Exam (elaborations)
_620a90043d82b_fundamentals_of_heat_and_mass_transfer___6th___solution_manual___gearteam
- Course
- Institution
_fundamentals_of_heat_and_mass_transfe
[Show more]
Content preview
PROBLEM 1.1KNOWN: Heat rate, q, through one-dimensional wall of area A, thickness L, thermal
conductivity k and inner temperature, T1.
FIND: The outer temperature of the wall, T2.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction in the x-direction, (2) Steady-state conditions,
(3) Constant properties.
ANALYSIS: The rate equation for conduction through the wall is given by Fourier’s law,
dT T −T
q cond = q x = q ′′x ⋅ A = -k ⋅ A = kA 1 2 .
dx L
Solving for T2 gives
q cond L
T2 = T1 − .
kA
Substituting numerical values, find
3000W × 0.025m
T2 = 415$ C -
0.2W / m ⋅ K × 10m2
T2 = 415$ C - 37.5$ C
T2 = 378$ C. <
COMMENTS: Note direction of heat flow and fact that T2 must be less than T1.
, PROBLEM 1.2
KNOWN: Inner surface temperature and thermal conductivity of a concrete wall.
FIND: Heat loss by conduction through the wall as a function of ambient air temperatures ranging from
-15 to 38°C.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction in the x-direction, (2) Steady-state conditions, (3)
Constant properties, (4) Outside wall temperature is that of the ambient air.
ANALYSIS: From Fourier’s law, it is evident that the gradient, dT dx = − q′′x k , is a constant, and
hence the temperature distribution is linear, if q′′x and k are each constant. The heat flux must be
constant under one-dimensional, steady-state conditions; and k is approximately constant if it depends
only weakly on temperature. The heat flux and heat rate when the outside wall temperature is T2 = -15°C
are
q′′x = − k
dT
=k
T1 − T2
= 1W m ⋅ K
25$ C − −15$ C
= 133.3W m 2 .
( ) (1)
dx L 0.30 m
q x = q′′x × A = 133.3 W m 2 × 20 m 2 = 2667 W . (2) <
Combining Eqs. (1) and (2), the heat rate qx can be determined for the range of ambient temperature, -15
≤ T2 ≤ 38°C, with different wall thermal conductivities, k.
3500
2500
Heat loss, qx (W)
1500
500
-500
-1500
-20 -10 0 10 20 30 40
Ambient air temperature, T2 (C)
Wall thermal conductivity, k = 1.25 W/m.K
k = 1 W/m.K, concrete wall
k = 0.75 W/m.K
For the concrete wall, k = 1 W/m⋅K, the heat loss varies linearily from +2667 W to -867 W and is zero
when the inside and ambient temperatures are the same. The magnitude of the heat rate increases with
increasing thermal conductivity.
COMMENTS: Without steady-state conditions and constant k, the temperature distribution in a plane
wall would not be linear.
, PROBLEM 1.3
KNOWN: Dimensions, thermal conductivity and surface temperatures of a concrete slab. Efficiency
of gas furnace and cost of natural gas.
FIND: Daily cost of heat loss.
SCHEMATIC:
ASSUMPTIONS: (1) Steady state, (2) One-dimensional conduction, (3) Constant properties.
ANALYSIS: The rate of heat loss by conduction through the slab is
T −T 7°C
q = k ( LW ) 1 2 = 1.4 W / m ⋅ K (11m × 8 m ) = 4312 W <
t 0.20 m
The daily cost of natural gas that must be combusted to compensate for the heat loss is
q Cg 4312 W × $0.01/ MJ
Cd = ( ∆t ) = ( 24 h / d × 3600s / h ) = $4.14 / d <
ηf 0.9 ×106 J / MJ
COMMENTS: The loss could be reduced by installing a floor covering with a layer of insulation
between it and the concrete.
, PROBLEM 1.4
KNOWN: Heat flux and surface temperatures associated with a wood slab of prescribed
thickness.
FIND: Thermal conductivity, k, of the wood.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction in the x-direction, (2) Steady-state
conditions, (3) Constant properties.
ANALYSIS: Subject to the foregoing assumptions, the thermal conductivity may be
determined from Fourier’s law, Eq. 1.2. Rearranging,
L W 0.05m
k=q′′x = 40
T1 − T2 m2 ( 40-20 ) C
k = 0.10 W / m ⋅ K. <
COMMENTS: Note that the °C or K temperature units may be used interchangeably when
evaluating a temperature difference.
The benefits of buying summaries with Stuvia:
Guaranteed quality through customer reviews
Stuvia customers have reviewed more than 700,000 summaries. This how you know that you are buying the best documents.
Quick and easy check-out
You can quickly pay through credit card or Stuvia-credit for the summaries. There is no membership needed.
Focus on what matters
Your fellow students write the study notes themselves, which is why the documents are always reliable and up-to-date. This ensures you quickly get to the core!
Frequently asked questions
What do I get when I buy this document?
You get a PDF, available immediately after your purchase. The purchased document is accessible anytime, anywhere and indefinitely through your profile.
Satisfaction guarantee: how does it work?
Our satisfaction guarantee ensures that you always find a study document that suits you well. You fill out a form, and our customer service team takes care of the rest.
Who am I buying these notes from?
Stuvia is a marketplace, so you are not buying this document from us, but from seller leonardmuriithi061. Stuvia facilitates payment to the seller.
Will I be stuck with a subscription?
No, you only buy these notes for $17.89. You're not tied to anything after your purchase.
Can Stuvia be trusted?
4.6 stars on Google & Trustpilot (+1000 reviews)
82388 documents were sold in the last 30 days
Founded in 2010, the go-to place to buy study notes for 14 years now