An expandable container is filled with 45 in.3 of air and is sitting in ice water that is 32°F. At what volume in in.3 will the
1 answer:
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Answer:
a. 0.4544 N
b.
Explanation:
For computing the normality and molarity of the acid solution first we need to do the following calculations
The balanced reaction
= 0.27264 g
= 0.006816 mol
Now
Moles of needed is
= 0.003408 mol
= 0.333984 g
Now based on the above calculation
a. Normality of acid is
= 0.4544 N
b. And, the acid solution molarity is
= 0.00005112
=
We simply applied the above formulas
Answer:
flow(m) = 54.45 kg/s
thermal efficiency u = 44.48%
Explanation:
Given:
- P_1 = P_8 = 10 KPa
- P_2 = P_3 = P_6 = P_7 = 800 KPa
- P_4 = P_5 = 10,000 KPa
- T_5 = 550 C
- T_7 = 500 C
- Power Output P = 80 MW
Find:
- The mass flow rate of steam through the boiler
- The thermal efficiency of the cycle.
Solution:
State 1:
P_1 = 10 KPa , saturated liquid
h_1 = 192 KJ/kg
v_1 = 0.00101 m^3 / kg
State 2:
P_2 = 800 KPa , constant volume process work done:
h_2 = h_1 + v_1 * ( P_2 - P_1)
h_2 = 192 + 0.00101*(790) = 192.80 KJ/kg
State 3:
P_3 = 800 KPa , saturated liquid
h_3 = 721 KJ/kg
v_3 = 0.00111 m^3 / kg
State 4:
P_4 = 10,000 KPa , constant volume process work done:
h_4 = h_3 + v_3 * ( P_4 - P_3)
h_4 = 721 + 0.00111*(9200) = 731.21 KJ/kg
State 5:
P_5 = 10,000 KPa , T_5 = 550 C
h_5 = 3500 KJ/kg
s_5 = 6.760 KJ/kgK
State 6:
P_6 = 800 KPa , s_5 = s_6 = 6.760 KJ/kgK
h_6 = 2810 KJ/kg
State 7:
P_7 = 800 KPa , T_7 = 500 C
h_7 = 3480 KJ/kg
s_7 = 7.870 KJ/kgK
State 8:
P_8 = 10 KPa , s_8 = s_7 = 7.870 KJ/kgK
h_8 = 2490 KJ/kg
- Fraction of steam y = flow(m_6 / m_3).
- Use energy balance of steam bleed and cold feed-water:
E_6 + E_2 = E_3
flow(m_6)*h_6 + flow(m_2)*h_3 = flow(m_3)*h_3
y*h_6 + (1-y)*h_3 = h_3
y*2810 + (1-y)*192.8 = 721
Compute y: y = 0.2018
- Heat produced by the boiler q_b:
q_b = h_5 - h_4 +(1-y)*(h_7 - h_8)
q_b = 3500 -731.21 + ( 1 - 0.2018)*(3480 - 2810)
Compute q_b: q_b = 3303.58 KJ/ kg
-Heat dissipated by the condenser q_c:
q_c = (1-y)*(h_8 - h_1)
q_c= ( 1 + 0.2018)*(2810 - 192)
Compute q_c: q_c = 1834.26 KJ/ kg
- Net power output w_net:
w_net = q_b - q_c
w_net = 3303.58 - 1834.26
w_net = 1469.32 KJ/kg
- Given out put P = 80,000 KW
flow(m) = P / w_net
compute flow(m) flow(m) = 80,000 /1469.32 = 54.45 kg/s
- Thermal efficiency u:
u = 1 - (q_c / q_b)
u = 1 - (1834.26/3303.58)
u = 44.48 %
Answer:
critical stress = 595 MPa
Explanation:
given data
fracture toughness = 74.6 MPa-
crack length = 10 mm
f = 1
solution
we know crack length = 10 mm
and crack length = 2a as given in figure attach
so 2a = 10
a = 5 mm
and now we get here with the help of plane strain condition , critical stress is express as
critical stress = ......................1
put here value and we get
critical stress =
critical stress = 595 MPa
so here stress is change by plane strain condition because when plate become thinner than condition change by plane strain to plain stress.
plain stress condition occur in thin body where stress through thickness not vary by the thinner section.
