Part B Now push the plate down and observe what happens on the shore. Repeat the action three to four more times, and watch the
1 answer:
You might be interested in
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:
Glass: Low-Loss dielectric
α = 8.42*10^-11 Np/m
β = 468.3 rad/m
λ = 1.34 cm
up = 1.34*10^8 m/s
ηc = 168.5 Ω
Tissue: Quasi-Conductor
α = 9.75 Np/m
β = 12.16 rad/m
λ = 51.69 cm
up = 0.52*10^8 m/s
ηc = 39.54 + j 31.72 Ω
Wood: Good conductor
α = 6.3*10^-4 Np/m
β = 6.3*10^-4 Np/m
λ = 10 km
up = 0.1*10^8 m/s
ηc = 6.28*( 1 + j )
Explanation:
Given:
Glass with µr = 1, εr = 5, and σ = 10−12 S/m at 10 GHz
Animal tissue with µr = 1, εr = 12, and σ = 0.3 S/m at 100 MHz.
Wood with µr = 1, εr = 3, and σ = 10−4 S/m at 1 kHz
Find:
Determine if the material is a low-loss dielectric, a quasi-conductor, or a good conductor, and then calculate α, β, λ, up, and ηc:
Solution:
- We need to determine the loss tangent to determine category of the medium as follows:
σ / w*εr*εo
Where, w is the angular speed of wave
εo is the permittivity of free space = 10^-9 / 36*pi
- Now we classify as follows:
- For σ / w*εr*εo < 0.01 --- Low-Loss dielectric and σ / w*εr*εo > 100 --- Good conducting material.
Glass: Low-Loss dielectric
Tissue: Quasi-Conductor
Wood: Good conductor
- Now we will use categorized material base equations from Table 17-1 as follows:
Glass: Low-Loss dielectric
α = (σ / 2)*sqrt(u / εr*εo) = (10^-12 / 2)*sqrt( 4*pi*10^-7/5*8.85*10^-12)
α = 8.42*10^-11 Np/m
β = w*sqrt (u*εr*εo) = 2pi*10^10*sqrt (4*pi*10^-7*5*8.85*10^-12)
β = 468.3 rad/m
λ = 2*pi / β = 2*pi / 468.3
λ = 1.34 cm
up = λ*f = 0.0134*10^10
up = 1.34*10^8 m/s
ηc = sqrt ( u / εr*εo ) = sqrt( 4*pi*10^-7/12*8.85*10^-12)
ηc = 168.5 Ω
Tissue: Quasi-Conductor
α = (σ / 2)*sqrt(u / εr*εo) = (0.3 / 2)*sqrt( 4*pi*10^-7/12*8.85*10^-12)
α = 9.75 Np/m
β = w*sqrt (u*εr*εo) = 2pi*100*10^6*sqrt (4*pi*10^-7*12*8.85*10^-12)
β = 12.16 rad/m
λ = 2*pi / β = 2*pi / 12.16
λ = 51.69 cm
up = λ*f = 0.5169*100*10^6
up = 0.52*10^8 m/s
ηc = sqrt ( u / εr*εo )*( 1 - j (σ / w*εr*εo))^-0.5
ηc = sqrt (4*pi*10^-7*12*8.85*10^-12)*( 1 - j 4.5)^-0.5
ηc = 39.54 + j 31.72 Ω
Wood: Good conductor
α = sqrt (pi*f*σ u) = sqrt( pi* 10^3 *4*pi* 10^-7 * 10^-4 )
β = α = 6.3*10^-4 Np/m
λ = 2*pi / β = 2*pi / 6.3*10^-4
λ = 10 km
up = λ*f = 10,000*1*10^3
up = 0.1*10^8 m/s
ηc = α*( 1 + j ) / б = 6.3*10^-4*( 1 + j ) / 10^-4
ηc = 6.28*( 1 + j )
Answer:
q = 1.73 W
Explanation:
given data
small end = 5 cm
large end = 10 cm
high = 15 cm
small end is held = 600 K
large end at = 300 K
thermal conductivity of asbestos = 0.173 W/mK
solution
first we will get here side of cross section that is express as
...............1
here x is distance from small end and S1 is side of square at small end
and S2 is side of square of large end and L is length
put here value and we get
S = 5 +
S = m
and
now we get here Area of section at distance x is
area A = S² ...............2
area A = m²
and
now we take here small length dx and temperature difference is dt
so as per fourier law
heat conduction is express as
heat conduction q = ...............3
put here value and we get
heat conduction q =
it will be express as
now we intergrate it with limit 0 to 0.15 and take temp 600 to 300 K
solve it and we get
q (30) = (0.173) × (600 - 300)
q = 1.73 W
