A ____________ is a term that originally was referring to a way to reproduce a technical drawing documenting an architectural or
1 answer:
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Answer:
Ts =Ta E)- 300(
569.5 K
5
Q12-W12 = -4014.26
Mol
AU2s = Q23= 5601.55
Mol
AUs¡ = Ws¡ = -5601.55
Explanation:
A clear details for the question is also attached.
(b) The P,V and T for state 1,2 and 3
P =1 bar Ti = 300 K and Vi from ideal gas Vi=
10
24.9x10 m
=
P-5 bar
Due to step 12 is isothermal: T1 = T2= 300 K and
VVi24.9 x 10x-4.9 x 10-3 *
The values at 3 calclated by Uing step 3l Adiabatic process
B-P ()
Since step 23 is Isochoric: Va =Vs= 4.99 m* and 7=
14
Ps-1x(4.99 x 103
P-1x(29x 10)
9.49 barr
And Ts =Ta E)- 300(
569.5 K
5
(c) For step 12: Isothermal, Since AT = 0 then AH12 = AU12 = 0 and
Work done for Isotermal process define as
8.314 x 300 In =4014.26
Wi2= RTi ln
mol
And fromn first law of thermodynamic
AU12= W12 +Q12
Q12-W12 = -4014.26
Mol
F'or step 23 Isochoric: AV = 0 Since volume change is zero W23= 0 and
Alls = Cp(L3-12)=5 x 8.311 (569.5 - 300) = 7812.18-
AU23= C (13-72) =5 x 8.314 (569.3 - 300) = 5601.53
Inol
Now from first law of thermodynamic the Q23
AU2s = Q23= 5601.55
Mol
For step 3-1 Adiabatic: Since in this process no heat transfer occur Q31= 0
and
AH
C,(T -Ts)=x 8.314 (300- 569.5)= -7842.18
mol
AU=C, (T¡-T)= x 8.314 (300
-5601.55
569.5)
mol
Now from first law of thermodynamie the Ws1
J
mol
AUs¡ = Ws¡ = -5601.55
Answer:
a.) 1.453MW/m2, b.) 2,477,933.33 BTU/hr c.) 22,733.33 BTU/hr d.) 1,238,966.67 BTU/hr
Explanation:
Heat flux is the rate at which thermal (heat) energy is transferred per unit surface area. It is measured in W/m2
Heat transfer(loss or gain) is unit of energy per unit time. It is measured in W or BTU/hr
1W = 3.41 BTU/hr
Given parameters:
thickness, t = 7.5mm = 7.5/1000 = 0.0075m
Temperatures 150 C = 150 + 273 = 423 K
50 C = 50 + 273 = 323 K
Temperature difference, T = 423 - 323 = 100 K
We are assuming steady heat flow;
a.) Heat flux, Q" = kT/t
K= thermal conductivity of the material
The thermal conductivity of brass, k = 109.0 W/m.K
Heat flux,
b.) Area of sheet, A = 0.5m2
Heat loss, Q = kAT/t
Heat loss,
Heat loss, Q = 726,666.667 * 3.41 = 2,477,933.33 BTU/hr
c.) Material is now given as soda lime glass.
Thermal conductivity of soda lime glass, k is approximately 1W/m.K
Heat loss,
Heat loss, Q = 6,666.67 * 3.41 = 22,733.33 BTU/hr
d.) Thickness, t is given as 15mm = 15/1000 = 0.015m
Heat loss,
Heat loss, Q = 363,333.33 * 3.41 = 1,238,966.67 BTU/hr