The rigging device which are used to move loads without the use of slings, but grip the load by biting down and using jaw tension to secure the load, is lifting clamps.
<h3>What are the rigging devices?</h3>
The rigging devices are used to lift the objects and items when the safety is required. This device is used in the industries.
Types of rigging devices
- Rigging hooks-These rigging device is used when the heavy load need to be lift.
- Lifting clamps-Lifting clamp are used to lift the device with jaw tension to secure the load. In this, there is no use of slings.
- Pulley and blocks-In the load is lifts with the help of block and pulley arrangement. This is a widely used rigging device.
Thus, the rigging device which are used to move loads without the use of slings, but grip the load by biting down and using jaw tension to secure the load, is lifting clamps.
Learn more about the rigging devices here;
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Answer:
The final temperature of water is 381.39 °C.
Explanation:
Given that
Mass of water = 5 kg
Heat transfer at constant pressure Q = 2960 KJ
Initial temperature = 240 °C
We know that heat transfer at constant pressure given as follows
We know that for water
Lets take final temperature of water is T
So
T=381.39 °C
So the final temperature of water is 381.39 °C.
Answer:
Explanation:
f = 50.0 Hz, L = 0.650 H, π = 3.14
C = 4.80 μF, R = 301 Ω resistor. V = 120volts
XL = wL = 2πfL
= 2×3.14×50* 0.650
= 204.1 Ohm
Xc= 1/wC
Xc = 1/2πfC
Xc = 1/2×3.14×50×4.80μF
= 1/0.0015072
= 663.48Ohms
1. Total impedance, Z = sqrt (R^2 + (Xc-XL)^2)= √ 301^2+ (663.48Ohms - 204.1 Ohm)^2
√ 90601 + (459.38)^2
√ 90601+211029.98
√ 301630.9844
= 549.209
Z = 549.21Ohms
2. I=V/Z = 120/ 549.21Ohms =0.218Ampere
3. P=V×I = 120* 0.218 = 26.16Watt
Note that
I rms = Vrms/Xc
= 120/663.48Ohms
= 0.18086A
4. I(max) = I(rms) × √2
= 0.18086A × 1.4142
= 0.2557
= 0.256A
5. V=I(max) * XL
= 0.256A ×204.1
=52.2496
= 52.250volts
6. V=I(max) × Xc
= 0.256A × 663.48Ohms
= 169.85volts
7. Xc=XL
1/2πfC = 2πfL
1/2πfC = 2πf× 0.650
1/2×3.14×f×4.80μF = 2×3.14×f×0.650
1/6.28×f×4.8×10^-6 = 4.082f
1/0.000030144× f = 4.082×f
1 = 0.000030144×f×4.082×f
1 = 0.000123f^2
f^2 = 1/0.000123048
f^2 = 8126.922
f =√8126.922
f = 90.14 Hz
Answer:
The maximum power that can be generated is 127.788 kW
Explanation:
Using the steam table
Enthalpy at 20 bar = 2799 kJ/kg
Enthalpy at 2 bar = 2707 kJ/kg
Change in enthalpy = 2799 - 2707 = 92 kJ/kg
Mass flow rate of steam = 5000 kg/hr = 5000 kJ/hr × 1 hr/3600 s = 1.389 kg/s
Maximum power generated = change in enthalpy × mass flow rate = 92 kJ/kg × 1.389 kg/s = 127.788 kJ/s = 127.788 kW
