Answer: 337J/kg
Explanation:
(a) For air, take k = 1.40, R = 287 J/kg ⋅ K, and cp = 1005 J/kg ⋅K. The adiabatic steady-flow energy equation (9.23) is used to compute the downstream velocity:+ = = + = +2 2 2p 21 1 1c T V constant 1005( 260) (75) 1005(207) V or .2 2 2
Ans2mV 335 s≈2 1 p 2 1 2 1 207 273 30Meanwhile, s s c ln(T /T ) R ln( p /p ) 1005 ln 287 ln ,260 273 140+ − = − = − ÷ ÷ +or s 2 − s1 = −105 + 442 ≈ 337 J/kg ⋅ K Ans.(a)
(b) For argon, take k = 1.67, R = 208 J/kg ⋅ K, and cp = 518 J/kg ⋅ K. Repeat part (a):2 2 2p 21 1 1c T V 518(260) (75) 518(207) V , solve .2 2 2 Ans+ = + = + 2mV 246 1 207 273 30s s 518 l n 208 ln 54 320 . (b)260 273 140
Answer:
A) m' = 351.49 kg/s
B) m'= 1036.91 kg/s
Explanation:
We are given;
Pressure Ratio;r_p = 12
Inlet temperature of compressor;T1 = 300 K
Inlet temperature of turbine;T3 = 1000 K
cp = 1.005 kJ/kg·K
k = 1.4
Net power output; W' = 70 MW = 70000 KW
A) Now, the formula for the mass flow rate using the total power output of the compressor and turbine is given as;
m' = W'/[cp(T3(1 - r_p^(-(k - 1)/k)) - T1(r_p^((k - 1)/k))
At, 100% efficiency, plugging in the relevant values, we have;
m' = 70000/(1.005(1000(1 - 12^(-(1.4 - 1)/1.4)) - 300(12^((1.4 - 1)/1.4)))
m' = 70000/199.1508
m' = 351.49 kg/s
B) At 85% efficiency, the formula will now be;
m' = W'/[cp(ηT3(1 - r_p^(-(k - 1)/k)) - (T1/η) (r_p^((k - 1)/k))
Where η is efficiency = 0.85
Thus;
m' = 70000/(1.005(0.85*1000(1 - 12^(-(1.4 - 1)/1.4)) - (300/0.85)(12^((1.4 - 1)/1.4)))
m' = 70000/(1.005*(432.09129 - 364.9189)
m'= 1036.91 kg/s
Answer:
Explanation:
Let assume that heating and boiling process occurs under an athmospheric pressure of 101.325 kPa. The heat needed to boil water is:
The heat liberated by the LP gas is:
A kilogram of LP gas has a minimum combustion power of . Then, the required mass is:
Answer:
The time required is 10.078 hours or 605 min
Explanation:
The formula to apply here is ;
K=(d²-d²₀ )/t
where t is time in hours
d is grain diameter to be achieved after heating in mm
d₀ is the grain diameter before heating in mm
Given
d=5.5 × 10^-2 mm
d₀=2.4 × 10^-2 mm
t₁= 500 min = 500/60 =25/3 hrs
t₂=?
n=2.2
First find K
K=(d²-d²₀ )/t₁
K={ (5.1 × 10^-2 mm)²-(2.4 × 10−2 mm)² }/ 25/3
K=(0.051²-0.024²) ÷25/2
K=0.000243 mm²/h
Re-arrange equation for K ,to get the equation for d as;
d=√(d₀²+ Kt) where now t=t₂
Ordinary Portland Cement (OPC)
Portland Pozzolana Cement (PPC)
Rapid Hardening Cement.
Extra Rapid Hardening Cement
Low Heat Cement.
Sulfates Resisting Cement.