Explanation:
A.
H = Aeσ^4
Using the stefan Boltzmann law
When we differentiate
dH/dT = 4AeσT³
dH/dT = 4(0.15)(0.9)(5.67)(10^-8)(650)³
= 8.4085
Exact error = 8.4085x20
= 168.17
H(650) = 0.15(0.9)(5.67)(10^-8)(650)⁴
= 1366.376watts
B.
Verifying values
H(T+ΔT) = 0.15(0.9)(5.67)(10)^-8(670)⁴
= 1542.468
H(T+ΔT) = 0.15(0.9)(5.67)(10^-8)(630)⁴
= 1205.8104
Error = 1542.468-1205.8104/2
= 168.329
ΔT = 40
H(T+ΔT) = 0.15(0.9)(5.67)(10)^-8(690)⁴
= 1735.05
H(T-ΔT) = 0.15(0.9)(5.67)(10^-8)(610)⁴
= 1735.05-1059.83/2
= 675.22/2
= 337.61
Answer:
true
Explanation:
if it is not true it is false
Answer:
See attachment for detailed answer.
Explanation:
Answer:
Explanation:
(c). looking for the radiation of the collector is given thus
C = 0.095 + 0.04 sin [360/365(n-100)] = 0.095 + 0.04 sin [360/365(1-100)]
C = 0.05535
∴ Diffuse radiation of the collector Idc = C*Ib + (1+cosσ/2)
Idc = 0.5535 * 908.7 (1+cos40/ 2) = 44.41 W/m²
Idc = 44.41 W/m²
Answer:
The heat transfer is 29.75 kJ
Explanation:
The process is a polytropic expansion process
General polytropic expansion process is given by PV^n = constant
Comparing PV^n = constant with PV^1.2 = constant
n = 1.2
(V2/V1)^n = P1/P2
(V2/0.02)^1.2 = 8/2
V2/0.02 = 4^(1/1.2)
V2 = 0.02 × 3.2 = 0.064 m^3
W = (P2V2 - P1V1)/1-n
P1 = 8 bar = 8×100 = 800 kPa
P2 = 2 bar = 2×100 = 200 kPa
V1 = 0.02 m^3
V2 = 0.064 m^3
1 - n = 1 - 1.2 = -0.2
W = (200×0.064 - 800×0.02)/-0.2 = -3.2/-0.2 = 16 kJ
∆U = 55 kJ/kg × 0.25 kg = 13.75 kJ
Heat transfer (Q) = ∆U + W = 13.75 + 16 = 29.75 kJ
