How the LED (light emitting diode) works?
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The expression of V(m³)=e^(t(s)) to make V in in³ and t in minutes is;
V(in³) = (¹/₆₁₀₂₄)a
We are given that;
Volume of microbial culture is observed to increase according to the formula;
V = e^(t)
where;
t is in seconds
V is in m³
We want to now express V in in³ and t in minutes.
Now, from conversions;
1 m³ = 61024 in³
Also; 1 second = 1/60 minutes
according to formula for exponential decay, we know that;
V = ae^(bt)
Thus, we have;
61024V = ae^(¹/₆₀b(t(h))
V(in³) = (¹/₆₁₀₂₄)a
Read more about subject of formula at; brainly.com/question/790938
Answer:
coupling is in tension
Force = -244.81 N
Explanation:
Diameter of Hose ( D1 ) = 35 mm
Diameter of nozzle ( D2 ) = 25 mm
water gage pressure in hose = 510 kPa
stream leaving the nozzle is uniform
exit speed and pressure = 32 m/s and atmospheric
<u>Determine the force transmitted by the coupling between the nozzle and hose </u>
attached below is the remaining part of the detailed solution
Inlet velocity ( V1 ) = V2 ( D2/D1 )^2
= 32 ( 25 / 35 )^2
= 16.33 m/s
Answer:
0.31
126.23 kg/s
Explanation:
Given:-
- Fluid: Water
- Turbine: P3 = 8MPa , P4 = 10 KPa , nt = 85%
- Pump: Isentropic
- Net cycle-work output, Wnet = 100 MW
Find:-
- The thermal efficiency of the cycle
- The mass flow rate of steam
Solution:-
- The best way to deal with questions related to power cycles is to determine the process and write down the requisite properties of the fluid at each state.
First process: Isentropic compression by pump
P1 = P4 = 10 KPa ( condenser and pump inlet is usually equal )
h1 = h-P1 = 191.81 KJ/kg ( saturated liquid assumption )
s1 = s-P1 = 0.6492 KJ/kg.K
v1 = v-P1 = 0.001010 m^3 / kg
P2 = P3 = 8 MPa( Boiler pressure - Turbine inlet )
s2 = s1 = 0.6492 KJ/kg.K .... ( compressed liquid )
- To determine the ( h2 ) at state point 2 : Pump exit. We need to determine the wok-done by pump on the water ( Wp ). So from work-done principle we have:
- From the following relation we can determine ( h2 ) as follows:
h2 = h1 + wp
h2 = 191.81 + 8.0699
h2 = 199.88 KJ/kg
Second Process: Boiler supplies heat to the fluid and vaporize
- We have already evaluated the inlet fluid properties to the boiler ( pump exit property ).
- To determine the exit property of the fluid when the fluid is vaporized to steam in boiler ( super-heated phase ).
P3 = 8 MPa
T3 = ? ( assume fluid exist in the saturated vapor phase )
h3 = hg-P3 = 2758.7 KJ/kg
s3 = sg-P3 = 5.7450 KJ/kg.K
- The amount of heat supplied by the boiler per kg of fluid to the water stream. ( qs ) is determined using the state points 2 and 3 as follows:
Third Process: The expansion ( actual case ). Turbine isentropic efficiency ( nt ).
- The saturated vapor steam is expanded by the turbine to the condenser pressure. The turbine inlet pressure conditions are similar to the boiler conditions.
- Under the isentropic conditions the steam exits the turbine at the following conditions:
P4 = 10 KPa
s4 = s3 = 5.7450 KJ/kg.K ... ( liquid - vapor mixture phase )
- Compute the quality of the mixture at condenser inlet by the following relation:
- Determine the isentropic ( h4s ) at this state as follows:
- Since, we know that the turbine is not 100% isentropic. We will use the working efficiency and determine the actual ( h4 ) at the condenser inlet state:
- We can now compute the work-produced ( wt ) due to the expansion of steam in turbine.
- The net power out-put from the plant is derived from the net work produced by the compression and expansion process in pump and turbine, respectively.
Answer: The mass flow rate of the steam would be 126.23 kg/s
- The thermal efficiency of the cycle ( nth ) is defined as the ratio of net work produced by the cycle ( Wnet ) and the heat supplied by the boiler to the water ( Qs ):
Answer: The thermal efficiency of the cycle is 0.31