitude of the acceleration is constant.

f) The mass of the particle does not affect the speed at which it travels.

Answer 4

Answer 5

Answer: b. Eddies

Answer 6

Answer: (2) b.

Answer 7

Answer: a. Convection

Answer 8

Answer: A) remarkable agreement with surface temperatures.

Answer 9

Ans: Approximately 300 mrad/yr

Answer 10

Answer: d.1500 N

Answer 11

(a) The partial pressures of the CO2 and H2O are both 0.25 atm. The equivalent mean hemi- spherical beam length, L, is






(b) To find the absorptivity to radiation from A1, the emittances of the H2O and CO2 must first be evaluated at T1 = 278 K.
Using the procedure shown above with:




Applying

(c) Repeating this procedure for T2 = 1390 K and pH2O L (Ts/TH2O) = 0.257

(d) The rate of heat flow from the gas to A1 is

The rate of heat flow from the gas to A2 is

The net rate of heat transfer from the gas is

Answer 12

(a) The path length for the hemisphere is

For a relative humidity of 0.5% , the partial pressure of the water vapor is

for 1 atm pressure (εH2O)pT=1 = 0.18
This must be corrected for the total pressure


Repeating this procedure for the CO2
Partial pressure


The emissivity of the mixture (ε m) is

where is determined by interpolating between values

Therefore

(b) To evaluate the rate of heat transfer, the emissivity and absorptivity must be evaluated at the surface temperature






The total absorptivity is the sum of the H2O and CO2 absorptivities

The rate of heat transfer is

Answer 13

(a)

The surface area of the sphere is

Since the air and the oven completely enclose the sphere, F12 = 1.0 and F1g = 1.0
the portion of the total radiation leaving the sphere that is received by the walls

where tm is the transmissivity of the air and the radio sity of the sphere, J1, is

Since the air temperature is the same as the sphere temperature, there will be no net heat transfer between the air and the sphere and the heat transfer from the sphere (q1) will be the same as the net heat transfer between the sphere and the walls (q12). Simplifying with the shape factors above and

Since the sphere is near the center of the cube, one half the beam length for a cube will be used

The partial pressure of the water vapor is

The emissivity of the water vapor (εm) can be calculated where Tm = 810 K

By Kirchoff’s radiation law

(b) The net radiation between the sphere and walls is 146.5 W from the walls to the sphere.

(c) The radiative heat transfer coefficient must satisfy the following equation

Answer 14

In steady state, the net rate of heat transfer from the heater must be equal to the power input

The net rate of heat transfer to the receiver in steady state must be zero: q2 = 0
Also, since the receiver and the surroundings are black

The shape factor F12 can be read

Since A1 = A2, F21 = F12 (This is also clear from the symmetry of the problem).
Since neither A1 nor A2 can see itself, F11 = F22 = 0
The shape factors from any given surface must sum to unity


The net rate transfer from a gray surface is


Combining Equations [2] and [4]

Substituting

Solving for T2

(b)


Applying



(c) The net rate of heat transfer between A1 and A2 is

(d) Applying the conservation of energy of both plates and the surroundings yields

The surroundings gain 300 watts from the plates.
(e) An estimate of the natural convection heat transfer will be made by treating the heater and receiver as a horizontal enclosed space, heated from below with the surface temperature calculated above.
for dry air at the average temperature of (459°C + 807°C)/2 = 633°C

The Rayleigh number is


where
[]
indicates that the quantity in the brackets should be taken to be zero if it is negative.

The rate of heat transfer by convection is given by

COMMENTS
The natural convection heat transfer rate is only about 4% of the total heat transfer rate. Therefore, the estimate of natural convection is probably adequate.

Answer 15

Let x be the fraction of the total satellite surface area (AT) which is covered by solar cells. By definition, the absorptivity is the fraction of the total irradiation absorbed by a
body


The rate of radiation from he satellite is the emissive power of the satellite from
multiplied by its area

For steady state

Solving for the fraction covered by solar cells

80.1% of the satellite can be covered by solar cells.

Answer 16

The tube and furnace can be thought of as two infinitely long concentric gray cylinders. The rate of radiative heat transfer is


The thermal circuit for this problem is shown below

where
Rci = Convective thermal resistance inside the tube
Rk = Conductive thermal resistance of the tube wall
Rr = Radiative thermal resistance
qc = Convective heat transfer to the tube wall interior

qk = Conductive heat transfer through the tube wall

q1–2 = Radiative heat transfer

For steady state, these three heat transfer rates must be equal

Solving for Tw1 from the first part of the equation

Substituting this into the second part of the equation to solve for T1 yields


Checking the units, then eliminating them for clarity

By trial and error

Answer 17

The shape factor F21 can be read off. For s/D = 50/13 = 3.85 and one row: F21 = 0.60.
Note that A1 = A2, therefore, F21 = F12.
The sum of the shape factors from a given surface must sum to unity

The rate of radiative heat transfer, between two gray surfaces connected by re-radiating surfaces is

where A2 f21 is note that A1 = A2 = A

The temperature changes in the material is given by

As T1 increases, the rate of heat transfer will decrease. Therefore, the equation above will be solved for
a chosen time increment and the temperature T1 will then be updated. This procedure will be repeated
until T1 = 760°C = 1033 K.
Let ∆t = 5 min = 300 s initially

Answer 18

(a) The total rate of radiation emission is the blackbody emissive power times the area

(b) For
The fraction of energy absorbed is the fraction, e of the total radiation below 0.3 m which can be read directly

(c) The distance between the house and the fireball center (L) is

The energy calculated in part (a) will spread evenly in all directions from the fireball. Therefore, the flux at the distance L = q1/AsL where AsL is the surface area of a sphere of radius L

However, the atmosphere will absorb 9.4% of this energy.

