(NUMBER 4.) solve for Lm, Pm, and Wm. I’ve attached a reference image with the solution method, I just need clarification.

Refrigeration and Air Conditioning Technology (MindTap Course List)
8th Edition
ISBN:9781305578296
Author:John Tomczyk, Eugene Silberstein, Bill Whitman, Bill Johnson
Publisher:John Tomczyk, Eugene Silberstein, Bill Whitman, Bill Johnson
Chapter42: Heat Gains And Heat Losses In Structures
Section: Chapter Questions
Problem 6RQ: The typical indoor design temperature used for heat gaincalculations is A. 65F. B.70 F. C.75 F. D....
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(NUMBER 4.) solve for Lm, Pm, and Wm. I’ve attached a reference image with the solution method, I just need clarification.
Po
TS
| W=LP₁ (KIC) =PL¹/₂²₁ 9 = Qin/2, N =
E'
c
7
+
IL
-
W = LP₁ (KIC) = PL¹/²₂ 9 = Q₁n/2, 19
L
E
ان
VF
3VS
1
c'
W=LD W €²₁ 9=W³P 2 - Qin
q=
1
1
q
E'
+ VF
M'L
L
1
W=LP 9 = C₁ q = W³P, 9 = Qin
q
드
E
VF
M'L
L
Transcribed Image Text:Po TS | W=LP₁ (KIC) =PL¹/₂²₁ 9 = Qin/2, N = E' c 7 + IL - W = LP₁ (KIC) = PL¹/²₂ 9 = Q₁n/2, 19 L E ان VF 3VS 1 c' W=LD W €²₁ 9=W³P 2 - Qin q= 1 1 q E' + VF M'L L 1 W=LP 9 = C₁ q = W³P, 9 = Qin q 드 E VF M'L L
With these in mind, the approximation equations are written in brackets next to the governing
equations. The solution for the toughness dominated regime can be found from simultaneously
satisfying Elasticity, Continuity, Propagation, and the Inlet Boundary Condition, that is
K₁=PL¹2,
W =
â=
One path to the algebraic solution is as follows:
W
LP
E
W =
O..
L
LP
E
W
t
W ĝ
t L
W g
t
L'
Lin
à.. W
Lt Ľ²
K₁=PL¹/¹²,
LP LP Qi
E tE Ľ²
Q
= →W= ::
t Ľ
QEt
P= → K₁=PL¹² :: Ke
Ľ²³
K. - QK¹ - L-(K)
QiE't
=
[5/2
QEt
LP
L₁, P₁ →W=W=
E
2
= ⇒P=
2/5
QEt
Ľ
â = Lin
QE't
Ľ
-L¹/2 =
QE't
LS/2
q=
1/5
2/5
QE't
L₁₂ =
⇒P=
- (CE₁) +P = OF ² = P = QE ( 2 E₁)" :-:-(5)
P =
QEt
Ľ
Et
EQt
K
1 K₁
EQEt EQ)
L
:.W₁ =
KQt
E'A
Transcribed Image Text:With these in mind, the approximation equations are written in brackets next to the governing equations. The solution for the toughness dominated regime can be found from simultaneously satisfying Elasticity, Continuity, Propagation, and the Inlet Boundary Condition, that is K₁=PL¹2, W = â= One path to the algebraic solution is as follows: W LP E W = O.. L LP E W t W ĝ t L W g t L' Lin à.. W Lt Ľ² K₁=PL¹/¹², LP LP Qi E tE Ľ² Q = →W= :: t Ľ QEt P= → K₁=PL¹² :: Ke Ľ²³ K. - QK¹ - L-(K) QiE't = [5/2 QEt LP L₁, P₁ →W=W= E 2 = ⇒P= 2/5 QEt Ľ â = Lin QE't Ľ -L¹/2 = QE't LS/2 q= 1/5 2/5 QE't L₁₂ = ⇒P= - (CE₁) +P = OF ² = P = QE ( 2 E₁)" :-:-(5) P = QEt Ľ Et EQt K 1 K₁ EQEt EQ) L :.W₁ = KQt E'A
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