As temperature increases what happens to the conductor
1 answer:
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Answer:
Explanation:
given data
types of drinking straws
- square cross-sectional shape
- round shape
solution
we know that both perimeter of the cross section are equal
so we can say that
perimeter of square = perimeter of circle
4 × S = π × D
here S is length and D is diameter
S = ....................1
and
ratio of flow rate through the square and circle is here
Answer:
a. T₂ = 837.2 K
b. ηt = 91.4%
Explanation:
Assuming k = 1.4
KNOWN: Air expands adiabatically through a turbine operating at steady state. Operating data are known.
FIND: Determine the exit temperature and the isentropic turbine efficiency.
ENGINEERING MODEL (See the pics):
(1) The control volume is at state.
(2) For the control volume, Qcv = 0 and kinetic and potential energy effects can be neglected.
(3) The air is modeled as an ideal gas with constant specific heats: k = 1.4.
ANALYSIS:
(a) Mass and energy rate balances reduce to give:
0 = - Wcv + m*(h₁ – h₂).
With h₁ – h₂ = cp*(T₁ – T₂)
then
cp = k*R/(k-1) = 1.4*(8.314/28.97)/(1.4 – 1) = 1.004 kJ/kg∙K and
T₂ = T₁ - (Wcv/(m*cp))
⇒ T₂ = 1040 K – (1120 kW)/[(5.5 kg/s)(1.004 kJ/kg∙K)(1 kJ/s/ 1 kW)
⇒ T₂ = 837.2 K
b) The isentropic efficiency is
ηt = (h₁ – h₂) / (h₁ – h₂s) = cp*(T₁ – T₂) / (cp*(T₁ – T₂s)).
To get T₂s we note that for an isentropic process of an ideal gas with constant specific heats
(T₂s / T₁) = (p₂ / p₁)∧((k-1)/k) ⇒ T₂s = T₁*(p₂ / p₁)∧((k-1)/k)
⇒ T₂s = (1040 K)*(120 / 278)∧((1.4-1)/1.4) = 818.1 K
Thus, the isentropic efficiency is
ηt = (1040 – 837.2)/(1040 – 818.1) = 0.914
ηt = 91.4%
