Answer:
using calculations Heat losses will be 4512 J
Answer:
See attached picture.
Explanation:
See attached picture for explanation.
Answer: 1.137*10^7 Btu/h.
Explanation:
Given data:
Efficiency of the plant = 4.5percent
Net power output of the plant = 150kw
Solution:
The required collection rate
QH = W/n
= 150/0.045 * 0.94782/ 1 /60 */60 Btu/h.
= 3333.333 *3412.152Btu/h.
= 11373840 Btu/h
= 1.137*10^7 Btu/h.
Answer:
a)temperature=69.1C
b)3054Kw
Explanation:
Hello!
To solve this problem follow the steps below, the complete procedure is in the attached image
1. draw a complete outline of the problem
2. to find the temperature at the turbine exit use termodinamic tables to find the saturation temperature at 30kPa
note=Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)
through prior knowledge of two other properties such as pressure and temperature.
3. Using thermodynamic tables find the enthalpy and entropy at the turbine inlet, then find the ideal enthalpy using the entropy of state 1 and the outlet pressure = 30kPa
4. The efficiency of the turbine is defined as the ratio between the real power and the ideal power, with this we find the real enthalpy.
Note: Remember that for a turbine with a single input and output, the power is calculated as the product of the mass flow and the difference in enthalpies.
5. Find the real power of the turbine
Answer:
The given blanks can be filled as given below
Voltmeter must be connected in parallel
Explanation:
A voltmeter is connected in parallel to measure the voltage drop across a resistor this is because in parallel connection, current is divided in each parallel branch and voltage remains same in parallel connections.
Therefore, in order to measure the same voltage across the voltmeter as that of the voltage drop across resistor, voltmeter must be connected in parallel.