The angle between this flux and the (normal to the) wall surface, σ, is given by

Therefore, the irradiation on the wall is

(d) By Kirchoff’s law, the absorptivity
Energy absorbed= class="w-image" />
(e) The radiative heat transfer coefficient (hr) is given by

Since Ts << Tf, the heat transfer coefficient will not vary much as the Tf changes. To estimate hr, let Tf = 500 K<br />
The Biot number for the wall is

Therefore, the internal thermal resistance of the oak is significant and the chart solution
will be used to estimate the surface temperature of the oak after 10 seconds: (L = s/2 =
(0.01 m)/2 = 0.005 m) The Fourier number is


(The center of the oak is still at the initial temperature after 10 s).
(b) for x/L = 1.0

where T(L,t) = the surface temperature of the wall after 10 seconds of exposure to the radiation.

Answer 19

The exterior surface area of the cylinder is

The mass of steel in the cylinder is

Performing an energy balance on the gas
heat input = rate of increase in enthalpy

Solving for the rate of change of the gas temperature

Case (a)
Initially

Finally

The rate of change of the gas temperature is essentially constant, therefore, the time required for the gas to reach 120°F is

Case (b)

Answer 20

The plate will absorb the solar radiation with the absorptivity

Law. However, it will radiate to its surroundings at longer infrared wavelengths with

The heat gain from solar radiation must equal the heat flux loss by radiation and convection at steady state

By trial and error

Answer 21

Since the space is evacuated, convection and conduction are negligible. Since the surfaces are simply infinite planes, equivalent conductance A1 f12 can be used.
The net rate of heat transfer from A1 to A2 is given by Equation (11.75)

Also, because of all the emittances and areas are identical
F12 = F23 =F34

Therefore Eb1 – Eb2 = Eb2 – Eb3 = Eb3 – Eb4
Eb2 = 0.5 (Eb1 + Eb3) and Eb3 = 0.5(Eb2 + Eb4)
Solving for Eb2

Answer 22

Since A2 completely surrounds A1 and A1 cannot see itself, F12 = 1.0 and F11 = 0
From Equation (11.46)

The shape factors from a given surface must sum to unity

Also

The net rate of heat transfer from A1 to A2 must equal the rate of heat transfer from the exterior sphere to the surroundings

The rate of heat transfer between the spheres is given by Equation (11.75)

where f12 is given in Equation (11.76) for concentric spheres.

Solving Equations [1] and [2] for T2

The rate of evaporation of the liquid air is

Answer 23

The gap between the plates can be considered to be a blackbody square cylindrical surface as shown in the sketch.

Also, since neither A1 nor A2 can see itself, F11 = F22 = 0.
From Equation (11.46) A1 F12 = A2 F21 → F12 = F21 (This is apparent from the symmetry of the problem).
The radiosities J1 and J2 are needed to calculate the rate of radiant heat transfer and can be determined using Equations (11.69) and (11.66)
From Equation (11.69) [1] A1 G1 = J1 A1 F11 + J2 A2 F21 + J3 A3 F31 = J2 A2 F21
[2] A2 G2 = J1 A1 F12 + J2 A2 F22 + J3 A3 F32 = J1 A1 F12
[3] A3 G3 = J1 A1 F13 + J2 A2 F23 + J3 A3 F33 = J1 A1 F13 + J2 A2 F23
From Equation (11.66) [4] J1 = ρ1 G1 + ε1 Eb1ρ
[5] J2 = ρ2 G2 + ε2 Eb2
[6] J3 = 0
Substituting Equations [4] and [5] into [1] and [2]

Substituting G1 from the first equation into the second and using F21 = F12 yields

The shape factor F12 can be determined from Figure 11.32: for x/D = y/D= 5 → F12 = 0.71
Also

From Equation [5]

Answer 24

All of the shape factors for the problem can be expressed in terms of F12 using Equation (11.46) and the fact that all shape factors from a given surface must sum to unity. Also

The net rate of heat transfer from surface 1 is given by Equation (11.67)

Where the radiosity (J1) and the irradiation (G1) can be calculated using Equations (11.69) and (11.66)

Substituting [4] into [1]

Substituting [2] and [3] into this Equation

where

Answer 25

The hydraulic diameter of the flattened tube is

The Reynolds number based on the hydraulic diameter is

(a) The Nusselt number for turbulent flow is given by Equation (7.61)

(b) The friction factor for turbulent flow in smooth tubes is given by Equation (7.57)

The pressure drop is given by Equation (7.13)

Answer 26

(a) The heat capacity rate of the water is

Since the heat capacity rate of the steam is essentially infinite, Cmin/Cmax = 0.
The effectiveness of the heat exchanger is given by Equation (10.22b) (Cc = Cmin).

For Cmin/Cmax = 0, the effectiveness for parallel or counterflow are the same and Equation (10.26) reduces to

(b) From Equation (10.3)

(c) Treating the first section as a separate heat exchanger and following the procedure of part (a)

This procedure must be repeated for each section. The results are tabulated below section (d).
(d) Following the procedure of section (b), the only value that changes is the overall heat transfer coefficient

Repeating parts (c) and (d) for each section yields

where x = distance from inlet to midpoint of section.
Plotting this data on a single graph