Answer:
A) strength decreases, chemical resistance decreases, and thermal insulation increases
Explanation:
Strength always decreases, chemical resistence decreases, and thermal condictivity must be reduced therefore themal insulation must increase.
Answer:
The air pressure in the tank is 53.9 
Solution:
As per the question:
Discharge rate, Q = 20 litres/ sec = 
(Since, 1 litre = 
)
Diameter of the bore, d = 6 cm = 0.06 m
Head loss due to friction, 
Height, 
Now,
The velocity in the bore is given by:
Now, using Bernoulli's eqn:
(1)
The velocity head is given by:
Now, by using energy conservation on the surface of water on the roof and that in the tank :
Answer:
Explanation:
t1 = 1000 F = 1460 R
t0 = 80 F = 540 R
T2 = 3600 R
The working substance has an available energy in reference to the 80F source of:
B1 = Q1 * (1 - T0 / T1)
B1 = 100 * (1 - 540 / 1460) = 63 BTU
The available energy of the heat from the heat wource at 3600 R is
B2 = Q1 * (1 - T0 / T2)
B2 = 100 * (1 - 540 / 3600) = 85 BTU
The reduction of available energy between the source and the 1460 R temperature is:
B3 = B2 - B1 = 85 - 63 = 22 BTU
Answer:
A) 
B) 
C) Second law efficiency 4.85%
exergy destruction for the cycle = 9.3237 kW
Explanation:
Given data:
degree celcius
degree celcius
Power to refrigerator = 9.8 kW
Cp = 3.35 kJ/kg degree C
b)
wil be max when COP maximum
taking surrounding temperature T_H = 20 degree celcius
we know that
c) second law efficiency
exergy destruction os given as 
= 9.8 - 0.473 = 9.3237 kW
Answer:
Explanation:
A plane wall of thickness 2L=40 mm and thermal conductivity k=5W/m⋅Kk=5W/m⋅K experiences uniform volumetric heat generation at a rateq
˙
q
q
˙
, while convection heat transfer occurs at both of its surfaces (x=-L, +L), each of which is exposed to a fluid of temperature T∞=20∘CT
∞
=20
∘
C. Under steady-state conditions, the temperature distribution in the wall is of the form T(x)=a+bx+cx2T(x)=a+bx+cx
2
where a=82.0∘C,b=−210∘C/m,c=−2×104C/m2a=82.0
∘
C,b=−210
∘
C/m,c=−2×10
4
C/m
2
, and x is in meters. The origin of the x-coordinate is at the midplane of the wall. (a) Sketch the temperature distribution and identify significant physical features. (b) What is the volumetric rate of heat generation q in the wall? (c) Determine the surface heat fluxes, q
′′
x
(−L)q
x
′′
(−L) and q
′′
x
(+L)q
x
′′
(+L). How are these fluxes related to the heat generation rate? (d) What are the convection coefficients for the surfaces at x=-L and x=+L? (e) Obtain an expression for the heat flux distribution q
′′
x
(x)q
x
′′
(x). Is the heat flux zero at any location? Explain any significant features of the distribution. (f) If the source of the heat generation is suddenly deactivated (q=0), what is the rate of change of energy stored in the wall at this instant? (g) What temperature will the wall eventually reach with q=0? How much energy must be removed by the fluid per unit area of the wall (J/m2)(J/m
2
) to reach this state? The density and specific heat of the wall material are 2600kg/m32600kg/m
3
and 800J/kg⋅K800J/kg⋅K, respectively.
