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const2013 [10]
3 years ago
9

In contrast to the leading-trailing drum brake system, the duo-servo drum brake system will:

Engineering
1 answer:
prohojiy [21]3 years ago
6 0

Leading/trailing shoe type drum brake This is called the servo effect (self-boosting effect) which realizes the powerful braking forces of drum brakes. ... This is because drum brakes generate the same braking force in either direction. Generally, this type is used for the rear brakes of passenger cars.

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What are the characteristics of carbon fibre?
nexus9112 [7]

Answer:

high tensile strenth

high stiffness

Explanation:

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3 years ago
Which one of the following types of gas is flammable? a. Neon b. Peroxide. c. Acetylene d. Carbon Dioxide ​
Delvig [45]

Answer:

D Carbon Dioxide

Explanation:

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4 years ago
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"Continuous duty" is defined as
larisa86 [58]

Answer:

operates at a substantially constant load for an indefinitely long time. ... Short-time duty: operates at a substantially constant load for a time that is definite, short, and specified. Varying duty: the loads and intervals of operation change.

Explanation:

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3 years ago
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An adiabatic air compressor compresses 10 L/s of air at 120 kPa and 20 degree C to 1000 kPa and 300 degree C.
Oksana_A [137]

Answer:

work=281.4KJ/kg

Power=4Kw

Explanation:

Hi!

To solve follow the steps below!

1. Find the density of the air at the entrance using the equation for ideal gases

density=\frac{P}{RT}

where

P=pressure=120kPa

T=20C=293k

R= 0.287 kJ/(kg*K)= gas constant ideal for air

density=\frac{120}{(0.287)(293)}=1.43kg/m^3

2.find the mass flow by finding the product between the flow rate and the density

m=(density)(flow rate)

flow rate=10L/s=0.01m^3/s

m=(1.43kg/m^3)(0.01m^3/s)=0.0143kg/s

3. Please use the equation the first law of thermodynamics that states that the energy that enters is the same as the one that must come out, we infer the following equation, note = remember that power is the product of work and mass flow

Work

w=Cp(T1-T2)

Where

Cp= specific heat for air=1.005KJ/kgK

w=work

T1=inlet temperature=20C

T2=outlet temperature=300C

w=1.005(300-20)=281.4KJ/kg

Power

W=mw

W=(0.0143)(281.4KJ/kg)=4Kw

5 0
3 years ago
7–53 Refrigerant-134a enters the condenser of a residential heat pump at 800 kPa and 35°C at a rate of 0.018 kg/s and leaves at
qwelly [4]

Answer:

a. The coefficient of power = 2.6364

b. The rate of heat absorption from the outside air is 1.96368kW

Explanation:

Given

First we need to get the enthalpy of R-34a.

When T = 35°C and P = 800kPa;

h1 = 271.24kj/kg

When x2 = 0 and P = 800kPa;

h1 = 95.48kj/kg

To calculate the COP, first we need to calculate the energy balance.

This is given as

Q = m(h1 - h2)

Where m = 0.018kg

Q = 0.018(271.24 - 95.48)

Q = 3.16368Kw

COP is then calculated as Q/W

Where W = Power consumption of the compressor = 1.2kW

COP = 3.16368Kw/1.2Kw

COP = 2.6364

Hence, the coefficient of power = 2.6364

b. The rate of heat absorption from the outside air is calculated as ∆Heat Rate

∆Heat Rate = Q - W

Where Q = Energy Balance = 3.16368Kw

W = Power consumption of the compressor = 1.2kW

∆Heat Rate = 3.16368Kw - 1.2kW

∆Heat Rate = 1.96368kW

Hence, The rate of heat absorption from the outside air is 1.96368kW

6 0
3 years ago
Read 2 more answers
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