Consider a flat plate that is 25 mm long, 30 mm wide, and 1 mm thick and a 50 mm long cylinder with the same volume as the plate
. Compare the heat transfer of the plate in parallel air flow to the heat transfer of the cylinder in cross flow under the same conditions. Both the plate and cylinder have a surface temperature of 80°C. The air is at 25°C with a velocity of 3 m/s. Note: Use the Churchill and Bernstein correlation to determine the Nusselt number for the cylinder case.
1 answer:
You might be interested in
Answer:
The mass flow rate of refrigerant is 0.352 kg/s
Explanation:
Considering the cycle of an ideal heat pump, provided in the attachment, we first find enthalpy at state B and D. For that purpose, we use property tables of refrigerant R134a:
<u>At State A</u>:
From table, we see the enthalpy and entropy value of saturated vapor at 0.2 MPa. Therefore:
ha = 244.5 KJ/kg
Sa = 0.93788 KJ/kg.k
<u>At State B</u>:
Since, the process from state A to B is isentropic. Therefore,
Sb = Sa = 0.93788 KJ/Kg
From table, we see the enthalpy value of super heated vapor at 1 MPa and Sb. Therefore:
hb = 256.85 KJ/kg (By interpolation)
<u>At State C</u>:
From table, we see the enthalpy and entropy value of saturated liquid at 1 MPa. Therefore:
hc = 107.34 KJ/kg
Now, from the diagram it is very clear that:
Heat Loss = m(hb = hc)
m = (Heat Loss)/(hb - hc)
where,
m = mass flow rate = ?
Heat Loss = (180,000 Btu/hr)(1.05506 KJ/1 Btu)(1 hr/3600 sec)
Heat Loss = 52.753 KW
Therefore,
m = (52.753 KJ/s)/(256.85 KJ/kg - 107.34 KJ/kg)
<u>m = 0.352 kg/s</u>
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:
Explanation:
The pictures below shows the whole explanation for the problem
